Instructor’s Resource Manual to accompany Electronic Devices Eighth Edition Thomas L. Floyd

Instructor’s Resource Manual to accompany

Electronic Devices Eighth Edition

Thomas L. Floyd

Upper Saddle River, New Jersey Columbus, Ohio

i

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Instructors of classes using Floyd, Electronic Devices, Eighth Edition, may reproduce material from the instructor’s manual for classroom use.

10 9 8 7 6 5 4 3 2 1

ii

ISBN-13: 978-0-13-242992-4 ISBN-10: 0-13-242992-6

Contents Solutions for End-of-Chapter Problems Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18

1 4 16 25 35 48 63 72 85 93 105 112 124 132 139 145 150 156

Application Activity Results

163

Summary of Multisim Circuit Files by Gary Snyder

185

Solutions Manual to Laboratory Exercises for Electronic Devices

205

Test Item File

275

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iv

Chapter 1 Introduction to Semiconductors Section 1-1 Atomic Structure 1.

An atom with an atomic number of 6 has 6 electrons and 6 protons.

2.

The third shell of an atom can have 2n2 = 2(3)2 = 18 electrons.

Section 1-2 Insulators, Conductors, and Semiconductors 3.

The materials represented in Figure 1−39 in the textbook are (a) insulator (b) semiconductor

4.

An atom with four valence electrons is a semiconductor.

5.

In a silicon crystal, each atom forms four covalent bonds.

(c)

conductor

Section 1-3 Current in Semiconductors 6.

When heat is added to silicon, more free electrons and holes are produced.

7.

Current is produced in silicon at the conduction band and the valence band.

Section 1-4 N-Type and P-Type Semiconductors 8.

Doping is the carefully controlled addition of trivalent or pentavalent atoms to pure (intrinsic) semiconductor material for the purpose of increasing the number of majority carriers (free electrons or holes).

9.

Antimony is a pentavalent (donor) material used for doping to increase free electrons. Boron is a trivalent (acceptor) material used for doping to increase the holes.

Section 1-5 The Diode 10.

The electric field across the pn junction of a diode is created by donor atoms in the n region losing free electrons to acceptor atoms in the p region. This creates positive ions in the n region near the junction and negative ions in the p region near the junction. A field is then established between the ions.

11.

The barrier potential of a diode represents an energy gradient that must be overcome by conduction electrons and produces a voltage drop, not a source of energy.

1

Chapter 1 Section 1-6 Biasing a Diode 12.

To forward-bias a diode, the positive terminal of a voltage source must be connected to the p region.

13.

A series resistor is needed to limit the current through a forward-biased diode to a value that will not damage the diode because the diode itself has very little resistance.

Section 1-7 Voltage-Current Characteristic of a Diode 14.

To generate the forward bias portion of the characteristic curve, connect a voltage source across the diode for forward bias and place an ammeter in series with the diode and a voltmeter across the diode. Slowly increase the voltage from zero and plot the forward voltage versus the current.

15.

A temperature increase would cause the barrier potential of a silicon diode to decrease from 0.7 V to 0.6 V.

Section 1-8 Diode Models 16.

(a) The diode is reverse-biased. (c) The diode is forward-biased.

17.

(a) (b) (c) (d)

VR = 5 V − 8 V = −3 V VF = 0.7 V VF = 0.7 V VF = 0.7 V

18.

(a) (b) (c) (d)

VR = 5 V − 8 V = −3 V VF = 0 V VF = 0 V VF = 0 V

19.

Ignoring rR′ : (a) VR ≅ 5 V − 8 V = −3 V

(b) The diode is forward-biased. (d) The diode is forward-biased.

100 V − 0.7 V = 174 mA 560 Ω + 10 Ω VF = IF rd′ + VB = (174 mA)(10 Ω) + 0.7 V = 2.44 V

(b) IF =

B

30 V 30 V = = 6.19 mA 4.85 kΩ Rtot 6.19 mA IF = = 3.1 mA 2 VF = IF rd′ + 0.7 V = (3.1 mA)(10 Ω) + 0.7 V = 0.731 V

(c) Itot =

2

Chapter 1 (d) Approximately all of the current from the 20 V source is through the diode. No current from the 10 V source is through the diode. 20 V − 0.7 V IF = = 1.92 mA 10 kΩ + 10 Ω VF = (1.92 mA)(10 Ω) + 0.7 V = 0.719 V

Section 1-9 Testing a Diode Since VD = 25 V = 0.5VS, the diode is open. The diode is forward-biased but since VD = 15 V = VS, the diode is open. The diode is reverse-biased but since VR = 2.5 V = 0.5VS, the diode is shorted. The diode is reverse-biased and VR = 0 V. The diode is operating properly.

20.

(a) (b) (c) (d)

21.

VA = VS1 = +25 V VB = VS1 − 0.7 V = 25 V − 0.7 V = +24.3 V VC = VS2 + 0.7 V = 8 V + 0.7 V = +8.7 V VD = VS2 = +8.0 V B

Multisim Troubleshooting Problems The solutions showing instrument connections for problems 22 through 30 are available from the Instructor Resource Center. The faults in the circuit files may be accessed using the password book (all lowercase). To access supplementary materials online, instructors need to request an instructor access code. Go to www.prenhall.com, click the Instructor Resource Center link, and then click Register Today for an instructor access code. Within 48 hours after registering, you will receive a confirming e-mail that includes an instructor access code. Once you have received your code, go to the site and log on for full instructions on downloading the materials you wish to use.

22.

Diode shorted

23.

Diode open

24.

Diode open

25.

Diode shorted

26.

No fault

27.

Diode shorted

28.

Diode leaky

29.

Diode open

30.

Diode shorted

3

Chapter 2 Diode Applications Section 2-1 Half-Wave Rectifiers 1.

See Figure 2-1.

Figure 2-1

2.

(a) PIV = V p = 5 V

3.

VAVG =

4.

(a) I F =

Vp

=

π

(b) I F =

200 V

π

(b) PIV = V p = 50 V

= 63.7 V

V( p )in − 0.7 V R V( p )in − 0.7 V R

=

5 V − 0.7 V 4.3 V = = 91.5 mA 47 Ω 47 Ω

=

50 V − 0.7 V 49.3 V = = 14.9 mA 3.3 kΩ 3.3 kΩ

5.

Vsec = nV pri = (0.2)120 V = 24 V rms

6.

Vsec = nV pri = (0.5)120 V = 60 V rms Vp(sec) = 1.414(60 V) = 84.8 V V p ( sec ) 84.8 V Vavg ( sec ) = = = 27.0 V PL ( p )

(V =

PL ( avg )

π

p ( sec )

π

− 0.7 V ) RL

(V =

avg ( sec )

RL

)

2

=

2

=

(84.1 V) 2 = 32.1 W 220 Ω

(27.0 V) 2 = 3.31 W 220 Ω

4

Chapter 2 Section 2-2 Full-Wave Rectifiers 7.

(a) VAVG =

Vp

(b) VAVG = (c) VAVG = (d) VAVG = 8.

=

π

2V p

π 2V p

π 2V p

π

5V

π =

= 1.59 V

2(100 V)

π

= 63.7 V

+ 10 V =

2(10 V)

π

+ 10 V = 16.4 V

− 15 V =

2( 40 V)

− 15 V = 10.5 V

π

(a) Center-tapped full-wave rectifier (b) Vp(sec) = (0.25)(1.414)120 V = 42.4 V V p ( sec ) 42.4 V = (c) = 21.2 V 2 2 (d) See Figure 2-2. VRL = 21.2 V − 0.7 V = 20.5 V

Figure 2-2

V p ( sec )

− 0.7 V

2

(e) IF =

RL

=

20.5 V = 20.5 mA 1.0 kΩ

(f) PIV = 21.2 V + 20.5 V = 41.7 V 9.

VAVG = VAVG =

120 V = 60 V for each half 2 Vp

π

Vp = πVAVG = π(60 V) = 186 V 10.

See Figure 2-3.

Figure 2-3

5

Chapter 2 π VAVG ( out )

=

π (50 V)

11.

PIV = Vp =

12.

PIV = Vp(out) = 1.414(20 V) = 28.3 V

13.

See Figure 2-4.

2

2

= 78.5 V

Figure 2-4

Section 2-3 Power Supply Filters and Regulators 14.

Vr(pp) = 0.5 V Vr ( pp ) 0.5 V = = 0.00667 r= VDC 75 V

15.

Vr(pp) =

V p (in ) fR L C

=

30 V = 8.33 V pp (120 Hz)(600 Ω)(50 μF)

⎛ ⎛ ⎞ 1 ⎞ 1 ⎟⎟V p (in ) = ⎜⎜1 − ⎟⎟30 V = 25.8 V VDC = ⎜⎜1 − ⎝ (240 Hz)(600 Ω)(50 μF) ⎠ ⎝ 2 fRLC ⎠ 16.

17.

⎛ Vr ( pp ) ⎞ 8.33 V ⎞ ⎟100 = ⎛⎜ %r = ⎜⎜ ⎟100 = 32.3% ⎟ ⎝ 25.8 V ⎠ ⎝ VDC ⎠ Vr(pp) = (0.01)(18 V) = 180 mV ⎛ 1 ⎞ ⎟⎟V p (in ) Vr(pp) = ⎜⎜ ⎝ fRLC ⎠ ⎛ 1 C = ⎜⎜ ⎝ fRLVr

18.

Vr ( pp ) =

⎞ ⎛ ⎞ 1 ⎟⎟V p (in ) = ⎜⎜ ⎟⎟18 V = 556 μF ⎝ (120 Hz)(1.5 kΩ)(180 mV) ⎠ ⎠

V p (in ) fRL C

=

80 V = 6.67 V (120 Hz)(10 kΩ)(10 μF)

⎛ ⎛ ⎞ 1 ⎞ 1 ⎟⎟V p (in ) = ⎜⎜1 − ⎟⎟80 V = 76.7 V VDC = ⎜⎜1 − ⎝ (240 Hz)(10 kΩ)(10 μF) ⎠ ⎝ 2 fRLC ⎠ Vr ( pp ) 6.67 V = = 0.087 r= VDC 76.7 V

6

Chapter 2 19.

V p (sec ) = (1.414)(36 V) = 50.9 V Vr(rect) = Vp(sec) − 1.4 V = 50.9 V − 1.4 V = 49.5 V ⎛ 1 ⎞ ⎛ ⎞ 1 Neglecting Rsurge, Vr(pp) = ⎜ ⎟V p ( rect ) = ⎜ ⎟ 49.5 V = 1.25 V ⎝ (120 Hz)(3.3 kΩ)(100 μ F) ⎠ ⎝ fRL C ⎠ V ⎛ 1 ⎞ ⎟⎟V p ( rect ) = V p ( rect ) − r ( pp ) = 49.5 V − 0.625 V = 48.9 V VDC = ⎜⎜1 − 2 ⎝ 2 fRLC ⎠

20.

Vp(sec) = 1.414(36 V) = 50.9 V See Figure 2-5.

Figure 2-5 21.

⎛ V − VFL ⎞ ⎛ 15.5 V − 14.9 V ⎞ ⎟⎟100% = ⎜ Load regulation = ⎜⎜ NL ⎟100% = 4% 14.9 V ⎝ ⎠ ⎝ VFL ⎠

22.

VFL = VNL − (0.005)VNL = 12 V − (0.005)12 V = 11.94 V

Section 2-4 Diode Limiting and Clamping Circuits 23.

See Figure 2-6.

Figure 2-6

7

Chapter 2 24.

Apply Kirchhoff’s law at the peak of the positive half cycle: (b) 25 V = VR1 + VR2 + 0.7 V 2VR = 24.3 V 24.3 V VR = = 12.15 V 2 Vout = VR + 0.7 V = 12.15 V + 0.7 V = 12.85 V See Figure 2-7(a). 11.3 V = 5.65 V 2 Vout = VR + 0.7 V = 5.65 V + 0.7 V = 6.35 V See Figure 2-7(b).

(c) VR =

4.3 V = 2.15 V 2 Vout = VR + 0.7 V = 2.15 V + 0.7 V = 2.85 V See Figure 2-7(c).

(d) VR =

Figure 2-7

8

Chapter 2 25.

See Figure 2-8.

Figure 2-8 26.

See Figure 2-9.

Figure 2-9 27.

See Figure 2-10.

Figure 2-10 28.

30 V − 0.7 V = 13.3 mA 2.2 kΩ (b) Same as (a).

(a) I p =

9

Chapter 2 30 V − (12 V + 0.7 V) = 7.86 mA 2.2 kΩ 30 V − (12 V − 0.7 V) (b) I p = = 8.5 mA 2.2 kΩ 30 V − (−11.3 V) = 18.8 mA (c) I p = 2.2 kΩ 30 V − (−12.7 V) (d) I p = = 19.4 mA 2.2 kΩ

29.

(a) I p =

30.

See Figure 2-11.

Figure 2-11 31.

(a) A sine wave with a positive peak at 0.7 V, a negative peak at −7.3 V, and a dc value of −3.3 V. (b) A sine wave with a positive peak at 29.3 V, a negative peak at −0.7 V, and a dc value of +14.3 V. (c) A square wave varying from +0.7 V to −15.3 V with a dc value of −7.3 V. (d) A square wave varying from +1.3 V to −0.7 V with a dc value of +0.3 V.

32.

(a) (b) (c) (d)

A sine wave varying from −0.7 V to +7.3 V with a dc value of +3.3 V. A sine wave varying from −29.3 V to +7.3 V with a dc value of +14.3 V. A square wave varying from −0.7 V to +15.3 V with a dc value of +7.3 V. A square wave varying from −1.3 V to +0.7 V with a dc value of −0.3 V.

Section 2-5 Voltage Multipliers 33.

VOUT = 2Vp(in) = 2(1.414)(20 V) = 56.6 V See Figure 2-12.

Figure 2-12

10

Chapter 2 34.

VOUT(trip) = 3Vp(in) = 3(1.414)(20 V) = 84.8 V VOUT(quad) = 4Vp(in) = 4(1.414)(20 V) = 113 V See Figure 2-13.

Figure 2-13

Section 2-6 The Diode Datasheet 35.

The PIV is specified as the peak repetitive reverse voltage = 100 V.

36.

The PIV is specified as the peak repetitive reverse voltage = 1000 V.

37.

IF(AVG) = 1.0 A 50 V RL(min) = = 50 Ω 1.0 A

Section 2-7 Troubleshooting 38.

If a bridge rectifier diode opens, the output becomes a half-wave voltage resulting in an increased ripple at 60 Hz.

39.

Vavg =

2V p

π

=

2(115 V)(1.414)

π

≅ 104 V

The output of the bridge is correct. However, the 0 V output from the filter indicates that the surge resistor is open or that the capacitor is shorted. 40.

(a) (b) (c) (d)

Correct Incorrect. Open diode. Correct Incorrect. Open diode.

11

Chapter 2 41.

120 V = 24 V rms 5 Vp(sec) = 1.414(24 V) = 33.9 V The peak voltage for each half of the secondary is V p ( sec ) 33.9 V = 17 V = 2 2 The peak inverse voltage for each diode is PIV = 2(17 V) + 0.7 V = 34.7 V The peak current through each diode is V p ( sec ) − 0.7 V 17.0 V − 0.7 V Ip = 2 = 49.4 mA = RL 330 Ω The diode ratings exceed the actual PIV and peak current. The circuit should not fail.

Vsec =

Application Activity Problems 42.

(a) (b) (c) (d) (e) (f)

Not plugged into ac outlet or no ac available at outlet. Check plug and/or breaker. Open transformer winding or open fuse. Check transformer and/or fuse. Incorrect transformer installed. Replace. Leaky filter capacitor. Replace. Rectifier faulty. Replace. Rectifier faulty. Replace.

43.

The rectifier must be connected backwards.

44.

−16 V with 60 Hz ripple

Advanced Problems 45.

46.

⎛ 1 ⎞ ⎟⎟V p (in ) Vr = ⎜⎜ ⎝ fRLC ⎠ ⎛ 1 ⎞ ⎞ ⎛ 1 ⎟⎟V p (in ) = ⎜⎜ ⎟⎟35 V = 177 μF C = ⎜⎜ ⎝ (120 Hz)(3.3 kΩ)(0.5 V) ⎠ ⎝ fRLVr ⎠ ⎛ 1 ⎞ ⎟⎟V p (in ) VDC = ⎜⎜1 − ⎝ 2 fRLC ⎠ ⎛ VDC 1 ⎞ ⎟ = ⎜⎜1 − V p (in ) ⎝ 2 fRLC ⎟⎠

1 V = 1 − DC 2 fRLC V p (in )

1 ⎛ V ⎞ 2 fRL ⎜1 − DC ⎟ ⎜ V p (in ) ⎟ ⎝ ⎠

=C

12

Chapter 2 1 1 = = 62.2 μF (240 Hz)(1.0 kΩ)(1 − 0.933) (240 Hz)(1.0 kΩ)(0.067) Then ⎛ 1 ⎞ ⎛ ⎞ 1 Vr = ⎜ ⎟V p ( in ) = ⎜ ⎟15 V = 2 V ⎝ (120 Hz)(1.0 kΩ)(62.2 μ F) ⎠ ⎝ fRL C ⎠ C=

47.

The capacitor input voltage is Vp(in) = (1.414)(24 V) − 1.4 V = 32.5 V V p (in ) 32.5 V Rsurge = = = 651 mΩ I surge 50 A The nearest standard value is 680 mΩ.

48.

See Figure 2-14. The voltage at point A with respect to ground is VA = 1.414(9 V) = 12.7 V Therefore, VB = 12.7 V − 0.7 V = 12 V Vr = 0.05VB = 0.05(12 V) = 0.6 V peak to peak ⎛ 1 ⎞ ⎛ ⎞ 1 ⎟⎟VB = ⎜⎜ ⎟⎟12 V = 245 μF C = ⎜⎜ ⎝ (120 Hz)(680 Ω)(0.6 V) ⎠ ⎝ fRLVr ⎠ B

B

The nearest standard value is 270 μF. Let Rsurge = 1.0 Ω. 12 V Isurge(max) = = 12 A 1.0 Ω 12 V = 17.6 mA IF(AV) = 680 Ω PIV = 2Vp(out) + 0.7 V = 24.7 V

Figure 2-14

13

Chapter 2 49.

See Figure 2-15. IL(max) = 100 mA 9V RL = = 90 Ω 100 mA Vr = 1.414(0.25 V) = 0.354 V Vr = 2(0.35 V) = 0.71 V peak to peak ⎞ ⎛ 1 ⎟⎟9 V Vr = ⎜⎜ ⎝ (120 Hz)(90 Ω)C ⎠ 9V = 1174 μF (120 Hz)(90 Ω)(0.71 V) Use C = 1200 μF. Each half of the supply uses identical components. 1N4001 diodes are feasible since the average current is (0.318)(100 mA) = 31.8 mA. Rsurge = 1.0 Ω will limit the surge current to an acceptable value. C=

Figure 2-15

50.

See Figure 2-16.

Figure 2-16 51.

VC1 = (1.414)(120 V) − 0.7 V = 170 V VC2 = 2(1.414)(120 V) − 2(0.7 V) = 338 V

14

Chapter 2 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 52 through 60 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 52.

Diode shorted

53.

Diode leaky

54.

Diode open

55.

Bottom diode open

56.

Reduced transformer turns ratio

57.

Open filter capacitor

58.

Diode leaky

59.

D1 open

60.

Load resistor open

15

Chapter 3 Special-Purpose Diodes Section 3-1 The Zener Diode 1.

See Figure 3-1.

Figure 3-1 2.

IZK ≅ 3 mA VZ ≅ −9 V

3.

ZZ =

4.

ΔIZ = 50 mA − 25 mA = 25 mA ΔVZ = ΔIZZZ = (+25 mA)(15 Ω) = +0.375 V VZ = VZ + ΔVZ = 4.7 V + 0.375 V = 5.08 V

5.

ΔT = 70°C − 25°C = 45°C (6.8 V)(0.0004/ °C) = 6.8 C + 0.12 V = 6.92 V VZ = 6.8 V + 45°C

ΔVZ 5.65 V − 5.6 V 0.05 V =5Ω = = ΔI Z 30 mA − 20 mA 10 mA

Section 3-2 Zener Diode Applications 6.

VIN(min) = VZ + IZKR = 14 V + (1.5 mA)(560 Ω) = 14.8 V

7.

ΔVZ = (IZ − IZK)ZZ = (28.5 mA)(20 Ω) = 0.57 V VOUT = VZ − ΔVZ = 14 V − 0.57 V = 13.43 V VIN(min) = IZKR + VOUT = (1.5 mA)(560 Ω) + 13.43 V = 14.3 V

8.

ΔVZ = IZZZ = (40 mA − 30 mA)(30 Ω) = 0.3 V VZ = 12 V + ΔVZ = 12 V + 0.3 V = 12.3 V V −VZ 18 V − 12.3 V = 143 Ω R = IN = 40 mA 40 mA

16

Chapter 3 9.

VZ ≅ 12 V + 0.3 V = 12.3 V See Figure 3-2.

Figure 3-2 10.

VZ(min) = VZ − ΔIZZZ = 5.1 V − (49 mA − 1 mA)(7 Ω) = 5.1 V − (48 mA)(7 Ω) = 5.1 V − 0.336 V = 4.76 V VR = 8 V − 4.76 V = 3.24 V V 3.24 V = 147 mA IT = R = R 22 Ω IL(max) = 147 mA − 1 mA = 146 mA VZ(max) = 5.1 V + (70 mA − 49 mA)(7 Ω) = 5.1 V + 147 mV = 5.25 V VR = 8 V − 5.25 V = 2.75 V 2.75 V = 125 mA IT = 22 Ω IL(min) = 125 mA − 70 mA = 55 mA

11.

% Load regulation =

12.

With no load and VIN = 6 V: V − VZ 6 V − 5.1 V = = 31 mA IZ ≅ IN 29 Ω R + ZZ VOUT = VZ − ΔIZZZ = 5.1 V − (49 mA − 31 mA)(7 Ω) = 5.1 V − 0.126 V = 4.97 V With no load and VIN = 12 V: V − VZ 12 V − 5.1 V = = 238 mA IZ ≅ IN 29 Ω R + ZZ VOUT = VZ + ΔIZZZ = 5.1 V + (238 mA − 49 mA)(7 Ω) = 5.1 V + 1.32 V = 6.42 V ΔVOUT 6.42 V − 4.97 V × 100% = % Line regulation = × 100% = 24.2% 12 V − 6 V ΔVIN

13.

% Load regulation =

VNL − VFL 8.23 V − 7.98 V × 100% = × 100% = 3.13% VFL 7.98 V

14.

% Line regulation =

ΔVOUT 0.2 V × 100% = × 100% = 4% ΔVIN 10 V − 5 V

15.

% Load regulation =

VNL − VFL 3.6 V − 3.4 V × 100% = 5.88% × 100% = VFL 3.4 V

VZ(max) − VZ(min) VZ(min)

× 100% =

17

5.25 V − 4.76 V × 100% = 10.3% 4.76 V

Chapter 3 Section 3-3 The Varactor Diode 16.

At 5 V, C = 20 pF At 20 V, C = 10 pF ΔC = 20 pF − 10 pF = 10 pF (decrease)

17.

From the graph, VR = 3 V @ 25 pF

18.

fr =

1 2π LCT

1 1 = 12.7 pF = 2 2 4π Lf r 4π ( 2 mH)(1 MHz)2 Since they are in series, each varactor must have a capacitance of 2CT = 25.4 pF

CT =

19.

2

Each varactor has a capacitance of 25.4 pF. Therefore, from the graph, VR must be slightly less than 3 V.

Section 3-4 Optical Diodes 24 V = 35.3 mA 680 Ω From the graph, the radiant power is approximately 80 mW.

20.

IF =

21.

See Figure 3-3. 5 V − 2.1 V R= = 97 Ω 30 mA The nearest standard 1% value is 97.6 Ω or the nearest standard 5% value is 91 Ω.

Figure 3-3

18

Chapter 3 22.

VF ≅ 2.2 V for IF = 20 mA Maximum LEDs/branch =

9V ≅ 4 2.2 V

Select 3 LEDs/branch: 48 = 16 3 9 V − 3(2.2 V) = 120 Ω RLIMIT = 20 mA Use sixteen 120 Ω resistors.

Number of branches =

23.

VF ≅ 2.5 V for IF = 30 mA Maximum LEDs/branch =

24 V ≅ 9.6 2.5 V

Select 5 LEDs/branch: 100 = 20 5 24 V − 5(2.5 V) RLIMIT = = 383 Ω 30 mA See Figure 3-4. Number of branches =

Figure 3-4 24.

IR =

10 V = 50 μA 200 kΩ

19

Chapter 3 25.

26.

(a) R =

VS 3V = = 30 kΩ I 100 μA

(b) R =

3V VS = = 8.57 kΩ 350 μA I

(c) R =

3V VS = = 5.88 kΩ 510 μA I

The microammeter reading will increase.

Section 3-5 Other Types of Diodes ΔV 125 mV − 200 mV − 75 mV = = = −750 Ω ΔI 0.25 mA − 0.15 mA 0.10 mA

27.

R=

28.

Tunnel diodes are used in oscillators.

29.

The reflective ends cause the light to bounce back and forth, thus increasing the intensity of the light. The partially reflective end allows a portion of the reflected light to be emitted.

Section 3-6 Troubleshooting 30.

(a) (b) (c) (d) (e)

All voltages are correct. V3 should be 12 V. Zener is open. V1 should be 120 V. Fuse is open. Capacitor C1 is open. R is open or D5 is shorted.

31.

(a) (b) (c) (d) (e) (f) (g) (h)

With D5 open, VOUT ≅ 30 V. With R open, VOUT = 0 V. With C leaky, VOUT has excessive 120 Hz ripple limited to 12 V. With C open, VOUT is full wave rectified voltage limited to 12 V. With D3 open, VOUT has 60 Hz ripple limited to 12 V. With D2 open, VOUT has 60 Hz ripple limited to 12 V. With T open, VOUT = 0 V. With F open, VOUT = 0 V.

Application Activity Problems 32.

(a) Faulty regulator

33.

Incorrect transformer secondary voltage

34.

LED open, limiting resistor open, faulty regulator, faulty bridge rectifier

35.

IL =

12 V = 12 mA; Vreg = 16 V − 12 V = 4 V 1 kΩ Preg = (4 V)(12 mA) = 48 mW

20

Chapter 3 Datasheet Problems 36.

From the datasheet of textbook Figure 3-7: (a) @ 25°C: PD(max) = 1.0 W for a 1N4738A (b) For a 1N4751A: @ 70°C; PD(max) = 1.0 W − (6.67 mW/°C)(20°C) = 1.0 W − 133 mW = 867 mW @ 100°C; PD(max) = 1.0 W − (6.67 mW/°C)(50°C) = 1.0 W − 333 mW = 667 mW (c) IZK = 0.5 mA for a 1N4738A (d) @ 25°C: IZM = 1 W/27 V = 37.0 mA for a 1N4750A (e) ΔZZ = 700 Ω − 7.0 Ω = 693 Ω for a 1N4740A

37.

From the datasheet of textbook Figure 3-24: (a) IF(max) = 200 mA (b) Cmax = 11 pF C 100 pF = 15.4 pF; range is 100 pF − 15.4 pF for an 836A. (c) C20 = 2 = CR 6.5

38.

From the datasheet of textbook 3-34: (a) 9 V cannot be applied in reverse across a TSMF1000 because VR(max) = 5 V. (b) When 5.1 V is used to forward-bias the TSMF1000 for IF = 20 mA, VF ≅ 1.3 V 5.1 V − 1.3 V 3.8 V = = 190 Ω R= 20 mA 20 mA (c) At 25°C maximum power dissipation is 190 mW. If VF = 1.5 V and IF = 50 mA, PD = 75 mW. The power rating is not exceeded. (d) For IF = 40 mA, radiant intensity is approximately 0.9 mW/sr. (e) For IF = 100 mA and θ = 20°, radiant intensity is 40% of maximum or (0.4)(25 mW/sr) = 10 mW/sr

39.

From the datasheet of textbook Figure 3-47: (a) With no incident light and a 10 kΩ series resistor, the typical voltage across the resistor is approximately VR = (1 nA)(1 kΩ) = 1 μV. (b) Reverse current is greatest at about 940 nm. (c) Sensitivity is maximum for λ ≅ 830 nm.

21

Chapter 3 Advanced Problems 40.

See Figure 3-5.

Figure 3-5 41.

VOUT(1) ≅ 6.8 V, VOUT(2) ≅ 24 V

42.

For a 10 kΩ load on each output: V 6.8 V IOUT(1) = OUT1 ≅ = 0.68 mA R1 10 kΩ V 24 V IOUT(2) = OUT2 ≅ = 2.4 mA 10 kΩ R2 VR1 ≅ 120 V − 6.8 V = 113.2 V 113.2 V − 0.68 mA = 112.5 mA IZ1 = 1 kΩ VR2 ≅ 120 V − 24 V = 96 V 96 V IZ2 = − 2.4 mA = 93.6 mA 1 kΩ IT = 0.68 mA + 2.4 mA + 112.5 mA + 93.6 mA = 209.2 mA The fuse rating should be 250 mA or 1/4 A.

43.

See Figure 3-6. Use a 1N4738A zener. IT = 35 mA + 31 mA = 66 mA 24 V − 8.2 V R= = 239 Ω 66 mA

Figure 3-6

22

Chapter 3 1 1 = 2 = 103.4 pF 2 4π Lf min 4π (2 mH)(350 kHz) 2 1 1 = 2 Cmin = = 17.5 pF 2 2 4π Lf max 4π (2 mH)(850 kHz) 2 To achieve this capacitance range, use an 826A varactor and change V2 to 30 V.

44.

Cmax =

45.

See Figure 3-7. From datasheet, VF = 2.1 V for red LED. R=

2

VD 12 V − 2.1 V = = 495 Ω I 20 mA

Use standard value of 510 Ω.

Figure 3-7 46.

See Figure 3-8.

Figure 3-8

23

Chapter 3 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 47 through 50 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 47.

Zener diode open

48.

Capacitor open

49.

Zener diode shorted

50.

Resistor open

24

Chapter 4 Bipolar Junction Transistors Section 4-1 BJT Structure 1.

Majority carriers in the base region of an npn transistor are holes.

2.

Because of the narrow base region, the minority carriers invading the base region find a limited number of partners for recombination and, therefore, move across the junction into the collector region rather than out of the base lead.

Section 4-2 Basic BJT Operation 3.

The base is narrow and lightly doped so that a small recombination (base) current is generated compared to the collector current.

4.

IB = 0.02IE = 0.02(30 mA) = 0.6 mA IC = IE − IB = 30 mA − 0.6 mA = 29.4 mA B

B

5.

The base must be negative with respect to the collector and positive with respect to the emitter.

6.

IC = IE − IB = 5.34 mA − 475 μA = 4.87 mA B

Section 4-3 BJT Characteristics and Parameters 7.

αDC =

I C 8.23 mA = = 0.947 I E 8.69 mA

8.

βDC =

25 mA IC = = 125 I B 200 μA

9.

IB = IE − IC = 20.5 mA − 20.3 mA = 0.2 mA = 200 μA I 20.3 mA βDC = C = = 101.5 I B 200 μ A

10.

IE = IC + IB = 5.35 mA + 50 μA = 5.40 mA 5.35 mA I αDC = C = = 0.99 I E 5.40 mA

11.

IC = αDCIE = 0.96(9.35 mA) = 8.98 mA

B

B

25

Chapter 4 VRC

5V = 5 mA RC 1.0 kΩ 5 mA I βDC = C = = 100 I B 50 μA

12.

IC =

13.

αDC =

14.

IB =

=

β DC

β DC + 1

=

100 = 0.99 101

VBB − VBE 3 V − 0.7 V = = 23 μA RB 100 kΩ IC = βDCIB = 200(23 μA) = 4.6 mA IE = IC + IB = 4.6 mA + 23 μA = 4.62 mA VCE = VCC − ICRC = 10 V − (4.6 mA)(1.0 kΩ) = 5.4 V B

B

B

15.

IC = does not change. For VCC = 10 V: VCE = VCC = ICRC = 10 V − (4.6 mA)(1.0 kΩ) = 5.4 V For VCC = 15 V: VCE = 15 V − (4.6 mA)(1.0 kΩ) = 10.7 V Δ VCE = 10.7 V − 5.4 V = 5.3 V increase

16.

IB =

VBB − VBE 4 V − 0.7 V 3.3 V = = = 702 μA RB 4.7 kΩ 4.7 kΩ V − VCE 24 V − 8 V = IC = CC = 34 mA RC 470 Ω IE = IC + IB = 34 mA + 702 μA = 34.7 mA 34 mA I βDC = C = = 48.4 I B 702 μA B

B

17.

(a) VBE = 0.7 V V −V 4.3 V IB = BB BE = = 1.1 mA RB 3.9 kΩ IC = βDCIB = 50(1.1 mA) = 55 mA VCE = VCC − ICRC = 15 V − (55 mA)(180 Ω) = 5.10 V VCB = VCE − VBE = 5.10 V − 0.7 V = 4.40 V B

B

(b) VBE = −0.7 V V −V − 3 V − (−0.7 V) − 2.3 V = IB = BB BE = = −85.2 μA RB 27 kΩ 27 kΩ IC = βDCIB = 125(−85.2 μA) = −10.7 mA VCE = VCC − ICRC = −8 V − (−10.7 mA)(390 Ω) = −3.83 V VCB = VCE − VBE = −3.83 V − (−0.7 V) = −3.13 V B

B

26

Chapter 4 18.

VCC 15 V = = 83.3 mA RC 180 Ω V −V 5 V − 0.7 V IB = BB BE = = 1.1 mA RB 3.9 kΩ IC = βDCIB = 50(1.1 mA) = 55 mA IC < IC(sat) Therefore, the transistor is not saturated.

(a) IC(sat) =

B

B

VCC 8V = = 20.5 mA RC 390 Ω V − VBE 3 V − 0.7 V = IB = BB = 85.2 μA RB 27 kΩ IC = βDCIB = 125(85.2 μA) = 10.7 mA IC < IC(sat) Therefore, the transistor is not saturated.

(b) IC(sat) =

B

B

19.

VB = 2 V VE = VB − VBE = 2 V − 0.7 V = 1.3 V V 1.3 V IE = E = = 1.3 mA RE 1.0 kΩ IC = αDCIE = (0.98)(1.3 mA) = 1.27 mA α DC 0.98 = βDC = = 49 1 − α DC 1 − 0.98 IB = IE − IC = 1.3 mA − 1.27 mA = 30 μA B

B

B

20.

(a) VB = VBB = 10 V VC = VCC = 20 V VE = VB − VBE = 10 V − 0.7 V = 9.3 V VCE = VC − VE = 20 V − 9.3 V = 10.7 V VBE = 0.7 V VCB = VC − VB = 20 V − 10 V = 10 V B

B

B

(b) VB = VBB = −4 V VC = VCC = −12 V VE = VB − VBE = −4 V − (−0.7 V) = −3.3 V VCE = VC − VE = −12 V − (−3.3) V = −8.7 V VBE = −0.7 V VCB = VC − VB = −12 V − (−4 V) = −8 V B

B

B

27

Chapter 4 21.

For βDC = 100: V − VBE 10 V − 0.7 V = IE = B = 930 μA RE 10 kΩ 100 β DC = αDC = = 0.990 1 + β DC 101 IC = αDCIE = (0.990)(930 μA) = 921 μA For βDC = 150: IE = 930 μA

β DC 150 = = 0.993 1 + β DC 151 IC = αDCIE = (0.993)(930 μA) = 924 μA ΔIC = 924 μA − 0.921 μA = 3 μA αDC =

22.

PD(max) = VCEIC PD(max) 1.2 W = 24 V VCE(max) = = 50 mA IC

23.

PD(max) = 0.5 W − (75°C)(1 mW/°C) = 0.5 W − 75 mW = 425 mW

Section 4-4 The BJT as an Amplifier 24.

Vout = AvVin = 50(100 mV) = 5 V

25.

Av =

26.

27.

Vout 10 V = = 33.3 Vin 300 mV

RC 560 Ω = = 56 re′ 10 Ω Vc = Vout = AvVin = 56(50 mV) = 2.8 V Av =

VBB − VBE 2.5 V − 0.7 V = = 18 μA RB 100 kΩ IC = βDC IB = 250(18 μA) = 4.5 mA V − VCE 9 V − 4 V = RC = CC = 1.1 kΩ IC 4.5 mA IB = B

B

28.

(a) DC current gain = βDC = 50 (b) DC current gain = βDC = 125

28

Chapter 4 Section 4-5 The BJT as a Switch 29.

VCC 5V = = 500 μA RC 10 kΩ I C(sat) 500 μA IB(min) = = = 3.33 μA β DC 150 VIN(min) − 0.7 V IB(min) = RB RBIB(min) = VIN(min) − 0.7 V VIN(min) = RBIB(min) + 0.7 V = (3.33 μA)(1.0 MΩ) + 0.7 V = 4.03 V IC(sat) =

B

B

30.

15 V = 12.5 mA 1.2 kΩ I C(sat) 12.5 mA IB(min) = = = 250 μA β DC 50 V − 0 .7 V 4.3 V RB(min) = IN = = 17.2 kΩ I B(min) 250 μA VIN(cutoff) = 0 V IC(sat) =

Section 4-6 The Phototransistor 31.

IC = βDCIλ = (200)(100 μA) = 20 mA

32.

Iλ = (50 lm/m2)(1 μA/lm/m2) = 50 μA IE = βDCIλ = (100)(50 μA) = 5 mA

33.

Iout = (0.30)(100 mA) = 30 mA

34.

I OUT = 0.6 I IN I 10 mA = 16.7 mA IIN = OUT = 0.6 0.6

Section 4-7 Transistor Categories and Packaging 35.

See Figure 4-1.

36.

(a) (b) (c) (d) (e)

Figure 4-1

Small-signal Power Power Small-signal RF

29

Chapter 4 Section 4-8 Troubleshooting 37.

With the positive probe on the emitter and the negative probe on the base, the ohmmeter indicates an open, since this reverse-biases the base-emitter junction. With the positive probe on the base and the negative probe on the emitter, the ohmmeter indicates a very low resistance, since this forward-biases the base-collector junction.

38.

(a) (b) (c) (d)

39.

(a) IB =

Transistor’s collector junction or terminal is open. Collector resistor is open. Operating properly. Transistor’s base junction or terminal open (no base or collector current).

5 V − 0.7 V = 63.2 μA 68 kΩ 9 V − 3.2 V IC = = 1.76 mA 3.3 kΩ 1.76 mA I βDC = C = = 27.8 I B 63.2 μA B

4.5 V − 0.7 V = 141 μA 27 kΩ 24 V − 16.8 V = 15.3 mA IC = 470 Ω 15.3 mA I βDC = C = = 109 I B 141 μA

(b) IB = B

Application Activity Problems 40.

Q1 OFF, Q2 ON IR2 = 0, PR2 = 0 mW IR1 = 0, PR2 = 0 mW 12 V − 0.7 V IR3 = IR4 = = 304 μA 1.2 kΩ + 36 kΩ PR3 = (304 μA)2(1.2 kΩ) = 110 μW PR4 = (304 μA)2(36 kΩ) = 3.3 mW 12 V − 0.176 V = 19 mA IR5 = 620 Ω PR5 = (19 mA)2(620 Ω) = 224 mW

30

Chapter 4 Q1 ON, Q2 OFF 12 V − 0.7 V IR2 = = 151 μA 75 kΩ PR2 = (151 μA)2(75 kΩ) = 1.7 mW (0.7 V)2 PR1 = = 0.49 μW 1.0 MΩ 12 V − 0.1 V IR4 ≅ = 9.9 mA 1.2 kΩ PR4 = (9.9 mA)2(1.2 kΩ) = 118 mW IR3 ≅ 0, PR3 = 0 mW IR5 = 0, PR5 = 0 mW 41.

IC(max) = 200 mA V 12 V = 60 Ω RL(min) = CC = I C(max) 200 mA

42.

See Figure 4-2.

Figure 4-2

Datasheet Problems 43.

From the datasheet of textbook Figure 4-20: (a) For a 2N3904, VCEO(max) = 40 V (b) For a 2N3904, IC(max) = 200 mA (c) For a 2N3904 @ 25°C, PD(max) = 625 mW (d) For a 2N3904 @ TC = 50°C, PD(max) = 625 mW − 5 mW/°C(25°C) = 625 mW − 125 mW = 500 mW (e) For a 2N3904 with IC = 1 mA, hFE(min) = 70

31

Chapter 4 44.

For an MMBT3904 with TA = 65°C: PD(max) = 350 mW − (65°C − 25°C)(2.8 mW/°C) = 350 mW − 40°C(2.8 mW/°C) = 350 mW − 112 mW = 238 mW

45.

For a PZT3904 with TC = 45°C: PD(max) = 1 W − (45°C − 25°C)(8 mW/°C) = 1 W − 20°C(8 mW/°C) = 1 W − 160 mW = 840 mW

46.

For the circuits of textbook Figure 4-66: 3 V − 0.7 V 2.3 V = (a) IB = = 6.97 mA 330 Ω 330 Ω Let hFE = 30 IC = 30(6.97 mA) = 209 mA VCC − VCE(sat) 30 V − 0.2 V IC(sat) = = 110 mA = RC 270 Ω The transistor is saturated since IC cannot exceed 110 mA. PD = (0.2 V)(110 mA) = 22 mW At 50°C, PD(max) = 350 mW − (50°C − 25°C)(2.8 mW/°C) = 280 mW No parameter is exceeded. B

(b) VCEO = 45 V which exceeds VCEO(max). 47.

For the circuits of textbook Figure 4-67: 5 V − 0.7 V 4.3 V = = 4.30 μA (a) IB = 10 kΩ 10 kΩ hFE(max) = 300 IC = 300(4.30 μA) = 129 mA 9V = 9 mA IC(sat) = 1.0 kΩ The transistor is saturated. B

3 V − 0.7 V 2.3 V = = 23 μA 100 kΩ 100 kΩ hFE(max) = 300 IC = 300(23 μA) = 6.90 mA 12 V = 21.4 mA IC(sat) = 560 Ω The transistor is not saturated.

(b) IB = B

48.

IB(min) = IB(max) =

IC hFE(max) IC hFE(min)

=

10 mA = 66.7 μA 150

=

10 mA = 200 μA 50

32

Chapter 4 49.

For the circuits of textbook Figure 4-69: 8 V − 0.7 V 7.3 V = (a) IB = = 107 μA 68 kΩ 68 kΩ hFE = 150 IC = 150(107 μA) = 16.1 mA VC = 15 V − (16.1 mA)(680 Ω) = 15 V − 10.95 V = 4.05 V VCE = 4.05 V − 0.7 V = 3.35 V PD = (3.35 V)(16.1 mA) = 53.9 mW At 40°C, PD(max) = 360 mW − (40°C − 25°C)(2.06 mW/°C) = 329 mW No parameters are exceeded. B

5 V − 0.7 V 4.3 V = = 915 μA 4.7 kΩ 4.7 kΩ hFE = 300 IC = 300(915 μA) = 274 mA 35 V − 0.3 V = 73.8 mA IC(sat) ≅ 470 Ω The transistor is in hard saturation. Assuming VCE(sat) = 0.3 V, PD = (0.3 V)(73.8 mA) = 22.1 mW No parameters are exceeded.

(b) IB = B

Advanced Problems α DC 1 − α DC βDC − βDCαDC = αDC βDC = αDC(1 + βDC) β DC αDC = (1 + β DC )

50.

βDC =

51.

IC = 150(500 μA) = 75 mA VCE = 15 V − (180 Ω)(75 mA) = 1.5 V Since VCE(sat) = 0.3 V @ IC = 50 mA, the transistor comes out of saturation.

52.

From the datasheet, βDC(min) = 15 (for IC = 100 mA) 150 mA IB(max) = = 10 mA 15 3 V − 0.7 V 2.3 V = = 230 Ω RB(min) = 10 mA 10 mA Use the standard value of 240 Ω for RB. To avoid saturation, the load resistance cannot exceed about 9 V −1 V = 53.3 Ω 150 mA See Figure 4-3. B

Figure 4-3

33

Chapter 4 53.

Since IB = 10 mA for IC = 150 mA, 9 V − 0.7 V 8.3 V = RB(min) = = 830 Ω 10 mA 10 mA Use 910 Ω. The load cannot exceed 53.3 Ω. See Figure 4-4. B

Figure 4-4 54.

RC(min) = Av re′ = 50(8 Ω) = 400 Ω (Use 430 Ω) 12 V − 5 V = 16.3 mA IC = 430 Ω Assuming hFE = 100, 16.3 mA = 163 μA IB = 100 4 V − 0.7 V = 20.3 kΩ (Use 18 kΩ) RB(max) = 163 μA See Figure 4-5. B

Figure 4-5

Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 55 through 62 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 55.

RB shorted

56.

RC open

57.

Collector-emitter shorted

58.

Collector-emitter open

59.

RE leaky

60.

Collector-emitter shorted

61.

RB open

62.

RC open

B

B

34

Chapter 5 Transistor Bias Circuits Section 5-1 The DC Operating Point 1.

The transistor is biased too close to saturation.

2.

IC = βDCIB = 75(150 μA) = 11.3 mA VCE = VCC − ICRC = 18 V − (11.3 mA)(1.0 kΩ) = 18 V − 11.3 V = 6.75 V Q-point: VCEQ = 6.75 V, ICQ = 11.3 mA

3.

IC(sat) ≅

4.

VCE(cutoff) = 18 V

5.

Horizontal intercept (cutoff): VCE = VCC = 20 V Vertical intercept (saturation): V 20 V IC(sat) = CC = 2 mA RC 10 kΩ

6.

B

VCC 18 V = = 18 mA RC 1.0 kΩ

VBB − 0.7 V RB VBB = IBRB + 0.7 V = (20 μA)(1.0 MΩ) + 0.7 V = 20.7 V IC = βDCIB = 50(20 μA) = 1 mA VCE = VCC − ICRC = 20 V − (1 mA)(10 kΩ) = 10 V IB = B

B

B

B

7.

See Figure 5-1. VCE = VCC − ICRC V − VCE 10 V − 4 V = RC = CC = 1.2 kΩ 5 mA IC I 5 mA = 0.05 mA IB = C = β DC 100 10 V − 0.7 V = 186 kΩ RB = 0.05 mA PD(min) = VCEIC = (4 V)(5 mA) = 20 mW B

B

Figure 5-1

35

Chapter 5 8.

VBB − VBE 1.5 V − 0.7 V = = 80 μA RB 10 kΩ V 8V IC(sat) = CC = = 20.5 mA RC 390 Ω IC = βDCIB = 75(80 μA) = 6 mA The transistor is biased in the linear region because 0 < IC < IC(sat). IB = B

B

9.

(a) IC(sat) = 50 mA (b) VCE(cutoff) = 10 V (c) IB = 250 μA IC = 25 mA VCE = 5 V B

10.

(a) IC ≅ 42 mA (b) Interpolating between IB = 400 μA and IB = 500 μA IB ≅ 450 μA (c) VCE ≅ 1.5 V See Figure 5-2. B

B

B

Figure 5-2

Section 5-2 Voltage-Divider Bias 11.

βDC(min)RE = 10R2 10 R2 47 kΩ βDC(min) = = = 69.1 RE 680 Ω

12.

IC(sat) =

VCC 15 V = = 6.88 mA RC + RE 2.18 kΩ VE(sat) = IC(sat)RE = (6.88 mA)(680 Ω) = 4.68 V VB = VE(sat) + 0.7 V = 4.68 V + 0.7 V = 5.38 V ⎛ R2 β DC RE ⎞ ⎜ ⎟ 15 V = 5.38 V ⎜R +R β R ⎟ 1 2 DC E ⎝ ⎠ B

36

Chapter 5

(R2 (R2 (R2 R2

β DC RE ) (15 V) = (5.38 V) (R1 + R2 β DC RE )

β DC RE ) (15 V) − (R2 β DC RE ) (5.38 V) = R1(5.38 V)

β DC RE ) (15 V − 5.38 V) = (22 kΩ)(5.38 V) (22 kΩ)(5.38 V) β DC RE = = 12.3 kΩ 15 V − 5.38 V 1 1 + = β DC RE 12.3 kΩ 1 1 + = 102 kΩ 12.3 kΩ

1 R2 1 R2 1 = 71.5 μS R2 R2 = 14 kΩ 13.

14.

⎛ R2 ⎞ 2 kΩ ⎞ ⎟⎟VCC = ⎛⎜ VB = ⎜⎜ ⎟ 15 V = 1.25 V ⎝ 24 kΩ ⎠ ⎝ R1 + R2 ⎠ VE = 1.25 V − 0.7 V = 0.55 V V 0.55 V IE = E = = 809 μA RE 680 Ω IC ≅ 809 μA VCE = VCC − ICRC − VE = 15 V − (809 μA)(1.5 kΩ + 680 Ω) = 13.2 V B

⎛ R2 β DC RE ⎞ ⎛ ⎞ 15 kΩ (110)(1.0 kΩ) ⎟ VCC = ⎜ ⎟9 V = 1.97 V VB = ⎜ ⎜R +R β R ⎟ ⎜ 47 kΩ + 15 kΩ (110)(1.0 kΩ) ⎟ 2 DC E ⎠ ⎝ 1 ⎝ ⎠ VE = VB − 0.7 V = 1.97 V − 0.7 V = 1.27 V V 1.27 V IC ≅ IE = E = = 1.27 mA RE 1.0 kΩ VC = VCC − ICRC = 9 V − (1.27 mA)(2.2 kΩ) = 6.21 V B

B

15.

See Figure 5-3.

Figure 5-3

16.

(a) RIN(base) = βDCRE = 150(560 Ω) = 84 kΩ ⎛ ⎞ 5.6 kΩ 84 kΩ ⎛ 5.25 kΩ ⎞ (−12 V) = ⎜ VB = ⎜ ⎟ (−12 V) = −1.65 V ⎜ 33 kΩ + 5.6 kΩ 84 kΩ ⎟⎟ ⎝ 38.25 kΩ ⎠ ⎝ ⎠ (b) RIN(base) = 50(560 Ω) = 28 kΩ ⎛ ⎞ 5.6 kΩ 28 kΩ ⎛ 4.67 kΩ ⎞ (−12 V) = ⎜ VB = ⎜ ⎟ (−12 V) = −1.49 V ⎜ 33 kΩ + 56 kΩ 28 kΩ ⎟⎟ ⎝ 37.7 kΩ ⎠ ⎝ ⎠ B

B

37

Chapter 5 17.

(a) VEQ = VB + 0.7 V = −1.49 V + 0.7 V = −0.79 V V −0.79 V ICQ ≅ IE = E = = −1.41 mA RE 560 Ω VCQ = VCC − ICRC = −12 V − (−1.41 mA)(1.8 kΩ) = −9.46 V VCEQ = VCQ − VEQ = −9.46 V − (−0.79 V) = −8.67 V (b) PD(min) = ICQVCEQ = (−1.41 mA)(−8.67 V) = 12.2 mW

18.

VB = −1.65 V

B

I1 =

VCC − VB −12 V − (1.65 V) = 314 μA = R1 33 kΩ

VB −1.65 V = 295 μA = R2 5.6 kΩ I B = I1 − I2 = 19 μA

I2 =

Section 5-3 Other Bias Methods 19.

Using Equation 5-7: −VEE − VBE 4.3 V −(−5 V) − 0.7 V = = IE = = 1.86 mA RE + RB / β DC 2.2 kΩ + 10 kΩ /100 2.2 kΩ + 0.1 kΩ IC ≅ IE = 1.86 mA I 1.86 mA IB = C ≅ = 18.6 μA 100 β VB = −IBRB = (18.6 μA)(10 kΩ) = −0.186 V VE = VB − 0.7 V = −0.186 − 0.7 V = −0.886 V VC = VCC − ICRC = 5 V − (1.86 mA)(1.0 kΩ) = 3.14 V B

B

B

B

B

20.

Assume VCE ≅ 0 V at saturation. VE = −0.886 V so VC(sat) = −0.886 VCC − VC(sat) 5 V − (−0.886 V) IC(sat) = = 5.89 mA = RC 1.0 kΩ V 4.11 V = 698 Ω RE(min) = RE = I C(sat) 5.89 mA

21.

At 100°C: VBE = 0.7 V − (2.5 mV/°C)(75°C) = 0.513 V −VEE − VBE 4.49 V −(−5 V) − 0.513 V = = IE = = 1.95 mA RE + RB / β DC 2.2 kΩ + 10 kΩ /100 2.3 kΩ At 25°C: I E = 1.86 mA (from problem 19) ΔI E = 1.95 mA − 1.86 mA = 0.09 mA

38

Chapter 5 22.

A change in βDC does not affect the circuit when RE >>RB/βDC. Since VEE − VBE IE = RE + RB / β DC In the equation, if RB/βDC is much smaller than RE, the effect of βDC is negligible. B

B

23.

Assume βDC = 100. VEE − VE 10 V − 0.7 V = = 16.3 mA IC ≅ IE = RE + RB / β 470 Ω + 10 kΩ /100 VCE = VEE − VCC − IC(RC + RE) = 20 V − 13.1 V = −6.95 V

24.

VB = 0.7 V VCC − VBE 3 V − 0.7 V = IC = = 1.06 mA RC + RB / β DC 1.8 kΩ + 33 kΩ / 90 VC = VCC − ICRC = 3 V − (1.06 mA)(1.8 kΩ) = 1.09 V

25.

IC = 1.06 mA from Problem 24. IC = 1.06 mA − (0.25)(1.06 mA) = 0.795 mA VCC − VBE IC = RC + RB / β DC V − VBE − I C RB / β DC 3 V − 0.7 V − (0.795 mA)(33 kΩ) / 90 = = 2.53 kΩ RC = CC IC 0.795 mA

26.

IC = 0.795 mA from Problem 25. VCE = VCC − ICRC = 3 V − (0.795 mA)(2.53 kΩ) = 0.989 V PD(min) = VCEIC = (0.989 V)(0.795 mA) = 786 μW

27.

See Figure 5-4. VCC − VBE 12 V − 0.7 V = = 7.87 mA IC = RC + RB / β DC 1.2 kΩ + 47 kΩ / 200 VC = VCC − ICRC = 12 V − (7.87 mA)(1.2 kΩ) = 2.56 V

B

Figure 5-4

28.

VBB = VCC; VE = 0 V V − 0.7 V 12 V − 0.7 V 11.3 V = = IB = CC = 514 μA RB 22 kΩ 22 kΩ IC = βDCIB = 90(514 μA) = 46.3 mA VCE = VCC − ICRC = 12 V − (46.3 mA)(100 Ω) = 7.37 V B

B

39

Chapter 5 29.

ICQ = 180(514 μA) = 92.5 mA VCEQ = 12 V − (92.5 mA)(100 Ω) = 2.75 V

30.

IC changes in the circuit with a common VCC and VBB supply because a change in VCC causes IB to change which, in turn, changes IC. B

31.

VBB − VBE 9 V − 0.7 V = = 553 μA RB 15 kΩ V 9V = 90 mA IC(sat) = CC = RC 100 Ω For βDC = 50: IC = βDCIB = 50(553 μA) = 27.7 mA VCE = VCC − ICRC = 9 V − (27.7 mA)(100 Ω) = 6.23 V For βDC = 125: IC = βDCIB = 125(553 μA) = 69.2 mA VCE = VCC − ICRC = 9 V − (69.2 mA)(100 Ω) = 2.08 V Since IC < IC(sat) for the range of βDC, the circuit remains biased in the linear region. IB = B

B

B

32.

IC(sat) =

VCC 9V = = 90 mA RC 100 Ω

At 0°C:

βDC = 110 − 110(0.5) = 55 V − VBE 9 V − 0.7 V = IB = CC = 553 μA RB 15 kΩ IC = βDCIB = 55(553 μA) = 30.4 mA B

B

VCE = VCC − ICRC = 9 V − (30.4 mA)(100 Ω) = 5.96 V At 70°C: βDC = 110 + 110(0.75) = 193 IB = 553 μA IC = βDCIB = 193(553 μA) = 107 mA IC > IC(sat), therefore the transistor is in saturation at 70°C. ΔIC = IC(sat) − IC(0°) = 90 mA − 30.4 mA = 59.6 mA ΔVCE ≅ VCE(0°) − VCE(sat) = 5.96 V − 0 V = 5.96 V B

B

Section 5-4 Troubleshooting 33.

The transistor is off; therefore, V1 = 0 V, V2 = 0 V, V3 = 8 V.

34.

V1 = 0.7 V, V2 = 0 V 8 V − 0.7 V 0.7 V − IB = = 221 μA − 70 μA = 151 μA 33 kΩ 10 kΩ IC = 200(151 μA) = 30.2 mA 8V = 3.64 mA, so VC ≅ VE = 0 V IC(sat) = 2.2 kΩ B

40

Chapter 5 If the problem is corrected, 10 kΩ ⎛ ⎞ V1 = ⎜ ⎟ 8 V = 1.86 V ⎝ 10 kΩ + 33 kΩ ⎠ V2 = VE = 1.86 V − 0.7 V = 1.16 V 1.16 V = 1.16 mA IE = 1.0 kΩ V3 = VC = 8 V − (1.16 mA)(2.2 kΩ) = 5.45 V 35.

(a) (b) (c) (d)

Open collector No problems Transistor shorted from collector-to-emitter Open emitter

36.

For βDC = 35: ⎛ 4.5 kΩ ⎞ VB = ⎜ ⎟(−10 V) = −3.1 V ⎝ 14.5 kΩ ⎠ For βDC = 100: ⎛ 5.17 kΩ ⎞ VB = ⎜ ⎟(−10 V) = −3.4 V ⎝ 15.17 kΩ ⎠ The measured base voltage at point 4 is within the correct range. VE = −3.1 V + 0.7 V = −2.4 V − 2.4 V = −3.53 mA IC ≅ IE = 680 Ω VC = −10 V − (−3.53 mA)(1.0 kΩ) = −6.47 V Allowing for some variation in VBE and for resistor tolerances, the measured collector and emitter voltages are correct. B

B

37.

(a) The 680 Ω resistor is open: Meter 1: 10 V Meter 2: floating ⎛ 5.6 kΩ ⎞ Meter 3: VB = ⎜ ⎟(−10 V) = −3.59 V ⎝ 15.6 kΩ ⎠ Meter 4: 10 V B

(b) The 5.6 kΩ resistor is open. 9.3 V IB = = 275 μA 10 kΩ + 35(680 Ω) IC = 35(275 μA) = 9.6 mA 10 V IC(sat) = = 5.95 mA 1680 Ω The transistor is saturated. Meter 1: 10 V Meter 2: (5.95 mA)(680 Ω) = 4.05 V Meter 3: 4.05 V + 0.7 V = 4.75 V Meter 4: 10 V − (5.95 mA)(1.0 kΩ) = 4.05 V B

41

Chapter 5 (c) The 10 kΩ resistor is open. The transistor is off. Meter 1: 10 V Meter 2: 0 V Meter 3: 0 V Meter 4: 10 V (d) The 1.0 kΩ resistor is open. Collector current is zero. Meter 1: 10 V Meter 2: 1.27 V − 0.7 V = 0.57 V ⎛ ⎞ 5.6 kΩ 680 Ω ⎟ (10 V) + 0.7 V = 0.57 V + 0.7 V = 1.27 V Meter 3: ⎜ ⎜ 10 kΩ + 5.6 kΩ 680 Ω ⎟ ⎝ ⎠ Meter 4: floating (e) A short from emitter to ground. Meter 1: 10 V Meter 2: 0 V Meter 3: 0.7 V (10 V − 0.7 V) 9.3 V = IB ≅ = 0.93 mA 10 kΩ 10 kΩ IC(min) = 35(0.93 mA) = 32.6 mA 10 V = 10 mA IC(sat) = 1.0 kΩ The transistor is saturated. Meter 4: ≅ 0 V B

(f) An open base-emitter junction. The transistor is off. Meter 1: 10 V Meter 2: 0 V ⎛ 5.6 kΩ ⎞ Meter 3: ⎜ ⎟ (10 V) = 3.59 V ⎝ 15.6 kΩ ⎠ Meter 4: 10 V

42

Chapter 5 Application Activity Problems 38.

With R1 open: VB = 0 V, VE = 0 V, VC = VCC = 9.1 V B

39.

Faults that will cause the transistor of textbook Figure 5-30(a) to go into cutoff: R1 open, R2 shorted, base lead or BE junction open.

40.

At 45°C: RTherm = 2.7 kΩ ⎛ RTherm ⎞ ⎛ 2.7 kΩ ⎞ VB ⎜ 9 V=⎜ ⎟ 9 V = 3.28 V ⎜ R1 + RTherm ⎟⎟ ⎝ 7.4 kΩ ⎠ ⎝ ⎠ VE = VB − 0.7 V = 2.58 V V 2.58 V = 5.49 mA IE = IC = E = R3 470 Ω VC = VOUT = 9 V − (5.49 mA)(1 kΩ) = 3.51 V B

At 48°C: RTherm = 1.78 kΩ ⎛ 1.78 kΩ ⎞ VB = ⎜ ⎟ 9 V = 2.47 V ⎝ 6.48 kΩ ⎠ VE = 2.47 V − 0.7 V = 1.77 V 1.77 V = 3.77 mA IE = IC = 470 Ω VC = VOUT = 9 V − (3.77 mA)(1 kΩ) = 5.23 V B

At 53°C: RTherm = 1.28 kΩ ⎛ 1.28 kΩ ⎞ VB = ⎜ ⎟ 9 V = 1.93 V ⎝ 5.98 kΩ ⎠ VE = 1.93 V − 0.7 V = 1.23 V 1.23 V = 2.62 mA IE = IC = 470 Ω VC = VOUT = 9 V − (2.62 mA)(1 kΩ) = 6.38 V B

41.

The following measurements would indicate an open CB junction: VC = VCC = +9.1 V VB normal VE ≅ 0 V B

Datasheet Problems 42.

For T = 45°C and R2 = 2.7 kΩ RIN(base) = 2.7 kΩ (30)(470 Ω) = 2.7 kΩ 14.1 kΩ = 2.27 kΩ min RIN(base) = 2.7 kΩ (300)(470 Ω) = 2.7 kΩ 141 kΩ = 2.65 kΩ max

2.27 kΩ ⎛ ⎞ ⎛ 2.27 kΩ ⎞ VB(min) = ⎜ ⎟9.1 V = ⎜ ⎟9.1 V = 2.62 V ⎝ 2.27 kΩ + 5.6 kΩ ⎠ ⎝ 7.87 ⎠ VE(min) = 2.62 V − 0.7 V = 1.92 V

43

Chapter 5 1.92 V = 4.09 mA 470 Ω VC(max) = 9.1 V − (4.09 mA)(1.0 kΩ) = 5.01 V 2.65 kΩ ⎛ ⎞ ⎛ 2.65 kΩ ⎞ VB(max) = ⎜ ⎟9.1 V = ⎜ ⎟9.1 V = 2.92 V ⎝ 2.65 kΩ + 5.6 kΩ ⎠ ⎝ 8.25 kΩ ⎠ VE(max) = 2.92 V − 0.7 V = 2.22 V 2.22 V = 4.73 mA So, IC ≅ IE = 470 Ω VC(min) = 9.1 V − (4.73 mA)(1.0 kΩ) = 4.37 V

So, IC ≅ IE =

For T = 55°C and R2 = 1.24 kΩ: RIN(base) = 1.24 kΩ (30)(470 Ω) = 1.24 kΩ 14.1 kΩ = 1.14 kΩ min RIN(base) = 1.24 kΩ (300)(470 Ω) = 1.24 kΩ 141 kΩ = 1.23 kΩ max

1.14 kΩ ⎛ ⎞ ⎛ 1.14 kΩ ⎞ VB(min) = ⎜ ⎟9.1 V = ⎜ ⎟9.1 V = 1.54 V ⎝ 1.14 kΩ + 5.6 kΩ ⎠ ⎝ 6.74 kΩ ⎠ VE(min) = 1.54 V − 0.7 V = 0.839 V 0.839 V = 1.78 mA So, IC ≅ IE = 470 Ω VC(max) = 9.1 V − (1.78 mA)(1.0 kΩ) = 7.32 V 1.23 kΩ ⎛ ⎞ ⎛ 1.23 kΩ ⎞ VB(max) = ⎜ ⎟9.1 V = ⎜ ⎟9.1 V = 1.64 V ⎝ 1.23 kΩ + 5.6 kΩ ⎠ ⎝ 6.83 kΩ ⎠ VE(max) = 1.64 V − 0.7 V = 0.938 V 0.938 V = 2.0 mA So, IC ≅ IE = 470 Ω VC(min) = 9.1 V − (2.0 mA)(1.0 kΩ) = 7.10 V 43.

At T = 45°C for minimum βDC: PD(max) = (5.01 V − 1.92 V)(4.09 mA) = (3.09 V)(4.09 mA) = 12.6 mW At T = 55°C for minimum βDC: PD(max) = (7.32 V − 0.839 V)(1.78 mA) = (6.48 V)(1.78 mA) = 11.5 mW For maximum beta values, the results are comparable and nowhere near the maximum. PD(max) = 625 mW − (5.0 m/°C)(30°C) = 475 mW No ratings are exceeded.

44.

For the datasheet of Figure 5-50 in the textbook: (a) For a 2N2222A, IC(max) = 1 A continuous (b) For a 2N2118A, VEB(max) = 6.0 V

45.

For a 2N2222A @ T = 100°C: PD(max) = 0.8 W − (4.57 mW/°C)(100°C − 25°C) = 0.8 W − 343 mW = 457 mW

46.

If IC changes from 1 mA to 500 mA in a 2N2219A, the percentage change in βDC is ⎛ 30 − 50 ⎞ ΔβDC = ⎜ ⎟100% = −40% ⎝ 50 ⎠

44

Chapter 5 Advanced Problems 47.

See Figure 5-5. V − VCEQ 15 V − 5 V RC = CC = 2 kΩ = 5 mA I CQ Assume βDC = 100. I 5 mA = 50 μA IBQ = CQ = β DC 100 V − VBE 15 V − 0.7 V = RB = CC = 286 kΩ I BQ 50 μA B

48.

Figure 5-5

See Figure 5-6. Assume βDC = 200. I 10 mA IBQ = CQ = = 50 μA β DC 200 Let RB = 1.0 kΩ 12 V − (50 μA)(1.0 kΩ) − 0.7 V 11.3 V = RE = = 1.13 kΩ 10 mA 10 mA 12 V − (−12 V + 11.3 V + 4 V) 8.7 V = = 870 Ω RC = 10 mA 10 mA 870 Ω and 1.13 kΩ are not standard values. RC = 820 Ω and RE = 1.2 kΩ give ICQ ≅ 9.38 mA, VCEQ ≅ 5.05 V. B

Figure 5-6 49.

See Figure 5-7. βDC(min) ≅ 70. Let RE = 1.0 kΩ. VE = IERE = 1.5 mA(1.0 kΩ) = 1.5 V VB = 1.5 V + 0.7 V = 2.2 V V − VCEQ − VE 9 V − 1.5 V − 3 V RC = CC = 3 kΩ = 1.5 mA I CQ B

R1 + R2 =

VCC 9V = 2.57 kΩ min = I CC(max) − I CQ 5 mA − 1.5 mA

Asssume βDCRE>>R2. The ratio of bias resistors equals the ratio of the voltages as follows. R1 6.8 V = = 3.09 R2 2.2 V R1 = 3.09R2 R1 + R2 = R2 + 3.09R2 = 2.57 kΩ 4.09R2 = 2.57 kΩ 2.57 kΩ R2 = = 628 Ω 4.09 Figure 5-7

45

Chapter 5 So, R2 ≅ 620 Ω and R1 = 1.92 kΩ ≅ 2 kΩ. From this, RIN(base) = 70(1.0 kΩ) = 70 kΩ >> R2 ⎛ 620 Ω ⎞ so, VB = ⎜ ⎟ 9 V = 2.13 V ⎝ 2.62 kΩ ⎠ VE = 2.13 V − 0.7 V = 1.43 V 1.43 V ICQ ≅ IE = = 1.43 mA 1.0 kΩ VCEQ = 9 V − (1.43 mA)(1.0 kΩ + 3 kΩ) = 3.28 V B

50.

See Figure 5-8. βDC ≅ 75. 10 mA = 133 μA IBQ = 75 V − VCE 5 V − 1.5 V = = 350 Ω (use 360 Ω) RC = CC I CQ 10 mA RB = B

VCE − 0.7 V 1.55 V − 0.7 V = = 6 kΩ (use 6.2 kΩ) I BQ 133 μA

5 V − 0.7 V = 9.71 mA 360 Ω + 6.2 kΩ / 75 VCEQ = VC = 5 V − (9.71 mA)(360 Ω) = 1.50 V

ICQ =

Figure 5-8

51.

The 2N3904 in textbook Figure 5-48 can be replaced with a 2N2222A and maintain the same voltage range from 45°C to 55°C because the voltage-divider circuit is essentially β independent and the βDC parameters of the two transistors are comparable.

52.

For the 2N2222A using the datasheet in textbook Figure 5-51 at IC = 150 mA and VCE = 1.0 V: At T = −55°C, hFE(min) = (0.45)(50) = 22.5 At T = 25°C, hFE(min) = (0.63)(50) = 31.5 At T = 175°C, hFE(min) = (0.53)(50) = 26.6

53.

If the valve interface circuit loading of the temperature conversion circuit changes from 100 kΩ to 10 kΩ, the Q-point will have a reduced VCEQ because the current through RC will consist of the same IC and a larger IL. ICQ is unaffected in the sense that the transistor collector current is the same, although the collector resistance current is larger. The transistor saturates sooner so that lower temperatures do not register as well, if at all.

54.

It is not feasible to operate the circuit from a 5.1 V dc supply and maintain the same range of output voltages because the output voltage at 60°C must be 6.478 V.

46

Chapter 5 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 55 through 60 are available from the Instructor’s Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 55.

RC open

56.

RB open

57.

R2 open

58.

Collector-emitter shorted

59.

RC shorted

60.

Base-emitter open

B

47

Chapter 6 BJT Amplifiers Section 6-1 Amplifier Operation 1.

Slightly greater than 1 mA minimum

2.

From the graph of Figure 6-4, the highest value of dc collector current is about 6 mA.

Section 6-2 Transistor AC Models 25 mV 25 mV = = 8.33 Ω IE 3 mA

3.

re′ =

4.

βac = hfe = 200

5.

IC = βDCIB = 130(10 μA) = 1.3 mA 1.3 mA I IE = C = = 1.31 mA α DC 0.99 25 mV 25 mV re′ = = = 19 Ω 1.31 mA IE

6.

βDC =

B

I C 2 mA = = 133 I B 15 μA 0.35 mA ΔI βac = C = = 117 3 μA ΔI B

Section 6-3 The Common-Emitter Amplifier 7.

See Figure 6-1.

Figure 6-1

48

Chapter 6 8.

⎛ R2 ⎞ ⎛ 4.7 kΩ ⎞ (a) VB = ⎜ ⎟15 V = 2.64 V ⎟VCC = ⎜ ⎝ 26.7 kΩ ⎠ ⎝ R1 + R2 ⎠ (b) VE = VB − 0.7 V = 2.64 − 0.7 V = 1.94 V V 1.94 V (c) IE = E = = 1.94 mA RE 1.0 kΩ (d) IC ≅ IE = 1.94 mA (e) VC = VCC − ICRC = 15 V − (1.94 mA)(2.2 kΩ) = 11.6 V B

B

9.

10.

ICC = IBIAS + IC V 2.64 V IBIAS = B = = 562 μA R2 4.7 kΩ ICC = 562 μA + 1.94 mA = 2.50 mA P = ICCVCC = (2.5 mA)(15 V) = 37.5 mW 4.7 kΩ ⎛ ⎞ (a) VB = ⎜ ⎟15 V = 2.64 V ⎝ 4.7 kΩ + 22 kΩ ⎠ VE = 2.64 V − 0.7 V = 1.94 V 1.94 V = 1.94 mA IE = 1.0 kΩ 25 mV 25 mV re′ ≅ = = 12.9 Ω IE 1.94 mA Rin(base) = β ac (re′ + RE ) = 100(1012.9 Ω) ≅ 101 kΩ B

(b) Rin = Rin(base) (c) Av = 11.

R1

R2 = 101 kΩ 22 kΩ 4.7 kΩ = 3.73 kΩ

RC 2.2 kΩ = = 2.17 ′ RE + re 12.02 Ω

(a) Rin(base) = β ac re′ = 100(12.9 Ω) = 1.29 kΩ (b) Rin = 1.29 kΩ 22 kΩ 4.7 kΩ = 968 Ω (c) Av =

12.

RC 2.2 kΩ = = 171 re′ 12.9 Ω

(a) Rin(base) = β ac re′ = 100(12.9 Ω) = 1.29 kΩ (b) Rin = 1.29 kΩ 22 kΩ 4.7 kΩ = 968 Ω (c) Av =

Rc RC RL 2.2 kΩ 10 kΩ = = = 140 re′ re′ 12.9 Ω

49

Chapter 6 13.

⎛ R2 β DC RE ⎞ ⎛ ⎞ 12 kΩ 75(1.0 kΩ) ⎟V = ⎜ ⎟18 V = 3.25 V (a) VB = ⎜ ⎜ R + R β R ⎟ CC ⎜ 47 kΩ + 12 kΩ 75(1.0 kΩ) ⎟ 2 DC E ⎠ ⎝ 1 ⎝ ⎠ (b) VE = VB − 0.7 V = 2.55 V V 2.55 V (c) IE = E = = 2.55 mA RE 1.0 kΩ (d) IC ≅ IE = 2.55 mA (e) VC = VCC − ICRC = 18 V − (2.55 mA)(3.3 kΩ) = 9.59 V (f) VCE = VC − VE = 9.59 V − 2.55 V = 7.04 V B

B

14.

From Problem 13, IE = 2.55 mA ⎛ 25 mV ⎞ ⎛ 25 mV ⎞ ⎟⎟ = 70⎜ (a) Rin(base) = β ac re′ ≅ β ac ⎜⎜ ⎟ = 686 Ω ⎝ 2.55 mA ⎠ ⎝ IE ⎠ (b) Rin = R1 R2 Rin (base ) = 47 kΩ 12 kΩ 686 Ω = 640 Ω

(c) Av =

RC RL 3.3 kΩ 10 kΩ = = 253 re′ 9.8 Ω

(d) Ai = βac = 70 (e) Ap = AvAi = (253)(70) = 17,710 15.

⎛ Rin ⎞ ⎛ ⎞ 640 Ω ⎟Vin = ⎜⎜ ⎟⎟12 μV Vb = ⎜⎜ ⎟ ⎝ 640 Ω + 600 Ω ⎠ ⎝ Rin + Rs ⎠ Attenuation of the input network is ⎛ Rin ⎞ ⎛ ⎞ 640 Ω ⎜ ⎟ ⎜ R + R ⎟ = ⎜⎜ 640 Ω + 600 Ω ⎟⎟ = 0.516 ⎝ ⎠ s ⎠ ⎝ in Av′ = 0.516 Av = 0.516(253) = 131 θ = 180°

16.

⎛ R2 β DC RE ⎞ ⎛ ⎞ 3.3 kΩ 150(100 Ω) ⎟V = ⎜ ⎟ 8 V = 1.47 V VB = ⎜ ⎜ R + R β R ⎟ CC ⎜ 12 kΩ + 3.3 kΩ 150(100 Ω) ⎟ 2 DC E ⎠ ⎝ 1 ⎝ ⎠ V − 0.7 V 1.47 V − 0.7 V = = 7.7 mA IE = B RE 100 Ω 25 mV 25 mV re′ = = = 3.25 Ω IE 7.7 mA 330 Ω RC = = 3.2 Av(min) = ′ RE + re 100 Ω + 3.25 Ω R 330 Ω = 102 Av(max) = C = ′ re 3.25 Ω B

50

Chapter 6 17.

Maximum gain is at Re = 0 Ω. RIN(base) = βDCRE = 150(100 Ω) = 15 kΩ ⎛ R2 RIN(base) ⎞ ⎞ 3.3 kΩ 15 kΩ) ⎟V = ⎛⎜ ⎟ 8 V = 1.47 V VB = ⎜ CC ⎜ 12 kΩ + 3.3 kΩ 15 kΩ) ⎟ ⎜R +R R ⎟ 1 2 IN(base) ⎝ ⎠ ⎝ ⎠ V − VBE 1.47 V − 0.7 V = = 7.7 mA IE = B RE 100 Ω 25 mV re′ ≅ = 3.25 Ω 7.7 mA R RL 330 Ω 600 Ω Av(max) = C = 65.5 = re′ 3.25 Ω B

Minimum gain is at Re = 100 Ω. R RL 212.9 Ω = = 2.06 Av(min) = C RE + re′ 103.25 Ω 18.

Rin = R1 R2 β ac re′ = 3.3 kΩ 12 kΩ 150(3.25 Ω) = 410 Ω Attenuation of the input network is Rin 410 Ω = = 0.578 Rin + Rs 410 Ω + 300 Ω Rc 330 Ω 1.0 kΩ = 76.3 = re′ 3.25 Ω Av′ = 0.5777 Av = 0.578(76.3) = 44.1

Av =

19.

See Figure 6-2. 25 mV re′ ≅ = 9.8 Ω 2.55 mA Re ≥ 10re′ Set Re = 100 Ω. The gain is reduced to RC 3.3 kΩ = Av = = 30.1 Re + re′ 109.8 Ω Figure 6-2

51

Chapter 6 Section 6-4 The Common-Collector Amplifier 20.

21.

⎛ R2 ⎞ 4.7 kΩ ⎞ ⎟⎟VCC = ⎛⎜ VB = ⎜⎜ ⎟ 5.5 V = 1.76 V ⎝ 14.7 kΩ ⎠ ⎝ R1 + R2 ⎠ V − 0.7 V 1.76 V − 0.7 V = IE = B = 1.06 mA RE 1.0 kΩ 25 mV re′ ≅ = 23.6 Ω 1.06 mA 1.0 kΩ RE = Av = = 0.977 RE + re′ 1.0 kΩ + 23.6 Ω B

Rin = R1 R2 β ac (re′ + RE ) ≅ R1 R2 β ac RE = 10 kΩ 4.7 kΩ 100 kΩ = 3.1 kΩ

⎛ R2 ⎞ ⎛ 4.7 kΩ ⎞ ⎟⎟VCC − 0.7 V = ⎜ VOUT = VB − 0.7 V = ⎜⎜ ⎟ 5.5 V − 0.7 V = 1.06 V ⎝ 14.7 kΩ ⎠ ⎝ R1 + R2 ⎠ B

22.

23.

The voltage gain is reduced because Av =

Re . Re + re′

⎛ R2 ⎞ ⎛ 4.7 kΩ ⎞ ⎟⎟VCC = ⎜ VB = ⎜⎜ ⎟ 5.5 V = 1.76 V ⎝ 14.7 kΩ ⎠ ⎝ R1 + R2 ⎠ V − VBE 1.76 V − 0.7 V = IE = B = 1.06 mA RE 1.0 kΩ 25 mV 25 mV = re′ ≅ = 23.6 Ω IE 1.06 mA B

Av =

RE re′ + RE

Av (re′ + RE RE

(RE (RE

RL RL

RL ) = RE

RL − Av (RE

RL

RL ) = Av re′

RL )(1 − Av ) = Av re′ Av re′ 0.9(23.6 Ω) RL ) = = = 212.4 Ω (1 − Av ) 1 − 0.9 RLRE = 212.4RL + 212.4RE RLRE − 212.4RL = 212.4RE 212.4 RE (212.4 Ω)(1000 Ω) RL = = 270 Ω = RE − 212.4 1000 Ω − 212.4 Ω

52

Chapter 6 24.

(a) VC1 = 10 V ⎛ R2 ⎞ ⎛ 22 kΩ ⎞ ⎟⎟VCC = ⎜ VB1 = ⎜⎜ ⎟ 10 V = 4 V ⎝ 55 kΩ ⎠ ⎝ R1 + R2 ⎠ VE1 = VB1 − 0.7 V = 4 V − 0.7 V = 3.3 V VC2 = 10 V VB2 = VE1 = 3.3 V VE2 = VB2 − 0.7 V = 3.3 V − 0.7 V = 2.6 V

′ = βDC1βDC2 = (150)(100) = 15,000 (b) β DC VE1 − 0.7 V 2.6 V = = 17.3 μA 100(1.5 kΩ) β DC2 RE 25 mV 25 mV re′1 ≅ = = 1.45 kΩ I E1 17.3 μ A V 2.6 V IE2 = E2 = = 1.73 mA RE 1.5 kΩ 25 mV 25 mV re′2 ≅ = = 14.5 Ω I E2 1.73 mV

(c) IE1 =

(d) Rin = R1 R2 Rin (base1) Rin(base1) = βac1βac2RE = (150)(100)(1.5 kΩ) = 22.5 MΩ Rin = 33 kΩ 22 kΩ 22.5 MΩ = 13.2 kΩ 25.

Rin(base) = βac1βac2RE = (150)(100)(1.5 kΩ) = 22.5 MΩ Rin = R2 R1 Rin (base ) = 22 kΩ 33 kΩ 22.5 MΩ = 13.2 kΩ Vin 1V = = 75.8 μA Rin 13.2 kΩ Vin 1V = Iin(base1) = = 44.4 nA Rin (base1) 22.5 MΩ Iin =

Ie ≅ βac1βac2Iin(base1) = (150)(100)(44.4 nA) = 667 μA I 667 μA Ai′ = e = = 8.8 I in 75.8 μA

Section 6-5 The Common-Base Amplifier 26.

The main disadvantage of a common-base amplifier is low input impedance. Another disadvantage is unity current gain.

53

Chapter 6 27.

28.

⎛ R2 ⎞ ⎛ 10 kΩ ⎞ ⎟⎟ VCC − VBE = ⎜ VE = ⎜⎜ ⎟ 24 V − 0.7 V = 6.8 V ⎝ 32 kΩ ⎠ ⎝ R1 + R2 ⎠ 6.8 V = 10.97 mA IE = 620 Ω 25 mV 25 mA = Rin(emitter) = re′ ≅ = 2.28 Ω IE 10.97 mA R 1.2 kΩ Av = C = = 526 re′ 2.28 Ω Ai ≅ 1 Ap = AiAv ≅ 526 (a) Common-base

(b) Common-emitter

(c) Common-collector

Section 6-6 Multistage Amplifiers 29.

Av′ = Av1Av2 = (20)(20) = 400

30.

Av′ ( dB) = 10 dB + 10 dB + 10 dB = 30 dB

20 log Av′ = 30 dB 30 = 1.5 log Av′ = 20 Av′ = 31.6 31.

⎛ R2 ⎞ 8.2 kΩ ⎛ ⎞ ⎟⎟ VCC − VBE = ⎜ (a) VE ⎜⎜ ⎟ 15 V − 0.7 V = 2.29 V ⎝ 33 kΩ + 8.2 kΩ ⎠ ⎝ R1 + R2 ⎠ V 2.29 V IE = E = = 2.29 mA RE 1.0 kΩ 25 mV 25 mV re′ ≅ = = 10.9 Ω IE 2.29 mA Rin(2) = R6 R5 β ac re′ = 8.2 kΩ 33 kΩ 175(10.9 Ω) = 1.48 kΩ Av1 =

RC Rin ( 2 )

=

3.3 kΩ 1.48 kΩ

re′ 10.9 Ω R 3.3 kΩ Av2 = C = = 303 re′ 10.9 Ω

= 93.6

(b) Av′ = Av1Av2 = (93.6)(303) = 28,361 (c) Av1(dB) = 20 log(93.6) = 39.4 dB Av2(dB) = 20 log(303) = 49.6 dB Av′ ( dB) = 20 log(28,361) = 89.1 dB

54

Chapter 6 32.

(a) Av1 = Av2 =

RC Rin ( 2 ) re′

=

3.3 kΩ 1.48 kΩ 10.9 Ω

= 93.6

RC R L 3.3 kΩ 18 kΩ = 256 = re′ 10.9 Ω

(b) Rin(1) = R1 R2 β ac re′ = 33 kΩ 8.2 kΩ 175(10.9 Ω) = 1.48 kΩ Attenuation of the input network is Rin (1) 1.48 kΩ = 0.95 = Rin (1) + R s 1.48 kΩ + 75 Ω

Av′ = (0.95)Av1Av2 = (0.95)(93.6)(256) = 22,764 (c) Av1(dB) = 20 log(93.6) = 39.4 dB Av2(dB) = 20 log(256) = 48.2 dB Av′ ( dB) = 20 log(22,764) = 87.1 dB

33.

⎛ R2 ⎞ ⎛ 22 kΩ ⎞ ⎟⎟ VCC = ⎜ VB1 = ⎜⎜ ⎟12 V = 2.16 V ⎝ 122 kΩ ⎠ ⎝ R1 + R2 ⎠ VE1 = VB1 − 0.7 V = 1.46 V V 1.46 V = 0.311 mA IC1 ≅ IE1 = E1 = R4 4.7 kΩ VC1 = VCC − IC1R3 = 12 V − (0.311 mA)(22 kΩ) = 5.16 V VB2 = VC1 = 5.16 V VE2 = VB2 − 0.7 V = 5.16 V − 0.7 V = 4.46 V V 4.46 V = 0.446 mA IC2 ≅ IE2 = E2 = R6 10 kΩ VC2 = VCC − IC2R5 = 12 V − (0.446 mA)(10 kΩ) = 7.54 V 25 mV 25 mV = re′2 ≅ = 56 Ω 0.446 mA IE2 Rin(2) = β ac re′2 = (125)(56 Ω) = 7 kΩ 25 mV 25 mV = re′1 ≅ = 80.4 Ω 0.311 mA I E1 R3 Rin (2)

=

22 kΩ 7 kΩ

= 66 80.4 Ω re′1 R 10 kΩ Av2 = 5 = = 179 ′ re 2 56 Ω Av′ = Av1Av2 = (66)(179) = 11,814

Av1 =

34.

(a) (b) (c) (d)

20 log(12) = 21.6 dB 20 log(50) = 34.0 dB 20 log(100) = 40.0 dB 20 log(2500) = 68.0 dB

55

Chapter 6 35.

⎛V ⎞ (a) 20 log ⎜⎜ 2 ⎟⎟ = 3 dB ⎝ V1 ⎠ ⎛V ⎞ 3 = 0.15 log ⎜⎜ 2 ⎟⎟ = ⎝ V1 ⎠ 20

⎛V ⎞ (b) 20 log ⎜⎜ 2 ⎟⎟ = 6 dB ⎝ V1 ⎠ ⎛V ⎞ 6 log ⎜ 2 ⎟ = = 0.3 ⎝ V1 ⎠ 20

V2 = 1.41 V1

V2 =2 V1

⎛V ⎞ (d) 20 log ⎜⎜ 2 ⎟⎟ = 20 dB ⎝ V1 ⎠ ⎛ V ⎞ 20 log ⎜⎜ 2 ⎟⎟ = =1 ⎝ V1 ⎠ 20

⎛V ⎞ (e) 20 log ⎜⎜ 2 ⎟⎟ = 40 dB ⎝ V1 ⎠ ⎛ V ⎞ 40 log ⎜⎜ 2 ⎟⎟ = =2 ⎝ V1 ⎠ 20

V2 = 10 V1

V2 = 100 V1

Section 6-7 The Differential Amplifier 36.

Determine IE for each transistor: VR 14.3 V I RE = E = = 6.5 mA RE 2.2 kΩ

I E(Q1) = I E(Q 2) =

I RE 2

= 3.25 mA

Determine IC for each transistor: I C(Q1) = α1 I E(Q1) = 0.980(3.25 mA) = 3.185 mA I C(Q 2) = α 2 I E(Q 2) = 0.975(3.25 mA) = 3.169 mA Calculate the collector voltages: VC(Q1) = 15 V − (3.185 mA)(3.3 kΩ) = 4.49 V VC(Q 2) = 15 V − (3.169 mA)(3.3 kΩ) = 4.54 V The differential output voltage is: VOUT = VC(Q 2) − VC(Q1) = 4.54 V − 4.49 V = 0.05 V = 50 mV 37.

V1 measures the differential output voltage. V2 measures the noninverting input voltage. V3 measures the single-ended output voltage. V4 measures the differential input voltage. I1 measures the bias current.

56

⎛V ⎞ (c) 20 log ⎜⎜ 2 ⎟⎟ = 10 dB ⎝ V1 ⎠ ⎛ V ⎞ 10 log ⎜⎜ 2 ⎟⎟ = = 0.5 ⎝ V1 ⎠ 20 V2 = 3.16 V1

Chapter 6 38.

Calculate the voltage across each collector resistor: VRC1 = (1.35 mA)(5.1 kΩ) = 6.89 V VRC2 = (1.29 mA)(5.1 kΩ) = 6.58 V The differential output voltage is: VOUT = VC(Q 2) − VC(Q1) = (VCC − VRC2 ) − (VCC − VRC1 ) = VRC1 − VRC2 = 6.89 V − 6.58 V = 0.31 V = 310 mV

39.

(a) Single-ended differential input, differential output (b) Single-ended, differential input, single-ended output (c) Double-ended differential input, single-ended output (d) Double-ended differential input, differential output

Section 6-8 Troubleshooting 40.

⎛ R1 ⎞ ⎛ 10 kΩ ⎞ ⎟⎟10 V − 0.7 V = ⎜ VE = ⎜⎜ ⎟10 V − 0.7 V = 1.05 V ⎝ 57 kΩ ⎠ ⎝ R1 + R2 ⎠ V 1.05 V = 1.05 mA IE = E = R4 1.0 kΩ VC = 10 V − (1.05 mA)(4.7 kΩ) = 5.07 V VCE = 5.07 V − 1.05 V = 4.02 V 4.02 V V ′ ≅ CE = rCE = 3.83 kΩ 1.05 mA IE With C2 shorted: RIN(2) = R6 β DC R8 = 10 kΩ 125(1.0 kΩ) = 9.26 kΩ Looking from the collector of Q1: ( rCE′ + R4 ) RIN(2) = (3.83 kΩ + 1.0 kΩ) 9.26 kΩ = 3.17 kΩ 3.17 kΩ ⎛ ⎞ VC1 = ⎜ ⎟10 V = 4.03 V ⎝ 3.17 kΩ + 4.7 kΩ ⎠

41.

Q1 is in cutoff. IC = 0 A, so VC2 = 10 V.

42.

(a) (b) (c) (d) (e) (f)

Reduced gain No output signal Reduced gain Bias levels of first stage will change. IC will increase and Q1 will go into saturation. No signal at the Q1 collector Signal at the Q2 base. No output signal.

57

Chapter 6 43.

re′ = 10.9 Ω Av1 = 93.6

Rin = 1.48 kΩ Av2 = 302

Test Point Input

DC Volts 0V

Q1 base Q1 emitter Q1 collector Q2 base Q2 emitter Q2 collector Output

2.99 V 2.29 V 7.44 V 2.99 V 2.29 V 7.44 V 0V

AC Volts (rms)

25 μA 20.8 μV 0V 1.95 mV 1.95 mV 0V 589 mV 589 mV

Application Activity Problems 44.

For the block diagram of textbook Figure 6-46 with no output from the power amplifier or preamplifier and only one faulty block, the power amplifier must be ok because the fault must be one that affects the preamplifier's output prior to the power amplifier. Check the input to the preamplifier.

45.

(a) (b) (c) (d) (e) (f)

46.

R7 = 220 Ω will bias Q2 off.

47.

(a) Q1 is in cutoff. (b) VC1 = VEE (c) VC2 is unchanged and at 5.87 V.

No output signal Reduced output signal No output signal Reduced output signal No output signal Increased output signal (perhaps with distortion)

58

Chapter 6 Datasheet Problems 48.

From the datasheet in textbook Figure 6-63: (a) for a 2N3947, βac(min) = hfe(min) = 100 (b) For a 2N3947, re′( min) cannot be determined since hre(min) is not given. (c) For a 2N3947, rc′( min) cannot be determined since hre(min) is not given.

49.

From the 2N3947 datasheet in Figure 6-63: (a) For a 2N3947, βac(max) = 700 h 20 × 10−4 = 40 Ω (b) For a 2N3947, re′( max) = re = hoe 50 μS (c) For a 2N3947, rc′( max) =

50.

hre + 1 20 × 10−4 + 1 = = 20 kΩ hoe 50 μS

For maximum current gain, a 2N3947 should be used.

Advanced Problems 51.

In the circuit of textbook Figure 6-62, a leaky coupling capacitor would affect the biasing of the transistors, attenuate the ac signal, and decrease the frequency response.

52.

See Figure 6-3.

Figure 6-3

59

Chapter 6 53.

For the 2nd stage: 30 V 30 V = IR6-7 = = 435 μA R6 + R7 69 kΩ VB2 = VCC − IR6-7R6 = 15 V − (435μA)(47 kΩ) = 15 V − 20.5 V = −5.5 V VE2 −5.5 V − 0.7 V = = −1.21 mA IE2 = R9 + R10 5.13 kΩ 25 mV re′2 = = 20.7 Ω 1.21 mA With R10 = 0 Ω for max gain: R8 6.8 kΩ = Av(2) = = 45.1 (unloaded) R9 + re′2 150.7 Ω With a 10 kΩ load: R & RL 6.8 kΩ & 10 kΩ 4.05 kΩ = 26.9 Av(2) = 8 = = 150.7 Ω 150.7 Ω R9 + re′2 To keep unloaded gain: 4.05 kΩ = 45.1 R9 + 20.7 Ω 4.05 kΩ = 45.1(R9 + 20.7 Ω) = 45.1R9 + 934 Ω 4.05 kΩ − 934 Ω = 69.1 Ω R9 = 45.1

54.

RC > (100)(330 Ω) = 33 kΩ To prevent cutoff, VC must be no greater than 12 V − (100)(1.414)(25 mV) = 8.46 V In addition, VC must fall no lower than 8.46 V − 3.54 V = 4.93 V to prevent saturation. RC = 100(RE + re′ ) 25 mV IE 12 V − ICRC = 8.46 V ICRC = 3.54 V I C (100( RE + re′) ) = 3.54 V re′ =

⎛ ⎛ 25 mV ⎞ ⎞⎟ ⎟ ≅ 3.54 V I C ⎜100⎜⎜ 330 Ω + ⎟⎟ ⎜ I C ⎠⎠ ⎝ ⎝ (33 kΩ)IC + 2.5 V = 3.54 V IC = 31.4 μA 25 mV = 797 Ω 31.4 μA RC = 100(330 Ω + 797 Ω) = 113 kΩ Let RC = 120 kΩ. VC = 12 V − (31.4 μA)(120 kΩ) = 8.23 V VC(sat) = 8.23 V − 3.54 V = 4.69 V re′ ≅

60

Chapter 6 RE(tot) RC RE(tot)

4.69 V 7.31 V = (0.642)(120 kΩ) = 77 kΩ. Let RE = 68 kΩ. =

VE = (31.4 μA)(68 kΩ) = 2.14 V VB = 2.14 V + 0.7 V = 2.84 V R2 2.84 V = = 0.310 R1 9.16 V R2 = 0.310R1. If R1 = 20 kΩ, R2 = 6.2 kΩ. B

The amplifier circuit is shown in Figure 6-4. From the design: ⎛ 6 .2 k Ω ⎞ VB = ⎜ ⎟12 V = 2.84 V ⎝ 26.2 kΩ ⎠ VE = 2.14 V 2.14 V IC ≅ IE = = 31.3 μA 68.3 kΩ 25 mV re′ = = 798 Ω 31.3 μA 120 kΩ = 106 or 40.5 dB Av = 795 Ω + 330 Ω VC = 12 V − (31.3 μA)(120 kΩ) = 8.24 V The design is a close fit. B

55.

Figure 6-4

See Figure 6-5. Rin = 120 kΩ 120 kΩ (100)(5.1 kΩ) = 53.6 kΩ minimum

Figure 6-5 56.

Figure 6-6

See Figure 6-6.

61

Chapter 6 57.

See Figure 6-7. 6 V − 0 .7 V = 10 mA IC = 510 Ω + 2 kΩ/100 25 mV re′ = = 2.5 Ω 10 mA 180 Ω Av = = 72.4 2.5 Ω This is reasonably close (≈3.3% off) and can be made closer by putting a 7.5 Ω resistor in series with the 180 Ω collector resistor. Figure 6-7

58.

Assuming βac = 200, 1 1 = C1 = 2π f c R 2π (100 Hz)(330 kΩ & 330 kΩ & (200 × 34 kΩ)) 1 = = 0.01 μF 2π (100 Hz)(161 kΩ) 1 1 C2 = = 2π f c R 2π (100 Hz)(22 kΩ + 47 kΩ & 22 kΩ & (200 × 5.13 kΩ)) 1 = = 0.043 μF 2π (100 Hz)(36.98 kΩ)

59.

IC ≅ IE VRC R RC RC R I Av = C ≅ = 40VRC ≅ = C C = re′ 25 mV/I E 25 mV/I C 25 mV 25 mV

Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 60 through 65 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 60.

C2 open

61.

C2 shorted

62.

RE leaky

63.

C1 open

64.

C2 open

65.

C3 open

62

Chapter 7 Power Amplifiers Section 7-1 The Class A Power Amplifier 1.

⎛ R2 ⎞ 330 Ω ⎞ ⎟⎟VCC = ⎛⎜ (a) VB = ⎜⎜ ⎟15 V = 3.72 V ⎝ 1.0 kΩ + 330 kΩ ⎠ ⎝ R1 + R2 ⎠ VE = VB − VBE = 3.72 − 0.7 V = 3.02 V VE 3.02 V = ICQ ≅ IE = = 68.4 mA RE1 + RE2 8.2 Ω + 36 Ω VCEQ = VCC − (IC)(RE1 + RE2 + RL) = 15 V − (68.4 mA)(8.2 Ω + 36 Ω + 100 Ω) = 5.14 V B

B

(b) Av =

100 Ω RL = = 11.7 RE1 + re′ 8.2 Ω + 0.37 Ω

Rin = βac(RE1 + re′ )

R1 R2

= 100 (8.2 Ω + 0.37 Ω)

330 Ω 1.0 kΩ = 192 Ω

⎛R ⎞ ⎛ 192 Ω ⎞ ⎟⎟ = 263 Ap = Av2 ⎜⎜ in ⎟⎟ = 11.7 2 ⎜⎜ ⎝ 100 Ω ⎠ ⎝ RL ⎠ The computed voltage and power gains are slightly higher if re′ is ignored. 2.

(a) If RL is removed, there is no collector current; hence, the power dissipated in the transistor is zero. (b) Power is dissipated only in the bias resistors plus a small amount in RE1 and RE2. Since the load resistor has been removed, the base voltage is altered. The base voltage can be found from the Thevenin equivalent drawn for the bias circuit in Figure 7-1.

Figure 7-1 Applying the voltage-divider rule and including the base-emitter diode drop of 0.7 V result in a base voltage of 1.2 V. The power supply current is then computed as ICC =

VCC − 1.2 V 15 V − 1.2 V = = 13.8 mA 1.0 kΩ R1

63

Chapter 7 Power from the supply is then computed as PT = ICCVCC = (13.8 mA)(15 V) = 207 mW (c) Av = 11.7 (see problem 1(b)). Vin = 500 mVpp = 177 mVrms. Vout = AvVin = (11.7)(177 mV) = 2.07 V V2 2.07 V 2 = 42.8 mW Pout = out = RL 100 Ω 3.

The changes are shown in Figure 7-2. The advantage of this arrangement is that the load resistor is referenced to ground.

Figure 7-2 4.

A CC amplifier has a voltage gain of approximately 1. Therefore, R 2.2 kΩ Ap = in = = 44 Rout 50 Ω

5.

(a) Stiff voltage divider: ⎛ R2 ⎞ ⎛ 510 Ω ⎞ VB = ⎜ ⎟VCC = ⎜ ⎟ 12 V = 5.14 V ⎝ 1190 Ω ⎠ ⎝ R1 + R2 ⎠ VE = VB − 0.7 V = 4.44 V VE 4.44 V = IE = = 55.7 mA RE1 + RE2 79.7 Ω IC ≅ IE = 55.7 mA VC = VCC − ICRC = 12 V − (55.7 mA)(100 Ω) = 6.43 V VCE = VC − VE = 6.43 V − 4.44 V = 1.99 V B

B

(b) Not stiff voltage divider: RIN(BASE) = βDC(RE1 + RE2) = 120(144 Ω) = 17.0 kΩ ⎛ R2 & RIN(BASE) ⎞ 4.7 kΩ & 17.0 kΩ ⎛ ⎞ V = 12 V VB = ⎜ ⎜ R1 + R2 & RIN(BASE) ⎟⎟ CC ⎜⎝ 12 kΩ + 4.7 kΩ & 17.0 kΩ ⎟⎠ ⎝ ⎠ B

⎛ 3.68 kΩ ⎞ =⎜ ⎟12 V = 2.8 V ⎝ 15.7 kΩ ⎠

64

Chapter 7 VE = VB − 0.7 V = 2.1 V VE 2.1 V = IE = = 14.6 mA RE1 + RE2 144 Ω IC ≅ IE = 14.6 mA VC = VCC − ICRC = 12 V − (14.6 mA)(470 Ω) = 5.14 V VCE = VC − VE = 5.14 V − 2.1 V = 3.04 V B

6.

The Q-point does not change because RL is capacitively coupled and does not affect the DC values.

7.

For the circuit in Figure 7-43(a): RIN(BASE) = βDC(RE1 + RE2) = (125)(79.7 ) = 9.96 kΩ Since RIN(BASE) > 10R2, it can be neglected. ⎛ R2 ⎞ ⎛ ⎞ ⎛ 510 Ω ⎞ 510 Ω VB = ⎜ ⎟VCC = ⎜ ⎟12 V = 5.14 V ⎟12 V = ⎜ ⎝ 680 Ω + 510 Ω ⎠ ⎝ 1190 Ω ⎠ ⎝ R1 + R2 ⎠ B

VE = VB − 0.7 V = 5.14 V − 0.7 V = 4.44 V V 4.44 V ICQ ≅ IE = E = = 55.7 mA RE 79.7 Ω VCQ = VCC − ICQRC = 12 V − (55.7 mA)(100 Ω) = 6.43 V VCEQ = VC − VE = 6.43 V − 4.44 V = 1.99 V Rc = RC RL = 100 Ω 100 Ω = 50 Ω B

Vce(cutoff) = VCEQ + ICQRc = 1.99 V + 55.7 mA(50 Ω) = 4.78 V Since VCEQ is closer to saturation, Ic is limited to V 1.99 V = 39.8 mA Ic(p) = CEQ = Rc 50 Ω Vout is limited to Vout(p) = VCEQ = 1.99 V For the circuit in 7-43(b): RIN(BASE) = βDC(RE1 + RE2) = (120)(142 Ω) = 17 kΩ Since RIN(BASE) < 10R2, it is taken into account. ⎛ R2 RIN(BASE) ⎞ ⎛ ⎞ 4.7 kΩ 17 kΩ ⎛ 3.68 kΩ ⎞ ⎟VCC = ⎜ VB = ⎜ ⎟⎟12 V = ⎜ ⎟12 V = 2.82 V ⎜ ⎜ R1 + R2 RIN(BASE) ⎟ 12 kΩ + 4.7 kΩ 17 kΩ ⎠ 15.68 kΩ ⎠ ⎝ ⎝ ⎝ ⎠ VE = VB − 0.7 V = 2.82 V − 0.7 V = 2.12 V ICQ ≅ IE = VE/RE = 2.12 V/142 Ω = 14.9 mA VCQ = VCC − ICQRC = 12 V − (14.9 mA)(470 Ω) = 5.0 V VCEQ = VCQ − VE = 5.0 V − 2.12 V = 2.88 V Rc = RC RL = 470 Ω 470 Ω = 235 Ω B

B

Vce(cutoff) = VCEQ + ICQRc = 2.88 V + 14.9 mA(235 Ω) = 6.38 V Since VCEQ is closer to saturation, Ic is limited to V 2.88 V = 12.3 mA Ic(p) = CEQ = 235 Ω Rc Vout is limited to Vout(p) = VCEQ = 2.88 V

65

Chapter 7 8.

⎛R ⎞ (a) Ap = Av2 ⎜⎜ in ⎟⎟ ⎝ RL ⎠ R RL 100 Ω 100 Ω 50 Ω R = 10.6 Av ≅ c = C = = RE1 RE1 4 .7 Ω 4.7 Ω Rin = R1 R2 Rin (base) = R1 R2 β ac RE1 Rin = 680 Ω 510 Ω (125)(4.7 Ω) = 680 Ω 510 Ω 588 Ω = 195 Ω

⎛ 195 Ω ⎞ ⎟⎟ = 219 Ap = (10.6)2 ⎜⎜ 100 Ω ⎝ ⎠ Av ≅

(b)

R RL 470 Ω 470 Ω 235 Ω Rc = 10.7 = C = = RE1 RE1 22 Ω 22 Ω

Rin = 12 kΩ 4.7 kΩ (120)(22 Ω) = 12 kΩ 4.7 kΩ 2.64 kΩ = 1.48 kΩ

⎛ 1.48 kΩ ⎞ ⎟⎟ = 361 Ap = (10.7)2 ⎜⎜ ⎝ 470 Ω ⎠ 9.

RIN(BASE) = βDCRE = 90(130 Ω) = 11.7 kΩ R2 RIN(BASE) = 1.0 kΩ 11.7 kΩ = 921 Ω

⎛ R2 RIN(BASE) ⎞ ⎛ 921 Ω ⎞ ⎟V = 24 V = 3.93 V VB = ⎜ ⎜ R1 + R2 RIN(BASE) ⎟ CC ⎜⎝ 5.62 kΩ ⎟⎠ ⎝ ⎠ VE = VB − 0.7 V = 3.93 V − 0.7 V = 3.23 V V 3.23 V = 24.8 mA ICQ ≅ IE = E = 130 Ω RE VC = VCC − ICQRC = 24 V − (24.8 mA)(560 Ω) = 10.1 V VCEQ = VC − VE = 10.1 V − 3.23 V = 6.87 V PD(min) = PDQ = ICQVCEQ = (24.8 mA)(6.87 V) = 170 mW B

B

10.

From Problem 9: ICQ = 24.8 mA and VCEQ = 6.87 V Vce(cutoff) = VCEQ + ICQRc = 6.87 V + (24.8 mA)(264 Ω) = 13.4 V 2 Rc = 0.5(24.8 mA)2(264 Ω) = 81.2 mW Pout = 0.5I CQ

η =

81.2 mW Pout Pout Pout = = = = 0.136 PDC VCC I CC VCC I CQ (24 V)(24.8 mA)

66

Chapter 7 Section 7-2 The Class B and Class AB Push-Pull Amplifiers 11.

(a) VB(Q1) = 0 V + 0.7 V = 0.7 V VB(Q2) = 0 V − 0.7 V = −0.7 V VE = 0 V V − (−VCC ) − 1.4 V 9 V − (−9 V) − 1.4 V = ICQ = CC = 8.3 mA R1 + R2 1.0 kΩ + 1.0 kΩ VCEQ(Q1) = 9 V VCEQ(Q2) = −9 V (b) Vout = Vin = 5.0 V rms (V ) 2 5.0 V 2 Pout = out = = 0.5 W RL 50 Ω

12.

VCC 9.0 V = = 180 mA RL 50 Ω Vce(cutoff) = 9 V Ic(sat) =

These points define the ac load line as shown in Figure 7-3. The Q-point is at a collector current of 8.3 mA (see problem 11) and the dc load line rises vertically through this point.

Figure 7-3 13.

Rin = β ac ( re′ + RL ) & R1 & R2 From Problem 11, ICQ = 8.3 mA so, IE ≅ 8.3 mA 25 mV re′ = =3Ω 8.3 mA Rin = 100(53 Ω) & 1.0 kΩ & 1.0 kΩ = 5300 Ω & 1.0 kΩ & 1.0 kΩ = 457 Ω

14.

The DC voltage at the output becomes negative instead of 0 V.

15.

(a) VB(Q1) = 7.5 V + 0.7 V = 8.2 V VB(Q2) = 7.5 V − 0.7 V = 6.8 V 15 V VE = = 7.5 V 2 V − 1.4 V 15 V − 1.4 V = = 6.8 mA ICQ = CC R1 + R2 1.0 kΩ + 1.0 kΩ VCEQ(Q1) = 15 V − 7.5 V = 7.5 V VCEQ(Q2) = 0 V − 7.5 V = −7.5 V

67

Chapter 7 (b) Vin = Vout = 10 Vpp = 3.54 V rms (V ) 2 (3.54 V) 2 PL = L = = 167 mW RL 75 Ω 16.

(a) Maximum peak voltage = 7.5 Vp. 7.5 Vp = 5.30 V rms 2 2 (V ) (5.30 V) PL(max) = L = = 375 mW RL 75 Ω (b) Maximum peak voltage = 12 Vp. 12 Vp = 8.48 V rms (VL ) 2 (8.48 V) 2 = 960 mW PL(max) = = RL 75 Ω

17.

(a) (b) (c) (d)

C2 open or Q2 open power supply off, open R1, Q1 base shorted to ground Q1 has collector-to-emitter short one or both diodes shorted

18.

Rin = β ac (re′ + RL ) & R1 & R2 From Problem 15: ICQ = 6.8 mA so, IE ≅ 6.8 mA 25 mV 25 mV re′ = = = 3.68 Ω 6.8 mA IE Rin = 200(78.7 Ω) & 1 kΩ & 1 kΩ = 485 Ω 485 Ω ⎞ ⎛ Vb = ⎜ ⎟1 V = 0.91 V rms ⎝ 485 Ω + 50 Ω ⎠

Section 7-3 The Class C Amplifier 19.

⎛t ⎞ PD(avg) = ⎜ on ⎟VCE(sat) I C(sat) = (0.1)(0.18 V)(25 mA) = 450 μW ⎝T ⎠

20.

fr =

21.

Vout(pp) = 2VCC = 2(12 V) = 24 V

22.

Pout =

1 2π LC

=

1 2π (10 mH)(0.001 μF)

= 50.3 kHz

2 0.5VCC 0.5(15 V) 2 = 2.25 W = Rc 50 Ω

⎛t ⎞ PD(avg) = ⎜ on ⎟VCE(sat) I C(sat) = (0.1)(0.18 V)(25 mA) = 0.45 mW ⎝T ⎠ Pout 2.25 W = η= = 0.9998 Pout + PD(avg) 2.25 W + 0.45 mW

68

Chapter 7 Section 7-4 Troubleshooting 23.

With C1 open, only the negative half of the input signal appears across RL.

24.

One of the transistors is open between the collector and emitter or a coupling capacitor is open.

25.

(a) (b) (c) (d)

No dc supply voltage or R1 open Diode D2 open Circuit is OK Q1 shorted from collector to emitter

Application Activity Problems 26.

For the block diagram of textbook Figure 7-34 with no signal from the power amplifier or preamplifier, but with the microphone working, the problem is in the power amplifier or preamplifier. It must be assumed that the preamp is faulty, causing the power amp to have no signal.

27.

For the circuit of Figure 7-35 with the base-emitter junction of Q2 open, the dc output will be approximately −15 V with a signal output approximately equal to the input.

28.

For the circuit of Figure 7-35 with the collector-emitter junction of Q5 open, the dc output will be approximately +15 V with a signal output approximately equal to the input (some distortion possible).

29.

On the circuit board of Figure 7-49, the vertically oriented diode has been installed backwards.

Datasheet Problems 30.

From the BD135 datasheet of textbook Figure 7-50: (a) βDC(min) = 40 @ IC = 150 mA, VCE = 2 V βDC(min) = 25 @ IC = 5 mA, VCE = 2 V (b) For a BD135, VCE(max) = VCEO = 45 V (c) PD(max) = 12.5 W @ TC = 25°C (d) IC(max) = 1.5 A

31.

PD = 10 W @ 50°C from graph in Figure 7-50.

32.

PD = 1 W @ 50°C. Extrapolating from the case temperature graph in Figure 7-50, since PD = 1.25 W @ 25°C ambient. This derating gives 1 W.

33.

As IC increases from 10 mA to approximately 125 mA, the dc current gain increases. As IC increases above approximately 125 mA, the dc current gain decreases.

34.

hFE ≅ 89 @ IC = 20 mA

69

Chapter 7 Advanced Problems 35.

36.

TC is much closer to the actual junction temperature than TA. In a given operating environment, TA is always less than TC.

24 V 24 V = = 55.8 mA 330 Ω + 100 Ω 430 Ω VCE(cutoff) = 24 V 1.0 kΩ ⎛ ⎞ VBQ = ⎜ ⎟ 24 V = 4.21 V Ω + Ω 1 . 0 k 4 . 7 k ⎝ ⎠ VEQ = 4.21 V − 0.7 V = 3.51 V 3.51 V = 35.1 mA IEQ ≅ ICQ = 100 Ω Rc = 330 Ω 330 Ω = 165 Ω IC(sat) =

VCQ = 24 V − (35.1 mA)(330 Ω) = 12.4 V VCEQ = 12.4 V − 3.51 V = 8.90 V 8.90 V = 89.1 mA Ic(sat) = 35.1 mA + 165 Ω Vce(cutoff) = 8.90 V + (35.1 mA)(165 Ω) = 14.7 V See Figure 7-4. Figure 7-4 37.

See Figure 7-5. 15 V = 174 mA IR1 ≅ IR2 = 86 Ω ⎛ 18 Ω ⎞ ⎟⎟ 15 V = 3.14 V VB ≅ ⎜⎜ ⎝ 86 Ω ⎠ B

VE = 3.14 V − 0.7 V = 2.44 V 2.44 V IE ≅ IC = = 503 mA 4.85 Ω VC = 15 V − (10 Ω)(503 mA) = 9.97 V VCE = 7.53 V 25 mV re′ = = 0.05 Ω 503 mA The ac resistance affecting the load line is Rc + Re + re′ = 10 Ω βac = βDC ≥ 100 7.53 V Ic(sat) = 503 mA + = 1.24 A 10.2 Ω Vce(cutoff) = 7.53 V + (503 mA)(10.2 Ω) = 12.7 V The Q-point is closer to cutoff so Pout = (0.5)(503 mA)2(10.2 Ω) = 1.29 W As loading occurs, the Q-point will still be closer to cutoff. The circuit will have Pout ≥ 1 W for RL ≥ 37.7 Ω. (39 Ω standard) 70

Figure 7-5

Chapter 7 38.

Preamp quiescent current: 30 V = 45 μA I1 = I2 = 660 kΩ 15 V − 0.7 V I3 = I4 = I5 = = 421 μA 34 kΩ 30 V I6 = I7 = = 435 μA 69 kΩ VB2 = 15 V − (435 μA)(47 kΩ) = −5.45 V −15 V − (−5.45 V − 0.7 V) = 1.73 mA I8 = I9 = I10 = 5.13 kΩ Itot = 45 μA + 421 μA + 435 μA + 1.73 mA = 2.63 mA Power amp quiescent current: I11 ≅ 0 15.7 V − 3(0.7 V) 13.6 V = I12 = = 13.6 mA 1.0 kΩ 1.0 kΩ −15 V − (−0.7 V) −14.3 V = I13 = = 65 mA 220 Ω 220 Ω Itot = 13.6 mA + 65 mA = 78.6 mA Signal current to load: Scope shows ≈ 9.8 V peak output. 0.707(9.8 V) = 866 mA IL = 8Ω Itot(sys) = 2.63 mA + 78.6 mA + 866 mA = 947 mA Amp. × hrs = 947 mA × 4 hrs = 3.79 Ah

Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 39 through 43 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 39.

Cin open

40.

RE2 open

41.

Q1 collector-emitter open

42.

D2 shorted

43.

Q2 drain-source open

71

Chapter 8 Field-Effect Transistors (FETs) Section 8-1 The JFET 1.

(a) (b)

A greater VGS narrows the depletion region. The channel resistance increases with increased VGS.

2.

The gate-to-source voltage of an n-channel JFET must be zero or negative in order to maintain the required reverse-bias condition.

3.

See Figure 8-1.

Figure 8-1 4.

See Figure 8-2.

Figure 8-2

Section 8-2 JFET Characteristics and Parameters 5.

VDS = VP = 5 V at point where ID becomes constant.

6.

VGS(off) = −VP = −6 V The device is on, because VGS = −2 V.

7.

By definition, ID = IDSS when VGS = 0 V for values of VDS > VP. Therefore, ID = 10 mA.

8.

Since VGS > VGS(off), the JFET is off and ID = 0 A.

72

Chapter 8 9.

10.

VP = −VGS(off) = −(−4 V) = 4 V The voltmeter reads VDS. As VDD is increased, VDS also increases. The point at which ID reaches a constant value is VDS = VP = 4 V. ⎞ ⎛ V ID = I DSS ⎜1 − GS ⎟ ⎜ VGS(off) ⎟ ⎠ ⎝

2

2

0V ⎞ ⎛ ID = 5 mA⎜1 − ⎟ = 5 mA ⎝ −8V⎠ 2

−1V ⎞ ⎛ ID = 5 mA⎜1 − ⎟ = 3.83 mA ⎝ −8V ⎠ 2

⎛ −2V⎞ ID = 5 mA⎜1 − ⎟ = 2.81 mA ⎝ −8V ⎠ 2

⎛ −3V⎞ ID = 5 mA⎜1 − ⎟ = 1.95 mA ⎝ −8V ⎠ 2

⎛ −4V⎞ ID = 5 mA⎜1 − ⎟ = 1.25 mA ⎝ −8V ⎠ 2

⎛ −5V⎞ ID = 5 mA⎜1 − ⎟ = 0.703 mA ⎝ −8V ⎠ 2

⎛ −6V⎞ ID = 5 mA⎜1 − ⎟ = 0.313 mA ⎝ −8V ⎠ 2

⎛ −7V⎞ ID = 5 mA⎜1 − ⎟ = 0.078 mA ⎝ −8V ⎠ 2

⎛ −8V ⎞ ID = 5 mA⎜1 − ⎟ = 0 mA ⎝ −8V ⎠ See Figure 8-3.

Figure 8-3

73

Chapter 8 11.

⎛ ⎞ V ID = I DSS ⎜1 − GS ⎟ ⎜ VGS(off) ⎟ ⎝ ⎠ V ID 1 − GS = VGS(off) I DSS

2

VGS ID = 1− VGS(off) I DSS

⎛ I D ⎞⎟ VGS = VGS(off) ⎜1 − ⎜ I DSS ⎟⎠ ⎝ ⎛ 2.25 mA ⎞⎟ VGS = −8 V ⎜1 − = −8 V(0.329) = −2.63 V ⎜ 5 mA ⎟⎠ ⎝

12.

13.

⎛ ⎞ ⎛ −4 V⎞ V ⎟⎟ = 1600 μS gm = g m 0 ⎜1 − GS ⎟ = 3200 μS⎜⎜1 − ⎜ VGS(off) ⎟ 8 V − ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ −2 V⎞ V ⎟⎟ = 1429 μS gm = g m 0 ⎜1 − GS ⎟ = 2000 μS⎜⎜1 − ⎜ VGS(off) ⎟ 7 V − ⎝ ⎠ ⎝ ⎠ gfs = gm = 1429 μS VGS 10 V = = 2000 MΩ I GSS 5 nA

14.

RIN =

15.

⎞ ⎛ V VGS = 0 V: ID = I DSS ⎜1 − GS ⎟ = 8 mA(1 − 0)2 = 8 mA ⎜ VGS(off) ⎟ ⎠ ⎝

2

2

−1V ⎞ ⎛ 2 2 VGS = −1 V: ID = 8 mA⎜1 − ⎟ = 8 mA(1 − 0.2) = 8 mA(0.8) = 5.12 mA − 5 V ⎝ ⎠ 2

⎛ −2V⎞ 2 2 VGS = −2 V: ID = 8 mA⎜1 − ⎟ = 8 mA(1 − 0.4) = 8 mA(0.6) = 2.88 mA ⎝ −5V⎠ 2

⎛ −3V ⎞ 2 2 VGS = −3 V: ID = 8 mA⎜1 − ⎟ = 8 mA(1 − 0.6) = 8 mA(0.4) = 1.28 mA − 5 V ⎝ ⎠ 2

⎛ −4V⎞ 2 2 VGS = −4 V: ID = 8 mA⎜1 − ⎟ = 8 mA(1 − 0.8) = 8 mA(0.2) = 0.320 mA ⎝ −5V⎠ 2

⎛ −5V⎞ 2 2 VGS = −5 V: ID = 8 mA⎜1 − ⎟ = 8 mA(1 − 1) = 8 mA(0) = 0 mA ⎝ −5V⎠

Section 8-3 JFET Biasing 16.

VGS = −IDRS = −(12 mA)(100 Ω) = −1.2 V

74

Chapter 8 17.

RS =

VGS −4V = = 800 Ω ID 5 mA

18.

RS =

VGS −3V = = 1.2 kΩ 2.5 mA ID

19.

(a) ID = IDSS = 20 mA (b) ID = 0 A (c) ID increases

20.

(a) VS = (1 mA)(1.0 kΩ) = 1 V VD = 12 V − (1 mA)(4.7 kΩ) = 7.3 V VG = 0 V VGS = VG − VS = 0 V − 1 V = −1 V VDS = 7.3 V − 1 V = 6.3 V

(b)

VS = (5 mA)(100 Ω) = 0.5 V VD = 9 V − (5 mA)(470 Ω) = 6.65 V VG = 0 V VGS = VG − VS = 0 V − 0.5 V = −0.5 V VDS = 6.65 V − 0.5 V = 6.15 V

(c) VS = (−3 mA)(470 Ω) = −1.41 V VD = −15 V − (3 mA)(2.2 kΩ) = −8.4 V VG = 0 V VGS = VG − VS = 0 V − (−1.41 V) = 1.41 V VDS = −8.4 V − (−1.41 V) = −6.99 V 21.

From the graph, VGS ≅ −2 V at ID = 9.5 mA. V −2V = 211 Ω RS = GS = ID 9.5 mA

22.

ID =

I DSS 14 mA = = 7 mA 2 2 VGS(off) − 10 V = VGS = = −2.93 V 3.414 3.414 V 2.93 V = 419 Ω (The nearest standard value is 430 Ω.) RS = GS = ID 7 mA

VDD − VD 24 V − 12 V = = 1.7 kΩ (The nearest standard value is 1.8 kΩ.) ID 7 mA Select RG = 1.0 MΩ. See Figure 8-4. RD =

Figure 8-4

75

Chapter 8 23.

RIN(total) = RG RIN =

RIN

VGS − 10 V = = 500 MΩ I GSS 20 nA

RIN(total) = 10 MΩ 500 MΩ = 9.8 MΩ 24.

For ID = 0, VGS = −IDRS = (0)(330 Ω) = 0 V For ID = IDSS = 5 mA VGS = −IDRS = −(5 mA)(330 Ω) = −1.65 V From the graph in Figure 8-69 in the textbook, the Q-point is VGS ≅ −0.95 V and ID ≅ 2.9 mA

25.

For ID = 0, VGS = 0 V For ID = IDSS = 10 mA, VGS = −IDRS = (10 mA)(390 Ω) = 3.9 V From the graph in Figure 8-70 in the textbook, the Q-point is VGS ≅ 2.1 V and ID ≅ 5.3 mA

26.

Since VRD = 9 V − 5 V = 4 V, VRD

4V = 0.85 mA RD 4.7 kΩ VS = IDRS = (0.85 mA)(3.3 kΩ) = 2.81 V ⎛ R2 ⎞ 2.2 MΩ ⎞ ⎟⎟VDD = ⎛⎜ VG = ⎜⎜ ⎟ 9 V = 1.62 V ⎝ 12.2 MΩ ⎠ ⎝ R1 + R2 ⎠ ID =

=

VGS = VG − VS = 1.62 V − 2.81 V = −1.19 V Q-point: ID = 0.85 mA, VGS = −1.19 V 27.

For ID = 0,

⎛ R2 ⎞ 2.2 MΩ ⎞ ⎟⎟VDD = ⎛⎜ VGS = VG = ⎜⎜ ⎟12 V = 4.8 V ⎝ 5.5 MΩ ⎠ ⎝ R1 + R2 ⎠ For VGS = 0 V, VS = 4.8 V V − VGS V 4 .8 V ID = S = G = 1.45 mA = RS RS 3.3 kΩ The Q-point is taken from the graph in Figure 8-72 in the textbook. ID ≅ 1.9 mA, VGS = −1.5 V

Section 8-4 The Ohmic Region 28.

RDS =

VDS 0.8 V = = 4 kΩ I D 0.20 mA

76

Chapter 8 29.

30.

31.

0.4 V = 2.67 kΩ 0.15 mA 0.6 V = 1.33 kΩ RDS2 = 0.45 mA Δ RDS = 2.67 kΩ − 1.33 kΩ = 1.34 kΩ

RDS1 =

⎛ ⎞ V −1 V ⎞ ⎛ gm = g m 0 ⎜1 − GS ⎟ = 1.5 mS ⎜1 − ⎟ ⎜ ⎟ ⎝ −3.5 V ⎠ ⎝ VGS(off) ⎠ = 1.5 mS(0.714) = 1.07 mS rds =

1 1 = = 935 Ω g m 1.07 mS

Section 8-5 The MOSFET 32.

See Figure 8-5.

Figure 8-5 33.

An n-channel D-MOSFET with a positive VGS is operating in the enhancement mode.

34.

An E-MOSFET has no physical channel or depletion mode. A D-MOSFET has a physical channel and can be operated in either depletion or enhancement modes.

35.

MOSFETs have a very high input resistance because the gate is insulated from the channel by an SiO2 layer.

Section 8-6 MOSFET Characteristics and Parameters 36.

K=

I D(on) (VGS − VGS(th) )

2

=

10 mA = 0.12 mA/V2 2 (−12 V + 3 V)

ID = K(VGS − VGS(off))2 = (0.12 mA/V2)(−6 V + 3 V)2 = 1.08 mA 2

37.

⎞ ⎛ V ID = I DSS ⎜1 − GS ⎟ ⎜ VGS(off) ⎟ ⎠ ⎝ ID 3 mA = 4.69 mA IDSS = = 2 2 −2V ⎞ ⎛ ⎞ ⎛ V ⎜1 − GS ⎟ ⎜1 − ⎟ ⎜ VGS(off) ⎟ ⎝ − 10 V ⎠ ⎝ ⎠

77

Chapter 8 38.

(a)

n channel

(b)

⎞ ⎛ V ID = I DSS ⎜1 − GS ⎟ ⎜ VGS(off) ⎟ ⎠ ⎝

2

⎛ −5V⎞ ID = 8 mA⎜1 − ⎟ = 0 mA ⎝ −5V⎠

2

⎛ −3V ⎞ ID = 8 mA⎜1 − ⎟ = 1.28 mA ⎝ −5V⎠

2

−1V ⎞ ⎛ ID = 8 mA⎜1 − ⎟ = 5.12 mA ⎝ −5V⎠

2

1V ⎞ ⎛ ID = 8 mA⎜1 − ⎟ = 11.5 mA ⎝ −5V⎠

2

3V ⎞ ⎛ ID = 8 mA⎜1 − ⎟ = 20.5 mA − 5V⎠ ⎝

2

5V ⎞ ⎛ ID = 8 mA⎜1 − ⎟ = 32 mA ⎝ −5V⎠

2

⎛ −4V⎞ ID = 8 mA⎜1 − ⎟ = 0.32 mA ⎝ −5V⎠

⎛ −2V⎞ ID = 8 mA⎜1 − ⎟ = 2.88 mA ⎝ −5V ⎠

2

2

0V ⎞ ⎛ ID = 8 mA⎜1 − ⎟ = 8 mA ⎝ −5V⎠

2V ⎞ ⎛ ID = 8 mA⎜1 − ⎟ = 15.7 mA − 5V⎠ ⎝ 4V ⎞ ⎛ ID = 8 mA⎜1 − ⎟ = 25.9 mA ⎝ −5V⎠

(c)

2

See Figure 8-6.

Figure 8-6

78

2

2

Chapter 8 Section 8-7 MOSFET Biasing 39.

40.

(a) (b) (c) (d)

Depletion Enhancement Zero bias Depletion

⎛ 10 MΩ ⎞ (a) VGS = ⎜ ⎟ 10 V = 6.8 V ⎝ 14.7 MΩ ⎠ ⎛ 1.0 MΩ ⎞ (b) VGS = ⎜ ⎟ (−25 V) = −2.27 V ⎝ 11 MΩ ⎠

This MOSFET is on. This MOSFET is off.

41.

Since VGS = 0 V for each circuit, ID = IDSS = 8 mA. (a) VDS = VDD − IDRD = 12 V − (8 mA)(1.0 kΩ) = 4 V (b) VDS = VDD − IDRD = 15 V − (8 mA)(1.2 kΩ) = 5.4 V (c) VDS = VDD − IDRD = −9 V − (−8 mA)(560 Ω) = −4.52 V

42.

(a) ID(on) = 3 mA @ 4 V, VGS(th) = 2 V ⎛ R2 ⎞ 4.7 MΩ ⎞ ⎟⎟VDD = ⎛⎜ VGS = ⎜⎜ ⎟ 10 V = 3.2 V ⎝ 14.7 MΩ ⎠ ⎝ R1 + R2 ⎠ K=

I D(on) (VGS − VGS(th) )

2

=

3 mA 3 mA = = 0.75 mA/V2 2 (4 V − 2 V) (2 V)2

ID = K(VGS − VGS(th))2 = (0.75 mA/V2)(3.2 V − 2 V)2 = 1.08 mA VDS = VDD − IDRD = 10 V − (1.08 mA)(1.0 kΩ) = 10 V − 1.08 V = 8.92 V (b) ID(on) = 2 mA @ 3 V, VGS(th) = 1.5 V ⎛ R2 ⎞ 10 MΩ ⎞ ⎟⎟VDD = ⎛⎜ VGS = ⎜⎜ ⎟ 5 V = 2.5 V ⎝ 20 MΩ ⎠ ⎝ R1 + R2 ⎠ K=

I D(on) (VGS − VGS(th) )

2

=

2 mA 2 mA = = 0.89 mA/V2 2 2 (3 V − 1.5 V) (1.5 V)

ID = K(VGS − VGS(th))2 = (0.89 mA/V2)(2.5 V − 1.5 V)2 = 0.89 mA VDS = VDD − IDRD = 5 V − (0.89 mA)(1.5 kΩ) = 5 V − 1.34 V = 3.66 V 43.

(a) VDS = VGS = 5 V V −V 12 V − 5 V ID = DD DS = = 3.18 mA RD 2.2 kΩ (b) VDS = VGS = 3.2 V V −V 8 V − 3.2 V = 1.02 mA ID = DD DS = RD 4.7 kΩ

44.

VDS = VDD − IDRD = 15 V − (1 mA)(8.2 kΩ) = 6.8 V VGS = VDS − IGRG = 6.8 V − (50 pA)(22 MΩ) = 6.799 V

79

Chapter 8 Section 8-8 The IGBT 45.

The input resistance of an IGBT is very high because of the insulated gate structure.

46.

With excessive collector current, the parasitic transistor turns on and the IGBT acts as a thyristor.

Section 8-9 Troubleshooting 47.

When ID goes to zero, the possible faults are: RD or RS open, JFET drain-to-source open, no supply voltage, or ground connection open.

48.

If ID goes to 16 mA, the possible faults are: The JFET is shorted from drain-to-source or VDD has increased.

49.

If VDD is changed to −20 V, ID will change very little or none because the device is operating in the constant-current region of the characteristic curve.

50.

The device is off. The gate bias voltage must be less than VGS(th). The gate could be shorted or partially shorted to ground.

51.

The device is saturated, so there is very little voltage from drain-to-source. This indicates that VGS is too high. The 1.0 MΩ bias resistor is probably open.

Application Activity Problems −500 mV −200 mV 0 mV 400 mV

52.

(a) (b) (c) (d)

53.

At VG2S = 6 V, ID ≅ 10 mA At VG2S = 1 V, ID ≅ 5 mA

54.

VGS1 = Vsensor = −400 mV VOUT = 9.048 V V − VOUT 12 V − 9.048 V = = 2.64 mA ID = DD R3 + R4 1120 Ω VGS1 = Vsensor = −300 mV VOUT = 7.574 V 12 V − 7.574 V = 3.95 mA ID = 1120 Ω VGS1 = Vsensor = −200 mV VOUT = 5.930 V 12 V − 5.930 V = 5.42 mA ID = 1120 Ω

80

Chapter 8 VGS1 = Vsensor = −100 mV VOUT = 4.890 V 12 V − 4.890 V ID = = 6.35 mA 1120 Ω VGS1 = Vsensor = 0 mV VOUT = 4.197 V 12 V − 4.197 V = 6.97 mA ID = 1120 Ω VGS1 = Vsensor = 100 mV VOUT = 3.562 V 12 V − 3.562 V = 7.53 mA ID = 1120 Ω VGS1 = Vsensor = 200 mV VOUT = 2.960 V 12 V − 2.960 V ID = = 8.07 mA 1120 Ω VGS1 = Vsensor = 300 mV VOUT = 2.382 V 12 V − 2.382 V = 8.59 mA ID = 1120 Ω See Figure 8-7.

Figure 8-7 55.

⎛ R2 ⎞ ⎛ 50 kΩ ⎞ VGS2 = ⎜ ⎟VDD = ⎜ ⎟12 V = 4 V ⎝ 150 kΩ ⎠ ⎝ R1 + R2 ⎠ From the graph in Figure 8-79 in the textbook for VG1S = 0 and VG2S = 4 V: ID ≅ 8 mA VOUT = 12 V − (8 mA)(1120 Ω) = 3.04 V

81

Chapter 8 Datasheet Problems 56.

The 2N5457 is an n-channel JFET.

57.

From the datasheet in textbook Figure 8-14: (a) For a 2N5457, VGS(off) = −0.5 V minimum (b) For a 2N5457, VDS(max) = 25 V (c) For a 2N5458 @ 25°C, PD(max) = 310 mW (d) For a 2N5459, VGS(rev) = −25 V maximum

58.

PD(max) = 310 mW − (2.82 mW/°C)(65°C − 25°C) = 310 mW − 113 mW = 197 mW

59.

gm0(min) = gfs = 2000 μS

60.

Typical ID = IDSS = 9 mA

61.

From the datasheet graph in textbook Figure 8-80: ID ≅ 1.4 mA at VGS = 0

62.

For a 2N3796 with VGS = 6 V, ID = 15 mA

63.

From the datasheet graph in textbook Figure 8-80: At VGS = +3 V, ID ≅ 13 mA At VGS = −2 V, ID ≅ 0.4 mA

64.

yfs = 1500 μS at f = 1 kHz and at f = 1 MHz for both the 2N3796 and 2N3797. There is no change in gfs over the frequency range.

65.

For a 2N3796, VGS(off) = −3.0 V typical

Advanced Problems 66.

For the circuit of textbook Figure 8-81: 2

⎛ ⎞ V ID = I DSS ⎜1 − GS ⎟ where VGS = IDRS ⎜ VGS(off) ⎟ ⎝ ⎠ From the 2N5457 datasheet: IDSS(min) = 1.0 mA and VGS(off) = −0.5 V minimum ID = 66.3 μA VGS = −(66.3 μA)(5.6 kΩ) = −0.371 V VDS = 12 V − (66.3 μA)(10 kΩ + 5.6 kΩ) = 11.0 V

82

Chapter 8 67.

For the circuit of textbook Figure 8-82: ⎛ 3.3 kΩ ⎞ VC = ⎜ ⎟ 9 V = (0.248)(9 V) = 2.23 V ⎝ 13.3 kΩ ⎠ From the equation, 2

⎛ VGS ⎞ ⎟ where VGS = VG − IDRS ID = I DSS ⎜ ⎜ 1 − VGS(off) ⎟ ⎝ ⎠ ID is maximum for IDSS(max) and VGS(off) max, so that IDSS = 16 mA and VGS(off) = −8.0 V ID = 3.58 mA VGS = 2.23 V − (3.58 mA)(1.8 kΩ) = 2.23 V − 6.45 V = −4.21 V

68.

From the 2N5457 datasheet: IDSS(min) = 1.0 mA and VGS(off) = −0.5 minimum ID(min) = 66.3 μA VDS(max) = 12 V − (66.3 μA)(15.6 kΩ) = 11.0 V and IDSS(max) = 5.0 mA and VGS(off) = −6.0 maximum ID(max) = 677 μA VDS(min) = 12 V − (677 μA)(15.6 kΩ) = 1.4 V

69.

VpH = +300 mV ID = (2.9 mA)(1 + 0.3 V/5.0 V)2 = (2.9 mA)(1.06)2 = 3.26 mA VDS = 15 V − (3.26 mA)(2.76 kΩ) = 15 V − 8.99 V = +6.01 V

70.

⎛ (1 mA)RS ⎞ ⎟ 1 mA = I DSS ⎜1 − ⎜ ⎟ V GS(off) ⎝ ⎠

2

⎛ (1 mA) RS ⎞ 1 mA = 2.9 mA⎜1 − ⎟ − 0.5 V ⎠ ⎝

2

2

⎛ (1 mA) RS ⎞ 0.345 = ⎜1 − ⎟ − 0 .5 V ⎠ ⎝ (1 mA)RS 0.587 = 1 − − 0.5 V (1 mA)RS 0.413 = − 0.5 V RS = 2.06 kΩ Use RS = 2.2 kΩ. Then ID = 963 μA VGS = VS = (963 μA)(2.2 kΩ) = 2.19 V So, VD = 2.19 V + 4.5 V = 6.62 V 9 V − 6.62 V RD = = 2.47 kΩ 963 μ A Use RD = 2.4 kΩ. So, VDS = 9 V −(963 μA)(4.6 kΩ) = 4.57 V

83

Chapter 8 71.

Let ID = 20 mA. 4V = 200 Ω RD = 20 mA Let VS = 2 V. 2V RS = = 100 Ω 20 mA I D(on) 500 mA = K= = 6.17 mA/V2 2 (VGS(on) − VGS(th) ) (10 V − 1 V)2 Let ID = 20 mA. 20 V = 3.24 6.17 mA/V 2 VGS − 1 V = 1.8 V VGS = 2.8 V VG = VS + 2.8 V = 4.8 V For the voltage divider: R1 7.2 V = = 1.5 R2 4.8 V Let R2 = 10 kΩ. R1 = (1.5)(10 kΩ) = 15 kΩ See Figure 8-8. (VGS − 1 V)2 =

Figure 8-8

Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 72 through 80 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 72.

RS shorted

73.

RD shorted

74.

RG shorted

75.

R1 open

76.

Drain-source open

77.

RD open

78.

R2 shorted

79.

Drain-source shorted

80.

R1 shorted

84

Chapter 9 FET Amplifiers and Switching Circuits Section 9-1 The Common-Source Amplifier Id = gmVgs = (6000 μS)(10 mV) = 60 μA Id = gmVgs = (6000 μS)(150 mV) = 900 μA Id = gmVgs = (6000 μS)(0.6 V) = 3.6 mA Id = gmVgs = (6000 μS)(1 V) = 6 mA

1.

(a) (b) (c) (d)

2.

Av = gmRd A 20 Rd = v = = 5.71 kΩ gm 3500 μS

3. 4.

⎛ R r′ ⎞ ⎛ (4.7 kΩ)(12 kΩ) ⎞ Av = ⎜⎜ D ds ⎟⎟ g m = ⎜ ⎟ 4.2 mS = 14.2 16.7 kΩ ⎝ ⎠ ⎝ RD + rds′ ⎠ Rd = RD rds′ = 4.7 kΩ 12 kΩ = 3.38 kΩ Av =

5.

g m Rd (4.2 mS)(3.38 kΩ) = = 2.73 1 + g m Rs 1 + (4.2 mS)(1.0 kΩ)

(a) N-channel D-MOSFET with zero-bias. VGS = 0 V. (b) P-channel JFET with self-bias. VGS = −IDRS = (−3 mA)(330 Ω) = −0.99 V (c) N-channel E-MOSFET with voltage-divider bias.

⎛ R2 ⎞ 4.7 kΩ ⎞ ⎟⎟VDD = ⎛⎜ VGS = ⎜⎜ ⎟12 V = 3.84 V ⎝ 14.7 kΩ ⎠ ⎝ R1 + R2 ⎠ 6.

(a) VG = 0 V, VS = 0 V VD = VDD − IDRD = 15 V − (8 mA)(1.0 kΩ) = 7 V (b) VG = 0 V VS = −IDRD = −(3 mA)(330 Ω) = −0.99 V VD = −VDD + IDRD = −10 V + (3 mA)(1.5 kΩ) = −5.5 V

⎛ R2 ⎞ 4.7 kΩ ⎞ ⎟⎟VDD = ⎛⎜ (c) VG = ⎜⎜ ⎟12 V = 3.84 V ⎝ 14.7 kΩ ⎠ ⎝ R1 + R2 ⎠ VS = 0 V VD = VDD − IDRD = 12 V − (6 mA)(1.0 kΩ) = 6 V

85

Chapter 9 7.

(a) n-channel D-MOSFET (b) n-channel JFET (c) p-channel E-MOSFET

8.

From the curve in Figure 9-16(a) in the textbook: Id(pp) ≅ 3.9 mA − 1.3 mA = 2.6 mA

9.

From the curve in Figure 9-16(b) in the textbook: Id(pp) ≅ 6 mA − 2 mA = 4 mA From the curve in Figure 9-16(c) in the textbook: Id(pp) ≅ 4.5 mA − 1.3 mA = 3.2 mA

10.

VD = VDD − IDRD = 12 V − (2.83 mA)(1.5 kΩ) = 7.76 V VS = IDRS = (2.83 mA)(1.0 kΩ) = 2.83 V VDS = VD − VS = 7.76 V − 2.83 V = 4.93 V VGS = VG − VS = 0 V − 2.83 V = −2.83 V

11.

12.

Av = gmRd = g m (RD RL ) = 5000 μS(1.5 kΩ 10 kΩ ) = 6.52 Vpp(out) = (2.828)(50 mV)(6.52) = 920 mV Av = gmRd Rd = 1.5 kΩ 1.5 kΩ = 750 Ω Av = (5000 μS)(750 Ω) = 3.75 Vout = AvVin = (3.75)(50 mV) = 188 mV rms

13.

(a) Av = gmRd = g m (RD RL ) = 3.8 mS(1.2 kΩ 22 kΩ ) = 3.8 mS(1138 Ω) = 4.32

14.

See Figure 9-1.

(b) Av = gmRd = g m (RD RL ) = 5.5 mS(2.2 kΩ 10 kΩ ) = 5.5 mS(1.8 kΩ) = 9.92

Figure 9-1 15.

ID =

I DSS 15 mA = = 7.5 mA 2 2

86

Chapter 9 16.

VGS = (7.5 mA)(220 Ω) = 1.65 V 2 I DSS 2(15 mA) = = 7.5 mS gm0 = 4V VGS(off) gm = (7.5 mS)(1 − 1.65 V/4 V) = 4.41 mS (4.41 mS)(820 Ω 3.3 kΩ) (4.41 mS)(657 Ω) g m Rd = 1.47 Av = = = 1 + g m RS 1 + (4.41 mS)(220 Ω) 1 + 0.97

17.

18.

19.

20.

Av = gmRd = (4.41 mS)(820 Ω 3.3 kΩ 4.7 kΩ) = (4.41 mS)(576 Ω) = 2.54

I DSS 9 mA = = 4.5 mA 2 2 VGS = −IDRS = −(4.5 mA)(330 Ω) = −1.49 V VDS = VDD − ID(RD + RS) = 9 V − (4.5 mA)(1.33 kΩ) = 3 V ID =

Av = gmRd = g m (RD RL ) = 3700 μS(1.0 kΩ 10 kΩ ) = 3700 μS(909 Ω) = 3.36 Vout = AvVin = (3.36)(10 mV) = 33.6 mV rms

⎛ R2 ⎞ 6.8 kΩ ⎞ ⎟⎟VDD = ⎛⎜ VGS = ⎜⎜ ⎟ 20 V = 5.48 V ⎝ 24.8 kΩ ⎠ ⎝ R1 + R2 ⎠ I D(on) 18 mA = K= = 0.32 mA/V2 2 2 (VGS − VGS(th) ) (10 V − 2.5 V) ID = K(VGS − VGS(th))2 = 0.32 mA/V2(5.48 V − 2.5 V)2 = 2.84 mA VDS = VDD − IDRD = 20 V − (2.84 mA)(1.0 kΩ) = 17.2 V

21.

RIN =

VGS − 15 V = = 600 MΩ I GSS 25 nA

Rin = 10 MΩ 600 MΩ = 9.84 MΩ 22.

Av = gmRd = 48 mS(1.0 kΩ 10 MΩ ) ≅ 4.8 Vout = AvVin = 4.8(10 mV) = 48 mV rms ID = IDSS = 15 mA VD = 24 V − (15 mA)(1.0 kΩ) = 9 V See Figure 9-2.

Figure 9-2

87

Chapter 9 23.

⎛ R2 ⎞ 47 kΩ ⎞ ⎟⎟VDD = ⎛⎜ VGS = ⎜⎜ ⎟18 V = 9 V ⎝ 94 kΩ ⎠ ⎝ R1 + R2 ⎠ I D(on) 8 mA = K= = 0.125 mA/V2 2 (VGS − VGS(th) ) (12 V − 4 V)2 ID(on) = K(VGS − VGS(th))2 = 0.125 mA/V2(9 V − 4 V)2 = 3.13 mA VDS = VDD − IDRD = 18 V − (3.125 mA)(1.5 kΩ) = 13.3 V Av = gmRD = 4500 μS(1.5 kΩ) = 6.75 Vds = AvVin = 6.75(100 mV) = 675 mV rms

Section 9-2 The Common-Drain Amplifier 24.

Rs = 1.2 kΩ 1 kΩ ≅ 545 Ω Av = RIN =

g m Rs (5500 μS)(545 Ω) = = 0.750 1 + g m Rs 1 + (5500 μS)(545 Ω) VGS − 15 V = = 3 × 1011 Ω I GSS 50 pA

Rin = 10 MΩ 3 × 1011 Ω ≅ 10 MΩ 25.

Rs = 1.2 kΩ 1 kΩ ≅ 545 Ω Av = RIN =

g m Rs (3000 μS)(545 Ω) = = 0.620 1 + g m Rs 1 + (3000 μS)(545 Ω) VGS − 15 V = = 3 × 1011 Ω I GSS 50 pA

Rin = 10 MΩ 3 × 1011 Ω ≅ 10 MΩ 26.

(a) Rs = 4.7 kΩ 47 kΩ = 4.27 kΩ Av =

g m Rs (3000 μS)(4.27 kΩ) = = 0.928 1 + g m Rs 1 + (3000 μS)(4.27 kΩ)

(b) Rs = 1.0 kΩ 100 Ω = 90.9 Ω Av =

27.

g m Rs (4300 μS)(90.9 Ω) = = 0.281 1 + g m Rs 1 + (4300 μS)(90.9 Ω)

(a) Rs = 4.7 kΩ 10 kΩ = 3.2 kΩ Av =

g m Rs (3000 μS)(3.2 kΩ) = = 0.906 1 + g m Rs 1 + (3000 μS)(3.2 kΩ)

(b) Rs = 100 Ω 10 kΩ = 99 Ω Av =

g m Rs (4300 μS)(99 Ω) = = 0.299 1 + g m Rs 1 + (4300 μS)(99 Ω)

88

Chapter 9 Section 9-3 The Common-Gate Amplifier 28.

Av = gmRd = 4000μS(1.5 kΩ) = 6.0

29.

Rin(source) =

30.

Av = gmRd = 3500μS(10 kΩ) = 35 ⎛ 1 ⎞ ⎛ ⎞ 1 ⎟ = 2.2 kΩ ⎜⎜ ⎟⎟ = 253 Ω Rin = RS ⎜⎜ ⎟ ⎝ 3500 μS ⎠ ⎝ gm ⎠

31.

XL = 2πfL = 2π(100 MHz)(1.5 mH) = 943 kΩ Av = gm(CG)XL = (2800 μS)(943 kΩ) = 2640 ⎛V ⎞ ⎛ 15 V ⎞ Rin = R3 & ⎜ GS ⎟ = 15 MΩ & ⎜ ⎟ ⎝ 2 nA ⎠ ⎝ I GSS ⎠

1 1 = = 250 Ω g m 4000 μS

= 15 MΩ & 500 MΩ = 14.6 MΩ

Section 9-4 The Class D Amplifier 2(9 V) 18 V = = 3600 5 mV 5 mV

32.

Av =

33.

Pout = (12 V)(0.35 A)= 4.2 W Pint = (0.25 V)(0.35 A) + 140 mW = 87.5 mW + 140 mW = 227.5 mW Pout 4.2 W η= = = 0.95 Pout + Pint 4.2 W + 227.5 mW

Section 9-5 MOSFET Analog Switching 34.

VG − Vp(out) = VGS(Th) Vp(out) = VG − VGS(Th) = 8 V − 4 V = 4 V Vpp(in) = 2 Vp(out) = 2 × 4 V = 8 V

35.

fmin = 2 × 15 kHz = 30 kHz

36.

37.

1 fC 1 1 = f= = 10 MHz RC (10 kΩ)(10 pF) R=

R=

1 1 = = 40 kΩ fC (25 kHz)(0.001 μ F)

89

Chapter 9 Section 9-6 MOSFET Digital Switching 38.

Vout = + 5 V when Vin = 0 Vout = 0 V when Vin = + 5 V

39.

(a) Vout = 3.3 V (c) Vout = 3.3 V

(b) (d)

Vout = 3.3 V Vout = 0 V

40.

(a) Vout = 3.3 V (c) Vout = 0 V

(b) (d)

Vout = 0 V Vout = 0 V

41.

The MOSFET has lower on-state resistance and can turn off faster.

Section 9-7 Troubleshooting 42.

(a) (b) (c) (d) (e)

VD1 = VDD; No signal at Q1 drain; No output signal VD1 ≅ 0 V (floating); No signal at Q1 drain; No output signal VGS1 = 0 V; VS = 0 V; VD1 less than normal; Clipped output signal Correct signal at Q1 drain; No signal at Q2 gate; No output signal VD2 = VDD; Correct signal at Q2 gate; No Q2 drain signal or output signal

43.

(a) Vout = 0 V if C1 is open. (b) Av1 = gmRd = 5000 μS(1.5 kΩ) = 7.5 g m Rd 7.5 = = 2.24 Av2 = 1 + g m Rs 1 + (5000 μS)(470 Ω) Av = Av1Av2 = (7.5)(2.24) = 16.8 Vout = AvVin = (16.8)(10 mV) = 168 mV (c) VGS for Q2 is 0 V, so ID = IDSS. The output is clipped. (d) No Vout because there is no signal at the Q2 gate.

Datasheet Problems 44.

The 2N3796 FET is an n-channel D-MOSFET.

45.

(a) (b) (c) (d)

46.

PD = 200 mW − (1.14 mW/°C)(55°C − 25°C) = 166 mW

47.

For a 2N3796 with f = 1 kHz, gm0 = 900 μS minimum

48.

At VGS = 3.5 V and VDS = 10 V, ID(min) = 9.0 mA, ID(typ) = 14 mA, ID(max) = 18 mA

49.

For a zero-biased 2N3796, ID(typ) = 1.5 mA

For a 2N3796, the typical VGS(off) = −3.0 V For a 2N3797, VDS(max) = 20 V At TA = 25°C, PD(max) = 200 mW For a 2N3797, VGS(max) = ±10 V

90

Chapter 9 50.

Av(max) = (1800 μS)(2.2 kΩ) = 3.96

Advanced Problems 51.

Rd(min) = 1.0 kΩ 4 kΩ = 800 Ω Av(min) = (2.5 mS)(800 Ω) = 2.0 Rd(max) = 1.0 kΩ 10 kΩ = 909 Ω Av(min) = (7.5 mS)(909 Ω) = 6.82

52.

IDSS(typ) = 2.9 mA 12 V RD + RS = = 4.14 kΩ 2.9 mA 1 1 = = 435 Ω g m 2300 μS If RS = 0 Ω, then RD ≅ 4 kΩ (3.9 kΩ standard) Av = (2300 μS)(3.9 kΩ) = 8.97 VDS = 24 V − (2.9 mA)(3.9 kΩ) = 24 V − 11.3 V = 12.7 V The circuit is a common-source zero-biased amplifier with a drain resistor of 3.9 kΩ.

53.

To maintain VDS = 12 V for the range of IDSS values: For IDSS(min) = 2 mA 12 V RD = = 6 kΩ 2 mA For IDSS(max) = 6 mA 12 V = 2 kΩ RD = 6 mA To maintain Av = 9 for the range of gm(yfs) values: For gm(min) = 1500 μS 9 RD = = 6 kΩ 1500 μS For gm(max) = 3000 μS 9 RD = = 3 kΩ 3000 μS A drain resistance consisting of a 2.2 kΩ fixed resistor in series with a 5 kΩ variable resistor will provide more than sufficient range to maintain a gain of 9 over the specified range of gm values. The dc voltage at the drain will vary with adjustment and depends on IDSS. The circuit cannot be modified to maintain both VDS = 12 V and Av = 9 over the full range of transistor parameter values. See Figure 9-3.

Figure 9-3

91

Chapter 9 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 54 through 62 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 54.

Drain-source shorted

55.

C2 open

56.

C1 open

57.

RS shorted

58.

Drain-source open

59.

R1 open

60.

RD open

61.

R2 open

62.

C2 open

92

Chapter 10 Amplifier Frequency Response Section 10-1 Basic Concepts 1.

If C1 = C2, the critical frequencies are equal, and they will both cause the gain to decrease at 40 dB/decade below fc.

2.

At sufficiently high frequencies, the reactances of the coupling capacitors become very small and the capacitors appear effectively as shorts; thus, negligible signal voltage is dropped across them.

3.

BJT: Cbe, Cbc, and Cce FET: Cgs, Cgd, and Cds

4.

Low-frequency response: C1, C2, and C3 High-frequency response: Cbc, Cbe, and Cce

5.

6. 7.

⎛ R2 ⎞ ⎛ 4.7 kΩ ⎞ ⎟⎟VCC − 0.7 V = ⎜ VE ≅ ⎜⎜ ⎟ 20 V − 0.7 V = 1.79 V ⎝ 37.7 kΩ ⎠ ⎝ R1 + R2 ⎠ V 1.79 V = 3.2 mA IE = E = RE 560 Ω 25 mV re′ = = 7.8 Ω 3.2 mA 2.2 kΩ 5.6 kΩ R Av = c = = 202 re′ 7.8 Ω Cin(miller) = Cbc(Av + 1) = 4 pF(202 + 1) = 812 pF

⎛ A +1⎞ 203 ⎞ ⎟ = 4 pF⎛⎜ Cout(miller) = Cbc ⎜⎜ v ⎟ = 4 pF ⎟ ⎝ 202 ⎠ ⎝ Av ⎠ ID = 3.36 mA using Eq. 9−2 and a programmable calculator. VGS = −(3.36 mA)(1.0 kΩ) = −3.36 V 2(10 mA) = 2.5 mS gm0 = 8V ⎛ 3.36 V ⎞ gm = (2.5 mS) ⎜1 − ⎟ = 1.45 mS 8V ⎠ ⎝ Av = gmRd = (1.45 mS) (1.0 kΩ 10 kΩ) = 1.32 Cgd = Crss = 3 pF Cin(miller) = Cgd(Av + 1) = 3 pF(2.32) = 6.95 pF ⎛ A +1⎞ 2.32 ⎞ ⎟ = 3 pF⎛⎜ Cout(miller) = C gd ⎜⎜ v ⎟ = 5.28 pF ⎟ ⎝ 1.32 ⎠ ⎝ Av ⎠

93

Chapter 10 Section 10-2 The Decibel 8.

Ap =

Pout 5W = = 10 Pin 0.5 W

⎛P Ap(dB) = 10 log⎜⎜ out ⎝ Pin

⎞ ⎟ = 10 log 10 = 10 dB ⎟ ⎠

Vout 1.2 V = 24 mV rms = Av 50 Av(dB) = 20 log(Av) = 20 log 50 = 34.0 dB

9.

Vin =

10.

⎛ 25 ⎞ The gain reduction is 20 log⎜ ⎟ = −8.3 dB ⎝ 65 ⎠

11.

12.

⎛ 2 mW ⎞ (a) 10 log⎜ ⎟ = 3.01 dBm ⎝ 1 mW ⎠ ⎛ 1 mW ⎞ (b) 10 log⎜ ⎟ = 0 dBm ⎝ 1 mW ⎠ ⎛ 4 mW ⎞ (c) 10 log⎜ ⎟ = 6.02 dBm ⎝ 1 mW ⎠ ⎛ 0.25 mW ⎞ (d) 10 log⎜ ⎟ = −6.02 dBm ⎝ 1 mW ⎠ ⎛ 4.7 kΩ ⎞ VB = ⎜ ⎟ 20 V = 1.79 V ⎝ 37.7 kΩ ⎠ 1.79 V = 3.20 mA IE = 560 Ω 25 mV re′ = = 7.81 Ω 3.2 mA 5.6 kΩ 2.2 kΩ = 202 Av = 7.81 Ω Av(dB) = 20 log(202) = 46.1 dB At the critical frequencies, Av(dB) = 46.1 dB − 3 dB = 43.1 dB B

Section 10-3 Low-Frequency Amplifier Response 13.

1 1 = = 318 Hz 2πRC 2π (100 Ω)(5 μF) 1 1 = (b) fc = = 1.59 kHz 2πRC 2π (1.0 kΩ)(0.1 μF)

(a) fc =

94

Chapter 10 14.

RIN(BASE) = βDCRE = 12.5 kΩ ⎛ R2 RIN(BASE) ⎞ ⎛ ⎞ 4.7 kΩ 12.5 kΩ ⎟ 9 V − 0.7 V = ⎜ VE = ⎜ ⎟ 9 V − 0.7 V = 1.3 V ⎜ ⎜ R1 + R2 RIN(BASE) ⎟ 12 kΩ + 4.7 kΩ 12.5 kΩ ⎟⎠ ⎝ ⎝ ⎠ VE 1.3 V = IE = = 13 mA RE 100 Ω 25 mV re′ = = 1.92 Ω 13 mA Rin(base) = β ac re′ = (125)(1.92 Ω) = 240 Ω Rin = 50 Ω + Rin(base) R1 R2 = 50 Ω + 240 Ω 12 kΩ 4.7 kΩ = 274 Ω For the input circuit: 1 1 = fc = = 581 Hz 2πRinC1 2π ( 274 Ω)(1 μF) For the output circuit: 1 1 = = 177 Hz fc = 2π ( RC + RL )C3 2π (900 Ω)(1 μF) For the bypass circuit: RTH = R1 R2 Rs = 12 kΩ 4.7 kΩ 50 Ω ≅ 49.3 Ω

fc =

1 2π (re′ + RTH / β DC

RC

RL

=

RE )C2

=

1 = 6.89 kHz 2π (2.31 Ω)(10 μF)

220 Ω 680 Ω

= 86.6 re′ 1.92 Ω Av(dB) = 20 log(86.6) = 38.8 dB The bypass circuit produces the dominant low critical frequency. See Figure 10-1. Av =

Figure 10-1

95

Chapter 10 15.

From Problem 14: Av(mid) = 86.6 Av(mid) (dB) = 38.8 dB For the input RC circuit: fc = 578 Hz For the output RC circuit: fc = 177 Hz For the bypass RC circuit: fc = 6.89 kHz The fc of the bypass circuit is the dominant low critical frequency. At f = fc = 6.89 kHz: Av = Av(mid) − 3 dB = 38.8 dB − 3 dB = 35.8 dB At f = 0.1fc: Av = 38.8 dB − 20 dB = 18.8 dB At 10fc (neglecting any high frequency effects): Av = Av(mid) = 38.8 dB

16.

At f = fc = XC = R ⎛X ⎞ θ = tan −1 ⎜ C ⎟ = tan −1 (1) = 45° ⎝ R ⎠ At f = 0.1fc, XC = 10R. θ = tan−1(10) = 84.3° At f = 10fc, XC = 0.1R. θ = tan−1(0.1) = 5.7°

17.

Rin(gate) = Rin = RG

VGS − 10 V = = 200 MΩ I GSS 50 nA Rin ( gate ) = 10 MΩ 200 MΩ = 9.52 MΩ

For the input circuit: 1 1 = fc = = 3.34 Hz 2πRinC1 2π (9.52 MΩ)(0.005 μF) For the output circuit: 1 1 = = 3.01 kHz fc = 2π ( RD + RL )C2 2π (560 Ω + 10 kΩ)(0.005 μF) The output circuit is dominant. See Figure 10-2. (Av is determined in Problem 18.)

Figure 10-2

96

Chapter 10 18.

2(15 mA) = 5 mS 6V Av(mid) = g m ( RD RL ) = 5 mS(560 Ω 10 kΩ) = 2.65 Av(mid) (dB) = 8.47 dB At fc: Av = 8.47 dB − 3 dB = 5.47 dB At 0.1fc: Av = 8.47 dB − 20 dB = −11.5 dB At 10fc: Av = Av(mid) = 8.47 dB (if 10fc is still in midrange) gm = gm0 =

Section 10-4 High-Frequency Amplifier Response 19.

From Problems 14 and 15: re′ = 1.92 Ω and Av(mid) = 86.6 Input circuit: Cin(miller) = Cbc(Av + 1) = 10 pF(87.6) = 876 pF Ctot = Cbe + Cin ( miller ) = 25 pF + 876 pF = 901 pF fc =

2π ( Rs

R1

1 R2

β ac re′) Ctot

=

1 = 4.32 MHz 2π ( 50 Ω 12 kΩ 4.7 kΩ 240 Ω ) 901 pF

Output circuit: ⎛ A + 1⎞ 87.6 ⎞ ⎟ = 10 pF⎛⎜ Cout(miller) = Cbc ⎜⎜ v ⎟ = 10.1 pF ⎟ ⎝ 86.6 ⎠ ⎝ Av ⎠ 1 1 = 94.9 MHz fc = = 2πRcCout ( miller ) 2π (166 Ω)(10.1 pF) Therefore, the dominant high critical frequency is determined by the input circuit: fc = 4.32 MHz. See Figure 10-3.

Figure 10-3

97

Chapter 10 20.

At f = 0.1fc = 458 kHz: Av = Av(mid) = 38.8 dB At f = fc = 4.58 MHz: Av = Av(mid) − 3 dB = 38.8 dB − 3 dB = 35.8 dB At f = 10fc = 45.8 MHz: Av = Av(mid) − 20 dB = 38.8 dB − 20 dB = 18.8 dB At f = 100fc = 458 MHz: The roll-off rate changes to −40 dB/decade at f = 94.6 MHz. So, for frequencies from 45.8 MHz to 94.6 MHz, the roll-off rate is −20 dB/decade and above 94.6 MHz it is −40 dB/decade. The change in frequency from 45.8 MHz to 94.6 MHz represents 94.6 MHz − 45.8 MHz × 100% = 11.8% 458 MHz − 45.8 MHz So, for 11.8% of the decade from 45.8 MHz to 458 MHz, the roll-off rate is −20 dB/decade and for the remaining 88.2% of the decade, the roll-off rate is −40 dB/ decade. Av = 18.8 dB − (0.118)(20 dB) − (0.882)(40 dB) = 18.8 dB − 2.36 dB − 35.3 dB = −18.9 dB

21.

Cgd = Crss = 4 pF Cgs = Ciss − Crss = 10 pF − 4 pF = 6 pF Input circuit: Cin(miller) = Cgd(Av + 1) = 4 pF(2.65 + 1) = 14.6 pF Ctot = C gs + Cin ( miller ) = 6 pF + 14.6 pF = 20.6 pF 1 1 = = 12.9 MHz 2π Rs Ctot 2π (600 Ω)(20.6 pF) Output circuit: ⎛ A + 1⎞ 2.65 + 1 ⎞ ⎟ = 4 pF⎛⎜ Cout(miller) = C gd ⎜⎜ v ⎟ = 5.51 pF ⎟ ⎝ 2.65 ⎠ ⎝ Av ⎠ 1 1 = fc = = 54.5 MHz 2πRd C out ( miller ) 2π (530 Ω)(5.51 pF) The input circuit is dominant. fc =

22.

From Problem 21: For the input circuit, fc = 12.9 MHz and for the output circuit, fc = 54.5 MHz. The dominant critical frequency is 12.9 MHz. At f = 0.1fc = 1.29 MHz: Av = Av(mid) = 8.47 dB, θ = 0° At f = fc = 12.9 MHz: Av = Av(mid) − 3 dB = 8.47 dB − 3 dB = 5.47 dB, θ = tan−1(1) = 45° At f = 10fc = 129 MHz: From 12.9 MHz to 54.5 MHz the roll-off is −20 dB/decade. From 54.5 MHz to 129 MHz the roll-off is −40 dB/decade. The change in frequency from 12.9 MHz to 54.5 MHz represents 54.5 MHz − 12.9 MHz × 100% = 35.8% 129 MHz − 12.9 MHz So, for 35.8% of the decade, the roll-off rate is −20 dB/decade and for 64.2% of the decade, the rate is −40 dB/decade. Av = 5.47 dB − (0.358)(20 dB) − (0.642)(40 dB) = −27.4 dB At f = 100fc = 1290 MHz: Av = −27.4 dB − 40 dB = −67.4 dB

98

Chapter 10 Section 10-5 Total Amplifier Frequency Response 23.

fcl = 136 Hz fcu = 8 kHz

24.

From Problems 14 and 19: fcu = 4.32 MHz and fcl = 6.89 kHz BW = fcu − fcl = 4.32 MHz − 6.89 kHz = 4.313 MHz

25.

ftot = (BW)Av(mid) f tot 200 MHz = BW = = 5.26 MHz Av ( mid ) 38 Therefore, fcu ≅ BW = 5.26 MHz

26.

6 dB/octave roll-off: At 2fcu: Av = 50 dB − 6 dB = 44 dB At 4fcu: Av = 50 dB − 12 dB = 38 dB 20 dB/decade roll-off: At 10fcu: Av = 50 dB − 20 dB = 30 dB

Section 10-6 Frequency Response of Multistage Amplifiers 27.

Dominant f cl′ = 230 Hz Dominant f cu′ = 1.2 MHz

28.

BW = 1.2 MHz − 230 Hz ≅ 1.2 MHz

29.

f cl′ =

400 Hz 21/2 − 1

=

400 Hz = 622 Hz 0.643

f cu′ = (800 kHz) 21/2 − 1 = 0.643(800 kHz) = 515 kHz BW = 515 kHz − 622 Hz ≅ 514 kHz 30.

31.

f cl′ =

50 Hz

f cl′ =

125 Hz

1/3

2

−1

=

50 Hz = 98.1 Hz 0.510

=

125 Hz = 194 Hz 0.643

21/2 − 1 f cu′ 2.5 MHz BW = 2.5 MHz − 194 Hz ≅ 2.5 MHz

99

Chapter 10 Section 10-7 Frequency Response Measurements 32.

fcl =

0.35 0.35 = = 350 Hz tf 1 ms

fcu =

0.35 0.35 = = 17.5 MHz tr 20 ns

33.

Increase the frequency until the output voltage drops to 3.54 V (3 dB below the midrange output voltage). This is the upper critical frequency.

34.

tr ≅ 3 div × 5 μs/div = 15 μs tf ≅ 6 div × 0.1 ms/div = 600 μs 0.35 0.35 = = 583 Hz fcl = tf 600 μs 0.35 0.35 = = 23.3 kHz tr 15 μs BW = 23.3 kHz − 583 Hz = 22.7 kHz fcu =

Application Activity Problems 35.

Q1 stage: 1 1 = = 1.77 Hz 2π ( R1 & R2 & β ac R4 )C1 2π (90 kΩ)1 μ F 1 1 = = 15.9 Hz fcl(bypass) = 2π R4 C2 2π (1 kΩ)10 μ F 1 1 = fcl(output) = = 4.42 Hz 2π ( R5 + R6 & R7 & β ac ( R9 + R10 )C3 2π (36 kΩ)1 μ F fcl(input) =

Q2 stage: 1 1 = = 4.42 Hz 2π ( R5 & R6 & R7 & β ac ( R9 + R10 )C3 2π (36 kΩ)1 μ F 1 1 fcl(bypass) = = = 7.76 Hz Ω π 2 (205 )100 μ F ⎛ R6 & R7 ⎞ 2π ⎜ R9 + ⎟C β ac ⎠ 4 ⎝ 1 1 = = 4.45 Hz fcl(output) = 2π ( R8 + RL )C5 2π (35.8 kΩ)1 μ F The dominant critical frequency of 15.9 Hz is set by the Q1 bypass circuit. fcl(input) =

36.

Changing to 1 μF coupling capacitors does not significantly affect the overall bandwidth because the upper critical frequency is much greater than the dominant lower critical frequency.

100

Chapter 10 37.

Increasing the load resistance on the output of the second stage has no effect on the dominant lower critical frequency because the critical frequency of the output circuit will decrease and the critical frequency of the first stage input circuit will remain dominant.

38.

The Q1 stage bypass circuit set the dominant critical frequency. 1 1 = = 15.9 Hz fcl(bypass) = 2π R4 C2 2π (1 kΩ)10 μ F This frequency is not dependent on βac and is not affected.

Datasheet Problems 39.

Cin(tot) = (25 + 1)4 pF + 8 pF = 112 pF

40.

BWmin =

41.

Cgd = Crss = 1.3 pF Cgs = Ciss − Crss = 5 pF − 1.3 pF = 3.7 pF Cds = Cd − Crss = 5 pF − 1.3 pF = 3.7 pF

fT Av ( mid )

=

300 MHz = 6 MHz 50

Advanced Problems 42.

From Problem 12: re′ = 7.81 Ω and IE = 3.2 mA VC ≅ 20 V − (3.2 mA)(2.2 kΩ) = 13 V dc The maximum peak output signal can be approximately 6 V. The maximum allowable gain for the two stages is 6V = 424 Av(max) = 1.414(10 mV) For stage 1: Rc = 2.2 kΩ 33 kΩ 4.7 kΩ (150)(7.81 Ω) = 645 Ω Av1=

645 Ω = 82.6 7.81 Ω

For stage 2: Rc = 2.2 kΩ 5.6 kΩ = 1.58 kΩ 1.58 kΩ = 202 7.81 Ω Av(tot) = (82.6)(202) = 16,685

Av1=

The amplifier will not operate linearly with a 10 mV rms input signal. The gains of both stages can be reduced or the gain of the second stage only can be reduced.

101

Chapter 10 One approach is leave the gain of the first stage as is and bypass a portion of the emitter resistance in the second stage to achieve a gain of 424/82.6 = 5.13. Rc = 5.13 Re + re′ R − 5.13re′ 1.58 kΩ − 40.1 Ω = Re = c = 300 Ω 5.13 5.13 Modification: Replace the 560 Ω emitter resistor in the second stage with an unbypassed 300 Ω resistor and a bypassed 260 Ω resistor (closest standard value is 270 Ω).

Av =

43.

From Problems 17, 18, and 21: Ctot = C gs + Cin ( miller ) = 20.6 pF ⎛ 2.65 + 1 ⎞ Cout(miller) = 4 pF⎜ ⎟ = 5.51 pF ⎝ 2.65 ⎠ Stage 1: 1 1 = 3.34 Hz fcl(in) = = 2πRin C1 2π (9.52 MΩ)(0.005 μF) 1 = 3.34 Hz since Rin(2) >> 560 Ω fcl(out) = 2π (9.52 MΩ)(0.005 μF) 1 fcu(in) = = 12.9 MHz 2π (600 Ω)(20.6 pF) 1 = 10.9 MHz fcu(out) = 2π (560 Ω)(20.6 pF + 5.51 pF) Stage 2: 1 1 = 3.34 Hz fcl(in) = = 2πRin C1 2π (9.52 MΩ)(0.005 μF) 1 = 3.01 kHz fcl(out) = 2π (10.6 kΩ)(0.005 μF) 1 fcu(in) = = 10.9 MHz 2π (560 Ω)(20.6 pF + 5.51 pF) 1 fcu(out) = = 54.5 MHz 2π (560 Ω 10 kΩ)(5.51 pF) Overall: fcl(in) = 3.34 kHz and fcu(in) = 10.9 MHz BW ≅ 10.9 MHz

102

Chapter 10 44.

Rin(1) = 22 kΩ (100)(320 Ω) = 13 kΩ ⎛ 13 kΩ ⎞ VB(1) = ⎜ ⎟ 12 V = 1.38, VE(1) = 0.681 V ⎝ 113 kΩ ⎠ 0.681 V = 2.13 mA, re′ = 11.7 Ω IE(1) = 320 Ω Rc(1) = 4.7 kΩ 33 kΩ 22 kΩ (100)(100 Ω) = 2.57 kΩ Av(1) =

2.57 kΩ = 23 112 Ω

Rin(2) = 22 kΩ (100)(1010 Ω) = 18 kΩ ⎛ 18 kΩ ⎞ VB(2) = ⎜ ⎟ 12 V = 4.24, VE(1) = 3.54 V ⎝ 51 kΩ ⎠ 3.54 V = 3.51 mA, re′ = 7.13 Ω IE(2) = 1.01 kΩ Rc(2) = 3 kΩ 10 kΩ = 2.31 kΩ 2.31 kΩ = 24 maximum 107.13 Ω 2.31 kΩ Av(2) = = 2.27 minimum 101 kΩ + 7.13 Ω

Av(2) =

Av(tot) = (23)(24) = 552 maximum Av(tot) = (23)(2.27) = 52.3 minimum This is a bit high, so adjust Rc(1) to 3 kΩ, then 3 kΩ 22 kΩ 33 kΩ 101 kΩ Av(1) = = 21.4 112 Ω Now, Av(tot) = (21.3)(24) = 513 maximum Av(tot) = (21.3)(2.27) = 48.5 minimum Thus, Av is within 3% of the desired specifications. Frequency response for stage 1: Rin = 22 kΩ 100 kΩ 32 kΩ = 11.5 kΩ fcl(in) =

1 = 1.38 Hz 2π (11.5 kΩ)(10 μF)

Remitter = 220 Ω (100 Ω + 11.7 Ω + (22 kΩ 100 kΩ /100) = 125 Ω fcl(bypass) =

1 = 12.7 Hz 2π (125 Ω)(100 μF)

Rout = 3 kΩ + (33 kΩ 22 kΩ (100)(107 Ω)) = 8.91 kΩ fcl(out) =

1 = 1.79 Hz 2π (8.91 kΩ)(10 μF)

103

Chapter 10 Frequency response for stage 2: fcl(in) = 1.79 Hz (same as fcl(out) for stage 1) Rout = 3 kΩ + 10 kΩ = 13 kΩ 1 fcl(out) = = 1.22 Hz 2π (13 kΩ)(10 μF) This means that CE(2) is the frequency limiting capacitance. Remitter 910 Ω (100 Ω + 7 Ω + (22 kΩ 33 kΩ 3 kΩ) / 100) = 115 Ω For f cl′ = 1 kHz: 1 = 1.38 μF 2π (115 Ω)(1 kHz) 1.5 μF is the closest standard value and gives 1 fcl(bypass) = = 922 Hz 2π (115 Ω)(1.5 μF) This value can be moved closer to 1 kHz by using additional parallel bypass capacitors in stage 2 to fine-tune the response.

CE(2) =

Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 45 through 48 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 45.

RC open

46.

Output capacitor open

47.

R2 open

48.

Drain-source shorted

104

Chapter 11 Thyristors Section 11-1 The Four-Layer Diode 1.

VA = VBE + VCE(sat) = 0.7 V + 0.2 V = 0.9 V V RS = VBIAS − VA = 25 V − 0.9 V = 24.1 V

IA =

2.

VRS RS

=

24.1 V = 24.1 mA 1.0 kΩ

VAK 15 V = = 15 MΩ I A 1 μA (b) From 15 V to 50 V for an increase of 35 V. (a) RAK =

Section 11-2 The Silicon-Controlled Rectifier (SCR) 3.

See Section 11-2 in the textbook.

4.

Neglecting the SCR voltage drop, 30 V − 0.7 V Rmax = = 2.93 kΩ 10 mA

5.

When the switch is closed, the battery V2 causes illumination of the lamp. The light energy causes the LASCR to conduct and thus energize the relay. When the relay is energized, the contacts close and 115 V ac are applied to the motor.

6.

See Figure 11-1.

Figure 11-1

105

Chapter 11 Section 11-3 SCR Applications 7.

Add a transistor to provide inversion of the negative half-cycle in order to obtain a positive gate trigger.

8.

D1 and D2 are full-wave rectifier diodes.

9.

See Figure 11-2.

Figure 11-2

Section 11-4 The Diac and Triac 10.

Vin(p) = 1.414Vin(rms) = 1.414(25 V) = 35.4 V 35.35 V Ip = Vin(p) = = 35.4 mA 1.0 kΩ 20 V = 20 mA Current at breakover = 1 .0 k Ω See Figure 11-3.

Figure 11-3

106

Chapter 11 11.

15 V = 3.19 mA 4.7 kΩ See Figure 11-4.

Ip =

Figure 11-4

Section 11-5 The Silicon-Controlled Switch (SCS) 12.

See Section 11-5 in the text.

13.

Anode, cathode, anode gate, and cathode gate

Section 11-6 The Unijunction Transistor (UJT) rB′1 2 .5 k Ω = = 0.385 ′ ′ rB1 + rB 2 2 .5 k Ω + 4 k Ω

14.

η=

15.

Vp = ηVBB + Vpn = 0.385(15 V) + 0.7 V = 6.48 V

16.

VBB − Vv V − VP < R1 < BB Iv Ip 12 V − 0.8 V 12 V − 10 V < R1 < 15 mA 10 μA 747 Ω < R1 < 200 kΩ

Section 11-7 The Programmable UJT (PUT) 17.

⎛ R3 (a) VA = ⎜⎜ ⎝ R 2 + R3 ⎛ R3 (b) VA = ⎜⎜ ⎝ R 2 + R3

⎞ 10 kΩ ⎞ ⎟⎟VB + 0.7 V = ⎛⎜ ⎟ 20 V + 0.7 V = 9.79 V ⎝ 22 kΩ ⎠ ⎠ ⎞ 47 kΩ ⎞ ⎟⎟VB + 0.7 V = ⎛⎜ ⎟ 9 V + 0.7 V = 5.2 V 94 ⎝ kΩ ⎠ ⎠

107

Chapter 11 18.

(a) From Problem 17(a), VA = 9.79 V at turn on. 9.79 V = 20.8 mA at turn on I= 470 Ω 10 V Ip = = 21.3 mA 470 Ω See Figure 11-5.

Figure 11-5

(b) From Problem 17(b), VA = 5.2 V at turn on. 5.2 V I = = 15.8 mA at turn on 330 Ω 10 V = 30.3 mA Ip = 330 Ω See Figure 11-6.

Figure 11-6

108

Chapter 11 19.

⎛ R3 ⎞ 10 kΩ ⎞ ⎟⎟ 6 V + 0.7 V = ⎛⎜ VA = ⎜⎜ ⎟ 6 V + 0.7 V = 3.7 V at turn on ⎝ 20 kΩ ⎠ ⎝ R 2 + R3 ⎠ VR1 ≅ VA = 3.7 V at turn on. See Figure 11-7.

Figure 11-7 20.

⎛ 15 kΩ ⎞ VA = ⎜ ⎟ 6 V + 0.7 V ⎝ 25 kΩ ⎠ = 4.3 V at turn on VR1 ≅ VA = 4.3 V See Figure 11-8.

Figure 11-8

Application Activity Problems 21.

The motor runs fastest at 0 V for the motor speed control circuit.

22.

If the rheostat resistance decreases, the SCR turns on earlier in the ac cycle.

23.

As the PUT gate voltage increases in the circuit, the PUT triggers on later in the ac cycle causing the SCR to fire later in the cycle, conduct for a shorter time, and decrease the power to the motor.

109

Chapter 11 Advanced Problems 24.

D1: R1: R2: Q1: R3:

15 V zener (1N4744) 100 Ω, 1 W 100 Ω, 1 W Any SCR with a 1 A minimum rating (1.5 A would be better) 150 Ω, 1 W

25.

See Figure 11-9.

Figure 11-9 26.

Vp = ηVBB + Vpn = (0.75)(12 V) + 0.7 V = 9.7 V Iv = 10 mA and Ip = 20 μA 12 V − 9.7 V = 115 kΩ R1 < 20 μA 12 V − 1 V R1 > = 1.1 kΩ 10 mA Select R1 = 51 kΩ as an intermediate value. During the charging cycle: V(t) = VF − (VF − V0 )e − t1 / R1C 9.7 V = 12 V − (12 V − 1 V)e −t1 / R1C −

t1 ⎛ 2 .3 V ⎞ = ln⎜ ⎟ R1C ⎝ 11 V ⎠

⎛ 2.3 V ⎞ 3 t1 = − R1C ln⎜ ⎟ = 1.56R1C = 79.8 × 10 C ⎝ 11 V ⎠

110

Chapter 11 During the discharging cycle (assuming R2 >> RB1): V(t) = VF − (VF − V0 )e −t 2 / R 2 C 1 V = 0 V − (0 V − 9.3 V)e −t 2 / R2C −

t2 ⎛ 1V ⎞ = ln⎜ ⎟ R2 C ⎝ 9.3 V ⎠

⎛ 1V ⎞ t2 = − R2 C ln⎜ ⎟ = 2.23R2C ⎝ 9.3 V ⎠ Let R2 = 100 kΩ, so t2 = 223 × 103C. Since f = 2.5 kHz, T = 400 μs T = t1 + t2 = 79.8 × 103C + 223 × 103C = 303 × 103C = 400 μs 400 μs = 0.0013 μF C= 303 × 10 3 See Figure 11-10. Figure 11-10

Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 27 through 29 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 27.

Cathode-anode shorted

28.

Gate-cathode open

29.

R1 shorted

111

Chapter 12 The Operational Amplifier Section 12-1 Introduction to Operational Amplifiers 1.

Practical op-amp: High open-loop gain, high input impedance. low output impedance, and high CMRR. Ideal op-amp: Infinite open-loop gain, infinite input impedance, zero output impedance, and infinite CMRR.

2.

Op amp 2 is more desirable because it has a higher input impedance, a lower output impedance, and a higher open-loop gain.

Section 12-2 Op-Amp Input Modes and Parameters 3.

(a) Single-ended differential input (b) Double-ended differential input (c) Common-mode

4.

CMRR (dB) = 20 log(250,000) = 108 dB

5.

⎛A CMRR (dB) = 20 log⎜⎜ ol ⎝ Acm

6.

CMRR = Acm =

⎞ 175,000 ⎞ ⎟⎟ = 20 log⎛⎜ ⎟ = 120 dB ⎝ 0.18 ⎠ ⎠

Aol Acm

Aol 90,000 = 0.3 = CMRR 300,000

8.3 μ A + 7.9 μ A = 8.1 μA 2

7.

IBIAS =

8.

Input bias current is the average of the two input currents. Input offset current is the difference between the two input currents. IOS = 8.3 μA − 7.9 μA = 400 nA

9.

Slew rate =

10.

Δt =

24 V = 1.6 V/μ s 15 μs

ΔVout 20 V = = 40 μ s slew rate 0.5 V/μs

112

Chapter 12 Section 12-4 Op-Amps with Negative Feedback 11.

(a) Voltage-follower (b) Noninverting (c) Inverting

12.

B=

Ri 1.0 kΩ = 9.90 × 10−3 = Ri + R f 101 kΩ

Vf = BVout = (9.90 × 10−3)5 V = 0.0495 V = 49.5 mV

1 1 = = 374 B 1.5 kΩ / 561.5 kΩ (b) Vout = Acl(NI)Vin = (374)(10 mV) = 3.74 V rms ⎛ 1.5 kΩ ⎞ (c) Vf = ⎜ ⎟ 3.74 V = 9.99 mV rms ⎝ 561.5 kΩ ⎠

13.

(a) Acl(NI) =

14.

(a) Acl(NI) =

15.

(a) 1 +

16.

(a) Acl(VF) = 1 ⎛ Rf (b) Acl(I) = − ⎜⎜ ⎝ Ri

1 B 1 (b) Acl(NI) = B 1 (c) Acl(NI) = B 1 (d) Acl(NI) = B

1 = 11 4.7 kΩ / 51.7 kΩ 1 = = 101 10 kΩ / 1.01 MΩ 1 = = 47.8 4.7 kΩ / 224.7 kΩ 1 = = 23 1.0 kΩ / 23 kΩ =

Rf

= Acl(NI) Ri Rf = Ri(Acl(NI) − 1) = 1.0 kΩ(50 − 1) = 49 kΩ Rf (b) = Acl(I) Ri Rf = −Ri(Acl(I)) = −10 kΩ(−300) = 3 MΩ (c) Rf = Ri(Acl(NI) − 1) = 12 kΩ(7) = 84 kΩ (d) Rf = −Ri(Acl(I)) = −2.2 kΩ(−75) = 165 kΩ ⎞ 100 kΩ ⎞ ⎟ = −⎛⎜ ⎟ = −1 ⎟ ⎝ 100 kΩ ⎠ ⎠ 1 1 = (c) Acl(NI) = = 22 47 kΩ ⎞ ⎛ Ri ⎞ ⎛ ⎟ ⎜ ⎟ ⎜ ⎜ Ri + R f ⎟ ⎝ 47 kΩ + 1.0 MΩ ⎠ ⎠ ⎝ ⎛ Rf ⎞ 330 kΩ ⎞ ⎟ = −⎛⎜ (d) Acl(I) = − ⎜⎜ ⎟ = −10 ⎟ ⎝ 33 kΩ ⎠ ⎝ Ri ⎠

113

Chapter 12 17.

(a) Vout ≅ Vin = 10 mV, in phase ⎛ Rf ⎞ ⎟Vin = −(1)(10 mV) = −10 mV, 180° out of phase (b) Vout = AclVin = − ⎜⎜ ⎟ ⎝ Ri ⎠ 1 1 (c) Vout = Vin = Vin = 10 mV = 223 mV, in phase 47 kΩ ⎞ ⎛ R1 ⎞ ⎛ ⎜ ⎟ ⎜⎜ ⎟⎟ ⎝ 1047 kΩ ⎠ ⎝ Ri + R f ⎠ ⎛ Rf ⎞ 330 kΩ ⎞ ⎟ Vin = −⎛⎜ (d) Vout = − ⎜⎜ ⎟ 10 mV = −100 mV, 180° out of phase ⎟ ⎝ 33 kΩ ⎠ ⎝ Ri ⎠

18.

Vin 1V = = 455 μA Rin 2.2 kΩ (b) If ≅ Iin = 455 μA (c) Vout = −IfRf = −(455 μA)(22 kΩ) = −10 V ⎛ Rf ⎞ 22 kΩ ⎞ ⎟ = −⎛⎜ (d) Acl(I) = − ⎜⎜ ⎟ = −10 ⎟ ⎝ 2 .2 k Ω ⎠ ⎝ Ri ⎠ (a) Iin =

Section 12-5 Effects of Negative Feedback on Op-Amp Impedances 19.

2.7 kΩ = 0.0048 562.5 kΩ Zin(NI) = (1 + Aol)Zin = [1 + (175,000)(0.0048)]10 MΩ = 8.41 GΩ Z out 75 Ω = Zout(NI) = = 89.2 mΩ 1 + Aol B 1 + (175,000)(0.0048)

(a) B =

1.5 kΩ = 0.031 48.5 kΩ Zin(NI) = (1 + AolB)Zin = [1 + (200,000)(0.031)]1 MΩ = 6.20 GΩ Z out 25 Ω = = 4.04 mΩ Zout(NI) = 1 + Aol B 1 + (200,000)(0.031)

(b) B =

56 kΩ = 0.053 1.056 MΩ Zin(NI) = (1 + AolB)Zin = [1 + (50,000)(0.053)]2 MΩ = 5.30 GΩ Z out 50 Ω = Zout(NI) = = 19.0 mΩ 1 + Aol B 1 + (50,000)(0.053)

(c) B =

114

Chapter 12 20.

(a) Zin(VF) = (1 + Aol)Zin = (1 + 220,000)6 MΩ = 1.32 × 1012 Ω = 1.32 TΩ Z out 100 Ω = = 455 μ Ω Zout(VF) = 1 + Aol 1 + 220,000 (b) Zin(VF) = (1 + Aol)Zin = (1 + 100,000)5 MΩ = 5 × 1011 Ω = 500 GΩ Z out 60 Ω = Zout(VF) = = 600 μ Ω 1 + Aol 1 + 100,000 (c) Zin(VF) = (1 + Aol)Zin = (1 + 50,000)800 kΩ = 40 GΩ Z out 75 Ω = = 1.5 mΩ Zout(VF) = 1 + Aol 1 + 500,000

21.

(a) Zin(I) ≅ Ri = 10 kΩ Ri 10 kΩ = B= = 0.0625 Ri + R f 160 kΩ Zout(I) =

Z out 40 Ω = = 5.12 mΩ 1 + Aol B 1 + (125,000)(0.0625)

(b) Zin(I) ≅ Ri = 100 kΩ 100 kΩ = 0.091 B= 1.1 MΩ Z out 50 Ω = = 7.32 mΩ Zout(I) = + + 1 Aol B 1 (75,000)(0.91) (c) Zin(I) ≅ Ri = 470 Ω 470 Ω B= = 0.045 10,470 Ω Z out 70 Ω = Zout(I) = = 6.22 mΩ 1 + Aol B 1 + (250,000)(0.045)

Section 12-6 Bias Current and Offset Voltage 22.

(a) Rcomp = Rin = 75 Ω placed in the feedback path. IOS = 42 μA − 40 μA = 2 μA (b) VOUT(error) = AvIOSRin = (1)(2 μA)(75 Ω) = 150 μV

23.

(a) Rc = Ri R f = 2.7 kΩ 560 kΩ = 2.69 kΩ (b) Rc = Ri R f = 1.5 kΩ 47 kΩ = 1.45 kΩ (c) Rc = Ri R f = 56 kΩ 1.0 MΩ = 53 kΩ See Figure 12-1. Figure 12-1

115

Chapter 12 24.

VOUT(error) = AvVIO = (1)(2 nV) = 2 nV

25.

VOUT(error) = (1 + Aol)VIO VOUT(error) 35 mV VIO = = 175 nV = Aol 200,000

Section 12-7 Open-Loop Response 26.

Acl = 120 dB − 50 dB = 70 dB

27.

The gain is ideally 175,000 at 200 Hz. The midrange dB gain is 20 log(175,000) = 105 dB The actual gain at 200 Hz is Av (dB) = 105 dB − 3 dB = 102 dB ⎛ 102 ⎞ Av = log −1 ⎜ ⎟ = 125,892 ⎝ 20 ⎠ BWol = 200 Hz

28.

29.

fc X C = f R Rf c (1.0 kΩ)(5 kHz) = = 1.67 kΩ XC = f 3 kHz (a)

(b)

(c)

(d)

(e)

Vout = Vin

Vout = Vin

Vout = Vin

Vout = Vin

Vout = Vin

1 ⎞ ⎟⎟ ⎠

2

⎞ ⎟⎟ ⎠

2

⎞ ⎟⎟ ⎠

2

⎞ ⎟⎟ ⎠

2

⎛ f ⎞ 1 + ⎜⎜ ⎟⎟ ⎝ fc ⎠

2

⎛ f 1 + ⎜⎜ ⎝ fc 1 ⎛ f 1 + ⎜⎜ ⎝ fc 1 ⎛ f 1 + ⎜⎜ ⎝ fc 1 ⎛ f 1 + ⎜⎜ ⎝ fc 1

=

1 ⎛ 1 kHz ⎞ 1+ ⎜ ⎟ ⎝ 12 kHz ⎠

=

1 ⎛ 5 kHz ⎞ 1+ ⎜ ⎟ ⎝ 12 kHz ⎠

=

= 0.707

2

1 ⎛ 20 kHz ⎞ 1+ ⎜ ⎟ ⎝ 12 kHz ⎠

=

= 0.923

2

1 ⎛ 12 kHz ⎞ 1+ ⎜ ⎟ ⎝ 12 kHz ⎠

=

= 0.997

2

= 0.515

2

1 ⎛ 100 kHz ⎞ 1+ ⎜ ⎟ ⎝ 12 kHz ⎠

2

116

= 0.119

Chapter 12 30.

(a) Aol =

(b) Aol =

(c) Aol =

(d) Aol =

Aol ( mid ) ⎞ ⎟ ⎟ ⎠

2

⎛ f 1+ ⎜ ⎜ f c ( ol ) ⎝ Aol ( mid )

⎞ ⎟ ⎟ ⎠

2

⎛ f 1+ ⎜ ⎜ f c ( ol ) ⎝ Aol ( mid )

⎞ ⎟ ⎟ ⎠

⎛ f 1+ ⎜ ⎜ f c ( ol ) ⎝ Aol ( mid )

⎛ f 1+ ⎜ ⎜ f c ( ol ) ⎝

31.

2

80,000 ⎛ 100 Hz ⎞ 1+ ⎜ ⎟ ⎝ 1 kHz ⎠

=

80,000 ⎛ 1 kHz ⎞ 1+ ⎜ ⎟ ⎝ 1 kHz ⎠

=

= 56,569

2

80,000 ⎛ 10 kHz ⎞ 1+ ⎜ ⎟ ⎝ 1 kHz ⎠

=

= 79,603

2

80,000 ⎛ 1 MHz ⎞ 1+ ⎜ ⎟ ⎝ 1 kHz ⎠

= 7960

2

2

= 80

⎛ f ⎞ 1 1 ⎛ 2 kHz ⎞ = = 1.59 kHz; θ = tan −1 ⎜⎜ ⎟⎟ = tan −1 ⎜ ⎟ = −51.5° 2πRC 2π (10 kΩ)(0.01 μF) ⎝ 1.59 kHz ⎠ ⎝ fc ⎠ ⎛ f ⎞ 1 1 ⎛ 2 kHz ⎞ = (b) fc = = 15.9 kHz; θ = tan −1 ⎜⎜ ⎟⎟ = tan −1 ⎜ ⎟ = −7.17° 2πRC 2π (1.0 kΩ)(0.01 μF) f ⎝ 15.9 kHz ⎠ ⎝ c⎠

(a) fc =

(c) fc =

32.

⎞ ⎟ ⎟ ⎠

2

=

⎛ f ⎞ 1 1 ⎛ 2 kHz ⎞ = = 159 Hz; θ = tan −1 ⎜⎜ ⎟⎟ = tan −1 ⎜ ⎟ = −85.5° 2πRC 2π (100 kΩ)(0.01 μF) ⎝ 159 Hz ⎠ ⎝ fc ⎠

⎛ f ⎞ ⎛ 100 Hz ⎞ (a) θ = tan −1 ⎜⎜ ⎟⎟ = tan −1 ⎜ ⎟ ⎝ 8.5 kHz ⎠ ⎝ fc ⎠ ⎛ f ⎞ ⎛ 400 Hz ⎞ (b) θ = tan −1 ⎜⎜ ⎟⎟ = tan −1 ⎜ ⎟ ⎝ 8.5 kHz ⎠ ⎝ fc ⎠ ⎛ f ⎞ ⎛ 850 Hz ⎞ (c) θ = tan −1 ⎜⎜ ⎟⎟ = tan −1 ⎜ ⎟ ⎝ 8.5 kHz ⎠ ⎝ fc ⎠ ⎛ f ⎞ ⎛ 8.5 kHz ⎞ (d) θ = tan −1 ⎜⎜ ⎟⎟ = tan −1 ⎜ ⎟ ⎝ 8.5 kHz ⎠ ⎝ fc ⎠

= −0.674° = −2.69° = −5.71° = −45.0°

⎛ f ⎞ ⎛ 25 kHz ⎞ (e) θ = tan −1 ⎜⎜ ⎟⎟ = tan −1 ⎜ ⎟ = −71.2° f ⎝ 8.5 kHz ⎠ ⎝ c⎠ ⎛ f ⎞ ⎛ 85 kHz ⎞ (f) θ = tan −1 ⎜⎜ ⎟⎟ = tan −1 ⎜ ⎟ = −84.3° ⎝ 8.5 kHz ⎠ ⎝ fc ⎠ See Figure 12-2.

117

Chapter 12

Figure 12-2 33.

(a) Aol(mid) = 30 dB + 40 dB + 20 dB = 90 dB

⎛ f ⎞ ⎛ 10 kHz ⎞ (b) θ1 = − tan −1 ⎜⎜ ⎟⎟ = − tan −1 ⎜ ⎟ = −86.6° f ⎝ 600 Hz ⎠ ⎝ c⎠ ⎛ f ⎞ ⎛ 10 kHz ⎞ θ2 = − tan −1 ⎜⎜ ⎟⎟ = − tan −1 ⎜ ⎟ = −11.3° ⎝ 50 kHz ⎠ ⎝ fc ⎠ ⎛ f ⎞ ⎛ 10 kHz ⎞ θ3 = − tan −1 ⎜⎜ ⎟⎟ = − tan −1 ⎜ ⎟ = −2.86° ⎝ 200 kHz ⎠ ⎝ fc ⎠ θtot = −86.6° − 11.3° − 2.86° − 180° = −281° 34.

(a) (b) (c) (d)

0 dB/decade −20 dB/decade −40 dB/decade −60 dB/decade

Section 12-8 Closed-Loop Frequency Response 35.

⎛ Rf ⎞ 68 kΩ ⎞ ⎟ = −⎛⎜ (a) Acl(I) = − ⎜⎜ ⎟ = −30.9; ⎟ ⎝ 2 .2 k Ω ⎠ ⎝ Ri ⎠ 1 1 (b) Acl(NI) = = = 15.7; B 15 kΩ / 235 kΩ

Acl(I) (dB) = 20 log(30.9) = 29.8 dB Acl(NI) (dB) = 20 log(15.7) = 23.9 dB

(c) Acl(VF) = 1; Acl(VF)(dB) = 20 log(1) = 0 dB These are all closed-loop gains. 36.

BWcl = BWol(1 + BAol(mid)) = 1500 Hz[1 + (0.015)(180,000)] = 4.05 MHz

118

Chapter 12 37.

Aol (dB) = 89 dB Aol = 28,184 Aclfc(cl) = Aolfc(ol) Aol f c ( ol ) (28,184)(750 Hz) = Acl = = 3843 f c ( cl ) 5.5 kHz Acl (dB) = 20 log(3843) = 71.7 dB

38.

39.

Aol f c ( ol )

(28,184)(750 Hz) = 3843 f c ( cl ) 5.5 kHz fT = Aclfc(cl) = (3843)(5.5 kHz) = 21.1 MHz Acl =

=

(a) Acl(VF) = 1 BW = fc(cl) =

fT 28 MHz = = 2.8 MHz 1 Acl

100 kΩ = −45.5 2.2 kΩ 2.8 MHz BW = = 61.6 kHz 45.5

(b) Acl(I) = −

12 kΩ = 13 1.0 kΩ 2.8 MHz BW = = 215 kHz 13

(c) Acl(NI) = 1 +

1 MΩ = −179 5.6 kΩ 2.8 MHz = 15.7 kHz BW = 179

(d) Acl(I) = −

40.

150 kΩ = 6.8 22 kΩ Aol f c ( ol ) (120,000)(150 Hz) fc(cl) = = 2.65 MHz = Acl 6.8 BW = fc(cl) = 2.65 MHz

(a) Acl =

1.0 MΩ = 100 10 kΩ Aol f c ( ol ) (195,000)(50 Hz) fc(cl) = = 97.5 kHz = Acl 100 BW = fc(cl) = 97.5 kHz

(b) Acl =

119

Chapter 12 Section 12-9 Troubleshooting 41.

(a) Faulty op-amp or open R1 (b) R2 open, forcing open-loop operation

42.

(a) (b) (c) (d)

43.

The gain becomes a fixed −100 with no effect as the potentiometer is adjusted.

Circuit becomes a voltage-follower and the output replicates the input. Output will saturate. No effect on the ac; may add or subtract a small dc voltage to the output. The voltage gain will change from 10 to 0.1.

Application Activity Problems 44.

The push-pull stage will operate nonlinearly if a diode is shorted, a transistor is faulty, or the opamp stage has excessive gain.

45.

If a 100 kΩ resistor is used for R2, the gain of the op amp will be reduced by a factor of 100.

46.

If D1 opens, the positive half of the signal will appear on the output through Q3 and Q4. The negative half is missing due to the open diode.

Datasheet Problems 47.

From the datasheet of textbook Figure 12-77: 470 Ω B= = 0.0099 47 kΩ + 470 Ω Aol = 200,000 (typical) Zin = 2.0 MΩ (typical) Zin(NI) = (1 + 0.0099)(200,000)(2 MΩ) = (1 + 1980)2 MΩ = 3.96 GΩ

48.

From the datasheet in Figure 12-77: R f 100 kΩ Zin(I) = Ri = = 1 kΩ = Acl 100

49.

Aol = 50 V/mV =

50.

Slew rate = 0.5 V/μs ΔV = 8 V − (−8 V) = 16 V 16 V Δt = = 32 μ s 0.5 V/μs

50 V 50,000 V = = 50,000 1 mV 1V

120

Chapter 12 Advanced Problems 51.

Using available standard values of Rf = 150 kΩ and Ri = 1.0 kΩ, 150 kΩ = 151 Av = 1 + 1.0 kΩ 1.0 kΩ B= = 6.62 × 10−3 151 kΩ Zin(NI) = (1 + (6.62 × 10−3)(50,000))300 kΩ = 99.6 MΩ The compensating resistor is Rc = Ri R f = 150 kΩ 1.0 kΩ = 993 Ω See Figure 12-3.

Figure 12-3 52.

See Figure 12-4. 2% tolerance resistors are used to achieve a 5% gain tolerance.

Figure 12-4 53.

From textbook Figure 12-78: fc = 10 kHz at Av = 40 dB = 100 In this circuit 33 kΩ Av = 1 + = 100.1 ≅ 100 333 Ω The compensating resistor is Rc = 33 kΩ 333 Ω = 330 Ω See Figure 12-5.

Figure 12-5

121

Chapter 12 54.

From textbook Figure 12-79: For a ±10 V output swing minimum, the load must be 600 Ω for a ±10 V and ≈ 620 Ω for −10 V. So, the minimum load is 620 Ω.

55.

For the amplifier, 100 kΩ = −50 Av = − 2 kΩ The compensating resistor is Rc = 100 kΩ 2 kΩ = 1.96 kΩ ≅ 2 kΩ See Figure 12-6.

Figure 12-6 56.

From textbook Figure 12-78 the maximum 741 closed loop gain with BW = 5 kHz is approximately 60 dB − (20 dB)log(5 kHz/1 kHz) = 60 dB − (20 dB)(0.7) = 46 dB Av(dB) = 20 log Av ⎛ Av ( dB) ⎞ 46 ⎟ = log −1 ⎛⎜ ⎞⎟ = 200 Av = log −1 ⎜⎜ ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠

Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 57 through 72 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 57.

Rf open

58.

Ri open

59.

Rf leaky

60.

Ri shorted

61.

Rf shorted

62.

Op-amp input to output open

63.

Rf leaky

122

Chapter 12 64.

Ri leaky

65.

Ri shorted

66.

Ri open

67.

Rf open

68.

Rf leaky

69.

Rf open

70.

Rf shorted

71.

Ri open

72.

Ri leaky

123

Chapter 13 Basic Op-Amp Circuits Section 13-1 Comparators 1.

Vout(p) = AolVin = (80,000)(0.15 mV)(1.414) = 17 V Since 12 V is the peak limit, the op-amp saturates. Vout(pp) = 24 V with distortion due to clipping.

2.

(a) Maximum negative (b) Maximum positive (c) Maximum negative

3.

⎛ R2 ⎞ ⎛ 18 kΩ ⎞ ⎟⎟(+ 10 V) = ⎜ VUTP = ⎜⎜ ⎟ 10 V = 2.77 V ⎝ 65 kΩ ⎠ ⎝ R1 + R2 ⎠ ⎛ R2 ⎞ ⎛ 18 kΩ ⎞ ⎟⎟(− 10 V) = ⎜ VLTP = ⎜⎜ ⎟ (−10 V) = −2.77 V ⎝ 65 kΩ ⎠ ⎝ R1 + R2 ⎠

4.

VHYS = VUTP − VLTP = 2.77 V − (−2.77 V) = 5.54 V

5.

See Figure 13-1.

Figure 13-1

124

Chapter 13 6.

⎛ R2 ⎞ ⎛ 18 kΩ ⎞ ⎟⎟ + Vout ( max ) = ⎜ (a) VUTP = ⎜⎜ ⎟ 11 V = 3.88 V ⎝ 51 kΩ ⎠ ⎝ R1 + R2 ⎠ VLTP = −3.88 V VHYS = VUTP − VLTP = 3.88 V − (−3.88 V) = 7.76 V

(

)

⎛ R2 ⎞ ⎛ 68 kΩ ⎞ ⎟⎟ + Vout ( max ) = ⎜ (b) VUTP = ⎜⎜ ⎟ 11 V = 3.43 V ⎝ 218 kΩ ⎠ ⎝ R1 + R2 ⎠ VLTP = −3.43 V VHYS = VUTP − VLTP = 3.43 V − (−3.43 V) = 6.86 V

(

7.

)

When the zener is forward-biased: 18 kΩ ⎛ ⎞ Vout = ⎜ ⎟ Vout − 0.7 V ⎝ 18 kΩ + 47 kΩ ⎠ Vout = (0.277)Vout − 0.7 V Vout(1 − 0.277) = −0.7 V − 0.7 V = −0.968 V Vout = 1 − 0.277 When the zener is reverse-biased: 18 kΩ ⎛ ⎞ Vout = ⎜ ⎟ Vout + 6.2 V ⎝ 18 kΩ + 47 kΩ ⎠ Vout = (0.277)Vout + 6.2 V Vout(1 − 0.277) = +6.2 V + 6.2 V Vout = = +8.57 V 1 − 0.277

8.

10 kΩ ⎛ ⎞ Vout = ⎜ ⎟ Vout ± (4.7 V + 0.7 V) ⎝ 10 kΩ + 47 kΩ ⎠ Vout = (0.175)Vout ± 5.4 V ± 5 .4 V Vout = = ±6.55 V 1 − 0.175 VUTP = (0.175)(+6.55 V) = +1.15 V VLTP = (0.175)(−6.55 V) = −1.15 V See Figure 13-2.

Figure 13-2

125

Chapter 13 Section 13-2 Summing Amplifiers 9.

(a) VOUT = −

(b) VOUT = − 10.

Rf Ri Rf Ri

(+1 V + 1.5 V) = −1(1 V + 1.5 V) = −2.5 V (0.1 V + 1 V + 0.5 V) = −

22 kΩ (1.6 V) = −3.52 V 10 kΩ

(a) VR1 = 1 V VR2 = 1.8 V 1V = 45.5 μA 22 kΩ 1.8 V IR2 = = 81.8 μA 22 kΩ If = IR1 + IR2 = 45.5 μA + 81.8 μA = 127 μA

(b) IR1 =

(c) VOUT = −IfRf = −(127 μA)(22 kΩ) = −2.8 V 11.

12.

⎛ Rf ⎞ ⎟Vin 5Vin = ⎜⎜ ⎟ ⎝ R ⎠ Rf =5 R Rf = 5R = 5(22 kΩ) = 110 kΩ

See Figure 13-3.

Figure 13-3 13.

⎡⎛ R f VOUT = − ⎢⎜⎜ ⎣⎢⎝ R1

⎞ ⎛R ⎞ ⎛R ⎞ ⎛R ⎞ ⎤ ⎟V1 + ⎜ f ⎟V2 + ⎜ f ⎟V3 + ⎜ f ⎟V4 ⎥ ⎟ ⎜R ⎟ ⎜R ⎟ ⎜R ⎟ ⎝ 4 ⎠ ⎦⎥ ⎠ ⎝ 2⎠ ⎝ 3⎠

⎡⎛ 10 kΩ ⎞ ⎛ 10 kΩ ⎞ ⎤ ⎛ 10 kΩ ⎞ ⎛ 10 kΩ ⎞ = − ⎢⎜ ⎟6 V ⎥ ⎟3 V + ⎜ ⎟3 V + ⎜ ⎟2 V + ⎜ ⎝ 180 kΩ ⎠ ⎦ ⎝ 91 kΩ ⎠ ⎝ 33 kΩ ⎠ ⎣⎝ 10 kΩ ⎠ = −(2 V + 0.91 V + 0.33 V + 0.33 V) = −3.57 V If =

VOUT 3.57 V = = 357 μA Rf 10 kΩ

126

Chapter 13 14.

Rf = 100 kΩ Input resistors: R1 = 100 kΩ, R2 = 50 kΩ, R3 = 25 kΩ, R4 = 12.5 kΩ, R5 = 6.25 kΩ, R6 = 3.125 kΩ

Section 13-3 Integrators and Differentiators 15.

dVout V 5V = − IN = − = −4.06 mV/μs (56 kΩ)(0.022 μ F) dt RC

16.

See Figure 13-4.

Figure 13-4 17.

18.

I=

CV pp T /2

=

(0.001 μF)(5 V) = 1 mA 10 μs / 2

⎛ V pp ⎞ ⎛ 2V ⎞ Vout = ± RC ⎜ ⎟ = ± (15 kΩ)(0.047 μ F) ⎜ ⎟ = ±2.82 V ⎝ 0.5 ms ⎠ ⎝T /2⎠ See Figure 13-5.

Figure 13-5

127

Chapter 13 19.

For the 10 ms interval when the switch is in position 2: V ΔVout 5V 5V = − IN = − =− = −50 V/s = −50 mV/ms RC (10 kΩ)(10 μF) 0.1 s Δt ΔVout = (−50 mV/ms)(10 ms) = −500 mV = −0.5 V For the 10 ms interval when the switch is in position 1: ΔVout V −5V −5V = − IN = − =− = +50 V/s = +50 mV/ms Δt RC (10 kΩ)(10 μF) 0.1 s ΔVout = (+50 mV/ms)(10 ms) = +500 mV = +0.5 V See Figure 13-6.

Figure 13-6

Section 13-4 Troubleshooting 20.

21.

⎛ R2 ⎞ ⎟⎟Vout ± (V Z + 0.7 V) VB = ⎜⎜ ⎝ R1 + R2 ⎠ ± (V Z + 0.7 V) VB = ⎛ R2 ⎞ ⎟⎟ 1 − ⎜⎜ ⎝ R1 + R2 ⎠ Normally, VB should be ± (4.3 V + 0.7 V) = ±10 V VB = 1 − 0.5 Since the negative portion of VB is only −1.4 V, zener D2 must be shorted: − (0 V + 0.7 V) = 1.4 V VB = 1 − 0.5 The output should be as shown in Figure 13-7. V2 has no effect on the output. This indicates that R2 is open.

Figure 13-7

128

Chapter 13 22.

Av =

2 .5 k Ω = 0.25 10 kΩ

The output should be as shown in Figure 13-8. An open R2 (V2 is missing) will produce the observed output, which is incorrect.

Figure 13-8 23.

The D2 input is missing (acts as a constant 0). This indicates an open 50 kΩ resistor.

Application Activity Problems 24.

The first thing that you should always do is visually inspect the circuit for bad contacts or loose connections, shorts from solder splashes or wire clippings, incorrect components, and incorrectly installed components. After careful inspection, you have found nothing wrong. Measurements are now necessary to isolate a component’s fault.

25.

An open decoupling capacitor can make the circuit more susceptible to power line noise.

26.

If a 1.0 kΩ resistor is used for R1, the inverting input would be increased, causing the pulse width to narrow for a given setting of the potentiometer.

Advanced Problems 27.

24 V = 39.2 μA 612 kΩ Minimum setting of R2: VINV = 12 V − (39.2 μA)(56 kΩ) = 9.8 V v = Vp sin θ v 9.8 V sin θ = = = 0.98 V p 10 V

IR1-2-3 =

⎛ v ⎞ = sin −1 (0.98) = 78.5° (on positive half cycle) ⎜ V p ⎟⎟ ⎝ ⎠ Angle from 78.5° to 90° Δθ = 90° − 78.5° = 11.5° Angle from 90° to next point at which v = 9.8 V: Δθ = 11.5° Angle from first point at which v = 9.8 V to second point at which v = 9.8 V on sine wave is θ = 11.5° + 11.5° = 23°

θ = sin−1 ⎜

129

Chapter 13 ⎛ 23° ⎞ min. duty cycle = ⎜ ⎟100 = 6.39% ⎝ 360° ⎠ See Figure 13.9(a). Maximum setting of R2: VINV = 12 V − (39.2 μA)(556 kΩ) = −9.8 V v −9.8 V = = −78.5° (on negative half cycle) sin θ = 10 V Vp

⎛ 360° − 23° ⎞ max. duty cycle = ⎜ ⎟100 = 93.6% ⎝ 360° ⎠ See Figure 13-9(b).

(a)

(b) Figure 13-9 28.

Let VINV = 4.8 V Let I1 = 39.2 μA VINV = 12 V − I1R1 −I1R1 = 4.8 V − 12 V I1R1 = 7.2 V 7.2 V R1 = = 184 kΩ 39.2 μ A Change R1 and R3 to 184 kΩ.

130

Chapter 13 29.

100 mV/μs = 5 V/RiC 5V RiC = = 50 μs 100 mV/μs For C = 3300 pF: 50 μs Ri = = 15.15 kΩ = 15 kΩ + 150 Ω 3300 pF For a 5 V peak-peak triangle waveform: 5V tramp up = tramp down = = 50 μs 100 mV/μs τ = 2(50 μs) = 100 μs fin = 1/100 μs = 100 kHz See Figure 13-10.

Figure 13-10

Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 30 through 39 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 30.

R1 open

31.

Op-amp inputs shorted together

32.

Op-amp + input to output shorted

33.

D1 shorted

34.

Top 10 kΩ resistor open

35.

Middle 10 kΩ resistor shorted

36.

Rf leaky

37.

Rf open

38.

C leaky

39.

C open

131

Chapter 14 Special-Purpose Op-Amp Circuits Section 14-1 Instrumentation Amplifiers R1 100 kΩ =1+ = 101 1.0 kΩ RG R 100 kΩ Av(2) = 1 + 2 = 1 + = 101 RG 1 .0 k Ω

1.

Av(1) = 1 +

2.

Acl = 1 +

3.

Vout = Acl (Vin ( 2 ) − Vin (1) ) = 202(10 mV − 5 mV) = 1.005 V

4.

Av = 1 +

2R 200 kΩ = 201 =1+ RG 1.0 kΩ

2R RG

2R = Av − 1 RG 2(100 kΩ) 200 kΩ 2R = = = 200.2 Ω ≅ 200 Ω RG = Av − 1 1000 − 1 999 50.5 kΩ Av − 1 50.5 kΩ Av = + 1 = 51.5 1.0 kΩ

5.

RG =

6.

Using the graph in textbook Figure 14-6, BW ≅ 300 kHz

7.

Change RG to 50.5 kΩ 50.5 kΩ = ≅ 2.2 kΩ RG = Av − 1 24 − 1

8.

RG =

50.5 kΩ 50.5 kΩ ≅ 2.7 kΩ = Av − 1 20 − 1

132

Chapter 14 Section 14-2 Isolation Amplifiers 9.

Av(total) = (30)(10) = 300

10.

(a) Av1 =

Rf1 Ri1 Rf 2

+1=

18 kΩ + 1 = 3.2 8 .2 k Ω

150 kΩ + 1 = 11 15 kΩ Ri 2 Av(tot) = Av1Av2 = (3.2)(11) = 35.2 Av2 =

(b) Av1 =

Rf1 Ri1 Rf 2

+1 =

+1=

330 kΩ + 1 = 331 1.0 kΩ

47 kΩ + 1 = 4.13 Ri 2 15 kΩ Av(tot) = Av1Av2 = (331)(4.13) = 1,367 Av2 =

11.

+1 =

Av2 = 11 (from Problem 10(a)) Av1Av2 = 100 Rf 1 100 + 1 = Av1 = = 9.09 11 Ri1 Rf1 = (9.09 − 1)Ri1 = (8.09)(8.2 kΩ) = 66 kΩ Change Rf (18 kΩ) to 66 kΩ. Use 68 kΩ ± 1% standard value resistor.

12.

Av1 = 331 (from Problem 10(b)) Av1Av2 = 440 Rf 2 440 = 1.33 + 1 = Av 2 = Ri 2 331 Change Rf (47 kΩ) to 3.3 kΩ. Change Ri (15 kΩ) to 10 kΩ.

13.

Connect pin 6 to pin 10 and pin 14 to pin 15. Make Rf = 0.

Section 14-3 Operational Transconductance Amplifiers (OTAs) I out 10 μA = 1 mS = Vin 10 mV

14.

gm =

15.

Iout = gmVin = (5000 μS)(100 mV) = 500 μA Vout = IoutRL = (500 μA)(10 kΩ) = 5 V

133

Chapter 14 16.

17.

I out Vin Iout = gmVin = (4000 μS)(100 mV) = 400 μA V 3.5 V = 8.75 kΩ RL = out = I out 400 μA + 12 V − (−12 V) − 0.7 V + 12 V − (−12 V) − 0.7 V 23.3 V = = IBIAS = = 106 μA 220 kΩ 220 kΩ RBIAS From the graph in Figure 14-57: gm = KIBIAS ≅ (16 μS/μA)(106 μA) = 1.70 mS V I R Av = out = out L = gmRL = (1.70 mS)(6.8 kΩ) = 11.6 Vin Vin

gm =

18.

The maximum voltage gain occurs when the 10 kΩ potentiometer is set to 0 Ω and was determined in Problem 17. Av(max) = 11.6 The minimum voltage gain occurs when the 10 kΩ potentiometer is set to 10 kΩ. + 12 V − (−12 V) − 0.7 V 23.3 V = = 101 μA IBIAS = 220 kΩ + 10 kΩ 230 kΩ gm ≅ (16 μS/μA)(101 μA) = 1.62 mS Av(min) = gmRL = (1.62 mS)(6.8 kΩ) = 11.0

19.

The VMOD waveform is applied to the bias input. The gain and output voltage for each value of VMOD is determined as follows using K = 16 μS/μA. The output waveform is shown in Figure 14-1. For VMOD = +8 V: + 8 V − (−9 V) − 0.7 V 16.3 V = = 418 μA IBIAS = 39 kΩ 39 kΩ gm = KIBIAS ≅ (16 μS/μA)(418 μA) = 6.69 mS V I R Av = out = out L = gmRL = (6.69 mS)(10 kΩ) = 66.9 Vin Vin Vout = AvVin = (66.9)(100 mV) = 6.69 V For VMOD = +6 V: + 6 V − ( −9 V) − 0.7 V 14.3 V = 367 μA IBIAS = = 39 kΩ 39 kΩ gm = KIBIAS ≅ (16 μS/μA)(367 μA) = 5.87 mS V I R Av = out = out L = gmRL = (5.87 mS)(10 kΩ) = 58.7 Vin Vin Vout = AvVin = (58.7)(100 mV) = 5.87 V

134

Chapter 14 For VMOD = +4 V: + 4 V − (−9 V) − 0.7 V 12.3 V = = 315 μA IBIAS = 39 kΩ 39 kΩ gm = KIBIAS ≅ (16 μS/μA)(315 μA) = 5.04 mS V I R Av = out = out L = gmRL = (5.04 mS)(10 kΩ) = 50.4 Vin Vin Vout = AvVin = (50.4)(100 mV) = 5.04 V For VMOD = +2 V: + 2 V − (−9 V) − 0.7 V 10.3 V = = 264 μA IBIAS = 39 kΩ 39 kΩ gm = KIBIAS ≅ (16 μS/μA)(264 μA) = 4.22 mS V I R Av = out = out L = gmRL = (4.22 mS)(10 kΩ) = 42.2 Vin Vin Vout = AvVin = (42.2)(100 mV) = 4.22 V For VMOD = +1 V: + 1 V − (−9 V) − 0.7 V 9.3 V = = 238 μA IBIAS = 39 kΩ 39 kΩ gm = KIBIAS ≅ (16 μS/μA)(238 μA) = 3.81 mS V I R Av = out = out L = gmRL = (3.81 mS)(10 kΩ) = 38.1 Vin Vin Vout = AvVin = (38.1)(100 mV) = 3.81 V

Figure 14-1 20.

+ 9V − (−9 V) − 0.7 V 17.3 V = = 444 μA 39 kΩ 39 kΩ VTRIG(+) = IBIASR1 = (444 μA)(10 kΩ) = +4.44 V VTRIG(−) = −IBIASR1 = (−444 μA)(10 kΩ) = −4.44 V

IBIAS =

135

Chapter 14 21.

See Figure 14-2.

Figure 14-2

Section 14-4 Log and Antilog Amplifiers 22.

(a) (b) (c) (d)

ln(0.5) = −0.693 ln(2) = 0.693 ln(50) = 3.91 ln(130) = 4.87

23.

(a) (b) (c) (d)

log10(0.5) = −0.301 log10(2) = 0.301 log10(50) = 1.70 log10(130) = 2.11

24.

Antilog x = 10x or ex, depending on the base used. INV ln = e1.6 = 4.95 INV log = 101.6 = 39.8

25.

The output of a log amplifier is limited to 0.7 V because the output voltage is limited to the barrier potential of the transistor’s pn junction.

26.

27.

⎛ V ⎞ Vout ≅ − (0.025 V)ln⎜⎜ in ⎟⎟ ⎝ I s Rin ⎠ ⎛ ⎞ 3V ⎟⎟ = −(0.025 V)ln(365.9) = −148 mV = − (0.025 V)ln⎜⎜ ⎝ (100 nA)(82 kΩ) ⎠ ⎛ Vin ⎞ ⎟⎟ Vout ≅ − (0.025 V)ln⎜⎜ I R EBO in ⎠ ⎝ ⎛ ⎞ 1.5 V ⎟⎟ = −(0.025 V)ln(531.9) = −157 mV = − (0.025 V)ln⎜⎜ ⎝ (60 nA)(47 kΩ) ⎠

136

Chapter 14 ⎛ Vin

28.

Vout = −(10 kΩ)(60 nA) e 29.

⎞

⎜ ⎟ ⎞ ⎛ V Vout = −Rf IEBO antilog⎜ in ⎟ = − R f I EBO e ⎝ 25 mV ⎠ ⎝ 25 mV ⎠ ⎛ 0.225 V ⎞ ⎜ ⎟ ⎝ 25 mV ⎠

= −(10 kΩ)(60 nA)e9 = −(10 kΩ)(60 nA)(8103) = −4.86 V

⎛ Vin ⎞ ⎞ ⎛ 1V ⎟⎟ = −(0.025 V) ln⎜⎜ ⎟⎟ Vout(max) ≅ − (0.025 V) ln⎜⎜ ⎝ (60 nA)(47 kΩ) ⎠ ⎝ I EBO Rin ⎠ = −(0.025 V)ln(354.6) = −147 mV ⎛ Vin ⎞ ⎞ ⎛ 100 mV ⎟⎟ = −(0.025 V) ln⎜⎜ ⎟⎟ Vout(min) ≅ − (0.025 V) ln⎜⎜ ⎝ (60 nA)(47 kΩ) ⎠ ⎝ I EBO Rin ⎠ = −(0.025 V) ln(35.5) = −89.2 mV The signal compression allows larger signals to be reduced without causing smaller amplitudes to be lost (in this case, the 1 V peak is reduced 85% but the 100 mV peak is reduced only 10%).

Section 14-5 Converters and Other Op-Amp Circuits 30.

(a) VIN = VZ = 4.7 V V 4.7 V IL = IN = = 4.7 mA Ri 1.0 kΩ

⎛ 10 kΩ ⎞ (b) VIN = ⎜ ⎟12 V = 6 V ⎝ 20 kΩ ⎠ Ri = 10 kΩ 10 kΩ + 100 Ω = 5.1 kΩ IL = 31.

VIN 6V = = 1.18 mA Ri 5.1 kΩ

See Figure 14-3.

Figure 14-3

137

Chapter 14 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 32 through 36 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 32.

RG leaky

33.

R open

34.

Rf open

35.

Zener diode open

36.

Lower 10 kΩ resistor open

138

Chapter 15 Active Filters Section 15-1 Basic Filter Responses 1.

(a) (b) (c) (d)

2.

BW = fc = 800 Hz

3.

fc =

4.

The roll-off is 20 dB/decade because this is a single-pole filter.

5.

BW = fc2 − fc1 = 3.9 kHz − 3.2 kHz = 0.7 kHz = 700 Hz f0 = f c1 f c 2 = (3.2 kHz)(3.9 kHz) = 3.53 kHz

1 1 = 48.2 Hz = 2πRC 2π (2.2 kΩ)(0.0015 μF) No, the upper response roll-off due to internal device capacitances is unknown.

Q=

6.

Band-pass High-pass Low-pass Band-stop

f0 3.53 kHz = = 5.04 700 Hz BW

f0 BW f0 = Q(BW) = 15(1 kHz) = 15 kHz

Q=

Section 15-2 Filter Response Characteristics 7.

(a) 2nd order, 1 stage R 1.2 kΩ =2−1=1 DF = 2 − 3 = 2 − 1.2 kΩ R4

Not Butterworth

(b) 2nd order, 1 stage R 560 Ω = 2 − 0.56 = 1.44 DF = 2 − 3 = 2 − 1.0 kΩ R4

(c) 3rd order, 2 stages, 1st stage (2 poles): R 330 Ω = 1.67 DF = 2 − 3 = 2 − 1.0 kΩ R4 2nd stage (1 pole): R DF = 2 − 6 = 1.67 Not Butterworth R7

139

Approximately Butterworth

Chapter 15 8.

(a) From Table 15-1 in the textbook, the damping factor must be 1.414; therefore, R3 = 0.586 R4 R3 = 0.586R4 = 0.586(1.2 kΩ) = 703 Ω Nearest standard value: 720 Ω (b)

R3 = 0.56 R4 This is an approximate Butterworth response (as close as you can get using standard 5% resistors).

(c) From Table 15-1, the damping factor of both stages must be 1, therefore R3 =1 R4 R3 = R4 = R6 = R7 = 1 kΩ (for both stages) 9.

(a) (b) (c) (d)

Chebyshev Butterworth Bessel Butterworth

Section 15-3 Active Low-Pass Filters 10.

High Pass 1st stage:

DF = 2 −

R3 1.0 kΩ =2− = 1.85 6 .8 k Ω R4

2nd stage: R7 6 .8 k Ω =2− = 0.786 5.6 kΩ R8 From Table 15-1 in the textbook: 1st stage DF = 1.848 and 2nd stage DF = 0.765 Therefore, this filter is approximately Butterworth. Roll-off rate = 80 dB/decade DF = 2 −

1

=

1

=

1

11.

fc =

12.

R = R1 = R2 = R5 = R6 and C = C1 = C2 = C3 = C4 Let C = 0.22 μF (for both stages). 1 1 = fc = 2 2 2πRC 2π R C 1 1 = = 3.81 kΩ R= 2πf c C 2π (190 Hz)(0.22 μF) Choose R = 3.9 kΩ (for both stages)

2π R1 R2 C1C 2

2π R5 R6 C 3 C 4

2π ( 4.7 kΩ)(6.8 kΩ)(0.22 μF)(0.1 μF)

140

= 190 Hz

Chapter 15 13.

See Figure 15-1.

Figure 15-1 14.

See Figure 15-2.

Figure 15-2

Section 15-4 Active High-Pass Filters 15.

Exchange the positions of the resistors and the capacitors. See Figure 15-3.

Figure 15-3

141

Chapter 15 16.

17.

1 2πRC 190 Hz = 95 Hz f0 = 2 1 1 R= = = 7615 Ω 2πf c C 2π (95 Hz)(0.22 μF) Let R = 7.5 kΩ. Change R1, R2, R5 and R6 to 7.5 kΩ. fc =

(a) Decrease R1 and R2 or C1 and C2. (b) Increase R3 or decrease R4.

Section 15-5 Active Band-Pass Filters 18.

(a) Cascaded high-pass/low-pass filters (b) Multiple feedback (c) State variable

19.

(a) 1st stage: 1 1 = = 3.39 kHz fc1 = 2πRC 2π (1.0 kΩ)(0.047 μF) 2nd stage: 1 1 fc2 = = 7.23 kHz = 2πRC 2π (1.0 kΩ)(0.022 μF) f0 =

f c1 f c 2 = (3.39 kHz)(7.23 kHz) = 4.95 kHz

BW = 7.23 kHz − 3.39 Hz = 3.84 kHz (b) f0 =

1 2π C

R1 + R3 1 47 kΩ + 1.8 kΩ = 449 Hz = R1 R3 R2 2π (0.022 μ F) (47 kΩ)(1.8 kΩ)(150 kΩ)

Q = πf0CR2 = π(449 Hz)(0.022 μF)(150 kΩ) = 4.66 f 449 Hz = 96.4 Hz BW = 0 = Q 4.66 (c) For each integrator: 1 1 = = 15.9 kHz fc = 2πRC 2π (10 kΩ)(0.001 μF) f0 = fc = 15.9 kHz ⎞ 1 ⎛ 560 kΩ 1⎛ R ⎞ 1 Q = ⎜⎜ 5 + 1⎟⎟ = ⎜ + 1⎟ = (56 + 1) = 19 3 ⎝ R6 ⎠ 3 ⎠ 3 ⎝ 10 kΩ f 15.9 kHz = 838 Hz BW = 0 = Q 19

142

Chapter 15 20.

⎞ 1 ⎛ R5 ⎜⎜ + 1⎟⎟ 3 ⎝ R6 ⎠ Select R6 = 10 kΩ. R 1 R + R6 Q= 5 + = 5 3R6 3 3R6 3R6Q = R5 + R6 R5 = 3R6Q − R6 = 3(10 kΩ)(50) − 10 kΩ = 1500 kΩ − 10 kΩ = 1490 kΩ 1 = 1.33 kHz f0 = 2π (12 kΩ)(0.01 μF) f 1.33 kHz = 26.6 Hz BW = 0 = 50 Q

Q=

Section 15-6 Active Band-Stop Filters 21.

See Figure 15-4.

Figure 15-4 22.

1 2πRC Let C remain 0.01 μF. 1 1 = = 133 kΩ R= 2πf 0 C 2π (120 Hz)(0.01 μF) Change R in the integrators from 12 kΩ to 133 kΩ. f0 = fc =

Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 23 through 31 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 23.

R4 shorted

24.

R3 open

25.

C3 shorted

26.

R5 open

143

Chapter 15 27.

R1 open

28.

R2 shorted

29.

R1 open

30.

C2 open

31.

R7 open

144

Chapter 16 Oscillators Section 16-1 The Oscillator 1.

An oscillator requires no input other than the dc supply voltage.

2.

Amplifier and positive feedback circuit

Section 16-2 Feedback Oscillator Principles 3.

Unity gain around the closed loop is required for sustained oscillation. Acl = AvB = 1 1 1 = B= = 0.0133 Av 75

4.

To ensure startup: Acl > 1 since Av = 75, B must be greater than 1/75 in order to produce the condition AvB > 1. For example, if B = 1/50, ⎛ 1 ⎞ AvB = 75⎜ ⎟ = 1.5 ⎝ 50 ⎠

Section 16-3 Oscillators with RC Feedback Circuits 5.

Vout 1 = Vin 3 2.2 V ⎛1⎞ Vout = ⎜ ⎟Vin = = 733 mV 3 ⎝3⎠

1 1 = = 1.28 kHz 2πRC 2π (6.2 kΩ)(0.02 μF)

6.

fr =

7.

R1 = 2R2 R 100 kΩ R2 = 1 = = 50 kΩ 2 2

8.

When dc power is first applied, both zener diodes appear as opens because there is insufficient output voltage. This places R3 in series with R1, thus increasing the closed-loop gain to a value greater than unity to assure that oscillation will begin.

9.

Rf = (Av − 1)(R3 + rds′ ) = (3 − 1)(820 Ω + 350 Ω) = 2.34 kΩ

145

Chapter 16 10.

fr =

11.

B=

1 = 10.6 kHz 2π (1.0 kΩ)(0.015 μF)

1 29 1 Acl = = 29 B Rf Acl = Ri Rf = AclRi = 29(4.7 kΩ) = 136 kΩ 1 = 628 Hz fr = 2π 6(4.7 kΩ)(0.022 μ F)

Section 16-4 Oscillators with LC Feedback Circuits 12.

(a) Colpitts: C1 and C3 are the feedback capacitors. 1 fr = 2π L1CT C1C 3 (100 μF)(1000 pF) = = 90.9 pF CT = C1 + C 3 1100 pF 1 = 236 kHz fr = 2π (5 mH)(90.9 pF)

(b) Hartley: fr =

1

2π LT C 2 LT = L1 + L2 = 1.5 mH + 10 mH = 11.5 mH 1 fr = = 68.5 kHz 2π (11.5 mH)(470 pF)

13.

47 pF = 0.1 470 pF The condition for sustained oscillation is 1 1 Av = = = 10 B 0.1

B=

Section 16-5 Relaxation Oscillators 14.

Triangular waveform. 1 ⎛ R2 ⎞ 1 ⎛ 56 kΩ ⎞ ⎜ ⎟= f= ⎟ = 1.61 kHz ⎜ 4 R1C ⎜⎝ R3 ⎟⎠ 4(22 kΩ)(0.022 μF) ⎝ 18 kΩ ⎠

146

Chapter 16 15.

Change f to 10 kHz by changing R1: 1 ⎛ R2 ⎞ ⎜ ⎟ f= 4 R1C ⎜⎝ R3 ⎟⎠ R1 =

16.

T=

1 ⎛ R2 ⎜ 4 fC ⎜⎝ R3

⎞ 1 ⎛ 56 kΩ ⎞ ⎟⎟ = ⎜ ⎟ = 3.54 kΩ ⎠ 4(10 kHz)(0.022 μF) ⎝ 18 kΩ ⎠

V p − VF

⎛ VIN ⎞ ⎟ ⎜ ⎜ RC ⎟ ⎠ ⎝ ⎛ R5 Vp = ⎜⎜ ⎝ R 4 + R5

⎞ 47 kΩ ⎞ ⎟⎟12 V = ⎛⎜ ⎟12 V = 3.84 V ⎝ 147 kΩ ⎠ ⎠ PUT triggers at about +3.84 V (ignoring the 0.7 V drop) Amplitude = +3.84 V − 1 V = 2.84 V ⎛ R2 ⎞ ⎛ 22 kΩ ⎞ ⎟⎟(−12 V) = ⎜ VIN = ⎜⎜ ⎟( −12 V) = −2.16 V ⎝ 122 kΩ ⎠ ⎝ R1 + R2 ⎠ 3.84 V − 1 V T= = 289 μs ⎛ ⎞ 2.16 V ⎜ ⎟ ⎝ (100 kΩ)(0.0022 μ F) ⎠

f=

1 1 = = 3.46 kHz T 289 μ s

See Figure 16-1.

Figure 16-1

147

Chapter 16 17.

18.

VG = 5 V. Assume VAK = 1 V. R5 = 47 kΩ ⎛ R5 ⎞ ⎟⎟ 12 V VG = ⎜⎜ ⎝ R 4 + R5 ⎠ Change R4 to get VG = 5 V. 5 V(R4 + 47 kΩ) = (47 kΩ)12 V R4(5 V) = (47 kΩ)12 V − (47 kΩ)5 V (12 V − 5 V)47 kΩ R4 = = 65.8 kΩ 5V V p − VF T= ⎛ VIN ⎞ ⎜ ⎟ ⎝ RC ⎠ ⎛ ⎞ 3V ⎛V ⎞ ⎟⎟10 μs + 1 V = 7.38 V Vp = ⎜ IN ⎟T + VF = ⎜⎜ ⎝ RC ⎠ ⎝ (4.7 kΩ)(0.001 μF) ⎠

Vpp(out) = Vp − VF = 7.38 V − 1 V = 6.38 V

Section 16-6 The 555 Timer as an Oscillator 19.

1 1 VCC = (10 V) = 3.33 V 3 3 2 2 VCC = (10 V) = 6.67 V 3 3

20.

f=

21.

f=

22.

1.44 1.44 = = 4.03 kHz ( R1 + 2 R2 )C ext (1.0 kΩ + 6.6 kΩ)(0.047 μF)

1.44 ( R1 + 2 R2 )C ext 1.44 1.44 = Cext = = 0.0076 μF ( R1 + 2 R2 ) f (1.0 kΩ + 6.6 kΩ)(25 kHz) R1 + R2 × 100% R1 + 2R2 dc(R1 + 2R2) = (R1 + R2)100 75(3.3 kΩ + 2R2) = (3.3 kΩ + R2)100 75(3.3 kΩ) + 150R2 = 100(3.3 kΩ) + 100R2 150R2 − 100R2 = 100(3.3 kΩ) − 75(3.3 kΩ) 50R2 = 25(3.3 kΩ) 25(3.3 kΩ) R2 = = 1.65 kΩ 50

Duty cycle (dc) =

148

Chapter 16 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 23 through 28 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 23.

Drain-to-source shorted

24.

C3 open

25.

Collector-to-emitter shorted

26.

R1 open

27.

R2 open

28.

R1 leaky

149

Chapter 17 Voltage Regulators Section 17-1 Voltage Regulation 1.

⎛ ΔV Percent line regulation = ⎜⎜ OUT ⎝ ΔVIN

2.

⎛ ΔV /V Percent line regulation = ⎜⎜ OUT OUT ΔVIN ⎝

3.

⎛V /V Percent load regulation = ⎜⎜ NL FL ⎝ ΔVFL

4.

⎞ 2 mV ⎞ ⎟⎟100% = ⎛⎜ ⎟100% = 0.0333% ⎝ 6V ⎠ ⎠ ⎞ 2 mV/8 V ⎞ ⎟⎟100% = ⎛⎜ ⎟100% = 0.00417%/V ⎝ 6V ⎠ ⎠

⎞ ⎛ 10 V − 9.90 V ⎞ ⎟⎟100% = ⎜ ⎟100% = 1.01% 9.90 V ⎠ ⎝ ⎠

From Problem 3, the percent load regulation is 1.01%. For a full load current of 250 mA, this can be expressed as 1.01% = 0.00404%/mA 250 mA

Section 17-2 Basic Linear Series Regulators 5.

See Figure 17-1.

Figure 17-1 6.

⎛ R ⎞ ⎛ 33 kΩ ⎞ VOUT = ⎜⎜1 + 2 ⎟⎟VREF = ⎜1 + ⎟ 2.4 V = 10.3 V R3 ⎠ ⎝ 10 kΩ ⎠ ⎝

7.

⎛ R ⎞ 5.6 kΩ ⎞ ⎛ VOUT = ⎜⎜1 + 2 ⎟⎟VREF = ⎜1 + ⎟ 2.4 V = 8.51 V 2.2 kΩ ⎠ R3 ⎠ ⎝ ⎝

150

Chapter 17 8.

9.

10.

For R3 = 2.2 kΩ: ⎛ R ⎞ 5.6 kΩ ⎞ ⎛ VOUT = ⎜⎜1 + 2 ⎟⎟VREF = ⎜1 + ⎟ 2.4 V = 8.5 V R3 ⎠ 2.2 kΩ ⎠ ⎝ ⎝ For R3 = 4.7 kΩ: ⎛ R ⎞ 5.6 kΩ ⎞ ⎛ VOUT = ⎜⎜1 + 2 ⎟⎟VREF = ⎜1 + ⎟ 2.4 V = 5.26 V R3 ⎠ 4.7 kΩ ⎠ ⎝ ⎝ The output voltage decreases by 3.24 V when R3 is changed from 2.2 kΩ to 4.7 kΩ. ⎛ R ⎞ 5.6 kΩ ⎞ ⎛ VOUT = ⎜⎜1 + 2 ⎟⎟VREF = ⎜1 + ⎟ 2.7 V = 9.57 V R 2.2 kΩ ⎠ ⎝ 3 ⎠ ⎝

0.7 V R4 0.7 V 0.7 mA = = 2.8 Ω R4 = I L (max) 250 mA IL(max) =

2 P = I L(max) R4 = (250 mA)22.8 Ω = 0.175 W, Use a 0.25 W.

11.

2 .8 Ω = 1.4 Ω 2 0.7 V 0.7 V = 500 mA IL(max) = = R4 1.4 Ω

R4 =

Section 17-3 Basic Linear Shunt Regulators 12.

Q1 conducts more when the load current increases, assuming that the output voltage attempts to increase. When the output voltage tries to increase due to a change in load current, the attempted increase is sensed by R3 and R4 and a proportional voltage is applied to the opamp’s noninverting input. The resulting difference voltage increases the op-amp output, driving Q1 more and thus increasing its collector current.

13.

ΔIC =

14.

ΔV R1 1V = 10 mA = R1 100 Ω

⎛ R ⎞ 10 kΩ ⎞ ⎛ VOUT = ⎜⎜1 + 3 ⎟⎟VREF = ⎜1 + ⎟ 5.1 V = 18.2 V R 3 .9 kΩ ⎠ ⎝ 4 ⎠ ⎝ V 18.2 V IL1 = OUT = = 18.2 mA RL1 1 kΩ V 18.2 V IL2 = OUT = = 15.2 mA RL2 1.2 kΩ ΔIL = 15.2 mA − 18.2 mA = −3.0 mA ΔIS = −ΔIL = 3.0 mA

151

Chapter 17 15.

IL(max) =

VIN 25 V = = 250 mA R1 100 Ω

2 PR1 = I L(max) R1 = (250 mA)2100 Ω = 6.25 W

Section 17-4 Basic Switching Regulators 16.

⎛t ⎞ VOUT = ⎜ on ⎟VIN ⎝T ⎠ ton = T − toff 1 1 T= = = 0.01 s = 10 ms f 100 Hz ⎛ 4 ms ⎞ VOUT = ⎜ ⎟ 12 V = 4.8 V ⎝ 10 ms ⎠

17.

f = 100 Hz, toff = 6 ms 1 1 T= = = 10 ms f 100 Hz ton = T − toff = 10 ms − 6 ms = 4 ms t 4 ms = 0.4 duty cycle = on = T 10 ms percent duty cycle = 0.4 × 100% = 40%

18.

The diode D1 becomes forward-biased when Q1 turns off.

19.

The output voltage decreases.

Section 17-5 Integrated Circuit Voltage Regulators 20.

21.

(a) (b) (c) (d)

7806: +6 V 7905.2: −5.2 V 7818: +18 V 7924: −24 V

⎛ R ⎞ 10 kΩ ⎞ ⎛ VOUT = ⎜⎜1 + 2 ⎟⎟VREF + I ADJ R2 = ⎜1 + ⎟ 1.25 V + (50 μA)(10 kΩ) R1 ⎠ ⎝ 1.0 kΩ ⎠ ⎝ = 13.7 V + 0.5 V = 14.3 V

152

Chapter 17 22.

⎡⎛ ⎤ R2(min) ⎞ ⎟VREF + I ADJ R2( min ) ⎥ VOUT(min) = − ⎢⎜⎜1 + ⎟ R1 ⎠ ⎣⎢⎝ ⎦⎥ R2(min) = 0 Ω VOUT(min) = − (1.25 V(1 + 0) + 0) = −1.25 V

⎡⎛ ⎤ R2(max) ⎞ ⎡ ⎤ 10 kΩ ⎞ ⎟VREF + I ADJ R2( max ) ⎥ = − ⎢1.25 V⎛⎜1 + VOUT(max) = − ⎢⎜⎜1 + ⎟ + (50 μA)(10 kΩ)⎥ ⎟ 470 Ω ⎠ R1 ⎠ ⎝ ⎣ ⎦ ⎣⎢⎝ ⎦⎥ = − (1.25 V(22.28) + 0.5 V ) = −28.4 V 23.

The regulator current equals the current through R1 + R2. VOUT 14.3 V IREG ≅ = 1.3 mA = R1 + R2 11 kΩ

24.

VIN = 18 V, VOUT = 12 V IREG(max) = 2 mA, VREF = 1.25 V V 1.25 V R1 = REF = = 625 Ω I REG 2 mA Neglecting IADJ: VR2 = 12 V − 1.25 V = 10.8 V V 10.8 V R2 = R 2 = = 5.4 kΩ I REG 2 mA For R1 use 620 Ω and for R2 use either 5600 Ω or a 10 kΩ potentiometer for precise adjustment to 12 V.

Section 17-6 Applications of IC Voltage Regulators 25.

VRext(min) = 0.7 V 0 .7 V 0.7 V = Rext = = 2.8 Ω I max 250 mA

26.

VOUT = +12 V 12 V IL = = 1200 mA = 1.2 A 10 Ω Iext = IL − Imax = 1.2 A − 0.5 A = 0.7 A Pext = Iext(VIN − VOUT) = 0.7 A(15 V − 12 V) = 0.7 A(3 V) = 2.1 W

27.

VRlim(min) = 0.7 V 0.7 V 0.7 V Rlim(min) = = = 0.35 Ω I ext 2A See Figure 17-2.

Figure 17-2

153

Chapter 17 28.

1.25 V = 2.5 Ω 500 mA See Figure 17-3. R=

Figure 17-3 29.

I = 500 mA 8V = 16 Ω R= 500 mA See Figure 17-4.

Figure 17-4 30.

VREF = 1.25 V The voltage divider must reduce the output voltage (12 V) down to the reference voltage (1.25 V). See Figure 17-38 in the text.

⎛ R1 ⎞ ⎟⎟VOUT VREF = ⎜⎜ ⎝ R1 + R2 ⎠ R1 V = REF R1 + R2 VOUT R1 = R1(VREF/VOUT) + R2(VREF/VOUT) R − R1 (VREF / VOUT ) R1 (1 + VREF / VOUT ) R2 = 1 = (VREF / VOUT ) (VREF / VOUT ) Let R1 = 10 kΩ. 10 kΩ(1 − 1.25 V/12 V) R2 = = 86 kΩ (1.25 V/12 V)

154

Chapter 17 Multisim Troubleshooting Problems The solutions showing instrument connections for Problems 31 through 34 are available from the Instructor Resource Center. See Chapter 1 for instructions. The faults in the circuit files may be accessed using the password book (all lowercase). 31.

R2 leaky

32.

Zener diode open

33.

Q2 collector-to-emitter open

34.

R1 open

155

Chapter 18 Communications Section 18-1 Basic Receivers 1.

See Figure 18-1.

Figure 18-1 2.

See Figure 18-2.

Figure 18-2 3.

fLO = 680 kHz + 455 kHz = 1135 kHz

4.

fLO = 97.2 MHz + 10.7 MHz = 107.9 MHz

5.

fRF = 101.9 MHz − 10.7 MHz = 91.2 MHz fIF = 10.7 MHz (always)

156

Chapter 18 Section 18-2 The Linear Multiplier Vout ≅ −2.5 V Vout ≅ −1.6 V Vout ≅ +1.0 V Vout ≅ +10 V

6.

(a) (b) (c) (d)

7.

Vout = KVXVY = 0.125(+3.5 V)(−2.9 V) = −1.27 V

8.

Connect the two inputs together.

9.

(a) Vout = KV1V2 = (0.1)(+2 V)(+1.4 V) = +0.28 V (b) Vout = KV1V2 = KV12 (0.1)(−3.2 V) 2 = +1.024 V − V1 − (6.2 V) (c) Vout = = = +2.07 V −3V V2

(d) Vout = V1 = 6.2 V = +2.49 V

Section 18-3 Amplitude Modulation 10.

fdiff = f1 − f2 = 100 kHz − 30 kHz = 70 kHz fsum = f1 + f2 = 100 kHz + 30 kHz = 130 kHz

11.

f1 =

12.

fc = 1000 kHz fdiff = 1000 kHz − 3 kHz = 997 kHz fsum = 1000 kHz + 3 kHz = 1003 kHz

13.

f1 =

14.

fc = 1.2 MHz by inspection fm = fc − fdiff = 1.2 MHz − 1.1955 MHz = 4.5 kHz

9 cycles = 9000 cycles/s = 9 kHz 1 ms 1 cycle = 1000 cycles/s = 1 kHz f2 = 1 ms fdiff = f1 − f2 = 9 kHz − 1 kHz = 8 kHz fsum = f1 + f2 = 9 kHz + 1 kHz = 10 kHz

18 cycles = 1.8 MHz 10 μs 1 cycle f2 = = 100 kHz 10 μs fdiff = f1 − f2 = 1.8 MHz − 100 kHz = 1.7 MHz fsum = f1 + f2 = 1.8 MHz + 100 kHz = 1.9 MHz fc = 1.8 MHz

157

Chapter 18 f diff + f sum

8.47 kHz + 853 KHz = 850 kHz 2 2 fm = fc − fdiff = 850 kHz − 847 kHz = 3 kHz =

15.

fc =

16.

fdiff(min) = 600 kHz − 3 kHz = 597 kHz fdiff(max) = 600 kHz − 300 Hz = 599.7 kHz fsum(min) = 600 kHz + 300 kHz = 600.3 kHz fsum(max) = 600 kHz + 3 kHz = 603 kHz See Figure 18-3.

Figure 18-3

Section 18-4 The Mixer 1 [cos( A − B ) − cos( A + B )] 2 Vin(1) = 0.2 V sin [2π(2200 kHz)t] Vin(2) = 0.15 V sin [2π(3300 kHz)t] Vin(1)Vin(2) = (0.2 V)(0.15 V) sin [2π(2200 kHz)t] sin [2π(3300 kHz)t] (0.2 V)(0.15 V) Vout = [cos 2π(3300 kHz − 2200 kHz)t − cos 2π (3300 kHz + 2200 kHz)t] 2 Vout = 15 mV cos [2π(1100 kHz)t] − 15 mV cos [2π(5500 kHz)t]

17.

(sin A)(sin B) =

18.

fIF = fLO − fc = 986.4 kHz − 980 kHz = 6.4 kHz

158

Chapter 18 Section 18-5 AM Demodulation 19.

See Figure 18-4.

Figure 18-4 20.

See Figure 18-5.

21.

See Figure 18-6.

Figure 18-5

Figure 18-6

Section 18-6 IF and Audio Amplifiers 22.

fc − fm = 1.2 MHz − 8.5 kHz = 1.1915 MHz fc + fm = 1.2 MHz + 8.5 kHz = 1.2085 MHz fc = 1.2 MHz fLO − fm = 455 kHz − 8.5 kHz = 446.5 kHz fLO + fm = 455 kHz + 8.5 kHz = 463.5 kHz fLO = 455 kHz

23.

The IF amplifier has a 450 kHz to 460 kHz passband. The audio/power amplifiers have a 10 Hz to 5 kHz bandpass.

159

Chapter 18 24.

C4 between pins 1 and 8 makes the gain 200. With R1 set for minimum input, Vin = 0 V. Vout(min) = AvVin(min) = 200(0 V) = 0 V With R1 set for maximum input, Vin = 10 mV rms. Vout(max) = AvVin(max) = 200(10 mV) = 2 V rms

Section 18-7 Frequency Modulation 25.

The modulating input signal is applied to the control voltage terminal of the VCO. As the input signal amplitude varies, the output frequency of the VCO varies proportionately.

26.

An FM signal differs from an AM signal in that the information is contained in frequency variations of the carrier rather than amplitude variations.

27.

Varactor

Section 18-8 The Phase-Locked Loop (PLL) 28.

See Figure 18-7.

Figure 18-7 29.

(a) The VCO signal is locked onto the incoming signal and therefore its frequency is equal to the incoming frequency of 10 MHz. (b) Vc =

Vi V o (250 mV)(400 mV) cos θ e cos(30° − 15°) = (0.050)(0.966) = 48.3 mV 2 2

30.

Δfo = +3.6 kHz, ΔVc = +0.5 V Δf o + 3.6 kHz K= = 7.2 kHz/V = ΔV c + 0.5 V

31.

K = 1.5 kHz/V, ΔVc = +0.67 V Δf o K= ΔV c Δfo = KΔVc = (1.5 kHz/V)(+0.67 V) = 1005 Hz

160

Chapter 18 32.

For a PLL to acquire lock the following conditions are needed: (1) The difference frequency, f0 − fi must fall within the filter’s bandwidth. (2) The maximum frequency deviation of the VCO frequency, Δfmax, must be sufficient to permit f0 to change to equal fi.

33.

The free-running frequency: 1 .2 1 .2 fo = = 233 kHz = 4 R1C1 4(3.9 kΩ)(330 pF) The lock range: 8f 8( 233 kHz) 1.864 MHz flock = ± o = ± =± = ±104 kHz VCC 18 V 18 V The capture range: fcap = ±

1 2π

⎛ 2πf lock ⎜⎜ ⎝ 3600 × C 2

= ±

1 2π

⎛ 2π (103.6 kHz) ⎞ 1 ⎜⎜ ⎟⎟ = ± 2π ⎝ 3600 × 0.22 μF ⎠

⎞ ⎟⎟ ⎠ ⎛ 650.9 kHz ⎞ ⎟⎟ = ±4.56 kHz ⎜⎜ ⎝ 792 μF ⎠

Section 18-9 Fiber Optics 34.

The light ray will be reflected because the angle of incidence (30°) is greater than the critical angle (15°).

35.

θC = cos−1(n2/n1) = cos−1(1.25/1.55) = 36.2°

There are no Multisim Troubleshooting Problems in this chapter.

161

Application Activity Results Chapter 2 1.

A comparison of Equations 2-5 and 2-7 shows that the output voltage of a bridge rectifier is nearly twice that of a center-tapped rectifier. Also, two diode drops contribute to the bridge output voltage, but only one diode drop contributes to the center-tapped output voltage.

2.

A comparison of Equations 2-6 and 2-8 shows that the PIV of a bridge rectifier is nearly half that of a center-tapped rectifier.

3.

V p ( out ) = V p ( sec ) − 1.4 V V p ( sec ) = V p ( out ) + 1.4 V = 16 V + 1.4 V = 17.4 V peak = 12.3 V rms

4.

Minimum VA = (12.3 V)(250 mA) = 3.075 VA Use the 12 V CT rated at 6 VA. The 12 V rms secondary voltage is within 10% of the specified 12.3 V determined in 3.

5.

The primary current is determined as follows: I p ( pri ) Vsec = I p ( sec ) V pri ⎛V ⎞ ⎛ 12 V ⎞ I p ( pri ) = ⎜ sec ⎟ I p ( sec ) = ⎜ ⎟ 354 mA = 35.4 mA ⎜ V pri ⎟ ⎝ 120 V ⎠ ⎝ ⎠ The fuse rating should be at least 20% greater than I p ( pri ) .

6.

7.

8.

r=

Vr ( pp )

VDC Vr ( pp ) = rVDC = 0.03(16 V) = 480 mV

⎛ 1 ⎞ Vr ( pp ) = ⎜ ⎟V p ( rectified ) ⎝ f RL C ⎠ ⎛ ⎞ ⎛ ⎞ 1 1 C= ⎜ V p ( rectified ) = ⎜ ⎟17 V = 0.004612 F ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ f RLVr ( pp ) ⎠ ⎝ (120 Hz)(64 Ω)(480 mV) ⎠ Use 4700 μF.

PL = VLIL = (16 V)(258 mA) = 4.13 W The rating should be at least 5 W.

163

Application Activity Results Troubleshooting

(a)

Open capacitor. Circuit is operating as full-wave rectifier with no filter.

(b)

No fault

(c)

Rectifier diode open. Circuit is operating as a half-wave rectifier.

(d)

Diode and filter capacitor open. Circuit is operating as a half-wave rectifier with no filter.

Chapter 3 1.

VOUT(max) = 12 V + 120 mV = 12.12 V (typical)

2.

ΔVOUT = 5 mV

3.

VF = 1.9 V 12 V − 1.9 V Rlimit = = 505 W 20 mA P = (10.1 V)20 mA = 202 mW. Use 0.25 W rating.

4.

⎛V ⎞ ⎛ 12.6 V ⎞ Ipri = ⎜ sec ⎟ I sec = ⎜ ⎟ 250 mA = 26.3 mA ⎜ V pri ⎟ ⎝ 120 V ⎠ ⎝ ⎠ Use the next highest standard fuse rating.

(12 V) 2 = 3.06 W 47 Ω

5.

PL =

6.

The printed circuit board and the schematic agree.

7.

Ireg = ILED + IL = 20 mA + 250 mA = 270 mA Preg = (16 V − 12 V)270 mA = 1.08 W

Chapter 4 1.

From the datasheet, VCE(sat) = 0.3 V. VCC − VCE(sat) 12 V − 0.3 V R3 = = 1170 Ω (use std. 1.2 kΩ) = 10 mA I C1

2.

From the datasheet, β DC(min) = hFE(min) = 75.

IB1 =

3.

R1 =

I C1

β DC(min)

=

10 mA = 133 μA 75

Vsensor − VBE 12 V − 0.7 V = = 85 Ω (use std. 75 Ω) I B1 133 μ A

164

Application Activity Results 4.

IC2 = IB2 =

VCC − VCE(sat) RL I C2

β DC(min)

12 V − 0.3 V = 18.9 mA 620 Ω 18.9 mA = = 252 μA 75 =

VCC − VBE R3 + R4 V − VBE 12 V − 0.7 V R3 + R4 = CC = = 44.8 kΩ I B2 252 μ A R4 = 44.8 kΩ − 1.2 kΩ = 43.6 kΩ (use lower std. value 39 kΩ or even 36 kΩ to ensure saturation)

IB2 =

5.

The value of VCE(sat) in the simulation is 0.126 V. The datasheet shows a value of 0.3 V at 150 mA. The simulated value is less because the collector current is much less than the current at which the datasheet value of 0.3 V is specified.

6.

Each circuit on the printed circuit board is in agreement with the schematic. See Figure AA-1 for the component identification.

Figure AA-1 Each of the four circuits is identical.

7.

The resistor values are in close agreement.

8.

See Figure AA-1 for pin labels.

165

Application Activity Results 9.

Connect VCC and ground. Apply +12V to input 1. Check output 1 for +12 V. Apply 0 V to input 1. Check output 1 for approximately 0 V (VCE(sat0). Repeat for each of the four circuits using the inputs 2-4 and outputs 2-4.

10.

Add a second circuit board with only two circuits installed for zones 5 and 6. The outputs are externally connected to the circuit outputs of the other board to form a single output to the alarm.

Chapter 5 1.

See Figure AA-2.

Figure AA-2

2.

RIN(BASE) = βDCRE = 300(470 Ω) = 141 kΩ The voltage divider is stiff for all values of RTherm. At 60°C:

At 65°C:

At 70°C:

⎛ RTherm ⎞ ⎛ 1.256 kΩ ⎞ VB = ⎜ ⎟VCC = ⎜ ⎟ 9 V = 1.9 V + R R ⎝ 5.965 kΩ ⎠ Therm ⎠ ⎝ 1 V − VBE 1.9 V − 0.7 V IE = B = = 2.6 mA RE 470 Ω IC = IE = 2.6 mA ⎛ RTherm ⎞ ⎛ 1.481 kΩ ⎞ VB = ⎜ ⎟VCC = ⎜ ⎟ 9 V = 2.16 V ⎝ 6.181 kΩ ⎠ ⎝ R1 + RTherm ⎠ V − VBE 2.16 V − 0.7 V = IE = B = 3.1 mA RE 470 Ω IC = IE = 3.1 mA ⎛ RTherm ⎞ ⎛ 1.753 kΩ ⎞ VB = ⎜ ⎟VCC = ⎜ ⎟ 9 V = 2.44 V ⎝ 6.453 kΩ ⎠ ⎝ R1 + RTherm ⎠ V − VBE 2.44 V − 0.7 V IE = B = = 3.7 mA RE 470 Ω IC = IE = 3.7 mA

166

Application Activity Results At 75°C:

At 80°C:

⎛ RTherm ⎞ ⎛ 2.084 kΩ ⎞ VB = ⎜ ⎟VCC = ⎜ ⎟ 9 V = 2.76 V ⎝ 6.784 kΩ ⎠ ⎝ R1 + RTherm ⎠ V − VBE 2.76 V − 0.7 V = IE = B = 4.4 mA RE 470 Ω IC = IE = 4.4 mA ⎛ RTherm ⎞ ⎛ 2.490 kΩ ⎞ VB = ⎜ ⎟VCC = ⎜ ⎟ 9 V = 3.12 V ⎝ 7.19 kΩ ⎠ ⎝ R1 + RTherm ⎠ V − VBE 3.12 V − 0.7 V = IE = B = 5.1 mA RE 470 Ω IC = IE = 5.1 mA VOUT = VCC − ICR2 = 9 V − (2.6 mA)(1.0 kΩ) = 6.4 V VOUT = VCC − ICR2 = 9 V − (3.1 mA)(1.0 kΩ) = 5.9 V VOUT = VCC − ICR2 = 9 V − (3.7 mA)(1.0 kΩ) = 5.3 V VOUT = VCC − ICR2 = 9 V − (4.4 mA)(1.0 kΩ) = 4.6 V VOUT = VCC − ICR2 = 9 V − (5.1 mA)(1.0 kΩ) = 3.9 V

3.

At 60°C: At 65°C: At 70°C: At 75°C: At 80°C:

4.

The calculated values agree reasonably well with the values measured in the simulation.

Chapter 6 1.

Second stage is set for maximum gain when R10 is completely bypassed and equal to 0 Ω. Rc1 = R5 & R6 & R7 & β DC R9 = 22 kΩ & 47 kΩ & 22 kΩ & 200(130 Ω) = 6.6 kΩ R 6.6 kΩ Av1 = c1 = = 6.6 R4 1.0 kΩ R8 6.8 kΩ = = 52.3 R9 130 Ω

2.

Av2 =

3.

Av(tot) = Av1Av2 = (6.6)(52.3) = 345 (This exceeds the specified maximum gain)

4.

I1 = I2 =

30 V = 45.5 μA 660 kΩ 14.3 V IE1 = IC1 = = 421 μA 34 kΩ 30 V I6 = I7 = = 435 μA 69 kΩ VB2 = VCC − I6R6 = 15 V − 20.4 V = −5.4 V −15 V − (−5.4 V − 0.7 V) IE2 = IC2 = = 1.7 mA 130 Ω + 5 kΩ ITOT = 45.5 μA + 421 μA + 435 μA + 1.7 mA = 2.6 mA PD = 30 V (2.6 mA) = 78 mW

167

Application Activity Results 5.

Vin(p) = 45 mV From the scope display, Vout(p) = 5.5 V. 5.5 V Av = = 122 45 mV

6.

The calculated maximum voltage gain is 345. The amplifier is not adjusted for maximum gain, but the value of 122 is within the specified range (90 to 170).

7.

The pc board agrees with the schematic.

8.

See Figure AA-3.

Figure AA-3

9.

There is a 400 mV peak signal at the collector of Q1 and the same signal appears at the base of Q2. There is no output signal. Either Q2 is faulty or C5 (the output coupling capacitor) is open.

10.

There is a 50 mV peak signal at the base of Q1 and a 300 mV peak signal at the collector of Q1. There is no signal at the base of Q2. The coupling capacitor C3 is open.

Chapter 7 1.

Assuming βDC for Q5 is 200, RIN(BASE) = βDCR4 = 200(220 Ω) = 44 kΩ RIN = RIN(BASE) = 44 kΩ

2.

Av =

R3 1.0 kΩ = = 4.5 R4 220 Ω

168

Application Activity Results 3.

The output waveform has a peak of approximately 10 V. Vout(rms) = 0.707(10 V) = 7.07 V 2 Vout (7.07 V) 2 ( rms ) = = 6.1 W Pout = RL 8.2 Ω

10.5 V = 4.2 2.5

4.

Av =

5.

The calculated gain is 4.5 and the measured gain is 4.2. They are close within less than 10%.

6.

The scope display shows the peak output of the power amplifier is approximately 9.8 V. 2 Vout (6.93 V) 2 ( rms ) Pout = = = 5.86 W 8.2 Ω RL

7.

From the scope display, Vin(peak) = 2 V and Vout(peak) = 9.8 V. 9.8 V Av = = 4.9 2V

8.

Av =

9.

The pc board agrees with the schematic.

10.

See Figure AA-4.

9.8 V = 245 40 mV

Figure AA-4

169

Application Activity Results 11.

An output transistor is faulty.

12.

The interconnections are correct.

Chapter 8 1.

From the graph, Vsensor = 100 mV for a pH of 8.

2.

From the graph, Vsensor = −400 mV for a pH of 3.

3.

From the datasheet y21s = gfs = 24 mS

4.

The maximum drain current is specified to be 30 mA regardless of VDS.

5.

From the datasheet graph, with VG2S = 1 V and VG1S = 0 V, ID is approximately 4.5 mA.

6.

For VOUT = 7 V, Vsensor is between −200 mV and −300 mV so the pH is between 4 and 5, making the solution acidic.

7.

See Figure AA-5.

Figure AA-5

8.

The pc board agrees with the schematic.

170

Application Activity Results 9.

See Figure AA-6.

Figure AA-6

10.

See Figure AA-6.

Chapter 9 1.

Channel 7: 174-180 MHz Channel 8: 180-186 MHz Channel 9: 186-192 MHz Channel 10: 192-198 MHz Channel 11: 198-204 MHz Channel 12: 204-210 MHz Channel 13: 210-216 MHz

2.

824-896 MHz and 1850-1990 MHz

3.

C2 provides an ac ground for the gate of Q1 (common gate).

4.

RIN(gate) =

5.

Rin = R3 & RIN(gate) = 10 MΩ & 20 × 109 Ω = 10 MΩ

6.

From the datasheet, gfs = 3000 μmhos = 3000 μS

7.

88 MHz is the lower end of the FM band.

VGS 20 V = = 20 × 109 Ω I GSS 1 nA

171

Application Activity Results 8.

(b)

Vout(rms) = 0.707Vout(p) = 0.707(21 mV) = 14.8 mV 21 mV = 2100 Av = 10 μ V Vout(rms) = 0.707Vout(p) = 0.707(100 mV) = 70.7 mV 70.7 mV Av = = 7070 10 μ V

(c)

9.

The pc board agrees with the schematic.

10.

It is a decoupling capacitor connected to the dc supply line for noise elimination.

11.

See Figure AA-7.

Figure AA-7

Chapter 10 1.

For stage 1 input circuit: 1 fcl(input) = 2π ( R1 & R2 & β ac R4 )C1 1 1 C1 = = = 13.2 nF 2π ( R1 & R2 & β ac R4 ) f cl ( input ) 2π (330 kΩ & 330 kΩ & 100(1.0 kΩ)193 Hz

2.

For stage 2 input circuit:

fcl(input) = C3 =

1 2π ( R5 + R6 & R7 & β ac ( R9 + R10 ) ) C3 1

2π ( R5 + R6 & R7 & β ac ( R9 + R10 ) ) f cl ( input )

172

=

1 = 23.4 nF 2π (35.2 kΩ)193 Hz

Application Activity Results 3.

Midrange gain is 33.4 dB. Gain at 60 Hz is 9.7 dB. 33.4 dB − 9.7 dB = 23.7 dB

4.

At 5 kHz from scope display: Vout = 0.707(2.6 V) = 1.84 V

5.

At 60 Hz from scope display: Vout = 0.707(240 mV) = 170 mV

6.

At 300 Hz: Vout = 0.707(1.84 V) = 1.30 V

7.

Av =

8.

Av =

2.6 V = 57.8 45 mV Av (dB) = 20 log Av = 20 log(57.8) = 35.2 dB 170 mV = 3.78 45 mV Av (dB) = 20 log Av = 20 log(3.78) = 11.6 dB

Chapter 11 1.

From the datasheet, VPRM = 400 V for the 2N6397.

2.

From the datasheet, the maximum on-state rms current is 12 A.

3.

Von(max) = 1.5 V IF = 50 mA Pmax = Von(max)IF = (1.5 V)(50 mA) = 750 mW

4.

The SCR conducts during the flat portion of the waveform.

5.

The SCR will conduct more it the control voltage is reduced.

6.

The motor speed will increase when the SCR conducts more as a result of a reduced control voltage.

7.

The pc board agrees with the schematic.

173

Application Activity Results 8.

See Figure AA-8.

Figure AA-8

9.

(a) (b) (c)

Test of board 1: Working properly Test of board 2: SCR is not triggering on. Test of board 3: SCR continuously on

10.

(a) (b) (c)

No problem PUT faulty or SCR gate open SCR shorted

Chapter 12 1.

The op-amp configuration is a noninverting amplifier.

2.

The maximum and minimum gains of the op-amp stage are

Av = 1 +

Rf Ri

150 kΩ = 151 1 kΩ 150 kΩ =1+ = 75 2 kΩ

Av (max) = 1 + Av (min) 3.

Maximum output voltage of the op-amp stage is Vout(max)⇒ = Av(max)Vin = (150)(50 mV) = 7.5 V

174

Application Activity Results 4.

The ideal maximum power to the speaker is Vout ( rms ) = 7.5 V

Pmax =

2 Vout ( rms )

Rspeaker

=

(7.5 V) 2 = 7.03 W 8Ω

5.

Noninverting input: pin 3; inverting input: pin 2; output: pin 6; ±15 V: pin 7; −15 V pin 4.

6.

From the datasheet: Maximum power consumption @ +15 V is 85 mW.

7.

From the datasheet, the typical output voltage swing is +14 V.

8.

V1(rms) = 50 mV input voltage V2(rms) = 0.707(10 V) = 7.07 V output of op-amp V3(rms) = 0.707(9 V) = 6.36 V voltage across the load

9.

Voltage gain of op-amp stage:

10.

Av =

Vout V2 7.07 V = = = 141.4 Vin V1 50 mV

Av =

Vout V3 6.36 V = = = 127.2 Vin V1 50 mV

Overall voltage gain:

11.

The pc board agrees with the schematic.

12.

See Figure AA-9.

Figure AA-9

175

Application Activity Results 13.

(a) (b) (c) (d)

Working properly No signal at C1. Top half of output clipped. Bottom half of output clipped.

14.

(a) (b) (c) (d)

No problem Op-amp or C1 faulty. Q3 or Q4 faulty. Q1 or Q2 faulty.

Chapter 13 1.

R1, R2, and R3 determine the reference voltage.

2.

IR1 = IR2 = IR3 =

3.

VR2/R3 = IR3(R2 + R3) = 39 μA(556 kΩ) = 21.7 V VREF(max) = −12 V + 21.7 V = +9.7 V

4.

The output pulse amplitude is limited to +5 V and 0 V by the +5V supply voltage.

5.

As R2 is adjusted, the reference voltage at which the comparator is triggered is varied from a maximum of +9.82 V to a minimum of −9.7 V. This changes the point on the input sine wave where the comparator changes state. The higher up on the sine wave that the trigger point occurs, the smaller the duty cycle.

6.

Noninverting input: pin 2; inverting input: pin 3; output: pin 7; +12 V: pin 8; −12 V: pin 4.

7.

Vp = 10 V Vrms = 0.707(10 V) = 7.07 V

8.

Vpulse = 5 V

9.

T = (2 div)(50 μs/div) = 100 μs 1 1 f= = = 10 kHz T 100 μ s

10.

(a)

24 V 24 V = = 39 μA R1 + R2 + R3 612 kΩ VR3 = IR3R3 = 39 μA(56 kΩ) = 2.18 V VREF(min) = −12 V + VR3 = −12 V + 2.18 V = −9.82 V

Each small division = 40 μs Pulse width = 1.5(40 μs) = 60 μs Period = 5 div(200 μs/div) = 1000 μs ⎛ 60 μ s ⎞ % Duty cycle = ⎜ ⎟100 = 6% (exceeds specs) ⎝ 1000 μ s ⎠

176

Application Activity Results (b)

Pulse width = 4.6 div(200 μs) = 920 μs Period = 5 div(200 μs/div) = 1000 μs ⎛ 920 μ s ⎞ % Duty cycle = ⎜ ⎟100 = 92% (exceeds spec) ⎝ 1000 μ s ⎠

(c)

Each small division = 4 μs Pulse width = 2(4 μs) = 8 μs Period = 5 div(20 μs/div) = 100 μs ⎛ 8 μs ⎞ % Duty cycle = ⎜ ⎟100 = 8% (exceeds spec) ⎝ 100 μ s ⎠

(d)

Pulse width = 4.6(20 μs) = 92 μs Period = 5 div(20 μs/div) = 100 μs ⎛ 92 μ s ⎞ % Duty cycle ⎜ ⎟100 = 92% (exceeds spec) ⎝ 100 μ s ⎠

11.

The pc board agrees with the schematic.

12.

See Figure AA-10.

Figure AA-10

13.

The connections are correct.

Chapter 14 1.

Pin 3 is connected to pin 11. From Table 1 on the on-line datasheet, the gain is 500.

177

Application Activity Results 2.

Table 1 in the on-line datasheet shows pin connections for various fixed gains. Also, the formula can be applied for pins 3 and 16, as shown in Figure AA-11.

Figure AA-11 Operating connections for G = 20.

3.

From the scope, Vout(pp) = (2.9 div)(50 mV/div) = 145 mV ⎛ 145 mV ⎞ Vout(rms) = 0.707 ⎜ ⎟ = 52.1 mV 2 ⎝ ⎠ Vout ( rms ) 51.2 mV Av = = = 512 100 μ V Vin ( rms ) The gain indicated by the pin connections is 500.

4.

The hysteresis is approximately 50 mV.

5.

The shield-guard driver output on the scope display is approximately 100 mV rms. This is the same as the common-mode signal.

6.

The pc board agrees with the schematic.

178

Application Activity Results 7.

See Figure AA-12.

Figure AA-12

Chapter 15 1.

RFID systems are used for tracking and identification of objects.

2.

The basic RFID system components are tag, reader, and processor.

3.

The tag stores an identification code and other information which it transmits to a reader in the form of a modulated signal.

4.

The reader receives the transmitted information from the tag and converts it to digital form.

5.

The band-pass filter eliminates frequencies other than the frequency transmitted by the tag.

6.

The low-pass filter eliminates the carrier frequency and passes only the modulating signal frequency.

7.

Gain of amplifier U2 and U3: Av = 1 +

8.

fo =

R5 R 100 kΩ = 46.5 =1+ 7 = 1 + 2.2 kΩ R6 R8

1 = 2π C

R1 + R3 1 15 kΩ + 1 kΩ = = 132 kHz R1 R2 R3 2π (200 pF) (15 kΩ)(39 kΩ)(1 kΩ)

This is within 6% of the measured center frequency.

179

Application Activity Results 9.

10.

11.

The low-pass filter is set for an approximate Butterworth response since the damping factor is ⎛R ⎞ ⎛ 560 Ω ⎞ DF = 2 − ⎜ 12 ⎟ = 2 − ⎜ ⎟ = 1.44 ⎝ 1000 Ω ⎠ ⎝ R11 ⎠ fc =

1 1 = = 15.9 kHz 2π RC 2π (1 kΩ)(10 nF)

⎛ R15 ⎞ ⎛ 15 kΩ ⎞ VREF = ⎜ ⎟VCC = ⎜ ⎟15 V = 1.96 V ⎝ 115 kΩ ⎠ ⎝ R14 + R15 ⎠ A positive reference is required because the output of the LP filter is a positive signal.

12.

The pc board agrees with the schematic.

13.

See Figure AA-13.

Figure AA-13

180

Application Activity Results Chapter 16 1.

Av =

C1 C4 100 nF = = = 10 C2 C5 10 nF

2.

CT =

C4C5 (100 nF)(10 nF) = = 9.1 nF 110 nF C4 + C5

fr =

1 1 = = 136 kHz 2π L1CT 2π (150 μ H)(9.1nF)

This calculated frequency is within 10% of the 125 kHz measured frequency. 3.

fr =

1.44 1.44 = = 11.6 kHz ( R7 + R8 )C3 (6.2 kΩ + 6.2 kΩ)10 nF

The measured frequency is 10 kHz. The calculated value differs by 16%. 4.

The forward resistance of the diode is a possible factor in the difference between the calculated and measured values in 2 and 3.

5.

Q2 operates as an analog switch that turns the carrier signal (125 kHz) on and off.

6.

First waveform: Output of ASK generator containing bursts of the 125 kHz carrier signal at 10 kHz. Second waveform: output of the RFID rectifier. Third waveform: The rectified signal after filtering out the 125 kHz carrier. Fourth waveform: Pulse output of the RFID reader.

7.

The pc board agrees with the schematic of the ASK test generator.

181

Application Activity Results 8.

See Figure AA-14.

Figure AA-14

Chapter 17 1.

From the datasheet, VO(min) = 8.65 V and VO(max) = 9.35 V.

2.

From the datasheet, the output voltage can change a maximum of 180 mV.

3.

IL(max) =

4.

5.

VO(max)

9.4 V = 28.3 mA 330 Ω R1 PD(max) = (28.3 mA)2330 Ω = 265 mW =

⎛ 1 kΩ ⎞ VR2(max) = ⎜ ⎟ 9.4 V = 7.07 V ⎝ 1330 Ω ⎠ (7.07 V) 2 = 50 mW PD(max) = 1 kΩ I pri I sec

=

Vsec V pri

⎛ (0.707)(34 V) ⎞ I pri = ⎜ ⎟100 mA = 20 mA 120 V ⎝ ⎠ Fuse rating should exceed 20 mA.

182

Application Activity Results 6.

The pc board agrees with the schematic.

7.

PD = (32.6 V − 9 V)100 mA = 2.36 W

8.

PD = (32.6 V − 30 V)100 mA = 236 mW

183

Summary of MultiSIM Circuit Files Chapter 1: Circuit E01-01

F01-40a F01-40b F01-40c F01-40d F01-41a F01-41b F01-41c F01-41d F01-42

TSP01-22 TSP01-23 TSP01-24 TSP01-25 TSP01-26 TSP01-27 TSP01-28 TSP01-29 TSP01-30

MultiSIM VR(1) = 9.29 V VD(1) = 0.713 V VR(2) = 4.99 V VD(2) = 4.99 V VR = 0.020 nV VR = 99.151 V VR = 14.01 V VD = 0.651 V VD = 25.000 V VD = 15.000 V VD = 2.500 V VR = 11.285 nV VA = 25.00 V VB = 24.255 V VC = 8.746 V VD = 8.000 V VR = 3.000 V VD = 99.994 V VD = 32.434 VD = -1.000 pV VD = 0.668 V VD = 0.192 nV VD = 2.577 V VD = 11.994 V VA = 25.00 V VB = 24.253 V VC = 8.000 V VD = 8.000 V

Chapter 2: Circuit E02-02a E02-02b E02-03 E02-05 E02-06 E02-07 E02-09

MultiSIM Half-wave VP = 4.3 V Half-wave VP = 99.2 V Half-wave VP = 84.2 V Full-wave VP = 24.3 V Full-wave VP = 15.74 V VL(max) = 14.7 V VL(min) = 13.8 V Half-wave VP(max) = 9.1 V VP(min) = 779.1 mV

185

Summary of MultiSIM Files Circuit E02-10 E02-11 E02-12 F02-73a F02-73b F02-74 F02-76 F02-77 F02-78 F02-79 F02-83a F02-83b F02-86 F02-88 TSE02-01 TSE02-02 TSE02-03 TSP02-52 TSP02-53 TSP02-54 TSP02-55 TSP02-56 TSP02-57 TSP02-58 TSP02-59

MultiSIM Clipped VP(max) = 5.7 V VP(min) = -5.7 V Negative Clamp VP(max) = 9.0 V VP(min) = -18.0 V Clipped VP(max) = 729 mV VP(min) = -45.5 V Half-wave VP = 4.2 V Half-wave VP = -49.3 V Half-wave VP = 83.964 V Full-wave VP = 20.476 V Full-wave VP = 32.296 V VL(max) = 47.6 V VL(min) = 46.6 V Half-wave VP(max) = 712.7 mV VP(min) = -10.0 V Clipped VP(max) = 721.9 mV VP(min) = -722.0 mV Clipped VP(max) = 722.0 mV VP(min) = -722.0 mV VSEC = 119.812 VAC VREC = 0.2 V VL = 59.1 V Full-wave VP = 16.211 V VFUSE = 0 V VSURGE = 0 V Not implemented Unrectified AC 10.0 VPP VL(max) = 38.4 V VL(min) = -49.3 V VL = 0 V Half-wave VP = 19.6 V Clipped Full-wave VP(max) = 59.8 V VP(min) = 1.6 V Full-wave VP(max) = 49.1 V VP(min) = 784.1 mV Positive Clipped VP(max) = 679.1 mV VP(min) = -909.1 mV Half-wave VP(min) = -30.0 V

186

Summary of MultiSIM Files Circuit TSP02-60

MultiSIM Half-wave VP(max) = 16.0 V VP(min) = 480.1 mV

Chapter 3: Circuit E03-05

E03-06 E03-08

F03-70

F03-73 F03-77 TSE03-01 TSE03-02 TSE03-03 TSE03-04 TSP03-47 TSP03-48 TSP03-49 TSP03-50

MultiSIM ID(1) = 4.749 µA VD(1) = 4.860 V ID(2) = 206 mA VD(2) = 5.137 V ID = 1.330 mA VR = 0.024 A VD = 11.930 V Clipped VP(max)(1) = 5.6 V VP (min)(1) = -3.8 V VP(max)(2) = 6.7 V VP (min)(1) = -15.5 V IR(1) = 150 mA VR(1) = 4.692 V IR(2) = 40 mA VR(2) = 5.116 V ID = 34 mA VOUT = 11.002 V VOUT = 10.439 V VC = 0 VDC VOUT = 0.000 VDC VD = 15.0 VDC with ripple VOUT = 15.0 VDC VD = 0 V Blown fuse VOUT = 8.000 VDC Spikey output VOUT = 12.0 V VSPIKE = 9.0 V VR > 26 VDC VOUT = 0.079 nVDC VR = 22.243 V

Chapter 4: Circuit E04-02

MultiSIM IB = 412 µA IC = 62 mA IE = 62 mA VCB = 2.94 V VCE = 3.821 V

187

Summary of MultiSIM Files Circuit E04-04 E04-11

F04-54 F04-55a F04-55b F04-56 F04-57

F04-58 F04-59 TSE04-01 TSE04-02 TSE04-03 TSP04-55 TSP04-56 TSP04-57 TSP04-58 TSP04-59 TSP04-60 TSP04-61 TSP04-62

MultiSIM IB = 217 µA IC = 9.838 mA IC(1) = 16 mA VCE(1) = 2.952 V IC(2) = 13 mA VCE(2) = 3.709 V IB = 667 µA IC = 32 mA IE = 33 mA VBC = -4.608 V VBE = 877 mV VCE = 5.458 V VBC = 3.254 V VBE = -0.834 V VCE = -4.088 V IB = 24 µA IC = 1.197 mA IE = 1.221 mA VB(1) = 10.000 V VC(1) = 20.000 V VE(1) = 9.228 V VB(2) = -4.000 V VC(2) = -12.000 V VE(2) = -3.216 V VC = 222 mV VB = 0.798 V VC = 0.389 V VB = 0.716V VC = 5.172 V VB = 0.598 V VC = 0.022 V VB = 2.998 V VC = 9.000 V VRB = 2.471 V VRC = 14.982 V VCE = -0.021 nV VCE = 10.0 V VB = 10.00 V VC = 20.00 V IC = 91 mA IC = 5.456 mA IB = 0.444 pA VC = 4.999 V IB = 4.350 µA IC = -0.021 pA

188

Summary of MultiSIM Files Chapter 5: Circuit E05-01 E05-02 E05-04 E05-08a E05-08b E05-10 F05-34 F05-35 F05-38 F05-39 F05-41 F05-42 TSE05-01 TSE05-02 TSE05-03 TSE05-04

MultiSIM VBE = 869 mV VCE = 7.187 V IC = 39 mA VBE = 814 mV VCE = 2.775 V IC = 4.615 mA VBE = -794 mV VCE = -3.082 V IC = 2.155 mA VBE = 0.802 V VCE = 10.385 V IC = 2.884 mA VBE = 0.806 V VCE = 10.101 V IC = 3.391 mA VBE = 0.767 V VCE = 2.366 V IC = 763 µA VCE = 10.771 V IC = 923 µA VCE = 6.001 V IC = 5.126 mA VB = 2.047 V VC = 6.237 V VE = 1.267 V VB = -1.666 V VC = -9.197 V VE = -0.881 V VBE = -0.846 V VCE = -7.182 V IC = 16 mA VBE = 0.774 V VCE = 1.169 V IC = 1.017 mA VB = 0.047 µV VC = 10.000 V VE = 4.700 nV VB = 3.197 V VC = 10.000 V VE = 0.000 V VB = 3.145 V VC = 5.061 V VE = 2.329 V VB = 3.197 µV VC = 10.000 V VE = 2.695 V

189

Summary of MultiSIM Files Circuit TSP05-55 TSP05-56 TSP05-57 TSP05-58 TSP05-59 TSP05-60

MultiSIM VB = 669 mV VC = 18 mV IB = -0.444 pA VB = 3.664 V VC = -2.846 V VC = -10.000 V VE = 2.438 V VB = 2.999 V

Chapter 6: Circuit E06-08

E06-09 E06-11 F06-51

F06-52

F06-53

F06-54

F06-57

F06-62

MultiSIM VC = 5.530 VDC Vb = 26.1 mVPP Vc = 221.9 mVPP DC values off, gain is OK Vb = 2.8 VPP Vc = 2.8 VPP DC values off, gain is OK Vb = 14.0 mVPP Vc = 930.5 mVPP VB = 2.570 VDC Vb = 99 mVAC VOUT ≈ 0 VDC Vout = 216 mVAC VB = 1.593 VDC Vb = 100 mVAC VOUT = 7.990 mVDC Vout = 395 mVAC VB = 1.593 VDC Vb = 100 mVAC VOUT = 7.990 mVDC Vout = 395 mVAC VB = 1.741 VDC Vb = 1.01 VAC VOUT = 88 mVDC Vout = 881 mVAC VC1 = 7.097 VDC VB1 = 2.859 VDC Vb1 = 0.051 VAC VE1 = 2.106 VDC VC2 = 6.204 VDC VB2 = 1.657 VDC VOUT2 = -1.862 VDC Vout2 = 5.949 VAC Vin = 0.100 V Vout = 3.28 V

190

Summary of MultiSIM Files Circuit TSE06-01

TSE06-02

TSE06-03

MultiSIM VB(1) = 1.680 VDC VC(1) = 5.772 VDC VE(1) = 908 mVDC Vb(1) = 0.100 VAC Vc(1) = 5.117 mVAC Ve(1) = 0.473 µVAC VB(2) = 1.680 VDC VC(2) = 5.732 VDC VE(2) = 908 mVDC Vb(2) = 5.132 VAC Vc(2) = 832 mVAC Ve(2) = 908 mVAC VOUT = -0.040 VDC Vout = 832 mVAC VB(1) = 1.680 VDC VC(1) = 5.772 VDC VE(1) = 908 mVDC Vb(1) = 0.100 VAC Vc(1) = 0.016 VAC Ve(1) = 0.469 µVAC VB(2) = 1.680 VDC VC(2) = 5.772 VDC VE(2) = 908 mVDC Vb(2) = 0.010 pVAC Vc(2) = 0.047 pVAC Ve(2) = 0.000 VAC VOUT = 1.721 pVDC Vout = 2.658 pVAC VB(1) = 1.680 VDC VC(1) = 5.772 VDC VE(1) = 908 mVDC Vb(1) = 0.100 VAC Vc(1) = 5.117 mVAC Ve(1) = 0.473 µVAC VB(2) = 1.680 VDC VC(2) = 5.772 VDC VE(2) = 908 mVDC Vb(2) = 0.010 VAC Vc(2) = 0.046 mVAC Ve(2) = 9.838 mVAC VOUT = -0.097 mVDC Vout = 0.046 VAC

191

Summary of MultiSIM Files Circuit TSE06-04

TSP06-60 TSP06-61 TSP06-62 TSP06-63 TSP06-64 TSP06-65

MultiSIM VB(1) = 1.680 VDC VC(1) = 5.772 VDC VE(1) = 908 mVDC Vb(1) = 0.100 VAC Vc(1) = 0.010 VAC Ve(1) = 0.473 µVAC VB(2) = 1.754 VDC VC(2) = 9.999 VDC VE(2) = 0.000 VDC Vb(2) = 0.010 VAC Vc(2) = 0.041 pVAC Ve(2) = 0.000 VAC VOUT = 0.012 nVDC Vout = 0.016 nVAC Vb = 0.100 VAC Vc = 0.215 VAC Vout = 0.000 VAC Vb = 0.099 VAC IE = 0.000 ADC Vb = 0.100 VAC Vout = 0.926 VAC Vin = 0.100 VAC Vb = 0.090 pV Vout = 0.000 V Vb = 0.100 VAC Vc = 0.114 VAC Vc(1) = 2.764 VAC Vb(2) = 0.013 pVAC

Chapter 7: Circuit E07-03 Lower than 40 VPP E07-04 Lower than 20 VPP F07-42 F07-44 F07-45 F07-46 TSE07-01 TSE07-02 TSE07-03

MultiSIM Vout = 37.9 VPP Vout = 13.733 VPP VCE = 6.296 VDC IC = 0.060 A Vout = 6.295 VPP for Vin = 100 mVrms Vout = 13.641 VPP Vout = 13.759 VPP VB = 2.596 VPP with output clipped on bottom VB = 0.839 VDC Vout(max) = 2.173 VP Vout(min) = -1.816 VP Vout = 2.043 VPP with nonlinear distortion 192

Summary of MultiSIM Files Circuit TSP07-39 TSP07-40 TSP07-41 TSP07-42 TSP07-43

MultiSIM IC = 0.060 ADC VCE = 6.281 VDC VB = 0.021 pVrms VB = 4.204 VDC VRE2 = 3.715 VDC Vout is negative pulses Vout = -6.820 VPP half-wave output VD2 = 6.885 pVDC Vout = 13.060 VPP VOUT is a 12.039 VP positive half-wave output

Chapter 8: Circuit E08-06

E08-09 E08-12 F08-66a F08-66b F08-66c F08-69 F08-70 F08-71 F08-72 F08-75a F08-75b F08-75c

MultiSIM ID = 8.078 mADC VD = 5.053VDC VGS = -3.116 VDC VS = 3.147 VDC ID = 4.827 mADC VD = 7.706VDC VGS = -2.700 VDC ID = 1.281 mADC VD = 7.129VDC VGS = -2.020 VDC VGS = -0.997 VDC VDS = 6.262VDC ID = 1.007 mADC VGS = -0.497 VDC VDS = 6.139VDC ID = 5.020 mADC VGS = -1.408 VDC VDS = 6.922VDC ID = 3.025 mADC ID = 2.854 mADC VGS = -0.856 VDC ID = 3.356 mADC VGS = 1.190 VDC ID = 0.857 mADC VGS = -1.186 VDC ID = 1.917 mADC VGS = -1.508 VDC IS = 8.000 mADC VDS = 4.000 VDC IS = 8.000 mADC VDS = 5.400 VDC IS = -8.000 mADC VDS = -4.520 VDC 193

Summary of MultiSIM Files Circuit F08-76a F08-76b F08-81 TSE08-01

TSE08-02

TSE08-03

TSP08-72 TSP08-73 TSP08-74 TSP08-75 TSP08-76 TSP08-77 TSP08-78 TSP08-79 TSP08-80

MultiSIM ID = 0.904 mADC VGS = 3.098 VDC ID = 5.316 mADC VGS = 4.762 VDC VDS = 10.940 VDC VGS = -0.381 VDC VGS = -0.080 VDC VD = 14.631 VDC VS = 0.081 VDC ID = 0.368 mADC VGS = 0.000 VDC VD = 0.000VDC VS = 0.000 VDC ID = 0.000 ADC VGS = -1.994 VDC VD = 14.999 VDC VS = 2.014 VDC ID = -1.776 µADC VD = 0.031 VDC VS = 2.547 pVDC VG = 0.345 µVDC VD = 6.000 VDC VS = 0.906 VDC VG = 0.063 mVDC VD = -2.988 VDC VS = -1.303 VDC VG = 0.000 VDC VD = 7.166 VDC VS = 1.287 VDC VG = 0.018 mVDC VD = 12.000 VDC VS = 0.000 VDC VG = 4.737 VDC VD = 0.000 VDC VG = 0.000 VDC VD = 10.000 VDC VG = 1.000 VDC VD = 4.307 VDC VS = 4.307 VDC VG = 0.000 VDC VD = 8.352 VDC VS = 8.311 VDC VG = 9.000 VDC

194

Summary of MultiSIM Files Chapter 9: Circuit E09-08 E09-09 E09-10 F09-56 F09-57a F09-57b F09-58 F09-59 F09-60 F09-62

F09-65a F09-65b TSE09-01

MultiSIM IS = 2.633 mADC VD = 6.318 VDC VRL = 1.712 Vrms VGS = -0.623 VDC Vin = 9.998 mVrms Vout = 9.142 mVrms Vin = 9.998 mVrms Vout = 0.096 Vrms ID = 2.834 mADC VDS = 4.914 VDC VS = 2.834 VDC ID = 1.843 mADC Vin = 0.050 Vrms Vout = 0.215 Vrms ID = 1.016 mADC Vin = 0.050 Vrms Vout = 0.496 Vrms ID = 5.847 mADC Vin = 0.050 Vrms Vout = 0.141 Vrms ID = 4.463 mADC VDS = 3.065 VDC VG = -1.458 VDC VD = 17.151 VDC VG = 5.484 VDC IS = 2.849 mADC IS = 0.015 mADC VD = 9.000 VDC Vd = 0.048 Vrms Vin = 9.996 mVrms ID = 0.279 mADC Vin = 0.050 Vrms Vout = 0.046 Vrms ID = 6.615 mADC Vin = 0.050 Vrms Vout = 0.014 Vrms Vin = 9.998 mVrms ID(1) = 3.125 mADC VD(1) = 7.313 VDC VG(2) = -0.135 mVDC Vg(2) = 0.075 Vrms ID(2) = 3.135 mADC Vout(2) = 0.562 Vrms

195

Summary of MultiSIM Files Circuit TSE09-02

TSE09-03

TSE09-04

TSP09-54 TSP09-55

TSP09-56

TSP09-57

TSP09-58 TSP09-59

TSP09-60

MultiSIM Vin = 9.998 mVrms ID(1) = 6.092 mADC VD(1) = 2.863 VDC VG(2) = 2.863 mVDC Vg(2) = 0.433 Vrms ID(2) = 6.115 mADC Vout(2) = 0.128 mVrms Vin = 9.997 mVrms ID(1) = 1.776 µADC VD(1) = 12.001 VDC VG(2) = 0.067 mVDC Vg(2) = 0.000 Vrms ID(2) = 3.125 mADC Vout(2) = 0.000 Vrms Vin = 9.998 mVrms ID(1) = 3.125 mADC VD(1) = 7.313 VDC VG(2) = -0.199 mVDC Vg(2) = 0.075 Vrms ID(2) = 0.000 ADC Vout(2) = 0.026 nVrms ID = 4.800 mADC VDS = 4.801 pVDC VS = 4.800 VDC ID = 1.842 mADC VS = 0.034 VDC Vin = 0.050 Vrms Vout = 0.069 Vrms ID = 1.000 ADC VG = 0.000 VDC Vin = 0.050 Vrms Vout = 6.203 pVrms ID = 0.013 ADC VGS = 0.080 mVDC Vin = 0.050 Vrms Vout = 0.208 Vrms ID = 9.008 nADC VDS = 9.000 VDC VGS = -2.941 µVDC VR1 = 20.000 VDC VG = 0.136 mVDC VD = 20.000 VDC ID = 0.028 nADC VRD = 24.00 VDC Vin = 0.010 Vrms IC = 0.240 µADC VD = 0.050 mVDC Vd = 0.080 µVrms Vout = 0.080 µVrms

196

Summary of MultiSIM Files Circuit TSP09-61

TSP09-62

MultiSIM ID = 9.876 mADC VDS = 3.186 VDC VGS = 17.991 VDC Vout = 0.024 Vrms ID = 0.279 mADC Vs = 0.047 Vrms Vin = 0.050 Vrms Vout = 0.000 Vrms

Chapter 10: Circuit E10-03 E10-05 E10-06 E10-07 E10-08 E10-11 F10-56 F10-57 F10-60 TSP10-45 TSP10-46 TSP10-47 TSP10-48

MultiSIM fC ≈ 209 Hz fC ≈ 50 Hz fC ≈ 30 Hz fC ≈ 16 Hz fC ≈ 17 Hz fC ≈ 1.78 MHz fC(1) ≈ 3.1 kHz fC(2) ≈ 25.0 kHz fC ≈ 93 kHz fC ≈ 2.95 kHz VRC = 19.765 VDC Vd = 202 mVPP Vout = 0.0 Vrms VB = 2.589 VDC VRD = 10.00 VDC ID = 0.018 ADC

Chapter 11: Circuit E11-03 E11-04 F11-53 TSP11-27 TSP11-28 TSP11-29

MultiSIM ISCR = 0.436 A Expected SCR operation Sine wave with missing piece from 90° to 180° with RPOT = 82% Expected SCR operation Cathode-anode shorted SCR never fires VR1 = 0.074 nV Vout is half-wave output

Chapter 12: Circuit E12-03

MultiSIM Vin = 0.100 Vrms Vout = 2.227 Vrms

197

Summary of MultiSIM Files Circuit E12-04 E12-05 E12-07 E12-12 F12-63 F12-64a F12-64b F12-64c F12-64d F12-66a F12-66b F12-66c F12-66d F12-67

F12-68a

F12-68b

F12-68c

F12-70a

MultiSIM Vin = 0.100 Vrms Vout = 9.997 Vrms Vin = 0.100 Vrms Vout = 2.299 Vrms Vin = 0.010 Vrms Vout = 9.997 Vrms BW(1) ≈ 48.9 kHz BW(2) ≈ 70.0 kHz Vin = 10.000 mVrms Vout = 3.742 Vrms Vin = 10.000 mVrms Vout = 0.110 Vrms Vin = 10.000 mVrms Vout = 1.010 Vrms Vin = 10.000 mVrms Vout = 0.478 Vrms Vin = 10.000 mVrms Vout = 10.003 Vrms Vin = 1.000 Vrms Vout = 1.000 Vrms Vin = 1.000 Vrms Vout = 1.000 Vrms Vin = 1.000 Vrms Vout = 0.223 Vrms Vin = 10.000 mVrms Vout = 9.998 Vrms Vin = 10.000 mVrms Is = 0.030 mArms Vout = 10.004 Vrms If = 0.030 mArms Vin = 10.000 mVrms Is = 0.100 nArms Vout = 2.084 Vrms If = 3.726 µArms Vin = 10.000 mVrms Is = 0.100 nArms Vout = 0.323 Vrms If = 6.685 µArms Vin = 10.000 mVrms Is = 0.100 nArms Vout = 0.189 Vrms If = 0.181 µArms Vin = 10.000 mVrms Is = 1.000 µArms Vout = 0.150 Vrms If = 1.001 µArms

198

Summary of MultiSIM Files Circuit F12-70b

F12-70c

TSE12-01 TSE12-02 TSE12-03 TSE12-04 TSP12-57 TSP12-58 TSO12-59 TSP12-60 TSP12-61 TSP12-62 TSP12-63 TSP12-64 TSP12-65 TSP12-66 TSP12-67 TSP12-68 TSP12-69 TSP12-70 TSP12-71 TSP12-72

MultiSIM Vin = 10.000 mVrms Is = 0.100 µArms Vout = 1.000 Vrms If = 0.110 µArms Vin = 10.000 mVrms Is = 0.021 mArms Vout = 0.213 Vrms If = 0.021 mArms Vin = 0.100 Vrms Vout = 2.227 Vrms Vin = 0.047 µVrms Vout is railed Vin = 0.100 Vrms Vout = 9.997 Vrms Vin = 0.100 Vrms Vout = 0.000 Vrms Vin = 28.3 VPP Vout is railed Vin = 2.83 VPP Vout = 2.83 VPP Vin = 28.3 mVPP Vout = 28.5 mVPP Vout is railed VRi ≈ 0 VPP Vin = 28.3 mVPP Vout = 28.3 mVPP Vin = 2.83 VPP Vout ≈ 0 VPP Vin = 14.1 VPP Vout = 14.1 mVPP Vin = 2.83 VPP Vout = 28.0 VPP Vin = 2.83 mVPP Vout is railed Vin = 2.83 VPP Vout ≈ 0 VPP Vin = 2.83 VPP Vout is railed Vin = 2.83 VPP Vout = 3.02 VPP Vin = 28.3 VPP Vout is railed Vin = 2.83 VPP Vout ≈ 0 VPP Vin = 2.83 VPP Vout ≈ 0 VPP Vin = 28.3 mVPP Vout = 3.43 VPP

199

Summary of MultiSIM Files Chapter 13: Circuit E13-01 E13-02 E13-03 E13-05 E13-06 E13-07 E13-08

E13-10 F13-61 F13-63a F13-63b F13-64 F13-65a F13-65b F13-66 F13-67 F13-68 F13-69 TSE13-01 TSE13-02 TSE13-03 TSE13-04 TSE13-05 TSE13-06 TSE13-07

MultiSIM VLTP = +1.627 V VUTP = +1.627 V VLTP = -2.551 V VUTP = +2.468 V VUTP = -2.391 V VLTP = +2.391 V Vout = -7.722 V, +7.684 V VOUT = -12 V VOUT = -7 V VOUT = -2.5 V V1 = 3.000 V V2 = 2.000 V V3 = 8.000 V VOUT = -8.838 V Integrator output ΔV/Δt = -4.816 V/200 μs VLTP = -2.637 V VUTP = +2.574 V VLTP = -3.848 V VUTP = +3.817 V VLTP = -3.628 V VUTP = +3.411 V VUTP = -1.206 V VLTP = +1.206 V Vout = -6.144 V, +6.142 V VOUT = -2.500 V VOUT = -3.520 V VOUT = -14.00 V VOUT = -3.572 V Integrator output ΔV/Δt = -4.058 V/1.0 ms VOUT = 20 VPP square wave V(-) = 2.000 V VOUT is railed VIN = 3 V, 2 V, 8 V V(-) = 6.682 V VOUT = 6.682 V Vin = 0 V to 5 V square wave Vout = 0 V Vin = 10 VPP triangle wave Vout = 0 V Vout(min) = -898 mV Vout(max) = 6.908 V 0 V output VUTP = -4.589 V VLTP = +4.589 V Vout = -7.806 V, +7.804 V 200

Summary of MultiSIM Files Circuit TSE13-08 TSE13-09 TSP13-30 TSP13-31 TSP13-32 TSP13-33 TSP13-34 TSP13-35 TSP13-36 TSP13-37 TSP13-38 TSP13-39

MultiSIM Vin = 1 V, 0.5 V, 0.2 V, 0.1 V V(-) = 0.450 V Vout is railed negative Vin = 1 V, 0.5 V, 0.2 V, 0.1 V V(-) = 0.500 V Vout is railed negative VLTP = -5.137 V VUTP = +10.649 V Vout is railed VLTP = -6.135 V VUTP = -9.472 V Vout = ±10 VP square wave Vin = 2.0 VP Vout = 2.0 VP Vout(min) = -711.0 mV Vout(max) = +5.700 V VOUT = -1.500 V V(-) = 1.000 V Vout is railed negative VOUT = -0.013 V V(-) = 2.432 V Vout is railed negative Vout = 2.804 V Vout = 8.928 mVPP square wave

Chapter 14: Circuit E14-08 F14-53 F14-60 F14-62a F14-62b TSP14-32 TSP14-33 TSP14-34 TSP14-35 TSP14-36

MultiSIM VOUT = -153.926 mV Vin(1) = 1.005 Vrms Vin (2) = 1.005 Vrms Vout = 2.010 Vrms VOUT = -162.307 mV IL = 4.659 mA IL = 1.176 mA Vout = 21.958 Vrms IQ = 0 A IQ = 27.756 nA IL = 6 mA IL = 594.058 µA

Chapter 15: Circuit E15-03 E15-06

MultiSIM fC ≈ 7.1 kHz fC(1) ≈ 634 Hz fC(2) ≈ 854 Hz

201

Summary of MultiSIM Files Circuit E15-07 E15-08 F15-43a F15-43b F15-43c F15-45 F15-46 F15-47a F15-47b F15-47c F15-48 TSP15-23 TSP15-24 TSP15-25 TSP15-26 TSP15-27 TSP15-28 TSP15-29 TSP15-30 TSP15-31

MultiSIM fC(1) ≈ 7.1 kHz fC(2) ≈ 7.3 kHz fC ≈ 60.1 Hz fC ≈ 11.2 kHz fC ≈ 162 kHz fC ≈ 49 kHz fC(1) ≈ 174 Hz fC(2) ≈ 200 Hz fC ≈ 1.27 kHz f1 ≈ 116 Hz f2 ≈ 218 Hz fC ≈ 159 Hz f1 ≈ 402 Hz f2 ≈ 500 Hz fC ≈ 445 Hz f1 ≈ 15.7 kHz f2 ≈ 16.2 kHz fPEAK ≈ 15.8 kHz f1 ≈ 1.3 kHz f2 ≈ 1.75 kHz fNOTCH ≈ 1.32 kHz IR4 = 0.000 A VR4 = 0.432 V IR3 = 0 A V+(2) = 1.321 pVrms fC ≈ 270 Hz fC ≈ 2.5 kHz V+(1) = 2.950 pV IR2 = 1.475 mA VS = 0.100 Vrms V on end of R1 = 0 V VC2 = 0.100 Vrms IC2 = 0.017 µArms Vout(2) 36.7 mVPP V-(3) = 0 V

Chapter 16: Circuit E16-01 E16-02 F16-58 F16-59 F16-60a F16-60b F16-62

MultiSIM Vout = 2.2 VPP @ 1.6 kHz Vout = 16.6 VPP @ 6.37 kHz Vout = 2.2 VPP @ 9.81 kHz Vout = 8.1 VPP @ 692 Hz Vout = 386 mVPP @ 264 kHz (Tank circuit is loaded) Vout = 4.02 VPP @ 58.9 kHz Vout = 11.7 VPP @ 1.56 kHz

202

Summary of MultiSIM Files Circuit F16-64 TSP16-23 TSP16-24 TSP16-25 TSP16-26 TSP16-27 TSP16-28

MultiSIM tH = 139 µs tL = 109 µs fOSC = 4.02 kHz RDS = 0 Ω VC3 open from oscilloscope trace VCE ≈ 2.557 pV VR1 = 661.758 mV Resistance of R2 out of range tH = 110.5 µs tL = 110.5 µs

Chapter 17: Circuit E17-03 E17-05 F17-45 F17-46 F17-47

F17-48 TSP17-31 TSP17-32 TSP17-33

TSP17-34

MultiSIM VOUT = 10.340 VDC VOUT = 10.340 VDC VOUT = 10.625 VDC VOUT = 8.752 VDC VR4 = 0.529 VDC IR4 = 0.189 ADC IQ2 = 0.191 mADC IL = 0.188 ADC VOUT = 9.419 VDC IL = 0.037 A VR3 = 2.496 VDC VOUT = 2.471 VDC VZ = 11.881 VDC VOUT = 11.999 VDC VR4 = 0.353 VDC IR4 = 0.126 ADC IQ2 < 2 µADC IL = 0.126 ADC VOUT = 9.419 VDC VR1 = 24.99 VDC IL = 0.019 mADC

203

Solutions Manual

Laboratory Exercises for Electronic Devices Eighth Edition

David M. Buchla Steven Wetterling

205

Instructor note: The B experiments include useful notes regarding the Anadigm Signal Processor (ASP) and the Programmable Analog Module (PAM). These are presented on an “as needed” basis with the heading Instructor note:.

Experiment 1: The Diode Characteristic Part 1: The Diode Characteristic Curve Step 1: Answers depend on the meter. A meter with a diode test will typically show about 0.6 V in one direction and an open in the other direction. Table 1-1 Listed Component Value R1 330 Ω R2 1.0 MΩ

Measured Value 331 Ω 1.05 MΩ

Table 1-2 VF 0.45 V 0.50 V 0.55 V 0.60 V 0.65 V 0.70 V 0.75 V Table 1-3 VS (measured) 5.0 V 10.0 V 15.0 V

VR1 (measured)

IF (computed)

3.8 mV

11 µA

16.3 mV

49 µA

83 mV

250 µA

230 mV

695 µA

690 mV

2.08 mA

1.85 V

5.59 mA

4.58 V

13.8 mA

VR2 (measured)

IR (computed)

454 mV

0.43 µA

890 mV

0.85 µA

1.3 V

1.24 µA

206

Step 8: Plot of the diode curve: 14 12

Current (mA)

10 8.0 6.0 4.0 2.0 -15 0

Note Scale change -10

-5.0 0.1

−2.0

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Voltage (V)

VI Curve for a 1N4001 Diode Plot 1-1

Step 9: The semilog plot shows that logarithm of forward current is proportional to the diode drop. 100

IF (mA)

10

1

0.1

0.01 0.4

0.45

0.5

0.55

0.6

0.65

0.7

0.75

0.8

VF (V)

Plot 1-2

Questions: Part 1 1. The input impedance is equal to the series 1 MΩ resistor, which can be seen from the voltage: divider rule. 2.

The larger resistor is needed to develop enough voltage to be easily measured.

207

Part 2: Plotting Diode Curves with an Oscilloscope Step 3: The barrier potential is reduced by the addition of heat, causing current to increase for a given voltage. Step 4: The threshold is about 1.5 V for a red LED. Step 5: The threshold is about 1.8 V for a green LED. Step 6: The zener forward voltage is about 0.7 V; the breakdown is at 5.0 V. Questions: Part 2 1. The dynamic resistance is measured by observing a small voltage change in the characteristic curve at the point it is being measured and dividing by a corresponding current change. 2.

The setup can be used to measure any two-terminal device because it enables a plot of current versus voltage.

Experiment 2 Diode Applications Part 1: Diode Rectifiers Step 1: Waveforms: 20 V

20 V

Vsec

V out

-20 V

-20 V 2.0 ms/div

2.0 ms/div

Plot 2-1 Table 2-1 Half-wave rectifier. Without Filter Capacitor

With Filter Capacitor

Computed

Measured

Compute

Measured

Vsec(rms)

Vsec(rms)

Vout(p)

Vout(p))

12.6 V ac 14.4 V rms 20.3 Vp

20.0 Vp

Measured VOUT(DC) 19.1 V dc

Step 4: Waveforms: 10 V

10 V

Vsec

Vout

-10 V

-10 V 2.0 ms/div

2.0 ms/div

Plot 2-2

208

Vr(pp) 1.4 Vpp

Ripple Frequency 60 Hz

Table 2-2 Full-wave rectifier circuit. Without Filter Capacitor

With Filter Capacitor

Computed

Measured

Compute

Measured

Vsec(rms)

Vsec(rms)

Vout(p)

Vout(p))

6.3 V ac

7.3 Vrms

10.2 Vp

9.8 Vp

Measured VOUT(DC) 9.5 V dc

Vr(pp) 0.35 Vpp

Ripple Frequency 120 Hz

Step 7: A second parallel load resistor increases the load current and the ripple voltage. Table 2-3 Bridge rectifier circuit. Without Filter Capacitor

With Filter Capacitor

Computed

Measured

Computed

Measured

Vsec(rms)

Vsec(rms)

Vout(p)

Vout(p))

VOUT(DC)

12.6 V ac 14.5 Vrms 20.5 Vp

19.5 Vp

19.0 V dc

Measured Vr(pp) 0.65 Vpp

Ripple Frequency 120 Hz

Step 10: The output voltage drops; ripple voltage is doubled; ripple frequency is 60 Hz. Questions: Part 1 1. The scope ground connection, if common with the circuit, will short the secondary if it is connected to one side, so two channels are necessary. 2.

An open diode will cause the ripple frequency to drop to 60 Hz (instead of 120 Hz).

Part 2: Diode Clipping Circuits Step 1: R2 and RL form a voltage divider; the voltage across RL is only slightly less than VS. Step 2: Waveforms: +3.0 V

Vs

0V

−3.0 V

+0.7 V

Vout 0 V

−2.5 V +2.3 V

VR2

−0.5 V

0V

Vertical 2 V/div Horiz = 0.2 ms/div

Plot 2-3

209

Step 3: The load has a waveform with a minimum of -2.72 V and a maximum of 0.68 V similar to the second waveform in Plot 2-3. The change in the lower part of the waveform is due to loading effects. Step 4: The positive clipping level follows changes in the power supply voltage. As the supply is increased, the positive clipping level moves up from about 0.7 V to the peak of the input waveform. Step 5: The negative portion of the waveform is clipped. The power supply controls the level, which varies from −0.7 V to the positive peak. Step 6: Reversing the power supply voltage moves the clipping level to negative voltages. Questions: Part 2 1. The clipping level stays the same when the load is reduced but the lower level of the waveform changes due to loading effects. 2.

At all times, the algebraic sum of the voltages around a closed path is zero. The voltage across R2 can be found directly by subtracting the load voltage from the source voltage.

Part 3: Diode Clamping Circuits Step 1: The output is an ac waveform that varies from approximately −0.6 V to +5.4 V. The output tracks changes in the input amplitude but the lower peak remains at −0.6 V (after a short delay for settling to the new level). Step 2: The power supply voltage adds to the output. See Plot 2-4 for output waveform. +7.26 V Vo ut +1.26 V 0V

2 V/div

Plot 2-4 Step 3: The circuit is now a negative clamping circuit. The peak is at approximately 0.6 V when the dc supply is zero and the clamping point moves more negative as the dc voltage increases. Questions: Part 3 1. The capacitor is polarized, so how it charges needs to be considered in placing it in the circuit. 2.

The output will “settle” faster due to the decreased time constant. There is also a small additional distortion on the bottom of the waveform when the diode conducts. 210

Experiment 3 Special-Purpose Diodes Part 1: The Zener Diode and Regulator Table 3-1 Resistor R1 R2 RL

+10 mA

Listed Value 220 Ω 1.0 kΩ 2.2 kΩ

Measured Value 219 Ω 1.00 kΩ

0 mA

2.22 kΩ

−10 mA −10 V

0V

+10 V

Plot 3-1 Table 3-2 Vs Vout

(measured)

2.0 V 4.0 V 6.0 V 8.0 V 10.0 V

IL

(computed)

VR1

(computed)

Is (computed)

IZ (computed)

1.82 V

0.82 mA

0.18 V

0.82 mA

0.0 mA

3.60 V

1.62 mA

0.40 V

1.81 mA

0.19 mA

4.70 V

2.12 mA

1.30 V

5.90 mA

3.78 mA

5.07 V

2.28 mA

2.93 V

13.3 mA

11.0 mA

5.24 V

2.26 mA

4.76 V

21.6 mA

19.3 mA

Step 4: As shown in Table 3-2, zener current increases for increasing source voltage. Table 3-3 RL Vout

(measured)

1.0 kΩ 750 Ω 500 Ω 250 Ω 100 Ω *

IL

(computed)

VR1

(computed)

Is (computed)

IZ (computed)

5.32 V

5.32 mA

6.68 V

30.4 mA

28.0 mA

5.31 V

7.08 mA

6.69 V

30.4 mA

28.0 mA

5.26 V

10.5 mA

6.72 V

30.5 mA

28.1 mA

5.08 V

20.3 mA

6.92 V

31.5 mA

29.2 mA

3.75 V

37.5 mA

8.25 V

37.5 mA

0.0 mA*

out of regulation with this load.

211

10

Voltage (V)

8 6 4 2 0

0

200

400

600

800

1000

Resistance (Ω )

Plot 3-2 Step 9: From the data taken, the zener was able to regulate as long as the load was at least 250 Ω. Step 11: The regulated output had approximately 1 mV of ripple. The ripple waveform showed only the tip of the positive waveform. Noise level was less than 1 mV. Although this circuit had the advantage of a larger filter capacitor, the ripple is hundreds of times smaller than the unregulated supply in step 9 of Experiment 2 (see Table 2-3 for a comparison.) Questions: Part 1 1. (a) The region between zener breakdown and 0.7 V is approximated by an open circuit. (b) Both the forward-bias region and the reverse-bias region are approximated by a short.

ΔVout 5.24 V − 5.07 V × 100% = ×100% = 8.5% 10 V − 8.0 V ΔVin

2.

Percent line regulation =

3.

For the worst case (1 kΩ and 100 Ω loads) the percentage load regulation was poor since the zener carried no current for the 100 Ω load.

Percent load regulation =

VNL – VFL 5.32 V – 3.75 V = × 100% = 42% VFL 3.75 V

Part 2: The Varactor Diode Step 1. Note: Measured resonant frequencies are shown in Table 3-4. Three varactor diodes were tested, all from the same manufacturer, with consistently lower than calculated resonant frequencies. The lower measured frequencies may be due to stray capacitance acting to reduce the measured frequency. With no stray capacitance, the calculated fr should be approximately 226 kHz with −4 V of bias.

212

Table 3-4

Resonant VBIAS Frequency, fr 130 kHz 0.0 V 146 kHz 1.0 V 158 kHz 2.0 V 168 kHz 4.0 V 180 kHz 8.0 V 15.0 V 195 kHz 190 180

fr (kHz)

170 160 150 140 130

0

2

4

6

8 10 VBIAS (V)

Table 3-5 Parameter

Resonant frequency, fr Upper critical frequency, fcu Lower critical frequency, fcl Bandwidth, BW Q

12

14

16

Measured Value 194.7 kHz 196.2 kHz 192.5 kHz 3.7 kHz 52

Questions: Part 2 1. Answers will vary. For the varactors tested, the resonant frequency was 168 kHz at 4 V of bias. This implies the total capacitance is 1 1 C= = = 60 pF 2 2 ( 2πf r ) L ( 2π (168 kHz ) ) 15 mH

Note: This is larger than the varactor capacitance alone. The reason is likely due to stray capacitance as discussed in step 1. (Typical varactor capacitance is 33 pF at −4 V).

213

2.

The higher resonant frequency that was observed with increasing bias implies a smaller capacitance with increasing bias.

3.

(a) It would have very little effect on the resonant frequency because the resonant frequency is primarily determined by the LC circuit. (b) It will affect the Q because R1 acts as a resistive load on the resonant circuit.

Part 3: Light-Emitting Diode and Photodiode Step 1: The red LED drops approximately 1.65 V when forward-biased (measured at 10 mA of current) and has the smallest ac resistance (approximately 50 Ω in the region tested). The yellow LED drops approximately 1.9 V when forward-biased. The green LED drops approximately 2.0 V and has the highest ac resistance (approximately 100 Ω in the region tested). Step 2: The reverse curve drops as light is increased. (Note: The optimum value of R1 depends on the diode and lighting conditions). Step 4: Measured data for a green LED is shown in the table and plot. Angle −90˚ −75˚ −60˚ −45˚ −30˚ −15˚ 0˚ +15˚ +30˚ +45˚ +60˚ +75˚ +90˚

Voltage 0.56 V 0.90 V 1.1 V 1.5 V 3.5 V 5.6 V 6.0 V 5.3 V 2.3 V 1.1 V 0.85 V 0.69 V 0.29 V

0 +15o o

o

6V

+30

5V

o

4V

+45

−15 o −30 o −45

o

3V o

+60

o

+75

o

+90

2V

−60 o

1V

−75 o −90 o

Figure 3-9 Directional data for a green LED

Questions: Part 3 18.0 V − 2.0 V 18.0 V − 2.0 V 1. (a) I = = 533 Ω = 23.5 mA (b) R = 680 Ω 30 mA

2.

Electrical: forward voltage drop at a specified forward current, reverse leakage current IR, and reverse breakdown voltage V(BR)R. Optical: peak emission wavelength (in nm), power output (in milliwatts or microwatts) PO, and spectral output.

3.

The photodiode is a current source; a larger resistor develops more voltage for a given current, making the circuit more sensitive to light.

214

Experiment 4: Bipolar Junction Transistors Part 1: The BJT Characteristic Curve Values shown are for a 2N3904 transistor:

Table 4-1 Resistor Listed Value 33 kΩ R1

Measured Value 32.8 kΩ

100 Ω

R2 Table 4-2 VCE (measured)

104 Ω

Base Current = 50 µA

Base Current = 100 µA

VR2 (measured) IC (computed)

2.0 V 4.0 V 6.0 V 8.0 V

Base Current = 150 µA

VR2 (measured) IC (computed) VR2 (measured) IC (computed)

0.900 V

8.65 mA

1.91 V

18.4 mA

2.93 V

28.2 mA

0.931 V

8.95 mA

2.01 V

19.3 mA

3.14 V

30.2 mA

0.955 V

9.18 mA

2.13 V

20.5 mA

3.38 V

32.5 mA

0.992 V

9.54 mA

2.23 V

21.4 mA

3.58 V

34.4 mA

Step 11: Plot of characteristic curves: 35

IB = 150 μA

30 25

IB = 100 μA

IC 20

(mA)

15 10

IB = 50 μA

5 0

0

1

2

3

4

5

6

7

VCE (V) Plot 4-1

Table 4-3

VCE 3.0 V 5.0 V

Current Gain, ßDC IB = 50 µA IB = 100 µA IB = 150 µA 176

189

193

192

200

209

215

8

Questions: Part 1 1. For the test transistor, the ßdc was not constant at all points (see Plot 4-1). This has an effect on linearity as the gain will change as the operating point changes.

2.

It would raise all of the curves.

3.

Answers vary. For the test transistor, maximum power dissipated was 8 V x 34 mA = 272 mW.

4.

(a)

IC IB IC β dc I I IB α = I C = I +C I = I C = I β dc + 1 E C B IB + B IB

β=

(b) Answers vary. The alpha for the test transistor was approximately 0.995. 5.

VCE would equal VCC. Without base current, there would be no collector current and the supply voltage would appear across the transistor.

Part 2: BJT Switching Circuits Table 4-4 Resistor Listed Value 10 kΩ RB 1.0 kΩ RC 10 kΩ RC1 330 Ω RE

Table 4-5 Quantity Computed Value VCE(cutoff) 12.0 V VCE(sat) 0.1 V VRC(sat) 9.9 V Isat 9.9 mA

Measured Value 9.88 kΩ 1.00 kΩ 9.86 kΩ 327 Ω

Measured Value 10.45 V* 0.08 V 9.87 V

* The LED and transistor appear to be open; thus the measurement is affected by meter loading. Answers vary.

Table 4-6

Quantity VIN (LED on) VOUT (LED on) VIN (threshold) VOUT (threshold)

Table 4-7

Measured Value

Quantity VIN (LED on) VOUT (LED on) VIN (upper threshold) VOUT (upper threshold) VIN (lower threshold) VOUT (lower threshold)

0.0 V 0.080 V 0.645 V 10.47 V*

216

Measured Value 0.0 V 2.735 V 3.16 V 10.47 V 1.03 V 2.735 V

Step 8: Transfer curve: 16 14

VOUT

(V)

12 10 8 6 4 2 0

0

2

4

6

8

VIN (V)

Plot 4-2 Questions: Part 2 1. Some advantages of transistor switching circuits: (1) Very fast (2) Electrically controlled with a low voltage or current (3) Reliable (4) Low cost (5) Can be adopted to interface high current loads with low current circuits.

2.

RB limits current to the base-emitter diode of Q1. Without it, current could destroy the transistor when the potentiometer is set near maximum.

3.

Measure the voltage drop across RB and apply Ohm’s law.

4.

The collector resistances are different for the two transistors.

Experiment 5 Transistor Bias Circuits Part 1: Three Bias Circuits Table 5-1

Table 5-2 DC Parameter VRB

Resistor RB

Listed Value 1.0 MΩ

Measured Value 994 kΩ

RC

2.0 kΩ

1.97 kΩ

IB

11.37 μA

RE

1.5 kΩ

1.47 kΩ

IC

2.27 mA

VRC VC

217

Computed Value 11.3 V

Measured Value Q1 Q2 Q3 11.4 V

11.3 V

11.3 V

4.48 V

3.79 V

3.25 V

5.08 V

7.52 V

8.18 V

8.74 V

6.91 V

Table 5-3 DC Parameter VRB

Computed Value 8.67 V

Measured Value Q1 Q2 Q3 8.99

9.21

8.27

Table 5-4

Resistor R1

Listed Value 33 kΩ

Measured Value 32.96 kΩ

R2

6.8 kΩ

6.70 kΩ

IB

8.73 μA

IC

1.75 mA

VRC

3.44 V

2.86

2.56

3.79

RE

680 Ω

676 Ω

VC

8.56 V

9.11

9.42

8.21

RC

2.0 kΩ

1.97 kΩ

Resistor RB

Listed Value 360 kΩ

Measured Value 363 kΩ

RC

2.0 kΩ

1.97 kΩ

Table 5-5 DC Computed Measured Value Parameter Value Q1 Q2 Q3 2.03 V 1.97 V 1.96 V 1.98 V VB

VE

1.33 V

IE ≈IC VRC

1.99 mA 3.97 V

3.86 V 3.74 V 3.88 V

VC

8.03 V

8.14 V 8.25 V 8.12 V

Table 5-7 DC Parameter IC

Computed Value

1.33 V 1.29 V 1.34 V

Table 5-6

Measured Value Q1 Q2 Q3

3.07 mA

VRC

6.13 V

5.73 V 5.23 V 6.27 V

VC

5.87 V

6.26 V 6.76 V 5.73 V

Questions: Part 1 1. The data for voltage-divider bias shows the least variation between the transistors.

2.

An emitter resistor will improve bias stability as it is a form of negative feedback.

Part 2: Emitter Bias and Two-Supply Voltage-Divider Bias Table 5-8

Resistor RB RC RE R2

Listed Value 4.7 kΩ 330 Ω 470 Ω 4.7 kΩ

Measured Value 4.62 kΩ 329 Ω 477 Ω 4.60 kΩ

218

Table 5-9 Quantity

VE IE ≈IC VC

Computed Value −1 V 17.0 mA 3.39 V

Table 5-10 Quantity Computed Value

VE IE ≈IC VC

Measured Value Q1 Q2 Q3 −1.33 V −1.10 V −1.24 V 3.74 V

3.57 V

3.67 V

−0.7 V

Measured Value Q1 Q2 Q3 −1.05V −0.93 V −1.02 V

17.7 mA 3.16 V

3.55 V

3.45V

3.52 V

Step 8: The LED fades out as the thermistor warms. The decrease in resistance as it warms causes the base voltage to drop and the transistor stops conducting. Questions: Part 2 1. An advantage to emitter bias is great stability. A disadvantage is the requirement for both a positive and negative supply voltage.

2.

For most troubleshooting work, it is valid to assume the emitter should be near −1 V. For precise design work, it is better to apply KVL.

3.

The thermistor can be moved to the other side of the voltage-divider (in series with R1). Alternatively, the thermistor could be changed for a positive temperature coefficient type

219

Experiment 6 BJT Amplifiers Part 1: The Common-Emitter Amplifier Table 6-1

Resistor R1 R2 RE1

100 Ω

100 Ω

RE2

330 Ω

333 Ω

RC

1.0 kΩ

1.00 kΩ

RL

10 kΩ

10.1 kΩ

Table 6-3 AC Quantity Vin = Vb

Ve re Vout = Vc Av Rin(tot) βac

Table 6-2 DC Computed Quantity Value VB 3.85 V

Listed Measured Value Value 10 kΩ 9.87 kΩ 4.7 kΩ 4.66 kΩ

Computed Value

Measured Value

300 mVpp

300 mVpp

290 mVpp

280 mVpp

Measured Value 3.74 V

VE

3.15 V

IE

7.02 mA

VC

4.98 V

4.87 V

VCE

1.83 V

1.93 V

3.04 V

3.6 Ω 8.8

8.43

2.63 Vpp

2.53 Vpp

2.9 kΩ*

2.89 kΩ

300 mVpp

300 mVpp

* Measured ßac of 300 used for calculation

Step 8: The gain drops to 2.2 when C2 is opened. Step 9: The gain drops to 4.5 when RL is replaced with a 1.0 kΩ resistor. Step 10: Transistor is cutoff since there is no path for base current. (Note that a measurement of VCE could mislead student to thinking transistor is near saturation; however, the power supply voltage is across the reverse-biased basecollector junction, not across RC). Step 11: Transistor is saturated. VC and VE are nearly the same and current is limited only by RC and RE. Maximum current is in the collector circuit. Questions: Part 1 1. Monitoring the output voltage ensures that the amplifier is performing normally during the test. If the output is clipped or distorted, the measurement is invalid.

2.

The load resistor has no effect on the input resistance because it is an isolated transistor, which looks like a current source. 220

3.

The unbypassed emitter resistor is used for gain stability and makes gain much less dependent on the particular transistor that is used. It also increases the input resistance of the amplifier.

4.

The gain is inversely proportional to the ac emitter resistance. When the bypass capacitor is open, the ac emitter resistance increases, and the gain goes down.

Part 2: The Common-Collector Amplifier Table 6-4

Measured Value

R1

Listed Value 33 kΩ

32.5 kΩ

Table 6-6 AC Quantity Vb

R2

10 kΩ

9.98 kΩ

Ve

RE

1.0 kΩ

1.00 kΩ

re

RL

1.0 kΩ

1.01 kΩ

Av

0.99

0.99

Rin(tot)

6.63 kΩ

7.03 kΩ

Table 6-5 DC Quantity VB VE IE VCE

Computed Value

Measured Value

2.82 V

3.08 V

3.52 V

3.77 V

Computed Measured Value Value 1.0 Vpp* 1.0 Vpp 298 mVpp 298 mVpp 3.0 Ω

Ap

6.5

6.9

8.48 mA

−3.52 V

−3.77 V

Step 6: There is no phase shift between the input and output signal. Table 6-7

Trouble

DC Predictions VB VE VCE

DC Measurements VB VE VCE

Effect of Trouble o Vout

R1 open

0.69 V

1.39 V

−1.39 V 0.66 V 1.37 V

−1.37 V

See note 1.

R2 open

12.0 V

12.0 V

−12.0 V 12. 0 V 12.0 V

−12.0 V

No output.

R1 shorted

12.0 V

12.0 V

−12.0 V 12. 0 V 12.0 V

−12.0 V

No output.

RE open open collector

2.82 V

0.0 V

0.0 V

0.0 V

No output.

10.2 V

10.9 V

−10.9 V 10.3 V 11.0 V

−11.0 V

See note 2.

open emitter

2.82 V

12.0 V

−12.0 V 2.82 V 12.0 V

−12.0 V

No output.

12. 0 V

0.0 V

Note 1: Little effect. The dc conditions can be computed by assuming base bias; however, these conditions change when an ac signal is applied due to clamping action of the transistor and input coupling capacitor. The output ac signal is maintained as a result. Note 2: Output reduced. The input impedance drops significantly when the collector is open. Although the voltage gain drops slightly (to about 0.88), the power gain drops significantly (measured = 0.38)

221

Step 8: Cutoff clipping is observed first. Step 9: As the resistance increases, the clipping level is observed to rise. Questions: Part 2 1. In the equivalent npn circuit, positive clipping occurs when the transistor has maximum conduction, hence it is saturation clipping.

2.

The output voltage is always smaller than the input voltage but the output current is larger than the input current, hence there is power gain.

Part 3: Multistage Amplifiers Table 6-8 Resistor

RA RB R1 R2 RE1 RE2 RC1 R3 R4 RE3 RE4 RC2 RL Table 6-10 AC Parameter re′(Q1) re′(Q2) Rout(Q1) Rin(Q2) Av (NL)(Q1) Av(NL)(Q2)

Listed Value 100 kΩ 2.0 kΩ 330 kΩ 330 kΩ 33 kΩ 1.0 kΩ 22 kΩ 47 kΩ 22 kΩ 4.7 kΩ 220 Ω 6.8 kΩ 10 kΩ

Table 6-9 DC Parameter VB(Q1) VE(Q1) IE(Q1) VC(Q1) VCE(Q1) VB(Q2) VE(Q2) IE(Q2) VC(Q2) VCE(Q2)

Measured Value 100 kΩ 1.98 kΩ 322 kΩ 321 kΩ 32.5 kΩ 1.00 kΩ 21.8 kΩ 46.7 kΩ 21.6 kΩ 4.68 kΩ

Computed Value

Measured Value

0.0 V

0.46 V

0.70 V

1.10 V

0.43 mA -5.69 V

-6.02 V

-6.39 V

-7.12 V

-5.43 V

-5.83 V

-6.13 V

-6.39 V

1.80 mA 2.74 V

3.31 V

8.87 V

9.70 V

219 Ω 6.71 kΩ 10.0 kΩ

Table 6-11 AC Parameter Av’ Rin(Q1) Rout(Q2) Vin(Q1) Vout(Q2)

Computed Value 58.1 Ω 13.9 Ω 21.8 kΩ 9.1 kΩ 20.4*

Computed Value 78* 63.8 kΩ

Measured Value 82 68.6 kΩ

6.71 kΩ 10 mV

6.71 kΩ

780 mVpp

820 mVpp

* calculated with 10 kΩ load resistor. AV′ = 20.4 x (9.1 kΩ/39.9 kΩ) x 28.0 x (10 kΩ/16.7 kΩ) = 78

28.0*

222

Table 6-12 Generator Setting 0.5 V 1.0 V 2.0 V 4.0 V 6.0 V 8.0 V 10.0 V 20.0 V

Vin 10 mV* 20 mV 40 mV 80 mV 120 mV 160 mV 200 mV 400 mV

Vout

Av

2.8 V

280

5.4 V

270

9.1 V

220

12.0 V

150

12.8 V

107

13.2 V

82.5

13.2 V

66

12.0 V

36

*Vin should be 2% of the generator setting. All voltages shown are peak-to-peak values. Some distortion is observed on the output for Vin = 20 V

400

300

Av

200

100

0

0

100

200

300

400

Vin (mVpp)

Plot 6-1

Questions: Part 3 1. The input and output impedances are ac quantities. The input impedance is determined, in part, by re’, a quantity that can not be measured directly.

2.

Capacitors can be considered “shorts” to ac for frequencies for which the reactance is small compared to the circuit impedance.

3.

“Looking” back from the load resistor, the circuit can be considered to be a current source (the transistor) in parallel with the collector resistor. Assuming the current source is an open, the ac and dc resistances are equivalent and are represented by the collector resistor.

4.

(a) Base bias is formed by removing R4 and increasing the value of R3 (the base resistor). To find the appropriate value, divide the emitter current (1.80 mA) by ß; this is the desired base current and current in R3. Find the resistance value by dividing the desired voltage across it (20.4 V) by the base current. For a ß of 100, the computed base resistor is 1.1 MΩ. (b) The disadvantage of this is that the amplifier would be sensitive to the ß of the particular transistor that is used.

5.

Both stages are CE amplifiers, therefore they each change the phase by 180˚. The result is that there is no phase shift to the output.

223

Experiment 7 Power Amplifiers Part 1: The Class-A Power Amplifier Table 7-1

Resistor R1 R2 RE

Listed Value 10 kΩ 22 kΩ

Measured Value

22 Ω, 2 W

23 Ω

Table 7-3 Quantity

Table 7-2 CC Amp (Q1,2)

9.9 kΩ

VB VE IE VCE re Av(NL) Av(FL)

21.9 kΩ

8Ω

Input resistance, Rin

6.81 kΩ

Output rms voltage, Vout 0.848 Vrms Input rms voltage, Vin Load power, PL Input power, Pin Power gain, Ap

Measured Value

8.25 V

8.05 V

6.85 V

6.75 V

311 mA 5.15 V

5.25 V

1 Ω (est.) 0.99

0.98

0.98

0.96

Table 7-4 Quantity

Value

Load resistance, RL

Computed Value

Value

Quiescent power, PQ

3.73 W

Load power, PL Efficiency (percentage)

90 mW 2.4%

0.883 Vrms 90 mW 0.113 mW 796

Questions: Part 1 1. (a) The efficiency will be higher because there is no wasted power in the emitter resistor. The disadvantage is that there is dc voltage across the speaker.

(b) The efficiency will be higher because a larger fraction of the input power will become ac power to the load. Although both the dc and ac power will rise, the ac rises in greater proportion. (This can be tested with the Multisim simulation that goes with this experiment.) 2.

The Darlington arrangement isolates the load more and increases the input resistance of the amplifier.

224

Part 2: The Class-B Power Amplifier Table 7-5

Listed Value 330 Ω

351 Ω

R1

10 kΩ

9.92 kΩ

R2

10 kΩ

9.90 kΩ

R3 R4

68 kΩ 2.7 kΩ

67.4 kΩ

Resistor RL

Measured Value

Vin Vout

Horiz = 0.1 ms/div Vertical = 5 V/div

2.68 kΩ Plot 7-1

Table 7-6 DC Computed Parameter Value VE 0V VB1 0.7 V

VB2 IR1 = ICQ Table 7-8 DC Parameter VB3

−0.7 V

Table 7-7 AC Parameter Vp(out)

Measured Value 0.015 V

Ip(out) P(out)

0.64 V

−0.60 V

Computed Value 7.5 Vp

Measured Value 7.2 V

23.1 mAp 86.5 mW

830 µA

Table 7-9 AC Parameter Av′

Computed Measured Value Value −6.11 V −6.15 V

VE3

−6.81 V

ICQ3

811 µA

−6.78 V

Questions: Part 2 1. Open D2, bad Q2, no −9 V voltage.

2.

Half the signal will be missing.

3.

Bypass R4.

225

Computed Value 3.21

Measured Value 3.14

Experiment 8 Field-Effect Transistors Part 1: JFET Characteristic Curve Table 8-1

Resistor R1 R2

Listed Value 10 kΩ

Measured Value 10.0 kΩ

100 Ω

100 Ω

Table 8-2

VG = 0 V VR2 ID

VDS

VG = −0.5 V VR2 ID

VG = −1.0 V VR2 ID

VG = −1.5 V VR2 ID

1.0 V

0.304 V 3.04 mA 0.248 V

2.48 mA 0.197 V

1.97 mA 0.152 V

1.52 mA

2.0 V

0.487 V 4.87 mA 0.388 V

3.88 mA 0.291 V

2.91 mA 0.201 V

2.01 mA

3.0 V

0.560 V 5.60 mA 0.443 V

4.43 mA 0.320 V

3.20 mA 0.214 V

2.14 mA

4.0 V

0.588 V 5.88 mA 0.458 V

4.58 mA 0.328 V

3.28 mA 0.218 V

2.18 mA

6.0 V

0.605 V 6.05 mA 0.469 V

4.69 mA 0.333 V

3.33 mA 0.222 V

2.22 mA

8.0 V

0.616 V 6.16 mA 0.469 V

4.69 mA 0.333 V

3.33 mA 0.223 V

2.23 mA

7 VG = 0 V

6 5

VG = −0.5 V

ID 4 (mA)

VG = −1.0 V

3

VG = −1.5 V

2 1 0

Table 8-3 Measured JFET Parameters VGS(off) = −3.22 V IDSS = 6.1 mA

0

1

2

3

4 5 VDS (V)

6

7

8

Plot 8-1 Step 12: At VDS = 3.22 V, the current for the test transistor reached 5.46 mA, 90% of IDSS. Questions: Part 1 1. In the constant-current region, read ID for each value of VGS. Plot VGS on the negative x-axis as a function of ID.

2.

(a) The transconductance is dependent on VGS. (b) Each characteristic curve was drawn with the same increase in VGS, but the corresponding drain currents do not follow a linear pattern.

226

Part 2:

JFET as a Voltage-Controlled Resistor

Table 8-4

Listed Resistor Value R1 56 kΩ

Table 8-5 DC Computed Quantity Value 6.11 V VB

Measured Value 56.3 kΩ

R2 RE

39 kΩ 6.2 kΩ

38.7 kΩ

VE

5.41 V

6.24 kΩ

IE

0.867 mA

RC

3.9 kΩ

3.88 kΩ

R3

100 kΩ

100.4 kΩ

VC VCE

Measured Value 6.01 V 5.39 V

11.64 V

11.68 V

6.23 V

6.29 V

Table 8-6

Predicted Gain Maximum Gain VGG = 0 V

Measured Vout

13.8

Measured Gain

4.92 Vpp

12.3*

Gain with VGG = −0.5 V

4.24 Vpp

10.6

Gain with VGG = −1.0 V

3.56 Vpp

8.9

Gain with VGG = −1.5 V

2.84 Vpp

7.1

Gain with VGG = −2.0 V

2.00 Vpp

5.0

Gain with VGG = −2.5 V

0.98 Vpp

2.45

Gain with VGG = −3.0 V Gain with VGG = −5.0 V

0.26 Vpp

0.65

0.24 Vpp

0.60

* Three FETs were tested. The smallest observed gain for 0 bias was 9.8 but all fell to a minimum of 0.6

Step 5: The output continues to drop as VGG is increased until VGS(off) is reached; then it remains at a gain of about 0.60. Table 8-7

Measured Vout

Measured Gain

Vin = 100 mVpp

1.06 Vpp

10.6

Vin = 400 mVpp Vin = 800 mVpp

3.40 Vpp

8.5

3.56 Vpp

4.45

Vin = 1.2 Vpp

2.84 Vpp

2.37

Questions: Part 2 1. The ac signal is coupled through C3 and rectified by D1. Negative excursions charge capacitor C4, which biases the FET. A larger signal produces a larger dc voltage on the gate, thus increasing the emitter resistance of Q1 and reducing the gain.

2.

R4 and C4 maintain a dc voltage on the gate and determine how fast the gate voltage can respond to an input signal change. 227

Part 3: JFET as a DC Amplifier Table 8-8

Table 8-9 VIN

Resistor R1

Listed Value 5.6 kΩ

Measured Value 5.63 kΩ

R2

620 Ω

614 Ω

200 mV

3.71 V

R3

1.0 MΩ

998 kΩ

100 mV

4.53 V

RS RD

180 Ω 3.9 kΩ

177 Ω 3.91 kΩ

0V

5.36 V

−100 mV −200 mV

6.21 V 7.03 V

−300 mV

7.81 V

−400 mV

8.80 V

300 mV

VD (measured) 2.99 V

9 8

V D (V)

7 6 5 4 3 2 -400

-200

0

200

400

VIN (mV)

Plot 8-2 Questions: Part 3 1. The transfer curve in Plot 8-2 indicates the output is a linear function of the input dc voltage.

2.

The bias is self-bias, with the voltage developed across RS making the source more positive than the gate.

3.

An increase in the power supply voltage will increase the drain current, which is partially offset by the self-bias. The drain voltage will increase, but should still have a linear response.

228

Experiment 9 FET Amplifiers and Switching Circuits Part 1: The Common-Source JFET Amplifier Table 9-1

Resistor RS RD

Listed Value 1.0 kΩ 3.3 kΩ

Measured Value

Table 9-2 Parameters for CS Amplifier Quantity DC values AC values

1.02 kΩ

Gate voltage, VG

3.26 kΩ

Source voltage, VS

1.73 V

0V

RG

1.0 MΩ

1.01 MΩ

Drain voltage, VD

9.24 V

RL

10 kΩ

9.9 kΩ

Drain current, ID Input voltage, Vin

1.77 mA

Output voltage, Vout Voltage gain, Av Phase difference

500 mVpp 2.26 Vpp 4.53 180˚

4.

The measured output voltage went from 2.26 Vpp to 2.44 Vpp when the 1.0 kΩ source resistor was replaced with 510 Ω. The transconductance, gm, increases because the bias current increases. Gain is Av = gmrd.

5.

When the load resistor was changed from 10 kΩ to 100 kΩ, the output voltage went from 2.44 Vpp to 3.06 Vpp due to the increase in ac drain resistance, rd. As before, gain is Av = gmrd. Table 9-3

Fault

Observation

C2 is open Source and drain reversed

No change to dc; gain drops to 1.6.

VDD drops to +12 V

Slight drop in gain (to 4.4).

RG open

Gate has small variations in dc level; ac ok but can drop out completely.

No difference; gain remains at 4.5.

Questions: Part 1 1. There is no effect because the input impedance is controlled only by gate resistor.

2. The BJT will significantly load a 100 kΩ source; the FET will not, so it is better suited for this application.

229

Part 2: The Common-Drain JFET Amplifier Table 9-4 Parameters for CD Self-Biased Amplifier Quantity DC values AC values

Gate voltage, VG

0V

Source voltage, VS

1.83 V

Drain voltage, VD

15.0 V

Drain current, ID

1.83 mA

Input voltage, Vin

2.0 Vpp

Output voltage, Vout Voltage gain, Av Phase difference

1.27 Vpp 0.64 0˚

Table 9-5 Parameters for CD Current-Source Biased Amplifier Quantity DC values AC values

Q1 gate voltage, VG Q1 source voltage, VS Q1 drain voltage, VD Q2 gate voltage, VG Q2 source voltage, VS Q2 drain voltage, VD Drain current, ID Input voltage, Vin Output voltage, Vout Voltage gain, Av Phase difference

0V 1.72 V 15.0 V

−15.0 V −13.3 0V 1.70 mA 2.01 Vpp 2.00 Vpp 0.996 0˚

Step 4: With a 10 kΩ load, the signal level drops to 1.74 Vpp (measured). When the signal was increased, the output was observed to have a 20 Vpp signal with no clipping (the limit of the signal generator). Step 5: The dc offset remained very close to 0 V. Student answers may differ as the offset will change if the transistors are not matched. Questions: Part 2 1. The current source acts like a very high load impedance on the CD amplifier.

2.

The input resistance is equal to RG = 1.0 MΩ. The loading observed by the 10 kΩ load resistor indicates that the output impedance is 1.5 kΩ.

230

Part 3: A Cascode Amplifier Table 9-6

Listed Value 5.6 kΩ 620 Ω

Measured Value 5.63 kΩ 618 Ω

R3

1.0 MΩ

R4

Table 9-7 Parameter

VG(Q1) VS(Q1)

Measured Value 8.63 V 10.37 V

0.96 MΩ

VG(Q2)

0.001 V

1.5 MΩ

1.56 MΩ

VS(Q2)

0.215 V

RG

1.0 MΩ

1.02 MΩ

RS

100 Ω

99.8 Ω

Resistor R1 R2

Table 9-8 (Data for 500 kHz) Parameter Measured Value 50 mVpp Vin

Vout Av(Q1/Q2)

Table 9-9 (Data for 50 kHz) Parameter Measured Value 50 mVpp Vin

Vout

1.16 Vpp

Av(Q1/Q2)

23.2

2.68 Vpp 53.6

Step 6: The output drops. At 10 Vdc, the output at 50 kHz drops to 1.06 Vpp. Questions: Part 3 1. The data shows that the source voltage is higher than the gate voltage for both FETs.

2.

Because the higher frequencies have less gain, a parallel capacitance path to ground is suspected. This is likely caused by the capacitance of the protoboard.

231

Experiment 10 Amplifier Frequency Response Part 1: Low-Frequency Response Table 10-1

Table 10-2

Listed Value 1.0 kΩ

Measured Value

Computed Value

Measured Value

13.1 V

13.2 V

RB R1

47 Ω 68 kΩ

617 Ω

13.8 V

13.8 V

9.91 kΩ

VE IE

R2

10 kΩ

67.1 kΩ

VC

8.21 V

8.15 V

RE1

10 Ω

565 Ω

VCE

-5.59 V

-5.55 V

RE2

560 Ω

10 Ω

re

11.9 Ω

RC

3.9 kΩ

3.87 kΩ

Av

127

100

RL

10 kΩ

9.88 kΩ

Vout

1.27 V

1.00 V

Resistor RA

Table 10-3 Capacitor

Req 3.5 kΩ*

C2

24 Ω

C3

13.9 kΩ

25 μF

5.58 kΩ

2.1 mA

Table 10-4

C1

Table 10-5 Computed value of C2

Parameter VB

Capacitor C1

-- fcritical -Computed Measured 45 Hz

41 Hz

C2

66 Hz

70 Hz

C3 Overall

52 Hz

52 Hz

163 Hz

122 Hz

Measured low frequency f = 330 Hz

Calculated capacitance to raise low frequency to 300 Hz is 25 μF as follows: C=

1 1 1 = = = 25 μF (use 22 μF) 2πfReq 2πf RE2 || ( RE1 + re' ) 2π(300 Hz) ( 560 Ω|| (10 Ω + 11.9 Ω ) )

(

)

Questions: Part 1 1. C2 has the greatest effect because the RC time constant is lower than the others.

2.

Because all three responses are similar, capacitors that are 5X larger are the simplest change to reduce the lower frequency cutoff by a factor of five.

232

Part 2: High-Frequency Response Table 10-6

Parameter VB VE

Computed Value

Measured Value

1.91 V

1.80 V

1.22 V

1.13 V

IE

2.12 mA

VC

6.67 V

7.35 V

VCE

5.46 V

6.22 V

re

12.3 Ω

Av Vout

124

111

2.48 Vpp

2.22 Vpp

Table 10-7

Step 4

Parameter Cin

Computed Value

Measured Value

11.3 nF

5

Req(in)

44.8 Ω

6

fc(in)

314 kHz

7

Cout

201 pF

8 9

Rc fc(out)

2.81 kΩ

10

fcu

148 kHz

281 kHz 159 kHz

Questions: Part 2 1. The unbypassed emitter resistor reduces the gain of the amplifier and increases the upper frequency response. The Miller capacitance is directly proportional to the gain, so the reduced gain increases the upper frequency cutoff as a result.

2.

(a) For a wide band amplifier, the upper critical frequency and the bandwidth are essentially identical. Rise time is measured by using a fast-rising pulse at the input and observing the (inverted) pulse on the output for an inverting amplifier. The time required for the output to change from 90% to 10% is measured and substituted into the equation. (b) The calculated rise time (based on a measured value of 159 kHz) is 2.2 μs. (Measured result was 2.12 μs).

233

Experiment 11 Thyristor Part 1: The SCR Table 11-1 Resistor Listed Value R1 1.0 kΩ R3 160 Ω R4 1.0 kΩ

R5

10 kΩ

Table 11-2

Measured Value 993 Ω 161 Ω

Transistor Parameter Latch VAK(off state) 13.5 V VAK(on state) 0.803 V

1.001 kΩ

VGate trigger

0.768 V

0.736 V

9.94 kΩ

VR4

3.42 V

3.83 V

IH(min)

3.42 mA

3.83 mA

SCR 13.5 V 0.769 V

Step 7: S1 turns on the SCR. S2 turns it off. The firing point is controlled by R2.

20 V

Vr4

Drop out

0V

Lower portion of the positive half of the sine-wave appears across the LED.

20 V

VAK 0V 0

4.0

8.0 12 Time (ms)

16

20

Plot 11-1 Questions: Part 1 1. The voltage across R4 is proportional to the conduction current. The SCR is on for a shorter time and the back of the SCR waveform will drop earlier.

2.

In the circuit in Figure 11-3, the capacitor is charged by the positive supply voltage so that the right side is positive. Connecting it to ground through SW2 causes the anode to be more negative than the cathode, dropping the SCR out of conduction.

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Part 2: The Unijunction Transistor 15

Table 11-3

Resistor R1

Listed Value 47 Ω

10

Measured Value

VE 5

(V)

0

R2

220 Ω

15

R3

15 kΩ

10

VB1 5

(V)

Table 11-4 Resistance setting of R4 5.0 kΩ

Total Measured Resistance Frequency

0 15 10

VB2 5

25 kΩ 45 kΩ

(V) 0 Horiz = 0.5 ms/div (all views)

65 kΩ

Plot 11-2

85 kΩ

Frequency (Hz)

400 300 200 100 0

0

20 40 60 80 Resistance (kΩ )

100

Plot 11-3

4.

The frequency decreases in a nonlinear manner as the total resistance increases.

5.

The period of the RC charging waveform decreases as the total resistance decreases. The amplitude is not changed.

6.

The frequency is unaffected. The amplitude varies directly with the power supply.

Questions: Part 2 1. 6.7 V

2.

The emitter voltage must be at least one diode drop (about 0.7 V) higher than VK, the voltage established inside the device by the intrinsic standoff ratio.

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Experiment 12-A Operational Amplifiers Part 1: The Differential Amplifier Table 12-1 Resistor Listed Value RB1 100 kΩ RB2 100 kΩ RE1 100 Ω

Table 12-2 DC Computed Measured Parameter Value Value −1.1 V –1 V VA 1.4 mA IT

Measured Value 102 kΩ 102 kΩ 99 Ω

IE1 = IE2

0.7 mA

RE2

100 Ω

100 Ω

VC(Q1)

+15.0 V

+15.0 V

RT

10 kΩ

10.2 kΩ

VC(Q2)

+8.0 V

+8.35 V

RC2

10 kΩ

10.3 kΩ

Table 12-3 AC Parameter Vb(Q1)

Computed Value 100 mVpp

Measured Value

VA

50 mVpp

50 mVpp

re(Q1) = re(Q2)

100 mVpp

35.7 Ω

Av(d)

36.8

Vc(Q2) Rin(tot)

3.68 Vpp

3.55 Vpp

35.1 kΩ

36.5 kΩ

Av(cm)

0.5

0.44

CMRR′

37.3

38.1

35.5

Step 9: The CMRR’ rises significantly due to a much smaller common mode gain. The observed common-mode output was approximately 5 mVpp for a 7.7 Vpp input signal. This translates to a measured CMRR’ of 95 dB. This result is better than expected based on previous tests with this circuit; student answers will likely be lower. Questions: Part 1 1. It assures that the measurement is made when the amplifier is operating in its linear region.

2.

1.45 mA

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Part 2: Op-Amp Specifications Table 12-4

Specified Value Step Parameter 2d Input offset voltage, VOS 3d Input bias current, IBIAS 3e

Input offset current, IOS

4b 4c 4d

Differential gain, Av(d) Common-mode gain, Acm CMRR’

5

Slew rate

Table 12-5 Resistor Listed Value Rf 1.0 MΩ Ri 10 kΩ

RC

10 kΩ

Minimum

Typical

Maximum

Measured Value

2.0 mV

6.0 mV

0.66 mV

80 nA

500 nA

98 nA

20 nA

200 nA

1 nA 1000 0.032

70 dB

90 dB

89.9 dB

0.5 V/µs

0.8 V/µs

Table 12-6

Measured Value 1.01 MΩ

Resistor R1

10.2 kΩ

R2

Listed Value 100 kΩ

Measured Value 102 kΩ

100 kΩ

102 kΩ

10.2 kΩ

Table 12-7

Listed Resistor Value RA 100 Ω RB 100 Ω RC RD

100 kΩ 100 kΩ

Measured Value 102 Ω 101 Ω 102 kΩ 102 kΩ

Questions: Part 2 1. The input bias current is the average of the input currents; the input offset current is the difference between the input currents when the output voltage is 0 V.

2.

The advantage is the rejection of common-mode signals; these are undesired and frequently represent cross-talk, or other form of interference.

237

Part 3: Basic Op-Amp Circuits Table 12-8

Resistor R1 Rf

Listed Value 1.0 kΩ 10 kΩ

Measured Value 1.02 kΩ 10.2 kΩ

Table 12-9

Table 12-10

Measured Value

Acl(I)

Computed Value 500 mVpp −10

Vout

−5.0 Vpp

Parameter Vin

Parameter Vin

500 mVpp

Computed Value 500 mVpp

Measured Value 500 mVpp

Acl(NI)

11

−4.95 Vpp

Vout

5.5 Vpp

V(–)

0 Vpp

V(–)

5.5 Vpp 500 mVpp

Rin

1.0 kΩ

Rin

5 MΩ

Step 3: The observed gain at 1 kHz was 149.3. As the frequency is raised, the gain is observed to go down. At 10 kHz, the measured gain was 98.6. Questions: Part 3 1. Amplifier 1 gain = 20 log (11) = 20.8 dB Amplifier 2 gain = 20 log (10) = 20.0 dB

2.

The voltage on the inverting terminal is not close to ground; it is nearly the same as the input voltage.

3.

The RC low-pass filters within the op-amp cause the gain to roll-off as frequency increases.

Experiment 12B: Programmable Analog Design Part 1: Introduction to AnadigmDesigner2 Instructor note: If lab time is critical and the students have laptops or other computer access to the web, Part 1 can be done outside of the lab time. Steps 1-10: These steps are tutorial in nature and do not require a response from students. Step 11: Observations: The traces are described in the procedure but are summarized as follows: The largest trace (violet) is the 6 Vpp signal from the oscillator. In the

238

center of the screen is an inverted and attenuated 2 Vpp signal (green) that represents the output of the amplifier. The output traces (yellow and blue) are at the top of the display. They are offset by 2 V and have opposite phases and represent the differential outputs from the Anadigm chip. Time per division is 10 μs, volts per division is 1V with no offset for all signals.

Plot 12-1 Waveforms for 3 V 50 kHz sine wave amplified by −1/3. Step 12: The file extension is .ad2. Questions: Part 1 1. (a) 0.25 V (b) −0.333

2.

A ground referenced sine wave is centered on 0 V, with each peak an equal amount away from 0 V.

3.

The peaks of a 1 V peak sine are +1 V and –1 V from the reference. With a reference of +2 V, the peaks are at +3 V and +1 V.

4.

(a) The differential signal is the difference between the two signals. Mathematically, this can be expressed as OutputA – OutputB for the two outputs. (b) The differential signal is the same.

5.

.ad2

239

Part 2: Downloading the Configuration File Instructor note: There are +5 V and ground points located on the PAM board edge that must not be shorted together. Explain to students to be careful when connecting a probe to the board and make sure that the probe’s ground clip does not contact any point except a marked ground terminal. The additional points are shown here (but we chose not to show them in the student manual). Input2P

Input1P

GND

+5V GND

Signal, power, and ground points on the PAM board edge. Step 3: The signal level is approximately 1 Vp, but depends on the setting of the gain on the PAM output buffer. For the several PAM units tested the amplitude was 0.85 V to 1.05 V. More recently produced PAMs will show an output close to 1 V indicating an input-to-output system gain close to +1. For the board tested the amplitude was 0.85 V. The output resembles the simulation waveform from the inverting amplifier.

Time per division is 10 μs, volts per division is 500 mV

Plot 12-2 Measured waveform.

The output resembles the output of the gain stage shown on the internal simulation but with slightly lower amplitude (see the green waveform in the simulation). Step 4: The simulated waveform is 2.0 Vpp at 50 kHz referenced to ground. The observed signal from the PAM is 1.67 Vpp at 50 kHz referenced to ground.

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Instructor note: The Anadigm ASP and the PAM operate using a mixture of single-ended signals and differential-signals which can be confusing. The lab manual explains this through Experiments 12, 13 and 14 culminating in Figure 14-11 that shows the full details of the input signal path, connections to the Anadigm ASP and the output signal path. Here is a preview of the relevant facts (see Figure 14-11 as you read this): 1) The inputs and outputs of the PAM at the board edge (INPUT1, INPUT2, OUT1, OUT2) are referenced to ground. 2) The input buffer and output buffer devices provided on the PAM board operate using +5 V and −5 V power supplies to be able to deal with signals that are above and below ground. 3) The Anadigm ASP operates using only 0 V and +5 V power supplies so it can only handle signals that are above ground level. 4) The designers of the Anadigm ASP decided to do all signal processing level shifted to +2.0 V which they call the “mid-range reference level”. 5) The input buffers and output buffers on the PAM take care of level shifting between ground referenced signals in the outside world and the +2.0 V mid-range referenced signals of the Anadigm ASP. This is done so that the student will not need to worry about or even know much about the level shifted signal processing that the Anadigm ASP performs. 6) The Anadigm ASP does all of its internal signal processing differentially, even though the wires in the AD2 design space and the waveforms in the simulator are shown as if they are signal-ended ground referenced signals. 7) The signal outputs of the Anadigm ASP are differential. The PAM provides output buffers that do two things: i) Convert the Anadigm ASP differential outputs to single-ended signals. ii) Level shift the Anadigm ASP outputs from +2V to ground reference. This feature allows the student to observe signals on the simulator display and at the OUT1 and OUT2 terminals on the PAM without confusion regarding differential vs. singleended and +2.0 V referenced vs. ground referenced. 8) There are test points provided on the top surface of the PAM to observe the signals out of the Anadigm ASP: For the Anadigm ASP Output1, TP4 is the positive output and TP3 is the negative output. For the Anadigm ASP Output2, TP2 is the positive output and TP1 is the negative output. The positive differential signal observed at TP4 will have ½ the amplitude of the total Output1 signal with the +2.0 V offset. 9) The signals into the Anadigm ASP are differential. The PAM provides input buffers that do two things: i) Allow the input of either differential signals or single-ended signals according the Input Termination DIP switch settings. ii) Level shift ground referenced input signals to the +2.0 V referenced signals for the Anadigm ASP to use. 10) The differential input buffers on the PAM have a gain of approximately +1 for differential input signals and a gain of approximately + ½ for single-ended input signals. This ½ gain characteristic is typical for differential amplifiers where only one leg of the input is driven by a signal.

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Step 5: Answers can vary depending on the particular PAM board. For the one tested, the required gain is: 2.0 Vpp Av = × 0.33 = 0.395 1.67 Vpp Step 6: Measured output amplitude is 1.0 Vp (2.0 Vpp). Step 7: Answers will vary, depending on the setting of the gain stage and the particular PAM version. With the gain stage set for a gain of 0.395, the simulated amplitude is 1.17 Vp (2.34 Vpp). The measured amplitude for the PAM tested drops to 0.79 Vp (1.58 Vpp).

Instructor note: The remaining steps in the experiment are observations only and require a set of computer speakers, so that students can hear the audio tone that is generated. If speakers are not available, these steps can be skipped. Questions: Part 2 1. Steps to change gain are: • Double-click the gain’s stage to open the CAM window • Change gain value in the CAM window • Click in another part of the window to update and verify the realized value • Close the CAM window • Download the new configuration

2.

Signals internal to the Anadigm IC; and some signals at the pins of the Anadigm IC that are not sent to the output of the PAM board

3.

Increase the gain of the gain stage by the ratio of the desired output signal to the observed output signal.

Instructor note: This procedure changes the gain at all frequencies. There are filter adjustments that can avoid this, but this answer is not expected from students.) 4.

The Anadigm IC resets, losing the configuration information.

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Experiment 13-A Basic Op-Amp Circuit Part 1: The Comparator and Schmitt Trigger +15 V Vin Vs = 3.0 Vpp 50 Hz

+15 V

+

2

−

Vout

4 1.0 µF

VREF

10 kΩ

7 1.0 µF 6 741C

3

Vin 0V 3.0 Vpp

Vout 0V 28 Vpp

−15 V

−15 V

Horiz =5.0 ms/div

Plot 13-1 Comparator waveform Step 2: Varying the potentiometer changes the duty cycle of the output from 0 to 100%. The sine wave shows the same response. 0V

0V

X = 1 V/div Y = 10 V/div

X = 1 V/div Y = 10 V/div

0V

0V

The vertical line moves along the x-axis as the potentiometer is varied.

Plot 13-2 Comparator transfer curve

Plot 13-3 Comparator transfer curve (inputs reversed)

Step 4: Varying the potentiometer causes the vertical line along the x-axis. Steps 6/7: When the potentiometer is set to maximum, the output is in positive saturation. As the potentiometer is varied (less resistance), the output waveform suddenly becomes a 50% duty cycle. Reducing the resistance more causes the output waveform to shift to the left.

243

Note that the positive threshold is different than the negative threshold. +15 V V in Vs 3.0 Vpp 50 Hz

10 kΩ

7 1.0 µF 6 741C

2

−

3

+ 4

Vout

1.0 µF

Vin 0V 3.0 Vpp

Vout 0V 28 Vpp

−15 V R1 100 kΩ

Horiz =5.0 ms/div

Plot 13-4 Schmitt trigger waveform Step 8: The potentiometer varies the hysteresis; the upper and lower thresholds move apart as resistance is increased. 0V

X = 1 V/div Y = 10 V/div 0V

Plot 13-5 Schmitt trigger transfer curve Questions: Part 1 1. The threshold voltage is a dc quantity that adds to or subtracts from the input plotted along the x-axis of the transfer curve. Varying the threshold shifts the vertical line along the x-axis.

2.

The offset control on the generator does not affect the threshold for the circuit but it adds a dc component to the input voltage. As a result, the duty cycle of the output waveform can be changed by the offset control. (Note: it is useful to show this with the transfer curve – it varies only the endpoints of the transfer curve, not the vertical line).

3.

The transfer curve is a characteristic that is independent of the input. On an analog oscilloscope, the transfer curve will show various intensities that depend 244

on the input because the waveform determines the time the beam spends in graphing any given point on the curve. 4.

A comparator has the same threshold for rising or falling signals; a Schmitt trigger has a different threshold for each.

Part 2: The Summing Amplifier Table 13-1

Resistor RA RB RC Rf

Listed Measured Value Value 20 kΩ 20.0 kΩ 10 kΩ 10.1 kΩ 5.1 kΩ 5.08 kΩ 3.9 kΩ 3.93 kΩ

QA

0V

QB QC −4.4 V Horiz = 1 ms/div Vertical = 1.0 V/div

Horiz = 1 ms/div Vertical = 5.0 V/div

Plot 13-6

Plot 13-7

Step 4: The output waveform is the positive portion of the input waveform with no offset. The small jump on the output is due to slew rate limitation as the output moves from negative saturation. Step 5: The output is a negative half-wave rectified signal. With D1 removed, the output appears to be a full-wave rectified signal with overshoot on the back side of every other pulse. This overshoot is frequency dependent, and disappears at a few hundred hertz. Step 6: The output is a positive full-wave rectified signal. The signals are shown in Plot 13-8.

245

Input R i2

Input R i3

Vout

Horiz = 2 ms/div Vertical = 5.0 V/div

Plot 13-8 Questions: Part 2 1. (a) To approximate the column values in the binary system (1,2,4, etc.), the three inputs are amplified in proportion to the column values they represent. (This is in effect a three bit D/A converter).

(b) Add an inverting amplifier to the output. 2.

Gains are −0.195, −0.39, and −0.76. The output is 4.5 V x (−0.195 −0.39 −0.76) = −6.05 V

3.

One of the inputs to the summing amplifier is a variable dc as shown: Rf

R1 R2

Vs

−

+15 V 10 k Ω

VREF

Vout

+

−15 V

4.

The inverted half-wave rectified signal is amplified twice as much as the sine wave input in order to cancel the negative excursion of the sine wave and add a positive-going signal to it.

5.

The resistors shown are representative and can have other values; the ratios are important. Rf

R1

10 k Ω

R2 15 kΩ

30 k Ω

−

Vout

+

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Part 3: The Integrator and Differentiator Table 13-2

VOUT Red ON Green ON +2.1 V

-2.0 V

VREF Threshold

VA 0V

0.001 V

Step 3: Duty cycle changes as R3 is varied; output B slope follows.

VB 0V Horiz = 0.5 ms/div Vertical = 2.0 V/div

Plot 13-9 Table 13-3

Trouble No Negative Power Supply Red LED open C1 open R5 open

Symptoms Red LED on; B goes to positive saturation A = -2 V to + sat; B = - sat w/small deviation B goes to a square wave (+ and - saturation) No change in A; B goes toward - saturation.

VB

VB

Vout

Vout

0V

0V

0V

0V

Horiz = 0.5 ms/div Vertical = 2.0 V/div

Horiz = 0.5 ms/div Vertical = 2.0 V/div

Plot 13-10 Questions: Part 3 1. VREF(MIN) = −714 mV

Plot 13-11

VREF(MAX) = +714 mV

2.

The LEDs drop a maximum of about 2.0 V at the op-amp's current limit.

3.

(a) R4 establishes a virtual ground at the inverting input through negative feedback and stabilizes the operating point. Without it, the output will saturate. (b) The output went to negative saturation.

4.

Differentiator circuit

247

Experiment 13-B Programmable Analog Design Part 1: The Comparator and Comparator with Hysteresis Vin 0V

Vout 0V

Vertical = 2.0 V/div (both channels) Horizontal = 5.0 ms/div

Plot 13-12 Vin 0V

Vin 0V

Vout 0V

Vout 0V

Vertical = 2.0 V/div (both channels) Horizontal = 5.0 ms/div

Vertical = 2.0 V/div (both channels) Horizontal = 5.0 ms/div

Plot 13-13

Plot 13-14

Step 8: As the offset voltage is increased, the output positive time increases and the negative time decreases; when the offset voltage is decreased, the opposite occurs.

x-axis = 1.0 V/div y-axis = 2.0 V/div

Plot 13-15 Comparator transfer curve.

248

Step 10: The vertical transition on the transfer curve moves to the right to approximately 2.0 V on the x-axis. (For reference, the plot is shown here, but not required in the experiment. Note that the 2.0 V result is because the internal differential signal is ½ the external single-ended signal.)

x-axis = 1.0 V/div y-axis = 2.0 V/div

Step 10 result Step 11: The end points on the transfer curve move left or right, following the offset control.

Vout 0V

Vin

Channel 1 ( Vin): Vertical = 200 mV/div Channel 2 ( Vout): Vertical = 2.0 V/div Horizontal = 5.0 ms/div

x-axis = 200 mV/div y-axis = 2.0 V/div persist ON

Plot 13-16 Comparator with hysteresis.

Plot 13-17 Schmitt trigger transfer curve.

Questions: Part 1 Instructor note: The questions in Part 1 are stand alone, i.e. they do not specifically apply to the experiments completed in the laboratory work for this section.

1.

In a single-ended signal, the signal is referenced to a static value, which is typically ground. In a differential signal, the signal is referenced to another signal.

2.

Hysteresis is a characteristic of a circuit in which the trigger level (or switching point) for a rising signal is different than that of a falling signal.

3.

Hysteresis helps reduce the susceptibility to noise for a comparator. 249

4.

In a triangle wave the change of voltage over time is linear; in a sine wave the change in voltage over time is not linear. The transfer curve would look similar, but with a sine wave, the ends of the transfer curve would be brighter and the center dimmer on an analog scope display.

Part 2: The Summing Amplifier and Peak Detector Step 5: The simulation shows the sum of the two signals. The sum shows a pulsating appearance; the envelope represents the difference in the two frequencies. Step 7: An audio tone is heard in the speaker that represents the difference frequency between the two oscillators. The peak detector is necessary to hear the tone.

Instructor note: The simulated and measured waveforms are virtually identical. The simulated waveform is shown in Plot 13-18 for reference; the light gray signal on top is from the peak detector. A. A magnified view of the signals is shown in the unnumbered plot. (Note the decay time for the peak detector.)

Both channels: 1 V/div, 200 μs/div

Both channels: 1 V/div, 1 ms/div

Plot 13-18 Summing amplifier and peak detector outputs.

Magnified view of summing amplifier and peak detector outputs

Step 8: The sound heard will be the difference frequency between the oscillators. Step 9: There is no sound when the peak detector is removed. Questions: Part 2 1. The decay time constant controls how fast the peak detector responds to changes. A smaller time constant would allow the peak signal to show as a group of spikes that follow the wave instead of the smooth envelope that was observed.

2.

The peak detector converts the envelope to a frequency that can be heard. Without it, there is no audible frequency in the signal.

3.

The peak detector follows the envelope which represents a “beat” or difference frequency between the oscillators.

250

Part 3: The Differentiator Instructor note: The ASP outputs are not precisely differential, meaning that there is a slight DC offset between the two parts of each output. The output buffer, a difference amplifier, exacerbates this by subtracting the two signals. The amount of dc offset varies with different PAM units. In an actual application a DC blocking capacitor to remove the offset is recommended; students can use AC coupling on the scope to see true differentiation.

Vin

Vin

0V

0V

Vout

Vout Both channels = 1.0 V/div, ground at center Horizontal: 10 μs/div (see not about dc offset)

Both channels = 1.0 V/div, ground at center Horizontal: 10 μs/div

Plot 13-19 Differential output of a triangle wave.

Plot 13-20 Differential output of a sine wave.

Step 4: The output waveform is not affected by small changes in the dc offset. Vin

Vin Vout

0V

0V

Vout

Input = 1.0 V/div, output = 2.0 V/div Horizontal: 10 μs/div

Input = 1.0 V/div, output = 2.0 V/div Horizontal: 10 μs/div

Plot 13-21 Differential output of a square wave.

Step 7 result with a rectifier filter.

Step 7: The output has only positive excursions from 0 to 4.3 V, as shown in the unnumbered plot above. Questions: Part 3 1. The dc level had no effect because the rate of change of dc (a constant) is zero.

2.

For a constant amplitude input, changing the frequency changes the slope. A higher frequency will have a larger slope and hence a larger amplitude derivative.

3.

The derivative of a sine wave has the same shape but lags the input by 90o. 251

Experiment 14-A Special-Purpose Op-Amp Circuits Part 1: The Instrumentation Amplifier Table 14-1

Resistor R1 R2

Table 14-3

Listed Value 10 kΩ 10 kΩ

Measured Value

Measured Value

9.87 kΩ

Parameter Oscillator frequency

9.87 kΩ

Vout(pp) from oscillator

7.6 Vpp

Vout(pp) from IA

170 mVpp

RG

470 Ω

463 Ω

R3

10 kΩ

9.91 kΩ

R4

10 kΩ

9.87 kΩ

R5

10 kΩ

9.88 kΩ

R6 R8

8.2 kΩ 100 kΩ

8.10 kΩ

R9

100 kΩ

100 kΩ

682 Hz * *

*Output amplitude depends on battery voltage

101 kΩ

Table 14-2

Step 3

Parameter Differential input voltage, Vin(d) Differential gain, Av(d)

43.5

Differential output Voltage, Vout(d) Common-mode input voltage, Vin(cm) 4

Common-mode gain, Av(cm) Common-mode output voltage, Vout(cm)

5

Computed Value 300 mVpp

CMRR′

Measured Value 300 mVpp 43.0

13.1 Vpp

13.0 Vpp

10 Vpp

10.0 Vpp 0.008 80 mVpp 74.6 dB

Step 8: The measured differential signal was a 170 mV, 682 Hz square wave (from the 555 timer). It was amplified by the IA but the 10 Vpp common-mode signal was almost completely eliminated from the output as viewed on an oscilloscope. Questions: Part 1 1. For a CMRR of 130 dB, the ratio of Av(d) to Av(cm) is 3.16 x 106. This implies that the common-mode gain for the experiment is 43/3.16 x 106 = 13.6 x 10-6. The expected output signal is 10 Vpp x 13.6 x 10-6 = 136 µVpp.

2.

The oscillator signal was a differential-mode signal but the signal generator was a common-mode signal.

3.

The reference ground for the 555 timer needs to be isolated from the ground for the IA. The simplest way to do this is power it from an independent source.

252

4.

The IA has CMRR, both inputs to the IA are balanced, and it has high input impedance. (The input impedance was 100 kΩ, but it can be much higher).

Part 2: The Log Amplifier and Antilog Amplifier Table 14-4

Resistor R1 R2 12

Listed Value 100 kΩ 100 kΩ

Measured Value 100 kΩ 100 kΩ

12

Vin

10 8

Vin(V)

Vin

10 8

Vin(V)

6

6

4

4

2

2

0

0

Vout is from −0.58 V to −0.51 V (both plots)

Horiz = 0.2 ms/div

Horiz = 0.2 ms/div

Plot 14-1

Plot 14-2

Step 5: The output from the antilog amp matches the input to the log amp, as it should. Step 6: When the log amp is warmed (by touch), the output goes down; when the antilog amp is warmed (by touch), the output increases. −0.4

Data for

−0.45

VOUT -0.513 V -0.532 V -0.548 V -0.559 V -0.566 V -0.572 V -0.577 V

−0.5

VOUT (V)

Table 14-5 Log Amp VIN +1.0 V +2.0 V +4.0 V +6.0 V +8.0 V +10.0 V +12.0 V

−0.55 −0.6 −0.65 −0.7 1

2

3

4

VIN (V)

Plot 14-3

253

6

8

10 12

Questions: Part 2 1. Advantages of integrated circuit IA’s include (a) less temperature sensitivity (b) greater dynamic range (5 or 6 decades) (c) gain control (d) single package design (e) high accuracy and linearity (f) very low bias current (max of 5 pA in the LOG100)

2.

The antilog amp transfer curve is also a straight line plotted on semilog paper; however, the input voltage should be plotted on the linear scale and the output voltage is plotted on the log scale.

3.

With the same resistor in the log and antilog circuit, the original signal applied to the log amp is restored by the antilog amp. If the antilog amp has a larger resistor, its transfer curve is shifted up, resulting in more gain.

4.

(a) Multiplication of the inputs (and multiplying by −1 because of the three inversions). (Note the summing amplifier added the logs; the antilog amplifier returns the product of the original numbers but with opposite sign). (b) −6.0 V

Experiment 14-B Programmable Analog Design Part 1: Single-Ended Signals into a Differential Signal Circuit Step 3: (a) 1.00 Vpp

(b) 0.50 Vpp

Step 4: Typical values for OUT1 = 0.445 Vpp to 0.468 Vpp. (Answers to these steps depend on the particular PAM.) Step 5: Typical values for OUT2 = 0.455 Vpp to 0.494 Vpp. Step 6: Typical values for gain settings to equalize outputs = 0.970 to 978. Step 7: Some trial and error may be required, because the ASP cannot set exact gains. The ASP outputs of 0.445 Vpp need to be multiplied by 2.25 to achieve 1.0 Vpp, but a small adjustment was needed for one of the PAMs tested; the final gain setting for this PAM was 2.16. Questions: Part 1 1. This reduces any errors caused by differences in the equipment or problems with calibration.

2.

Any unshielded wire will capture nearby signals.

3.

Ground paths are sometimes called “returns”. If several signals use the aluminum structure for the return path, signals can combine, increasing the noise for any given signal. In short, this is a bad idea.

4.

Small differences in the actual gain are generally due to component variations.

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Part 2: Instrumentation CAMs Zero-crossing Detector

0V Vin Vout Ch 1 (Vin): Vertical 500 mV/div Ch 2 (Vout): Vertical 2.0 V/div Horizontal: 250 μs/div

Plot 14-4 Sine wave input and zero-crossing detector output. The output is at a level of – 5 V and pulses to +5.0 V when the zero crossing occurs.

For comparison, the simulation data is shown. The simulation shows the pulse is 2 μs wide.

Step 4: The delay is approximately 5 μs when the output is positive, which is significant at 50 kHz, representing 25% of the period. It is positive for about 2 μs, in good agreement with the simulation. Step 5: The zero crossing pulse is closer to the actual zero crossing as the frequency is increased.

255

Peak Detector Step 8: The signal at OUT1 is a half-wave signal that is ½ of the input peak. The signal at OUT2 is a full-wave signal that is also ½ of the input peak. Step 9: The half-wave signal returns to the base line between peaks, but the full-wave signal does not. Step 10: Neither the half-wave nor full-wave signal return to the baseline between peaks. The half-wave signal drops about 50%; the full-wave drops about 25%. Step 11: The signals approach the dc level of the peak as the frequency is raised. Gain-Limiter Steps 13 and 14: The simulated waveform and the measured waveform are virtually identical. The waveforms shown below are from the simulation, with the top waveform offset by +10 (1 div) and the bottom waveform offset by −10.

Plot 14-5 Simulated limiter waveform. (Plot 14-6 will be the same.)

256

Comparator Application – Tank Experiment

Plot 14-7 Simulated comparator waveform. The simulator waveform shows the trigger points at +40 mV and at −40 mV.

Vout Vin 0V

Ch 1 (Vin): Vertical 50 mV/div Ch 2 (Vout): Vertical 2.0 V/div Horizontal: 50 μs/div

Plot 14-8 Actual comparator waveform. Step 17: The trigger points on the actual waveform are at +80 mV and −80 mV.

257

Questions: Part 2 1. (a) One method is to rectify the input signal and combine it with a small negative dc voltage to offset the output and force it to cross 0 V. The rectified signal is twice the frequency of the input, which is then passed on to a zero-crossing detector. The configuration for this solution is shown. Note that the small dc voltage is generated with the Voltage CAM set to +3 V and a GainInv stage with a gain set to 0.02.

Instructor note: You can also generate a small dc voltage using the PeriodicWave CAM by cycling through a constant voltage. This CAM is discussed in Experiment 16-B, part 3.

Output of circuit for Question 1. Circuit for Question 1 (b) This circuit could be used to measure the time an arbitrary wave is above 0 V. 2.

When the time constant is short, the peak detector tends to follow the positive halfcycle if the input.

3.

(a) A long time constant causes the change from the peak value to decay slowly, acting as a low-pass filter. (b) A higher frequency causes the output to reset to the peak value sooner than a low frequency and tends to smooth the output as a result.

4.

Using a high-gain amplifier causes the signal to rapidly reach the limiter cutoff values, so the output signal looks like a square wave.

5.

Without hysteresis, the decision point for turning the valve on or off could be crossed several times because of noise in the system. Thus, the valve would tend to chatter.

258

Experiment 15-A Active Filters Part 1: Four-pole Low-Pass Filter Table 15-2 Component

Listed Value 8.2 kΩ 0.01 µF 10 kΩ 1.5 kΩ 22 kΩ 27 kΩ

RA1, RB1, RA2, RB2 CA1, CB1, CA2, CB2 Ri1 Rf1 Ri2 Rf2 Table 15-3 Frequency

VRL

A1

Measured Values B1 A2

8.17 kΩ

8.37 kΩ

8.00 kΩ 8.22 kΩ

0.01 µF

0.01 µF

0.01 µF

0.01 µF

10.1 kΩ 1.5 kΩ 21.9 kΩ 26.8 kΩ

3.0

2.55 V

1000 Hz

2.52 V

1500 Hz

2.45 V

2000 Hz

1.72 V

3000 Hz

0.45 V

4000 Hz

0.14 V

8000 Hz

0.009 V

VRL (V)

500 Hz

2.0

1.0 0.0 0

1.0

2.0

3.0

4.0

5.0

Frequency (kHz)

Plot 15-1

10

VRL (V)

1.0

0.1

0.01 10

B2

100

1.0 k Frequency (Hz)

Plot 15-2 259

10 k

6.0

7.0

8.0

Questions: Part 1 1. (a) The cutoff frequency is approximately 2.0 kHz.

(b) Answers vary; for the tested circuit, the average R is 8.19 kΩ, the average C is 0.010 µF. The computed cutoff frequency from these values is 1.94 kHz. 2.

(a) The measured voltage gain in the band pass is 2.55. (b) The desired value for a four-pole filter (from Table 15-1) is 2.574 (This is the product of the two gains: 1.152 x 2.235 = 2.574).

3.

The output is reduced by a factor of 104 (−80 dB for this filter), which is approximately 250 µV.

4.

The constructed filter is very close to the theoretical roll-off.

5.

(a) The gain computed from the actual resistors in the first stage is 1.149. (b) The gain computed from the actual resistors in the second stage is 2.227. The product of these gains results in an overall gain of 2.559 (compare to question 2).

Part 2: State-Variable Filter Table 15-4

Component R1 R2 R3 R4 R5 R6 R7 C1 C2

Listed Value 10 kΩ 10 kΩ 10 kΩ 1.0 kΩ 100 kΩ 1.0 kΩ 1.0 kΩ 0.1 μF 0.1 μF

Measured Value 10.0 kΩ 10.1 kΩ 9.91 kΩ 1.01 kΩ 101 kΩ 1.00 kΩ

Table 15-5 Quantity Computed Measured Center frequency, f0 = 1.59 kHz 1.62 kHz Vpp(center) = 13.3 Vpp 1.649 kHz Upper cutoff, fcu = 1.60 kHz Lower cutoff, fcl = Bandwidth, BW = 0.047 kHz 0.049 kHz 33.7 32.4 Q=

1.02 kΩ 0.0979 µF 0.1006 µF

260

Table 15-6

40 mVpp

200 Hz

75 mVpp

500 Hz

200 mVpp

1.0 kHz

0.48 Vpp

1.5 kHz 2.0 kHz

3.0 Vpp 1.14 Vpp

2.5 kHz

0.54 Vpp

3.0 kHz

0.37 Vpp

4.0 kHz

0.24 Vpp

5.0 kHz

180 mVpp

10 kHz 20 kHz

82 mVpp 41 mVpp

1.0

VRL (V)

Frequency 100 Hz

10

Output voltage Vout(pp)

0.1

0.01 100

1.0 k

10 k Frequency (Hz)

Plot 15-3 Questions: Part 2 1. If the Q of the circuit is made smaller, the output can be increased.

2.

(a) Advantages: high Q, easily tuned, and both high- and low-pass outputs are available. (b) The change of R6 to 100 kΩ, as noted in step 6, will eliminate the peaking, but greatly decreases the Q of the circuit so it affects the response of the band-pass filter.

3.

(a) To double the frequency, reduce R4 and R7 or reduce C1 and C2 by half. (b) To lower the Q, increase the value of R6.

4.

(a) A band reject or notch filter. (b) Rejection of an interfering noise frequency.

261

100 k

Experiment 15-B Programmable Analog Design Part 1: Single-Pole Low-Pass Filter using the Bilinear Filter CAM The simulation scope is shown. The two signals are nearly identical, so Channel 1, which represents the input, is offset by 10 units (1 div) to show both signals.

Plot 15-4 Input and output waveforms from a single-pole low-pass filter. Table 15-7 Simulated input and output voltage and time.

f 500 Hz 2.5 kHz 5.0 kHz 10 kHz 50 kHz

T 2000 μs 400 μs 200 μs 100 μs 20 μs

Δt 32 μs 29 μs 25 μs 18 μs 5.0 μs

Phase shift

Input Peak Voltage, Vp

5.8o 26.1o 45.0o 64.8o 90.0o

1.50 1.50 1.50 1.50 1.50

Output Peak Voltage, Vp

1.49 1.34 1.06 0.67 0.15

Table 15-8 Measured input and output voltage and time.

f 500 Hz 2.5 kHz 5.0 kHz 10 kHz 50 kHz

T 2000 μs 400 μs 200 μs 100 μs 20 μs

Δt 38.0 μs 30.3 μs 25.6 μs 18.5 μs 5.45 μs

Phase shift

6.8o 27.3o 46.1o 66.4o 98.1o

Input Peak Voltage, Vp

3.02 3.02 3.02 3.02 3.02

Output Peak Voltage, Vp

1.67 1.47 1.18 0.750 0.17

Step 9: The shape of the responses is nearly identical but the measured outputs are slightly larger than the simulation. The attenuation in the PAM input buffer required the larger specified input signal, and may account for the differences. Questions: Part 1 1. At the zero crossing, the signals are changing at the maximum rate, so it is easier to read time precisely. Also, the signals are not as sensitive to alignment on the scope.

262

2.

Yes. In general, low-pass filters are constructed as RC networks with the output taken across C, causing a phase lag. With RL networks, the output is taken across R, also producing a phase lag.

3.

The response of the simulation is nearly identical to theory for the points at 0.1fc, fc, and 10fc. The measured responses are higher at all frequencies. This may be accounted for by the PAM input buffer amplifier.

Part 2: Single-Pole Low-Pass Filter Using AnadigmFilter Step 9: Both filters have identical responses at all frequencies tested. Table 15-9 Simulated input and output voltage for a five-pole low-pass filter.

f 500 Hz 2.5 kHz 5.0 kHz 10 kHz 50 kHz

Input Peak Voltage, Vp

1.00 1.00 1.00 1.00 1.00

Output Peak Voltage, Vp

1.00 1.00 0.71 0.0316 0.00

Step 14: At the cutoff frequency of 5 kHz, the measured output is 0.73 Vp (1.46 Vpp). Step 15: At 10 kHz, the measured output is 0.377 Vp (73.5 mVpp). Questions: Part 2 1. The filter has a very sharp roll-off. At one decade, the predicted output is too small to measure (it should be approximately 10 μV). 10

(a) See Plot 15-5. (b) The frequency as read on the plot for Vout = 10 mV is approximately 13 kHz. This agrees with the calculated value for a 5 pole filter of 13.3 kHz.

1.0

Vout (V)

2.

0.1

0.01 100

1.0 k

10 k Frequency (Hz)

Plot 15-5

263

100 k

Experiment 16-A Oscillators Part 1: The Wien-Bridge Oscillator Table 16-2 Computed

Table 16-1

Listed Value 10 kΩ

Measured Value

R2

10 kΩ

10 kΩ

C1

0.01 μF 0.01 μF 0.01 μF 0.01 μF

Component R1

C2

fr

10 kΩ

1.59 kHz

Measured 1.46 kHz

Step 2: The output saturates on both positive and negative peaks. Freeze spray causes circuit to change and even stop oscillating. Circuit is temperature sensitive. Table 16-3 Vout(pp)

V(+)(pp)

V(–)(pp)

(pin 6)

(pin 3)

(pin 2)

4.0 Vpp

1.5 Vpp

1.5 Vpp

VG −1.0 Vdc

Step 5: The phase shift is 0o. Step 6: Very little effect with freeze spray. The output is much more stable. Table 16-4 Vout(pp)

V(+)(pp)

V(–)(pp)

(pin 6)

(pin 3)

(pin 2)

4.6 Vpp

1.56 Vpp

1.56 Vpp

VG −1.29 Vdc

Questions: Part 1 1. (a) Feedback fraction is very close to 1/3.

(b) The measured result agrees with theory. 2.

The extra diode causes C3 to charge for a smaller part of the cycle decreasing VG. This causes the FET resistance to drop (temporarily). The op-amp's gain (and output voltage) increase until the charge on C3 is returned to the proper level for a stable output.

3.

The diode causes the negative half-cycle of the output to charge the capacitor and bias the FET with a negative bias voltage.

4.

The frequency is halved to 790 Hz.

264

Part 2: The Hartley and Colpitts Oscillators Table 16-5

Resistor R1 R2 RE1 RE2 RC

Listed Measured Value Value 9.98 kΩ 10 kΩ 3.31 kΩ 3.3 kΩ 50 Ω 50 Ω * 1.01 kΩ 1.0 kΩ 2.69 kΩ 2.7 kΩ

* Set to the center position

Table 16-6 DC Computed Parameter Value VB 3.01 V VE 2.31 V IE 2.18 mA VC 6.11 V Table 16-8 Hartley Computed Oscillator Value frequency 969 kHz amplitude

Table 16-7 AC Computed Parameter Value Vb 100 mVpp re′ 61.5 Ω Av 43.9 Vc 4.39 Vpp

Measured Value 2.97 V 2.32 V 6.07 V

Table 16-9 Colpitts Computed Oscillator Value frequency 1.06 MHz amplitude

Measured Value 961 kHz 5.5 Vpp

Measured Value 100 mVpp 39.0 3.9 Vpp

Measured Value 1.04 MHz 7.4 Vpp

Step 6: Both the amplitude and the frequency of the oscillator decreased. Questions: Part 2 1. The amount of feedback decreased.

2.

The two conditions are positive feedback and a loop gain equal or greater than 1.

3.

In the Hartley oscillator, an inductor is used to provide positive feedback from the tank circuit. In a Colpitts oscillator, a capacitor is used to supply positive feedback.

265

Part 3: The 555 Timer Table 16-10

Table 16-11

Listed Value 8.2 kΩ

Measured Value 8.14 kΩ

R2

10 kΩ

Cext

0.01 μF

9.91 kΩ 10.3 nF

Component R1

Component Frequency

Computed Value 4.96 kHz

Duty cycle

0.65

Measured Value 5.15 kHz 0.65

Pin 3 0V

Pin 2 0V Vertical = 2 V/div Horiz = 50μ s/div

Plot 16-1 Step 5: The capacitor waveform shows the charge and very fast discharge. Frequency rises to 12.5 kHz, with only negative triggers on the output. Steps 6 and 7: Student answers will vary. One possible circuit is to change both RA and RB to 4.7 kΩ (no change to C). This will give a calculated frequency of 10.1 kHz. Questions: Part 3 1. The value of R1 needs to be larger and R2 smaller. R1 will need to be increased by twice the amount of reduction to R2. For example if R2 is reduced to 5.1 kΩ, then R1 should be a resistor that is near 2 x 4.9 kΩ larger than 8.2 kΩ, which is 18 kΩ.

2.

The shape will remain the same, but the waveform will go between 5 V and 10 V (1/3 and 2/3 of VCC.)

3.

For any real value of R1, the fraction (R1+R2)/(R1+2R2) > 0.5.

4.

(a) The trigger points are 1/3 and 2/3 of VCC. The capacitor charges and discharges between these two levels. (b) It has no effect on the frequency because the frequency is determined solely by the time constants.

5.

(a) The maximum output source or sink current is 200 mA but depends on the acceptable output voltage levels for high and low. (b) The high output voltage drops and the low output voltage rises.

266

Experiment 16-B Programmable Analog Design Part 1: Ring Oscillators Instructor Note: Values for frequency and amplitude will vary for different PAMs. Step 3: Frequency = 470 kHz Amplitude = 3.68 Vp (7.36 Vpp)

0V

Vertical: 2.00 V/div Horizontal 500 ns/div

Plot 16-2 Three-stage ring oscillator. Step 4: Frequency = 285 kHz

Amplitude = 4.04 Vp (8.08 Vpp)

0V

Vertical: 2.00 V/div Horizontal 1.0 μs/div

Plot 16-3 Seven-stage ring oscillator. Step 6: Frequency = 6.50 kHz

Amplitude = 1.05 Vp (2.10 Vpp)

0V

Vertical: 500 mV/div Horizontal 50 μs/div

Plot 16-4 Waveform from the special ring oscillator. 267

Step 7: As the integration time constant is increased, the frequency increases. An integration time constant of 4.0 produces a 12 kHz square wave. Step 8: The last circuit used most of the available resources. The LUT and counter are shown as available but only 2 capacitors (in CAB2) are available. Questions: Part 1 1. Each time a signal goes through the loop it represents either a logical LOW or logical HIGH. Thus each pass represents ½ the period of the wave.

2.

The measured frequency in step 4 is 285 kHz with a period of 3.50 μs. This time represents fourteen total delays (two passes through seven stages). Thus the delay for one stage is approximately 3.50 μs/14 = 250 ns. Note that this is an average for all stages and ignores other delays in the system such as the input and output buffers on the PAM unit.

3.

Make one inverting stage and the rest non-inverting stages. (This can be implemented as one instance of a GainInv CAM and three instances of a GainLimiter CAM.)

4.

In addition to showing how much of the available resources are used, it also indicates the power consumption.

Part 2: The Sine Wave Oscillator

0V

Vertical: 500 mV/div Horizontal: 500 ns/div

Plot 16-5 Oscillator signal at OUT1 of PAM.

268

0V

Vertical: 1.0 V/div Horizontal: 25 μs/div

Plot 16-6 A 7 kHz sine wave using the internal 250 kHz clock.

0V

Vertical: 1.0 V/div Horizontal: 5.0 μs/div

Plot 16-7 A 41 kHz sine wave using the internal 250 kHz clock (from a single-shot display.) Questions: Part 2 1. (a) At 5 samples per cycle of a sine wave, the reconstructed wave is recognizable but the steps are clearly visible between sample points. A smoothing filter gives a better result. (Note: Typical simulator software tools such as PSPICE, Multisim, or Saber) use at least 20 samples per cycle.)

(b) The Nyquist criteria requires more than two samples per cycle as the minimum sampling rate to reconstruct the frequency of the sine wave. 2.

The intent of the output stage smoothing filter is to make the steps in the output look more like a sine wave. Because the filter frequency on the PAM5002 is set to approximately 432 kHz, it is not effective for the 41 kHz signal used in the experiment (the signal is more than a factor of 10 below the filter fc).

3.

The signal is 800 MHz but looks more like a triangle wave than a sine wave, even after the 432 MHz smoothing filter.

269

Part 3: The Arbitrary Waveform Generator Instructor note: ClockA = 2000 kHz (500 ns period). With 24 samples, the period is T = (500ns)(24 samples) = 12,000 ns = 12 μs. This implies f = 83.3 kHz. If a student reports an incorrect value, check that he or she has set the counter reset value to = 23, not 24 or some other value.

0V

Vertical: 500 mV/div Horizontal: 5.0 μs/div

Vertical: 500 mV/div Horizontal: 5.0 μs/div

Plot 16-8 Simulated waveform. Frequency: Simulated 83.3 kHz

Plot 16-9 Measured waveform Measured 83.3 kHz

Step 6: The output filter on the PAM has a cutoff frequency at 432 kHz that affects the higher harmonics in the signal. This causes rounding of the edges as shown in Plot 16-9. Step 7: The frequency drops to 10.4 kHz, so the measured waveform has steeper changes and approaches the ideal wave. (The simulator shows the ideal wave.) Step 11: The frequency measured on the simulator is 1.64 kHz. The period is 0.61 ms. Step 12: The frequency measured from the PAM is 1.64 kHz. The period is 0.61 ms. Step 13: The frequency goes to 13.2 kHz. This is 8 times faster, which is expected because the sample clock was increased by 8 times. Step 14: The dc voltage on the simulator is −3.46 V. The measured voltage on the PAM (using a DMM) is −3.55 V. The measured voltage in the PAM is after the output buffer. Small variations in component values can affect the gain of the buffer. Questions: Part 3 1. There are 152 steps read at 250 kHz. This implies 4 μs per step. T = (152 steps)(4 μs per step) = 608 μs, which is a frequency of 1.64 kHz.

2.

Examples are CD and MP3 players.

3.

The clock determines how fast the lookup table is sampled, which determines how fast values are presented to the DAC.

270

Experiment 17 Voltage Regulators Part 1: The Series Regulator Table 17-1

Table 17-2

Listed Value 2.7 kΩ 330 Ω

Measured Value

R3

1.0 kΩ

1.00 kΩ

R4*

1 kΩ

1.00 kΩ

R5

1.2 kΩ

1.19 kΩ

RL

330 Ω

328 Ω

Resistor R1 R2

Parameter VOUT(min)

2.69 kΩ

VOUT(max)

329 Ω

Computed Value

Measured Value

8.30 V

8.54 V

15.3 V

15.09 V

*potentiometer; record maximum resistance

Table 17-4

Table 17-3

VIN +18.0 V

VOUT (measured) +10.0 V

Step Quantity 6 Line regulation

+17.0 V

9.94 V

VNL

+16.0 V

9.88 V

+15.0 V

9.82 V

+14.0 V

9.75 V

7

8

VFL Load regulation

Measured Value 0.63%

+10.0 V 9.96 V 0.4%

Vripple(in)

700 mVpp

Vripple(out)

45 mVpp

Questions: Part 1 1. Increase the size of C1. (A second 1000 µF capacitor in parallel with C1 halved the ripple).

2.

The power delivered to the load is 0.91 W. The efficiency is 45%.

Part 2: IC Regulators Step 1: The regulated output had approximately 1 mV of ripple. The ripple waveform showed only the tip of the positive waveform. Noise level was less than 1 mV. Although a larger filter capacitor was specified here, the ripple is significantly smaller than the unregulated supply in Step 9 of Experiment 2. Step 2: The LED brightness was the same for all input voltages because of the current source as indicated by the data in Table 17-5. Step 3: The current was nearly identical to the current in the single LED as indicated by the data in Table 17-6.

271

Table 17-5 Data for one LED

Table 17-6 Data for two LEDs

Voltage

Current

Voltage

Current

10 V

18.61 mA

10 V

18.61 mA

12 V

18.65 mA

12 V

18.61 mA

14 V

18.67 mA

14 V

18.61 mA

16 V

18.69 mA

16 V

18.61 mA

Questions: Part 2 1. There is a small additional current from the common (“ground”) lead on the regulator that adds to the current sourced by R1.

2.

Change R1 to 680 Ω (nearest standard value to 666 Ω).

Experiment 18 Communications Circuits Part 1: The IF Amplifier Table 18-1 Resistor R1 R2 R3 RE1 RE2 RL Table 18-3 AC Parameter Vb Vc Av Vout(tot)

Listed Value 56 kΩ

Measured Value 56.1 kΩ

4.7 kΩ 10 kΩ 220 Ω 470 Ω 10 kΩ

4.62 kΩ

Table 18-2 DC Parameter VB

Computed Value 0.697 V

Measured Value 0.697 V

VE

0.0 V*

0.0999 V

10.1 kΩ

IE

0 mA

222 Ω

VC

9.00 V

8.99 V

VCE 9.00 V 8.99 V *assuming VE cannot be negative

467 Ω 10.1 kΩ

Table 18-4 AC Parameter fc

Measured Value 300 mVpp* 8.0 mVpp** 26.7 1.2 mVpp

* Signal gen set to approx 500 mVpp ** Different transformers were tested. Results varied more than usual. Typical results shown.

Measured Value* 455.8 kHz

fcu

461.4 kHz

fcl

444.6 kHz

BW

16.8 kHz

Q

27.1

*frequency counter suggested for frequency measurements

Questions: Part 1 1. Adding a second probe doubles the loading effect. If probe loading is not a problem, no effect will be observed when the second probe is connected to the circuit.

2.

The resonant circuit has highest reactance at resonance; thus, amplifier gain is also highest here.

272

4.

The voltage across Rin is 3X larger than that dropped across R3, hence Rin = 3 x 10 kΩ = 30 kΩ.

5.

Answers (a) and (e) could account for 0 V on the collector.

Part 2: The Phase-Locked Loop Table 18-5 Listed Value 2.0 kΩ

Measured Value 1.99 kΩ

R2

1.0 kΩ

1.00 kΩ

R3

1.0 kΩ

1.01 kΩ

C1

2200 pF

2200 pF

C2

0.1 μF

0.1 µF

C3

1000 pF

1000 pF

Component R1*

* plus 10 kΩ potentiometer in circuit

Table 18-6 Step 2 and 3

Quantity free-running frequency, f0(min)

Computed Value 11.54 kHz

Measured Value 11.65 kHz

free-running frequency, f0(max)

68.5 kHz

68.3 kHz

4

lower capture frequency

5

upper capture frequency

Range =

± 1.9 kHz

lower lock frequency 6

8

Range =

23.8 kHz 28.1 kHz 17.5 kHz

± 8.3 kHz

34.6 kHz

frequency in (multiplier)

2.5 kHz

2.5 kHz

frequency out (multiplier)

25.0 kHz

25.0 kHz

upper lock frequency

Step 7: Pin 7 shows a capacitor charge and discharge waveform that goes between 8 and 10 V. The frequency is twice the input frequency. Questions: Part 2 1. A decrease in either R1 or C1 by a factor of ten will multiply the free-running frequency by ten.

2.

The transistor limits the input signal to the logic circuit between 0 and 5 V.

3.

The capture range is the range of frequencies over which the VCO can establish synchronization. After a signal is captured, the VCO can track it over the lock range.

4.

It must be multiplied by two.

5.

(a) ±12 V (b) The power supply voltage is inversely proportional to the lock range.

273

Test Item File

Electronic Devices Eighth Edition

Thomas L. Floyd

Contents 275

Chapter 1

Introduction to Semiconductors............................................................. 277

Chapter 2

Diode Applications ................................................................................ 282

Chapter 3

Special‐Purpose Diodes ......................................................................... 291

Chapter 4

Bipolar Junction Transistors .................................................................. 299

Chapter 5

Transistor Bias Circuits.......................................................................... 308

Chapter 6

BJT Amplifiers....................................................................................... 316

Chapter 7

Power Amplifiers ................................................................................... 325

Chapter 8

Field‐Effect Transistors (FETs) ............................................................. 333

Chapter 9

FET Amplifiers and Switching Circuits ................................................ 341

Chapter 10

Amplifier Frequency Response.............................................................. 349

Chapter 11

Thyristors ............................................................................................... 356

Chapter 12

The Operational Amplifier..................................................................... 363

Chapter 13

Basic Op‐Amp Circuits.......................................................................... 371

Chapter 14

Special‐Purpose Op‐Amp Circuits......................................................... 379

Chapter 15

Active Filters.......................................................................................... 386

Chapter 16

Oscillators .............................................................................................. 393

Chapter 17

Voltage Regulators................................................................................. 401

Chapter 18

Communications .................................................................................... 411

Answer Key ............. .....................................................................................................415

276

Chapterȱ1 Introduction to Semiconductors 1) Aȱmoleculeȱisȱtheȱsmallestȱparticleȱofȱanȱelementȱthatȱretainsȱtheȱcharacteristicsȱofȱthatȱelement. Diff:ȱ2

2) Aȱconductingȱgermaniumȱdiodeȱhasȱaȱpotentialȱofȱaboutȱ0.7ȱVȱacrossȱit. Diff:ȱ2

3) Siliconȱdopedȱwithȱimpuritiesȱisȱusedȱinȱtheȱmanufactureȱofȱsemiconductorȱdevices. Diff:ȱ2

4) Reverseȱbiasȱpermitsȱfullȱcurrentȱthroughȱaȱpnȱjunction. Diff:ȱ2

5) SemiconductorȱmaterialȱofȱtheȱpȬtypeȱhasȱfewȱfreeȱelectrons. Diff:ȱ2

6) Theȱapplicationȱofȱaȱdcȱvoltageȱtoȱcontrolȱdiodeȱconductionȱisȱcalled A) oscillation.

B) amplification.

C) bias.

D) aȱpnȱjunction.

Diff:ȱ2

7) Holesȱareȱtheȱmajorityȱcarriersȱin A) anȱnȬtypeȱsemiconductor.

B) aȱpȬtypeȱsemiconductor.

C) aȱpnȱjunctionȱsemiconductor.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

8) Semiconductorȱmaterialsȱareȱthoseȱwith A) conductiveȱpropertiesȱthatȱareȱinȱbetweenȱthoseȱofȱaȱconductorȱorȱanȱinsulator. B) conductiveȱpropertiesȱthatȱareȱveryȱgood. C) noȱconductiveȱproperties. D) EitherȱAȱorȱB. Diff:ȱ2

9) Thereȱisȱcurrentȱthroughȱtheȱjunctionȱofȱaȱforward-biasedȱdiode.ȱȱThisȱcurrentȱisȱcalled A) forwardȱcurrent.

B) reverseȱbreakdownȱcurrent.

C) avalancheȱcurrent.

D) reverseȱleakageȱcurrent.

Diff:ȱ2

277

10) Aȱtypicalȱvalueȱofȱreverseȱbreakdownȱvoltageȱinȱaȱdiodeȱis A) 0ȱV.

B) 0.3ȱV.

C) 0.7ȱV.

D) 50ȱVȱorȱlarger.

Diff:ȱ2

11) Theȱsmallȱcurrentȱwhenȱaȱdiodeȱisȱreverse-biasedȱisȱcalled A) forward-biasȱcurrent.

B) reverseȱbreakdownȱcurrent.

C) conventionalȱcurrent.

D) reverse-leakageȱcurrent.

Diff:ȱ2

12) Asȱtheȱforwardȱcurrentȱthroughȱaȱforward-biasedȱdiodeȱdecreases,ȱtheȱvoltageȱacrossȱthe diode A) increases.

B) immediatelyȱdropsȱtoȱ0ȱV.

C) isȱrelativelyȱconstant.

D) increasesȱandȱthenȱdecreases.

Diff:ȱ2

13) WhichȱstatementȱbestȱdescribesȱaȱpȬtypeȱsemiconductor? A) Siliconȱwithȱpentavalentȱimpurity. B) Siliconȱwithȱtrivalentȱimpurityȱatomsȱadded. C) Aȱmaterialȱwhereȱholesȱareȱtheȱminorityȱcarriers. D) Pureȱintrinsicȱsilicon. Diff:ȱ3

14) Theȱresistanceȱofȱaȱforward-biasedȱdiodeȱis A) perfectlyȱlinear.

B) infinite.

C) minimalȱbelowȱtheȱkneeȱofȱtheȱcurve.

D) minimalȱaboveȱtheȱkneeȱofȱtheȱcurve.

Diff:ȱ2

15) Aȱsiliconȱdiodeȱmeasuresȱaȱhighȱvalueȱofȱresistanceȱwithȱtheȱmeterȱleadsȱinȱbothȱpositions.ȱȱThe trouble,ȱifȱany,ȱis A) theȱdiodeȱisȱopen.

B) theȱdiodeȱisȱshortedȱtoȱground.

C) theȱdiodeȱisȱinternallyȱshorted.

D) nothing;ȱtheȱdiodeȱisȱgood.

Diff:ȱ2

16) Theȱforwardȱvoltageȱacrossȱaȱconductingȱsiliconȱdiodeȱisȱabout A) 1.3ȱV.

B) 0.3ȱV.

C) 0.7ȱV.

Diff:ȱ2

278

D) -0.3ȱV.

17) Anȱunknownȱtypeȱofȱdiodeȱisȱinȱaȱcircuit.ȱȱTheȱforwardȱvoltageȱmeasuredȱacrossȱitȱisȱfoundȱto beȱ0.3ȱV.ȱȱTheȱdiodeȱis A) aȱsiliconȱdiode.

B) aȱgermaniumȱdiode.

C) aȱtransistor.

D) shorted.

Diff:ȱ2

18) Aȱreverse-biasedȱdiodeȱhasȱtheȱ________ȱconnectedȱtoȱtheȱpositiveȱsideȱofȱtheȱsource,ȱandȱthe ________ȱconnectedȱtowardsȱtheȱnegativeȱsideȱofȱtheȱsource. A) cathode,ȱanode

B) cathode,ȱbase

C) base,ȱanode

D) anode,ȱcathode

Diff:ȱ3

19) TheȱboundaryȱbetweenȱpȬtypeȱmaterialȱandȱn-typeȱmaterialȱinȱaȱdiodeȱisȱcalled A) theȱcathode.

B) theȱcontrolȱgrid.

C) theȱpnȱjunction.

D) theȱanode.

Diff:ȱ2

20) Theȱatomicȱnumberȱofȱanȱatomȱrefersȱtoȱthe A) numberȱofȱprotonsȱinȱtheȱnucleus.

B) numberȱofȱelectronsȱinȱaȱchargedȱatom.

C) netȱelectricalȱchargeȱofȱtheȱatom.

D) numberȱofȱneutronsȱinȱtheȱnucleus.

Diff:ȱ1

21) Electronsȱorbitingȱtheȱnucleusȱofȱanȱatomȱareȱgroupedȱintoȱenergyȱbandsȱknownȱas A) slots.

B) tracks.

C) shells.

D) tunnels.

Diff:ȱ2

22) Valenceȱelectronsȱhaveȱ________ȱenergyȱlevelȱofȱallȱtheȱelectronsȱinȱorbitȱaroundȱtheȱnucleusȱof aȱgivenȱatom. A) theȱlowest

B) theȱhighest

C) theȱsame

D) Noneȱofȱtheȱabove.

Diff:ȱ2

23) Theȱdifferenceȱinȱenergyȱlevelsȱthatȱexistsȱbetweenȱtheȱvalenceȱbandȱandȱtheȱconductionȱband isȱcalled A) covalentȱgap.

B) semiconductorȱregion.

C) sparkȱgap.

D) energyȱgap.

Diff:ȱ3

279

24) Theȱvalenceȱelectronȱofȱaȱcopperȱatomȱexperiencesȱwhatȱkindȱofȱattractionȱtowardȱtheȱnucleus? A) None

B) Weak

C) Strong

D) Impossibleȱtoȱsay

Diff:ȱ2

25) Whatȱmustȱbeȱusedȱinȱseriesȱwithȱaȱforward-biasedȱdiodeȱtoȱpreventȱdamageȱdueȱtoȱexcessive current? A) Ammeter

B) Resistor

C) NCȱswitch

D) Nothingȱisȱrequired

Diff:ȱ2

26) Reverseȱbiasȱisȱaȱconditionȱthatȱessentiallyȱ________ȱcurrentȱthroughȱtheȱdiode. A) prevents

B) allows

C) increases

D) amplifies

Diff:ȱ2

27) Theȱkneeȱvoltageȱofȱaȱdiodeȱisȱapproximatelyȱequalȱtoȱthe A) appliedȱvoltage.

B) barrierȱpotential.

C) breakdownȱvoltage.

D) reverseȱvoltage.

Diff:ȱ2

28) Aȱnonconductingȱdiodeȱisȱ________ȱbiased. A) forward

B) inverse

C) poorly

D) reverse

C) 0.7ȱV.

D) 0.79ȱV.

Diff:ȱ2

29) Onȱdiodeȱcheck,ȱaȱshortedȱdiodeȱwillȱmeasure A) 0ȱV.

B) 0.3ȱV.

Diff:ȱ2

30) Howȱmuchȱforwardȱdiodeȱvoltageȱisȱthereȱwithȱtheȱideal -diodeȱapproximation? A) 0ȱV

B) 0.7ȱV

C) Moreȱthanȱ0.7ȱV

D) 1ȱV

Diff:ȱ1

31) AȱDMMȱmeasuresȱ0.13ơ̇̄ȱinȱbothȱdirectionsȱwhenȱtestingȱaȱdiode.ȱTheȱdiodeȱis A) constructedȱofȱSiȱandȱisȱgood.

B) open.

C) operatingȱnormally.

D) shorted.

Diff:ȱ3

280

32) Ifȱtheȱpositiveȱleadȱofȱanȱohmmeterȱisȱplacedȱonȱtheȱcathodeȱandȱtheȱnegativeȱleadȱisȱplacedȱon theȱanode,ȱwhichȱofȱtheȱfollowingȱreadingsȱwouldȱindicateȱaȱdefectiveȱdiode? A) 0ơ̇̄

B) Q ̛

C) 1ȱM̛

D) 400ȱk̛

Diff:ȱ3

33) Whatȱisȱtheȱmaximumȱnumberȱofȱelectronsȱthatȱcanȱexistȱinȱtheȱshellȱclosestȱtoȱtheȱnucleusȱof anȱatom? A) 1

B) 2

C) 4

D) 8

C) silicon.

D) copper.

Diff:ȱ2

34) Allȱofȱtheȱfollowingȱareȱsemiconductorsȱexcept A) carbon.

B) germanium.

Diff:ȱ2

35) Aȱreverse-biasedȱsiliconȱdiodeȱisȱconnectedȱinȱseriesȱwithȱaȱ12ȱVȱsourceȱandȱaȱresistor.ȱThe voltageȱacrossȱtheȱdiodeȱis A) 0ȱV.

B) 0.3ȱV.

C) 0.7ȱV.

D) 12ȱV.

Diff:ȱ2

36) Aȱreverse-biasedȱsiliconȱdiodeȱisȱconnectedȱinȱseriesȱwithȱaȱ12ȱVȱsourceȱandȱaȱresistor.ȱThe voltageȱacrossȱtheȱresistorȱis A) 0ȱV.

B) 0.3ȱV.

C) 0.7ȱV.

D) 12ȱV.

Diff:ȱ2

37) Siliconȱandȱgermaniumȱcontainȱ________ȱvalenceȱelectrons. A) one

B) two

C) four

D) eight

Diff:ȱ1

38) Germaniumȱhasȱlimitedȱuseȱinȱmodernȱelectronicsȱdueȱto A) shortagesȱofȱrawȱmaterials. B) higherȱforwardȱvoltageȱdropȱwhenȱcomparedȱtoȱSi. C) highȱtemperatureȱinstability. D) filamentȱwarm-upȱtime. Diff:ȱ2

39) Aȱdiodeȱisȱoperatedȱinȱreverseȱbias.ȱAsȱtheȱreverseȱvoltageȱisȱdecreased,ȱtheȱdepletionȱregion A) narrows.

B) hasȱaȱconstantȱwidth.

C) widens.

D) isȱnotȱrelatedȱtoȱreverseȱvoltage.

Diff:ȱ2

281

Chapterȱ2 DiodeȱApplications 1) Aȱdiodeȱconductsȱcurrentȱwhenȱforward-biasedȱandȱblocksȱcurrentȱwhenȱreverse-biased. Diff:ȱ2

2) Theȱlargerȱtheȱrippleȱvoltage,ȱtheȱbetterȱtheȱfilter. Diff:ȱ2

3) Clampingȱcircuitsȱuseȱcapacitorsȱandȱdiodesȱtoȱaddȱaȱdcȱlevelȱtoȱaȱwaveform. Diff:ȱ2

4) Oneȱofȱtheȱadvantagesȱofȱusingȱtransformerȱcouplingȱinȱaȱhalf-waveȱrectifierȱisȱthatȱitȱallows theȱacȱsourceȱtoȱbeȱdirectlyȱconnectedȱtoȱtheȱload. Diff:ȱ2

5) TheȱPIVȱratingȱofȱaȱdiodeȱinȱaȱfull-waveȱbridgeȱrectifierȱisȱmoreȱthanȱthatȱrequiredȱforȱa full-waveȱcenter-tappedȱconfiguration. Diff:ȱ2

6) Theȱdiodeȱinȱaȱhalf-waveȱrectifierȱconductsȱforȱ________ȱofȱtheȱinputȱcycle. A) 0°

B) 45°

C) 90°

D) 180°

Diff:ȱ2

7) Aȱfull-waveȱbridgeȱrectifierȱusesȱ________ȱdiode(s)ȱinȱaȱbridgeȱcircuit. A) 1

B) 2

C) 3

D) 4

Diff:ȱ2

8) Aȱsiliconȱdiodeȱisȱconnectedȱinȱseriesȱwithȱaȱ10ȱk̛ resistorȱandȱaȱ12ȱVȱbattery.ȱȱIfȱtheȱcathodeȱof theȱdiodeȱisȱconnectedȱtoȱtheȱpositiveȱterminalȱofȱtheȱbattery,ȱtheȱvoltageȱfromȱtheȱanodeȱtoȱthe negativeȱterminalȱofȱtheȱbatteryȱis A) 0ȱV.

B) 0.7ȱV.

C) 11.3ȱV.

Diff:ȱ3

282

D) 12ȱV.

9) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱvoltmeterȱacrossȱtheȱtransformerȱsecondaryȱreadsȱ0ȱV,ȱthe probableȱtroubleȱisȱthat A) oneȱofȱtheȱdiodesȱisȱopen. B) theȱfilterȱcapacitorȱisȱopen. C) theȱtransformerȱsecondaryȱisȱopen. D) theȱinductorȱisȱopen. E) Noȱtroubleȱexists;ȱeverythingȱisȱnormal. Diff:ȱ3

10) Referȱtoȱtheȱfigureȱabove.ȱȱInȱservicingȱthisȱpowerȱsupply,ȱyouȱnoticeȱthatȱtheȱrippleȱvoltageȱis higherȱthanȱnormalȱandȱthatȱtheȱrippleȱfrequencyȱhasȱchangedȱtoȱ60ȱHz.ȱȱTheȱprobableȱtrouble isȱthat A) theȱfilterȱcapacitorȱhasȱopened.

B) theȱinductorȱhasȱopened.

C) aȱdiodeȱhasȱshorted.

D) aȱdiodeȱhasȱopened.

Diff:ȱ3

11) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱoscilloscopeȱtraceȱindicatesȱtheȱoutputȱfrom A) aȱhalf-waveȱfilteredȱrectifier. B) aȱfull-waveȱrectifierȱwithȱnoȱfilterȱandȱanȱopenȱdiode. C) aȱfull-waveȱfilteredȱrectifier. D) aȱfull-waveȱfilteredȱrectifierȱwithȱanȱopenȱdiode. Diff:ȱ3

283

12) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱtraceȱonȱthisȱoscilloscopeȱindicatesȱtheȱoutputȱfrom A) aȱhalf-waveȱrectifierȱwithȱnoȱfilter. B) aȱfull-waveȱrectifierȱwithȱnoȱfilter. C) aȱfull-waveȱfilteredȱrectifier. D) aȱfull-waveȱfilteredȱrectifierȱwithȱanȱopenȱdiode. Diff:ȱ3

13) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱisȱtheȱoutputȱfrom A) aȱhalf-waveȱrectifierȱwithȱnoȱfilter. B) aȱfull-waveȱrectifierȱwithȱnoȱfilterȱandȱanȱopenȱdiode. C) aȱfull-waveȱfilteredȱrectifier. D) aȱfull-waveȱfilteredȱrectifierȱwithȱanȱopenȱdiode. Diff:ȱ3

14) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱtraceȱshowsȱtheȱoutputȱfrom A) aȱhalf-waveȱrectifierȱwithȱnoȱfilter. B) aȱfull-waveȱrectifierȱwithȱnoȱfilterȱandȱanȱopenȱdiode. C) aȱhalf-waveȱrectifierȱwithȱanȱopenȱdiode. D) aȱfull-waveȱfilteredȱrectifierȱwithȱanȱopenȱdiode. Diff:ȱ2

15) Referȱtoȱtheȱfigureȱabove.ȱȱTheseȱcircuitsȱareȱknownȱas A) amplifiers.

B) clippers.

C) clampers.

Diff:ȱ2

284

D) rectifiers.

FigureȱI

FigureȱII 16) WhichȱofȱtheȱcircuitsȱinȱFigureȱIȱwillȱproduceȱtheȱsignalȱinȱFigureȱIIȱ(a)? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

17) WhichȱofȱtheȱcircuitsȱinȱFigureȱIȱwillȱproduceȱtheȱsignalȱinȱFigureȱIIȱ(b)? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

18) WhichȱofȱtheȱcircuitsȱinȱFigureȱIȱwillȱproduceȱtheȱsignalȱinȱFigureȱIIȱ(c)? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

19) WhichȱofȱtheȱcircuitsȱinȱFigureȱIȱwillȱproduceȱtheȱsignalȱinȱFigureȱIIȱ(d)? A) (a)

B) (b)

C) (c)

Diff:ȱ2

285

D) (d)

20) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱThisȱrectifierȱarrangement A) willȱproduceȱaȱpositiveȱoutputȱvoltage.

B) willȱproduceȱaȱnegativeȱoutputȱvoltage.

C) isȱincorrectlyȱconnected.

D) AȱorȱCȱabove.

Diff:ȱ2

21) Referȱtoȱ(d)ȱinȱtheȱfigureȱabove.ȱȱThisȱrectifierȱarrangement A) willȱproduceȱaȱpositiveȱoutputȱvoltage.

B) willȱproduceȱaȱnegativeȱoutputȱvoltage.

C) isȱincorrectlyȱconnected.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

22) Aȱsiliconȱdiodeȱhasȱaȱvoltageȱtoȱgroundȱofȱ117ȱVȱfromȱtheȱanode.ȱȱTheȱvoltageȱtoȱgroundȱfrom theȱcathodeȱisȱ117.7ȱV.ȱȱTheȱdiodeȱis A) conducting.

B) shorted.

C) forward-biased.

D) reverse-biased.

Diff:ȱ2

286

23) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱprobableȱtrouble,ȱifȱany,ȱindicatedȱbyȱtheseȱvoltagesȱis A) oneȱofȱtheȱdiodesȱisȱopen. B) anȱopenȱtransformerȱprimary. C) anȱopenȱtransformerȱsecondary. D) theȱfilterȱcapacitorȱisȱopen. E) theȱinductorȱisȱopen. Diff:ȱ3

24) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱvoltmeterȱacrossȱtheȱtransformerȱsecondaryȱreadsȱ0ȱV,ȱthe probableȱtrouble,ȱifȱany,ȱwouldȱbe A) oneȱofȱtheȱdiodesȱisȱopen.

B) anȱopenȱtransformerȱprimary.

C) theȱinductorȱisȱopen.

D) theȱinductorȱisȱshorted.

Diff:ȱ3

25) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱoscilloscopeȱtraceȱindicatesȱtheȱoutputȱfromȱaȱfiltered full-waveȱrectifierȱwithȱanȱopenȱdiode? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ3

26) Theȱrippleȱfrequencyȱofȱaȱbridgeȱrectifierȱis A) theȱsameȱasȱtheȱinputȱfrequency.

B) doubleȱtheȱinputȱfrequency.

C) fourȱtimesȱtheȱinputȱfrequency.

D) one-halfȱtheȱinputȱfrequency.

Diff:ȱ2

287

27) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱcorrectȱdiodeȱarrangementȱtoȱsupplyȱaȱpositiveȱoutputȱvoltage is A) (a).

B) (b).

C) (c).

D) (d).

Diff:ȱ2

28) Withȱaȱhalf-waveȱrectifiedȱvoltageȱacrossȱaȱloadȱresistor,ȱloadȱcurrentȱexistsȱforȱwhatȱpartȱofȱa cycle? A) 0ȱdegrees

B) 90ȱdegrees

C) 180ȱdegrees

D) 360ȱdegrees

Diff:ȱ2

29) Whenȱaȱ60ȱHzȱsinusoidalȱsignalȱvoltageȱisȱappliedȱtoȱtheȱinputȱofȱaȱhalf -waveȱrectifier,ȱthe outputȱfrequencyȱis A) 120ȱHz.

B) 60ȱHz.

C) 30ȱHz.

D) 90ȱHz.

Diff:ȱ1

30) Theȱaverageȱvalueȱofȱtheȱhalf-waveȱrectifiedȱoutputȱvoltageȱisȱapproximatelyȱ________ȱofȱVp. A) 31.8%

B) 63.6%

C) 70.7%

D) 100%

Diff:ȱ2

31) Usingȱaȱpracticalȱforward-biasedȱdiode,ȱifȱtheȱvoltageȱatȱtheȱanodeȱwereȱ10ȱV,ȱtheȱvoltageȱat theȱcathodeȱwouldȱequal A) 10ȱV.

B) 9.3ȱV.

C) 10.7ȱV.

D) 10.3ȱV.

Diff:ȱ3

32) Ifȱinputȱfrequencyȱisȱ60ȱHz,ȱtheȱoutputȱfrequencyȱofȱaȱbridgeȱrectifierȱis A) 30ȱHz.

B) 60ȱHz.

C) 120ȱHz.

Diff:ȱ1

288

D) 240ȱHz.

33) Toȱreduceȱsurgeȱcurrent,ȱ________ȱshouldȱbeȱaddedȱtoȱaȱpowerȱsupplyȱcircuit. A) additionalȱfilterȱcapacitance

B) aȱlargerȱfuse

C) aȱvaractorȱtuningȱcircuit

D) aȱsurge-limitingȱresistor

Diff:ȱ3

34) Theȱdcȱcurrentȱthroughȱeachȱdiodeȱinȱaȱbridgeȱrectifierȱequals A) twiceȱtheȱdcȱloadȱcurrent.

B) halfȱtheȱdcȱloadȱcurrent.

C) theȱloadȱcurrent.

D) one-fourthȱtheȱdcȱloadȱcurrent.

Diff:ȱ3

35) Theȱpeakȱinverseȱvoltageȱacrossȱaȱnonconductingȱdiodeȱinȱanȱunfilteredȱbridgeȱrectifierȱequals approximately A) halfȱtheȱpeakȱsecondaryȱvoltage. B) twiceȱtheȱpeakȱsecondaryȱvoltage. C) theȱpeakȱvalueȱofȱtheȱsecondaryȱvoltage. D) fourȱtimesȱtheȱpeakȱvalueȱofȱtheȱsecondaryȱvoltage. Diff:ȱ3

36) Theȱidealȱdcȱoutputȱvoltageȱofȱaȱcapacitor-inputȱfilterȱequalsȱthe A) rmsȱvalueȱofȱtheȱrectifiedȱvoltage. B) peakȱvalueȱofȱtheȱrectifiedȱvoltage. C) averageȱvalueȱofȱtheȱrectifiedȱvoltage. D) peak-to-peakȱvalueȱofȱtheȱsecondaryȱvoltage. Diff:ȱ3

37) Aȱfilteredȱfull-waveȱrectifierȱvoltageȱhasȱaȱsmallerȱrippleȱthanȱdoesȱaȱhalf-waveȱrectifier voltageȱforȱtheȱsameȱloadȱresistanceȱandȱcapacitorȱvaluesȱbecause A) ofȱtheȱshorterȱtimeȱbetweenȱpeaks. B) ofȱtheȱlongerȱtimeȱbetweenȱpeaks. C) theȱlargerȱtheȱripple,ȱtheȱbetterȱtheȱfilteringȱaction. D) Noneȱofȱtheȱabove. Diff:ȱ3

38) Asȱtheȱloadȱresistanceȱinȱaȱfilteredȱpowerȱsupplyȱvaries,ȱtheȱoutputȱvoltage A) remainsȱconstant.

B) isȱunaffected.

C) varies.

D) doesȱnotȱchange.

Diff:ȱ1

289

39) Theȱvoltageȱregulationȱstageȱinȱaȱpowerȱsupply A) isȱlocatedȱprecedingȱtheȱtransformerȇsȱprimary. B) followsȱtheȱfilterȱstage. C) isȱconnectedȱtoȱtheȱinputȱofȱtheȱrectifier(s). D) isȱinsideȱtheȱtransformer. Diff:ȱ2

40) Aȱvoltageȱregulatorȱcompensatesȱforȱchangesȱin A) theȱinputȱvoltage.

B) temperature.

C) theȱloadȱconditions.

D) Allȱofȱtheȱabove.

Diff:ȱ2

41) Anotherȱnameȱforȱaȱdiodeȱlimiterȱis A) bridger.

B) clipper.

C) clamper.

D) dcȱrestorer.

Diff:ȱ1

42) Aȱdiodeȱclamperȱwill A) clipȱoffȱaȱportionȱofȱtheȱinputȱsignal. B) eliminateȱtheȱpositiveȱorȱnegativeȱalternationȱofȱaȱsignal. C) addȱanȱacȱvoltageȱtoȱaȱsignal. D) addȱaȱdcȱvoltageȱtoȱaȱsignal. Diff:ȱ3

43) Voltageȱmultipliersȱuseȱ________ȱactionȱtoȱincreaseȱpeakȱrectifiedȱvoltagesȱwithoutȱincreasing theȱinputȱtransformerȱvoltageȱrating. A) clipping

B) clamping

C) charging

D) cropping

Diff:ȱ3

44) Allȱofȱtheȱfollowingȱdiodeȱinformationȱisȱprovidedȱbyȱaȱmanufacturerȇsȱdataȱsheetȱexcept A) frequencyȱresponse.

B) PIVȱratings.

C) mechanicalȱdata.

D) temperatureȱparameters.

Diff:ȱ4

290

Chapterȱ3 Special-PurposeȱDiodes 1) Theȱregulatingȱabilityȱofȱaȱzenerȱdiodeȱdependsȱuponȱitsȱabilityȱtoȱoperateȱinȱaȱbreakdown condition. Diff:ȱ2

2) Darkȱcurrentȱisȱtheȱamountȱofȱthermallyȱgeneratedȱforwardȱcurrentȱinȱaȱphotodiodeȱinȱthe absenceȱofȱlight. Diff:ȱ2

3) Aȱ________ȱdiodeȱmaintainsȱaȱconstantȱvoltageȱacrossȱitȱwhenȱoperatingȱinȱtheȱbreakdown condition. A) silicon

B) germanium

C) zener

D) Noneȱofȱtheȱabove.

Diff:ȱ2

4) Aȱtunnelȱdiodeȱhasȱ________ȱcharacteristic(s). A) anȱextremelyȱwideȱdepletionȱregion

B) aȱnegativeȱresistance

C) aȱlinear

D) extremelyȱlightȱdoping

Diff:ȱ2

5) Typically,ȱtheȱmaximumȱV FȱforȱanȱLEDȱisȱbetween A) 0ȱVȱandȱ0.7ȱV.

B) 1ȱVȱandȱ1.4ȱV.

C) 1.2ȱVȱandȱ3.2ȱV.

D) 3.2ȱVȱandȱ10ȱV.

Diff:ȱ2

6) AnȱOLEDȱproducesȱlightȱbyȱaȱprocessȱcalledȱ________. A) Electrophosphorescence.

B) Organicȱlightȱemission.

C) Avalancheȱlightȱemission.

D) Photonicȱlightȱemission.

Diff:ȱ2

7) Whenȱaȱzenerȱdiodeȱisȱforward-biased,ȱit A) operatesȱlikeȱaȱrectifierȱdiode.

B) exhibitsȱaȱvoltageȱdropȱofȱVZȱ+ȱ0.7ȱV.

C) conductsȱnoȱcurrent.

D) emitsȱvisibleȱlight.

Diff:ȱ2

291

8) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱV in increases,ȱV R will A) increase.

B) decrease.

C) remainȱtheȱsame.

D) Unableȱtoȱdetermine.

Diff:ȱ3

9) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱzenerȱshorts A) VinȱwillȱbeȱequalȱtoȱVR.

B) VRL willȱbeȱequalȱtoȱzero.

C) thereȱisȱnoȱcurrentȱthroughȱRL.

D) Allȱofȱtheȱabove.

Diff:ȱ3

10) Referȱtoȱtheȱfigureȱabove.ȱȱMeasurementsȱshowȱthatȱV RL hasȱincreased.ȱȱWhichȱofȱthe followingȱfaults,ȱifȱany,ȱcouldȱhaveȱcausedȱthisȱproblem? A) Rȱopens.

B) Theȱzenerȱshorts.

C) Vinȱhasȱdecreased.

D) Theȱzenerȱopens.

Diff:ȱ3

11) Referȱtoȱtheȱfigureȱabove.ȱȱAsȱlongȱasȱtheȱzenerȱisȱinȱregulation A) thereȱisȱnoȱcurrentȱthroughȱRL.

B) thereȱisȱnoȱcurrentȱthroughȱtheȱzener.

C) VRL ȱwillȱbeȱequalȱtoȱ0.7ȱV.

D) theȱcurrentȱ(IRL )ȱremainsȱconstant.

Diff:ȱ3

12) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱloadȱcurrentȱincreases,ȱIR willȱ________ȱandȱI Zȱwillȱ________. A) remainȱtheȱsame,ȱincrease

B) decrease,ȱremainȱtheȱsame

C) increase,ȱremainȱtheȱsame

D) remainȱtheȱsame,ȱdecrease

Diff:ȱ3

13) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱV in decreases,ȱIR will A) increase.

B) decrease.

C) remainȱtheȱsame.

D) Unableȱtoȱdetermine.

Diff:ȱ2

292

14) Referȱtoȱtheȱfigureȱabove.ȱIfȱtheȱloadȱcurrentȱincreasesȱinȱtheȱzenerȱregulator A) theȱseriesȱcurrentȱincreases.

B) theȱseriesȱcurrentȱremainsȱtheȱsame.

C) theȱzenerȱcurrentȱincreases.

D) BothȱBȱandȱCȱabove.

Diff:ȱ3

15) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱTheȱsymbolȱisȱfor A) aȱzenerȱdiode. B) anȱLED. C) aȱSchottkyȱdiode. D) aȱphotodiode. E) aȱtunnelȱdiode. Diff:ȱ2

16) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱTheȱsymbolȱisȱfor A) aȱzenerȱdiode. B) anȱLED. C) aȱSchottkyȱdiode. D) aȱphotodiode. E) aȱtunnelȱdiode. Diff:ȱ2

17) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱTheȱsymbolȱisȱfor A) aȱzenerȱdiode. B) anȱLED. C) aȱSchottkyȱdiode. D) aȱphotodiode. E) aȱtunnelȱdiode. Diff:ȱ2

293

18) Referȱtoȱ(d)ȱinȱtheȱfigureȱabove.ȱȱTheȱsymbolȱisȱfor A) aȱzenerȱdiode. B) anȱLED. C) aȱSchottkyȱdiode. D) aȱphotodiode. E) aȱtunnelȱdiode. Diff:ȱ2

19) Referȱtoȱ(e)ȱinȱtheȱfigureȱabove.ȱȱTheȱsymbolȱisȱfor A) aȱzenerȱdiode. B) anȱLED. C) aȱSchottkyȱdiode. D) aȱphotodiode. E) aȱtunnelȱdiode. Diff:ȱ2

20) Aȱvaractorȱisȱaȱdiodeȱthat A) variesȱitsȱresistanceȱwithȱtemperature. B) changesȱitsȱcapacitanceȱwithȱvoltage. C) emitsȱlightȱwhenȱforward-biased. D) switchesȱveryȱfast. E) exhibitsȱanȱincreaseȱinȱreverseȱcurrentȱwithȱlightȱintensity. Diff:ȱ2

21) AȱSchottkyȱdiodeȱisȱaȱdiodeȱthat A) variesȱitsȱresistanceȱwithȱtemperature. B) changesȱitsȱcapacitanceȱwithȱvoltage. C) emitsȱlightȱwhenȱforward-biased. D) switchesȱveryȱfast. E) exhibitsȱanȱincreaseȱinȱreverseȱcurrentȱwithȱlightȱintensity. Diff:ȱ2

294

22) Aȱphotodiodeȱisȱaȱdiodeȱthat A) variesȱitsȱresistanceȱwithȱtemperature. B) changesȱitsȱcapacitanceȱwithȱlightȱintensity. C) emitsȱlightȱwhenȱforward-biased. D) switchesȱveryȱfast. E) exhibitsȱanȱincreaseȱinȱreverseȱcurrentȱwithȱlightȱintensity. Diff:ȱ2

23) TheȱcorrectȱwayȱtoȱilluminateȱanȱLEDȱisȱto A) placeȱtheȱLEDȱinȱreverseȱbiasȱdirectlyȱacrossȱaȱ5ȱVȱdcȱsupply. B) placeȱtheȱLEDȱinȱforwardȱbiasȱdirectlyȱacrossȱaȱ5ȱVȱdcȱsupply. C) placeȱtheȱLEDȱinȱseriesȱwithȱaȱresistorȱandȱaȱ5ȱVȱdcȱsupplyȱwithȱtheȱanodeȱtowardȱthe positiveȱsideȱofȱtheȱsupply. D) placeȱtheȱLEDȱinȱseriesȱwithȱaȱresisitorȱandȱaȱ5ȱVȱdcȱsupplyȱwithȱtheȱcathodeȱtowardȱthe positiveȱsideȱofȱtheȱsupply. Diff:ȱ3

24) Aȱdiodeȱwithȱaȱnegative-resistanceȱcharacteristicȱisȱneeded.ȱTheȱcorrectȱselectionȱwouldȱbeȱa A) tunnelȱdiode.

B) Gunnȱdiode.

C) varactorȱdiode.

D) Schottkyȱdiode.

Diff:ȱ2

25) Aȱ6.2ȱVȱzenerȱisȱratedȱatȱ1ȱwatt.ȱȱTheȱmaximumȱsafeȱcurrentȱtheȱzenerȱcanȱcarryȱis approximately A) 1.61ȱA.

B) 161ȱmA.

C) 16.1ȱmA.

D) 1.61ȱmA.

Diff:ȱ2

26) AnȱLEDȱisȱforward-biased.ȱȱTheȱdiodeȱshouldȱbeȱon,ȱbutȱnoȱlightȱisȱshowing.ȱȱAȱpossible problemȱmightȱbe A) thatȱtheȱdiodeȱisȱopen. B) thatȱtheȱseriesȱresistorȱisȱtooȱsmall. C) none;ȱtheȱdiodeȱshouldȱbeȱoffȱifȱforward-biased. D) thatȱtheȱpowerȱsupplyȱvoltageȱisȱtooȱhigh. Diff:ȱ3

27) Theȱbestȱtypeȱofȱdiodeȱtoȱuseȱinȱaȱtuningȱcircuitȱis A) anȱLED.

B) aȱSchottkyȱdiode.

C) aȱGunnȱdiode.

D) aȱvaractor.

Diff:ȱ2

295

28) Whatȱisȱtrueȱaboutȱtheȱbreakdownȱvoltageȱinȱaȱzenerȱdiode? A) Itȱdecreasesȱwhenȱcurrentȱincreases.

B) Itȱdestroysȱtheȱdiode.

C) Itȱoccursȱonlyȱinȱtheȱforwardȱregion.

D) Itȱisȱapproximatelyȱconstant.

Diff:ȱ2

29) Twoȱtypesȱofȱreverseȱbreakdownȱinȱaȱzenerȱdiodeȱare A) avalancheȱandȱzener.

B) avalancheȱandȱreverse.

C) avalancheȱandȱforward.

D) chargeȱandȱdischarge.

Diff:ȱ2

30) Forȱzenerȱdiodes,ȱtheȱtemperatureȱcoefficientȱis A) alwaysȱpositive. B) alwaysȱnegative. C) negativeȱforȱbreakdownȱvoltagesȱlessȱthanȱ5ȱVȱandȱpositiveȱforȱbreakdownȱvoltages greaterȱthanȱ6ȱV. D) alwaysȱzero. Diff:ȱ2

31) Dataȱsheetsȱforȱzenerȱdiodesȱusuallyȱspecifyȱtheȱzenerȱvoltageȱatȱaȱparticularȱtestȱcurrent designated A) IS.

B) IZK.

C) IZM.

D) IZT.

Diff:ȱ1

32) Whenȱtheȱsourceȱvoltageȱincreasesȱinȱaȱzenerȱregulator,ȱtheȱcurrentȱthatȱremainsȱapproximately constantȱisȱthe A) seriesȱcurrent.

B) zenerȱcurrent.

C) loadȱcurrent.

Diff:ȱ3

33) Ifȱtheȱloadȱresistanceȱdecreasesȱinȱaȱzenerȱregulator,ȱtheȱzenerȱcurrent A) decreases. B) staysȱtheȱsame. C) increases. D) equalsȱtheȱsourceȱvoltageȱdividedȱbyȱtheȱseriesȱresistance. Diff:ȱ2

296

D) totalȱcurrent.

34) Ifȱtheȱloadȱresistanceȱdecreasesȱinȱaȱzenerȱregulator,ȱtheȱseriesȱcurrent A) decreases. B) staysȱtheȱsame. C) increases. D) equalsȱtheȱsourceȱvoltageȱdividedȱbyȱtheȱseriesȱresistance. Diff:ȱ2

35) Theȱvaractorȱisȱusually A) forward-biased.

B) reverse-biased.

C) unbiased.

D) operatedȱinȱtheȱbreakdownȱregion.

Diff:ȱ2

36) Theȱcapacitanceȱofȱaȱvaractorȱdiode A) remainsȱconstantȱasȱtheȱbiasȱvoltageȱvaries. B) decreasesȱasȱtheȱreverseȱbiasȱvoltageȱincreases. C) increasesȱasȱtheȱreverseȱbiasȱvoltageȱincreases. D) isȱusuallyȱ1000ȱΐFȱorȱmore. Diff:ȱ3

37) Aȱphotodiodeȱisȱnormally A) reverse-biased.

B) forward-biased.

C) notȱbiased.

D) usedȱtoȱregulateȱvoltage.

Diff:ȱ2

38) Whenȱtheȱlightȱincreases,ȱtheȱreverseȱminorityȱcarrierȱcurrentȱinȱaȱphotodiode A) decreases.

B) increases.

C) isȱunaffected.

D) reversesȱdirection.

Diff:ȱ1

39) LEDȱcolorȱisȱdeterminedȱby A) theȱappliedȱLEDȱvoltage. B) theȱimpuritiesȱaddedȱduringȱdopingȱtoȱcontrolȱwavelength. C) theȱLEDȱcurrent. D) theȱmaterialȱusedȱforȱtheȱincandescentȱfilament. Diff:ȱ2

297

40) Toȱdisplayȱtheȱdigitȱ8ȱinȱaȱseven-segmentȱindicator A) onlyȱCȱmustȱbeȱlighted.

B) Gȱmustȱbeȱoff.

C) onlyȱFȱmustȱbeȱon.

D) allȱsegmentsȱmustȱbeȱon.

Diff:ȱ2

41) Modernȱtrafficȱsignalsȱandȱlargeȱvideoȱscreensȱareȱimplementedȱwith A) PINȱdiodes.

B) infraredȱphotodiodes.

C) highȱintensityȱLEDȱarrays.

D) hotȱfilamentȱbulbs.

Diff:ȱ2

42) AȱSchottkyȱdiode A) hasȱaȱforwardȱvoltageȱdropȱofȱaboutȱ2ȱV. B) hasȱnoȱlimitȱonȱmaximumȱcurrent. C) hasȱaȱmetal-to-semiconductorȱjunction. D) cannotȱoperateȱproperlyȱatȱhighȱfrequencies. Diff:ȱ2

43) TheȱPINȱdiode,ȱwhenȱreverse-biased,ȱactsȱlikeȱaȱnearlyȱconstant A) resistance.

B) capacitance.

C) voltageȱsource.

D) currentȱsource.

Diff:ȱ3

44) Whichȱofȱtheȱfollowingȱhasȱaȱnegative-resistanceȱregion? A) Tunnelȱdiode

B) Rectifierȱdiode

C) Schottkyȱdiode

D) LED

Diff:ȱ2

45) TheȱradiationȱpatternȱforȱanȱLED A) showsȱhowȱdirectionalȱtheȱlightȱis. B) showsȱtheȱcurrentȱvs.ȱvoltageȱcurveȱofȱtheȱLED. C) visuallyȱshowsȱtheȱcolorȱofȱtheȱlight. D) showsȱtheȱwavelengthȱofȱtheȱlight. Diff:ȱ3

46) Allȱofȱtheȱfollowingȱareȱcharacteristicsȱofȱaȱlaserȱdiodeȱexcept A) emitsȱmonochromaticȱlight.

B) emitsȱincoherentȱlight.

C) emitsȱlightȱofȱaȱsingleȱcolor.

D) containsȱaȱpnȱjunction.

Diff:ȱ3

298

Chapterȱ4 BipolarȱJunctionȱTransistorsȱ 1) BJTȱtransistorsȱhaveȱtwoȱpnȱjunctions. Diff:ȱ2

2) AȱBJTȱtransistorȱhasȱtheȱbase-emitterȱjunctionȱreverse-biasedȱforȱproperȱoperation. Diff:ȱ2

3) TheȱratioȱI E/I CȱisȱΆ dc. Diff:ȱ2

4) ProperȱoperationȱofȱaȱBJTȱrequiresȱthatȱtheȱbase-collectorȱjunctionȱshouldȱbeȱreverse-biased. Diff:ȱ2

5) TheȱformulaȱforȱICȱisȱICȱ=ȱIEȱ- IB. Diff:ȱ2

6) Whenȱaȱtransistorȱisȱcheckedȱout-of-circuit,ȱaȱveryȱlowȱresistanceȱreadingȱshouldȱbeȱobtained betweenȱtheȱCȱandȱEȱleadsȱofȱaȱgoodȱtransistor. Diff:ȱ2

7) Phototransistorsȱcomeȱinȱbothȱtwoȱandȱthreeȱterminalȱconfigurations. Diff:ȱ3

8) AȱBJTȱhasȱanȱI B ofȱ75ȱΐAȱandȱaȱΆ dc ofȱ100.ȱȱTheȱvalueȱofȱI C is A) 175ȱΐA.

B) 75ȱmA.

C) 10ȱmA.

D) 7.5ȱmA.

Diff:ȱ2

9) AȱcertainȱtransistorȱhasȱanȱI Cȱ= 12ȱmAȱandȱanȱI B = 125ȱΐA.ȱȱΆ dc is A) 150.

B) 15.

C) 96.

D) 12.

Diff:ȱ2

10) NormalȱoperationȱofȱanȱNPNȱBJTȱrequiresȱtheȱbaseȱtoȱbeȱ________ȱwithȱrespectȱtoȱtheȱemitter, andȱ________ȱwithȱrespectȱtoȱtheȱcollector. A) positive,ȱnegative

B) positive,ȱpositive

C) negative,ȱpositive

D) negative,ȱnegative

Diff:ȱ2

299

11) Aȱtransistorȱamplifierȱhasȱanȱinputȱvoltageȱofȱ67ȱmVȱandȱanȱoutputȱvoltageȱofȱ2..48ȱV.ȱȱThe voltageȱgainȱis A) 67.

B) 37.

C) 27.

D) 17.

Diff:ȱ2

12) Aȱ22ȱmVȱsignalȱisȱappliedȱtoȱtheȱbaseȱofȱaȱproperlyȱbiasedȱtransistorȱthatȱhasȱanȱ rȇeȱ=ȱ7ơ̇̄ and anȱRCȱ=ȱ1.2ȱk̛.ȱȱTheȱoutputȱvoltageȱatȱtheȱcollectorȱis A) 22ȱmV.

B) 17.1ȱV.

C) 7ȱV.

D) 3.77ȱV.

Diff:ȱ2

13) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱcircuitȱisȱoperating A) inȱcutoff. B) inȱsaturation. C) inȱtheȱactiveȱregion. D) incorrectlyȱbecauseȱtheȱbiasȱvoltagesȱareȱwrong. Diff:ȱ3

14) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱvalueȱofȱIB is A) 8.6ȱmA.

B) 860ȱΐA.

C) 1ȱmA.

D) 0.7ȱΐA.

Diff:ȱ2

15) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱvalueȱofȱV BB wereȱincreasedȱtoȱ10ȱV,ȱtheȱtransistorȱwouldȱbe operatingȱin A) cutoff.

B) saturation.

C) theȱactiveȱregion.

D) Cannotȱbeȱdetermined.

Diff:ȱ3

16) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱthisȱtransistorȱisȱoperatingȱinȱsaturation,ȱtheȱvalueȱofȱI C(sat) is A) 9.4ȱmA.

B) 4.26ȱmA.

C) 28.6ȱmA.

Diff:ȱ3

300

D) 42.6ȱmA.

17) Referȱtoȱtheȱfigureȱabove.ȱȱAssumeȱthatȱthisȱcircuitȱisȱoperatingȱinȱcutoff.ȱTheȱmeasurement,ȱif any,ȱthatȱwouldȱconfirmȱthisȱassumptionȱis A) VBEȱ=ȱ0.7ȱV. B) VCEȱ=ȱ8ȱV. C) VCEȱ=ȱ20ȱV. D) VCC ȱ=ȱ20ȱV. E) Noneȱofȱtheȱabove. Diff:ȱ3

18) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱvalueȱofȱIC atȱcutoffȱis A) 0ȱmA.

B) 2.13ȱmA.

C) 10.65ȱΐA.

D) 10ȱmA.

Diff:ȱ2

19) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱvalueȱofȱtheȱcollectorȱresistorȱisȱincreasedȱtoȱ6.8ȱk ̛,ȱtheȱnew valueȱof IC(sat)ȱis A) 2.13ȱmA.

B) 0.68ȱmA.

C) 1.47ȱmA.

Diff:ȱ2

301

D) 0ȱmA.

20) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱcollectorȱresistorȱvalueȱisȱchangedȱtoȱ4.7ȱk̛ȱandȱΆ dcȱ=ȱ200, IC(sat)ȱwould be A) 4.26ȱmA.

B) 8ȱmA.

C) 4.26ȱΐA.

D) 8.426ȱmA.

Diff:ȱ2

21) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱmeasuredȱvoltageȱfromȱtheȱcollectorȱtoȱgroundȱwereȱ0ȱV,ȱthe transistorȱisȱoperatingȱin A) saturation.

B) cutoff.

C) theȱactiveȱregion.

D) Notȱenoughȱdataȱtoȱdetermine.

Diff:ȱ2

22) Referȱtoȱtheȱfigureȱabove.ȱThisȱcircuitȱisȱsaturated.ȱToȱgetȱtheȱcircuitȱtoȱoperateȱcloseȱtoȱits linearȱrange A) RBȱshouldȱbeȱdecreased.

B) RC shouldȱbeȱdecreased.

C) Vinȱshouldȱbeȱincreased.

D) RB shouldȱbeȱincreased.

Diff:ȱ3

23) Aȱ35ȱmVȱsignalȱisȱappliedȱtoȱtheȱbaseȱofȱaȱproperlyȱbiasedȱtransistorȱwithȱanȱrȇeȱ=ȱ8ơ̇̄ȱandȱRC = 1ȱk̛.ȱȱTheȱoutputȱsignalȱvoltageȱatȱtheȱcollectorȱis A) 3.5ȱV.

B) 28.57ȱV.

C) 4.375ȱV.

Diff:ȱ2

302

D) 4.375ȱmV.

24) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱvalueȱofȱVCE is A) 9.9ȱV.

B) 9.2ȱV.

C) 0.7ȱV.

D) 19.3ȱV.

Diff:ȱ2

25) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱminimumȱvalueȱofȱI B thatȱwillȱproduceȱsaturationȱis A) 0.25ȱmA.

B) 5.325ȱΐA.

C) 1.064ȱΐA

Diff:ȱ2

303

D) 10.64ȱΐA

26) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱvoltageȱVCE wasȱmeasuredȱandȱfoundȱtoȱbeȱ20ȱV.ȱȱThe transistorȱis operatingȱin A) saturation.

B) cutoff.

C) theȱactiveȱregion.

D) Notȱenoughȱdataȱtoȱdetermine.

Diff:ȱ2

27) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱV CE isȱmeasuredȱandȱisȱequalȱtoȱnearlyȱzero,ȱtheȱtransistorȱis operatingȱin A) cutoff.

B) theȱactiveȱregion.

C) saturation.

D) Cannotȱbeȱdetermined.

Diff:ȱ3

28) InȱanȱNPNȱtransistor,ȱtheȱmajorityȱcarriersȱinȱtheȱbaseȱare A) freeȱelectrons.

B) holes.

C) neither.

D) both.

Diff:ȱ1

29) TheȱbaseȱofȱanȱNPNȱtransistorȱisȱthinȱand A) heavilyȱdoped.

B) lightlyȱdoped.

C) metallic.

D) dopedȱbyȱaȱpentavalentȱmaterial.

Diff:ȱ2

30) Inȱaȱtransistor,ȱtheȱrelationȱofȱtheȱthreeȱtransistorȱcurrentsȱis A) ICȱ=ȱIEȱ+ȱVC/RC.

B) IC = IB - 2IE.

C) IEȱ=ȱICȱ+ȱIB.

D) IC = IE + IB.

Diff:ȱ2

30 304

31) AȱtransistorȱhasȱaȱΆ DCȱofȱ250ȱandȱaȱbaseȱcurrent,ȱI B,ȱofȱ20ȱΐA.ȱȱTheȱcollectorȱcurrent,ȱIC,ȱequals A) 500ȱΐA.

B) 5ȱmA.

C) 50ȱmA.

D) 5ȱA.

Diff:ȱ2

32) Inȱaȱbipolarȱjunctionȱtransistor,ȱcollectorȱcurrentȱisȱcontrolledȱby A) collectorȱvoltage.

B) baseȱcurrent.

C) collectorȱresistance.

D) Allȱofȱtheȱabove.

Diff:ȱ2

33) MostȱofȱtheȱelectronsȱinȱtheȱbaseȱofȱanȱNPNȱtransistorȱflow A) outȱofȱtheȱbaseȱlead.

B) intoȱtheȱcollector.

C) intoȱtheȱemitter.

D) intoȱtheȱbaseȱsupply.

Diff:ȱ2

34) Whenȱaȱtransistorȱisȱoperatedȱinȱtheȱactiveȱregion,ȱchangesȱinȱtheȱcollectorȱsupplyȱvoltageȱV CC A) produceȱchangesȱinȱcollectorȱcurrent. B) produceȱchangesȱinȱbaseȱvoltage. C) haveȱlittleȱorȱnoȱeffectȱonȱcollectorȱcurrent. D) produceȱchangesȱinȱemitterȱvoltage. Diff:ȱ2

35) Aȱbipolarȱjunctionȱtransistorȱhasȱ________ȱregionsȱofȱoperation. A) 1

B) 2

C) 3

D) 4

Diff:ȱ2

36) Theȱregionȱinȱaȱtransistorȱthatȱhasȱtoȱdissipateȱtheȱmostȱheatȱisȱthe A) emitter.

B) base.

C) collector.

D) anode.

C) hj-fj.

D) Ά ac.

Diff:ȱ2

37) TheȱsymbolȱhFEȱisȱtheȱsameȱas A) Ά DC.

B) ΅DC.

Diff:ȱ2

38) VCEȱapproximatelyȱequalsȱ________ȱwhenȱaȱtransistorȱswitchȱisȱinȱsaturation. A) VC

B) VB

C) 0.2ȱV

Diff:ȱ2

305

D) 0.7ȱV

39) Whenȱaȱtransistorȱswitchȱisȱon,ȱtheȱcollectorȱcurrentȱisȱlimitedȱby A) theȱbaseȱcurrent.

B) theȱcollectorȱresistance.

C) theȱbaseȱvoltage.

D) theȱbaseȱresistance.

Diff:ȱ2

40) Theȱsignalȱoutputȱvoltageȱ(V out)ȱisȱaȱfunctionȱofȱthe A) currentȱfromȱbaseȱtoȱcollector. B) voltageȱdropȱfromȱbaseȱtoȱcollector. C) powerȱbeingȱdissipatedȱbyȱtheȱbaseȱsupplyȱvoltage. D) changingȱcollectorȱcurrentȱ(I C)ȱthroughȱtheȱcollectorȱresistorȱR C. Diff:ȱ2

41) Theȱsignalȱvoltageȱgainȱofȱanȱamplifier,ȱAV,ȱisȱdefinedȱas Vout . A) AVȱ=ȱ Vin B) AVȱ=ȱȱIBȱxȱRB. C) AVȱ=ȱȱȱ D) AVȱ=ȱ

rȇe . RC

Rc . ȱRL

Diff:ȱ2

42) Whenȱtransistorsȱareȱusedȱinȱdigitalȱcircuitsȱtheyȱusuallyȱoperateȱinȱthe A) activeȱregion.

B) breakdownȱregion.

C) saturationȱandȱcutoffȱregions.

D) linearȱregion.

Diff:ȱ1

43) Besidesȱoperatingȱasȱanȱamplifier,ȱtheȱBJTȱisȱoftenȱappliedȱasȱa A) variableȱinductor.

B) switch.

C) voltageȱcontrolledȱcapacitance.

D) varactor.

Diff:ȱ2

306

44) Anȱopenȱbaseȱresistorȱ(RB)ȱinȱaȱtransistorȱswitchȱwillȱresultȱinȱthe A) transistorȱalwaysȱbeingȱON. B) transistorȱalwaysȱbeingȱOFF. C) transistorȱoperatingȱinȱtheȱactiveȱregion. D) transistorȱbeingȱinstantlyȱdestroyed. Diff:ȱ3

45) Aȱtransistorȱcollectorȱcharacteristicȱcurveȱisȱaȱgraphȱshowing A) emitterȱcurrentȱ(I E)ȱversusȱcollector-emitterȱvoltageȱ(VCE)ȱforȱspecifiedȱvaluesȱofȱbase currentȱ(IB). B) collectorȱcurrentȱ(IC)ȱversusȱcollector-emitterȱvoltageȱ(VCE)ȱforȱspecifiedȱvaluesȱofȱbase currentȱ(IB). C) collectorȱcurrentȱ(IC)ȱversusȱcollector-emitterȱvoltageȱ(VC)ȱforȱspecifiedȱvaluesȱofȱbase currentȱ(IB). D) collectorȱcurrentȱ(IC)ȱversusȱcollector-emitterȱvoltageȱ(VCC )ȱforȱspecifiedȱvaluesȱofȱbase currentȱ(IB). Diff:ȱ3

46) Theȱbase-to-emitterȱjunctionȱofȱaȱcertainȱtransistorȱisȱcheckedȱwithȱaȱDMMȱonȱdiodeȱcheckȱin theȱforwardȱbiasȱdirection.ȱIfȱtheȱDMMȱindicatesȱ0.700,ȱtheȱtransistorȱis A) definitelyȱdefective.

B) siliconȱandȱmeasuringȱnormal.

C) germaniumȱandȱmeasuringȱnormal.

D) openȱbetweenȱtheȱbaseȱandȱemitter.

Diff:ȱ2

47) Ifȱtheȱcollectorȱresistanceȱinȱaȱtransistorȱamplifierȱisȱopen,ȱtheȱdcȱvoltageȱatȱtheȱcollectorȱwillȱbe closestȱto A) VBB.

B) VCC .

C) VCC /2.

Diff:ȱ2

307

D) 0ȱV.

Chapterȱ5 TransistorȱBiasȱCircuits 1) BiasingȱaȱBJTȱamplifierȱmeansȱestablishingȱdcȱoperatingȱvoltagesȱforȱproperȱoperation. Diff:ȱ2

2) Aȱtransistorȱoperatingȱinȱsaturationȱhasȱveryȱlittleȱcollectorȱcurrent. Diff:ȱ2

3) Voltage-dividerȱbiasingȱisȱrarelyȱusedȱdueȱtoȱinstability. Diff:ȱ2

4) Negativeȱfeedbackȱinȱtheȱcollector-feedbackȱcircuitȱprovidesȱmoreȱstableȱoperation. Diff:ȱ2

5) TheȱcorrectȱformulaȱforȱfindingȱtheȱdcȱcurrentȱgainȱisȱΆ DC = IC/I B. Diff:ȱ2

6) Aȱcertainȱtransistorȱinȱaȱfixed-biasȱcircuitȱhasȱtheȱfollowingȱvalues:ȱȱI B = 50ȱΐA,ȱΆ DCȱ=ȱ125, VCC ȱ=ȱ18ȱV,ȱandȱR Cȱ=ȱ1.2ȱk̛.ȱȱVCȱis A) 0ȱV.

B) 7.5ȱV.

C) 10.5ȱV.

D) 18ȱV.

Diff:ȱ2

7) Addingȱanȱemitterȱresistorȱtoȱaȱbase-biasȱcircuitȱproducesȱaȱbiasȱcircuitȱcalled A) emitterȱbias.

B) base-emitterȱbias.

C) emitter-feedbackȱbias.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

8) Anȱindicationȱofȱcutoffȱisȱthat A) ICȱ=ȱIC(sat).

B) VCE = 0ȱV.

C) VBE = 0.7ȱV.

Diff:ȱ2

308

D) VCEȱ=ȱVCC.

9) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱtransistorȱisȱbiasedȱforȱ________ȱoperation. A) saturation

B) linear

C) cutoff

D) EitherȱAȱorȱCȱabove.

Diff:ȱ2

10) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱvoltageȱatȱtheȱbaseȱofȱthisȱsiliconȱtransistorȱis A) 0.3ȱV.

B) 0ȱV.

C) 12ȱV.

D) 11.3ȱV.

E) 0.7ȱV.

Diff:ȱ2

11) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱΆ DC = 100,ȱtheȱminimumȱvalueȱofȱI B thatȱwouldȱcauseȱthis transistorȱto saturateȱis A) 100ȱΐA.

B) 50ȱΐA.

C) 1ȱmA.

D) 0.1ȱmA.

Diff:ȱ3

12) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱV C increasesȱtoȱ9ȱV,ȱwhichȱofȱtheȱfollowingȱwouldȱmakeȱtheȱdc collectorȱvoltageȱreturnȱtoȱ6ȱV? A) IncreaseȱtheȱvalueȱofȱR B B) ReplaceȱtheȱtransistorȱwithȱoneȱwithȱaȱlowerȱΆ dc C) DecreaseȱtheȱvalueȱofȱR B D) IncreaseȱVCC Diff:ȱ3

309

13) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱvalueȱofȱR C thatȱwillȱproduceȱaȱvalueȱofȱV Cȱ=ȱ10ȱVȱis A) 2.2ȱk̛.

B) 2ȱk̛.

C) 1ȱk̛.

D) 500ơ̇̄.

Diff:ȱ3

14) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱtransistorȱwereȱreplacedȱwithȱaȱtransistorȱwhoseȱȱ Ά dcȱ= 200, theȱchangeȱthatȱmightȱoccurȱis A) VCȱwouldȱincreaseȱtoȱnearȱ20ȱV.

B) VC wouldȱdecreaseȱtoȱnearȱ0ȱV.

C) IBȱwouldȱincreaseȱsignificantly.

D) VC wouldȱchangeȱaȱsmallȱamount.

Diff:ȱ3

15) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR1 = 10ȱk̛,ȱtheȱdcȱvoltageȱpresentȱatȱtheȱbaseȱofȱthisȱsilicon transistorȱis A) 0.7ȱV.

B) 9.1ȱV.

C) 10ȱV.

Diff:ȱ3

16) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱpurposeȱforȱR 1 andȱR2 isȱto A) establishȱaȱdcȱbaseȱvoltage. B) stabilizeȱtheȱoperatingȱpointȱwithȱnegativeȱfeedback. C) developȱtheȱoutputȱvoltage. D) maintainȱVBEȱatȱ0.7ȱV. Diff:ȱ3

17) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱpurposeȱofȱRC isȱto A) establishȱaȱdcȱbaseȱvoltage. B) stabilizeȱtheȱoperatingȱpointȱwithȱnegativeȱfeedback. C) developȱtheȱoutputȱvoltage. D) maintainȱVBEȱatȱ0.7ȱV. Diff:ȱ2

310

D) 20ȱV.

18) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱpurposeȱofȱRE isȱto A) establishȱaȱdcȱbaseȱvoltage. B) stabilizeȱtheȱoperatingȱpointȱwithȱnegativeȱfeedback. C) developȱtheȱoutputȱvoltage. D) maintainȱVBEȱatȱ0.7ȱV. Diff:ȱ2

19) Twoȱimportantȱyetȱeasilyȱmeasuredȱquantitiesȱthatȱcanȱhelpȱdetermineȱifȱaȱtransistorȱamplifier isȱoperatingȱcorrectlyȱare A) Ά dcȱandȱIB.

B) IC andȱVC.

C) VC andȱVBE.

D) VBEȱandȱIE.

Diff:ȱ2

20) Saturationȱandȱcutoffȱareȱoperatingȱconditionsȱthatȱareȱveryȱusefulȱwhenȱoperatingȱthe transistorȱas A) aȱlinearȱamplifier.

B) aȱswitch.

C) aȱcurrentȱamplifier.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

21) Forȱlinearȱoperation,ȱitȱisȱusualȱtoȱsetȱtheȱQ-pointȱsoȱthat A) VCEȱ=ȱV CC.

B) VCE = VE.

C) VCE = V CC /4.

Diff:ȱ2

311

D) VCEȱ=ȱVCC /2.

22) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱTheȱmostȱprobableȱcauseȱofȱtrouble,ȱifȱany,ȱfromȱtheseȱvoltage measurementsȱwouldȱbe A) theȱbase-emitterȱjunctionȱisȱopen.

B) RE isȱopen.

C) aȱshortȱfromȱcollectorȱtoȱemitter.

D) Thereȱareȱnoȱproblems.

Diff:ȱ3

23) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱTheȱmostȱprobableȱcauseȱofȱtrouble,ȱifȱany,ȱfromȱtheseȱvoltage measurementsȱis A) theȱbase-emitterȱjunctionȱisȱopen.

B) RE isȱopen.

C) aȱshortȱfromȱcollectorȱtoȱemitter.

D) Thereȱareȱnoȱproblems.

Diff:ȱ3

24) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱTheȱmostȱprobableȱcauseȱofȱtrouble,ȱifȱany,ȱfromȱtheseȱvoltage measurementsȱis A) theȱbase-emitterȱjunctionȱisȱopen.

B) RE isȱopen.

C) aȱshortȱfromȱcollectorȱtoȱemitter.

D) Thereȱareȱnoȱproblems.

Diff:ȱ3

25) Referȱtoȱ(d)ȱinȱtheȱfigureȱabove.ȱȱTheȱmostȱprobableȱcauseȱofȱtrouble,ȱifȱany,ȱfromȱtheseȱvoltage measurementsȱis A) theȱbase-emitterȱjunctionȱisȱopen.

B) RE isȱopen.

C) aȱshortȱfromȱcollectorȱtoȱemitter.

D) Thereȱareȱnoȱproblems.

Diff:ȱ3

312

26) Theȱmostȱsuitableȱbiasingȱtechniqueȱusedȱisȱthe A) base-bias.

B) emitter-bias.

C) voltage-dividerȱbias.

D) collector-bias.

Diff:ȱ2

27) Improperȱbiasingȱcanȱcauseȱdistortionȱinȱanȱamplifierȇs A) inputȱsignal.

B) outputȱsignal.

C) powerȱdissipation.

D) frequencyȱresponse.

Diff:ȱ2

28) Onȱaȱdcȱloadȱline,ȱtheȱareaȱbetweenȱsaturationȱandȱcutoffȱisȱcalledȱthe A) saturationȱzone.

B) depletionȱregion.

C) linearȱregion.

D) breakdownȱregion.

Diff:ȱ3

29) ThreeȱdifferentȱQ-pointsȱareȱshownȱonȱaȱdcȱloadȱline.ȱTheȱupperȱQ -pointȱrepresentsȱthe A) minimumȱcurrentȱgain.

B) intermediateȱcurrentȱgain.

C) maximumȱcurrentȱgain.

D) cutoffȱpoint.

Diff:ȱ2

30) Ifȱaȱtransistorȱoperatesȱatȱtheȱmiddleȱofȱtheȱdcȱloadȱline,ȱaȱdecreaseȱinȱtheȱcurrentȱgainȱwill moveȱtheȱQ-point A) offȱtheȱloadȱline.

B) nowhere.

C) up.

D) down.

Diff:ȱ2

31) Theȱinputȱresistance,ȱR IN,ȱofȱaȱvoltage-dividerȱbiasedȱNPNȱtransistorȱisȱ________ȱbyȱaȱfactor ofȱBeta. A) stepped-up

B) stepped-down

C) notȱaffected

D) Noneȱofȱtheȱabove.

Diff:ȱ3

32) Voltage-dividerȱbiasȱprovides A) anȱunstableȱQ-point. B) aȱstableȱQ-point. C) aȱQ-pointȱthatȱeasilyȱvariesȱwithȱchangesȱinȱtheȱtransistorȇsȱcurrentȱgain. D) BothȱAȱandȱCȱabove. Diff:ȱ1

313

33) Forȱtransistorsȱusingȱvoltage-dividerȱbias,ȱtheȱbaseȱcurrentȱshouldȱbe A) muchȱlargerȱthanȱtheȱcurrentȱthroughȱtheȱvoltageȱdivider. B) aboutȱone-halfȱtheȱcollectorȱcurrent. C) muchȱsmallerȱthanȱtheȱcurrentȱthroughȱtheȱvoltageȱdivider. D) Betaȱtimesȱlargerȱthanȱtheȱcollectorȱcurrent. Diff:ȱ3

34) Baseȱbiasȱprovides A) aȱveryȱstableȱQ-point. B) aȱveryȱunstableȱQ-point. C) noȱcurrentȱgain. D) zeroȱcurrentȱinȱtheȱbaseȱandȱcollectorȱcircuits. Diff:ȱ2

35) Aȱcircuitȱwithȱaȱfixedȱemitterȱcurrentȱisȱcalled A) base-bias.

B) emitter-bias.

C) grid-bias.

D) one-supplyȱbias.

Diff:ȱ2

36) Forȱaȱproperlyȱdesignedȱemitter-biasȱcircuit,ȱchangesȱinȱcurrentȱgain A) doȱnotȱaffectȱtheȱQ-point.

B) severelyȱaffectȱtheȱQ-point.

C) doȱnotȱoccurȱinȱtheȱtransistor.

D) affectȱtheȱcollectorȱvoltage.

Diff:ȱ3

37) AȱlinearȱamplifierȱshouldȱhaveȱtheȱQ-pointȱlocated A) closeȱtoȱsaturation. B) closeȱtoȱcutoff. C) inȱtheȱdistortionȱregion. D) approximatelyȱhalf-wayȱbetweenȱsaturationȱandȱcutoff. Diff:ȱ2

38) TheȱQ-pointȱofȱaȱtwoȱsupplyȱemitter-biasȱcircuitȱisȱnotȱaffectedȱby A) VCC.

B) collectorȱresistance.

C) emitterȱresistance.

D) currentȱgain.

Diff:ȱ3

314

39) Theȱemitterȱresistorȱinȱaȱvoltage-dividerȱbiasȱcircuitȱisȱopen.ȱTheȱcollectorȱvoltageȱwillȱequal approximately A) VCC.

B) 0ȱV.

C) one-halfȱVCC.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

40) Ifȱtheȱbase-emitterȱjunctionȱopensȱinȱaȱvoltage-dividerȱbiasedȱcircuit,ȱtheȱemitterȱvoltageȱwill measure A) 0ȱV.

B) 0.7ȱVȱlessȱthanȱtheȱbase.

C) 0.7ȱVȱmoreȱthanȱtheȱbase.

D) aȱvoltageȱnearlyȱequalȱtoȱV CC.

Diff:ȱ2

41) Ifȱtheȱcollectorȱresistorȱdecreasesȱtoȱzeroȱinȱaȱbase-biasedȱcircuit,ȱtheȱloadȱlineȱwillȱbecome A) horizontal.

B) vertical.

C) useless.

D) flat.

Diff:ȱ3

42) Theȱfirstȱstepȱinȱanalyzingȱemitter-biasedȱcircuitsȱisȱtoȱfindȱthe A) baseȱcurrent.

B) emitterȱvoltage.

C) transistorȱpower.

D) collectorȱcurrent.

Diff:ȱ2

43) TheȱmainȱdifferenceȱbetweenȱanȱNPNȱandȱPNPȱtransistorȱamplifierȱisȱthat A) PNPȱtransistorsȱneedȱmoreȱheatȱsinking. B) NPNȱtransistorsȱareȱlinearȱamplifiers,ȱwhileȱPNPȱamplifiersȱareȱnonlinear. C) PNPȱtransistorsȱcannotȱamplifyȱsineȱwaves. D) PNPȱtransistorsȱrequireȱoppositeȱbiasȱpolaritiesȱasȱcomparedȱtoȱNPN. Diff:ȱ2

315

Chapterȱ6 BJTȱAmplifiers 1) Aȱcommon-emitterȱamplifierȱhasȱveryȱhighȱinputȱimpedance,ȱhighȱvoltageȱgain,ȱandȱhigh currentȱgain. Diff:ȱ2

2) AȱhighȱinputȱimpedanceȱamplifierȱcouldȱbeȱimplementedȱwithȱaȱDarlingtonȱpair. Diff:ȱ2

3) Aȱcommon-collectorȱamplifierȱisȱalsoȱknownȱasȱanȱemitterȱfollower. Diff:ȱ2

4) Aȱcommon-baseȱamplifierȱhasȱaȱhighȱcurrentȱgain. Diff:ȱ2

5) TheȱSziklaiȱpairȱusesȱtoȱpnpȱtransistors. Diff:ȱ2

6) Aȱcertainȱtransistorȱhasȱaȱdcȱemitterȱcurrentȱofȱ25ȱmA.ȱȱTheȱvalueȱofȱrȇe is A) 25ơ̇̄.

B) 2.5ơ̇̄.

C) 1.2ơ̇̄.

D) 1ơ̇̄.

C) 5ȱV.

D) 0ȱV.

Diff:ȱ2

7) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱvalueȱofȱVC is A) 20ȱV.

B) 10ȱV.

Diff:ȱ2

8) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱanȱemitter-bypassȱcapacitorȱwereȱadded,ȱtheȱvoltageȱgain A) wouldȱnotȱchange.

B) wouldȱdecrease.

C) wouldȱincrease.

D) wouldȱdecreaseȱtoȱzero.

Diff:ȱ2

316

9) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR 2 opened,ȱVCE wouldȱbe A) 0ȱV.

B) 20ȱV.

C) 10ȱV.

D) 4.8ȱV.

Diff:ȱ3

10) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR 2 opened,ȱtheȱvalueȱofȱIC wouldȱbe A) 6ȱmA.

B) 6.67ȱmA.

C) 8ȱmA.

D) 10ȱmA.

Diff:ȱ3

11) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR C opened,ȱVE would A) increase.

B) decrease.

C) remainȱtheȱsame.

D) beȱundetermined.

Diff:ȱ2

12) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱemitterȱcollectorȱshorted,ȱtheȱvoltageȱV Cȱwouldȱbe A) 0ȱV.

B) 20ȱV.

C) 16.67ȱV.

D) 3.33ȱV.

Diff:ȱ2

13) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱcollectorȱopenedȱinternally,ȱtheȱvoltageȱonȱtheȱcollectorȱwould A) increase.

B) decrease.

C) remainȱtheȱsame.

D) beȱundetermined.

Diff:ȱ3

14) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱV E = 0,ȱtheȱtroubleȱmightȱbeȱthat A) REȱisȱopen.

B) RC isȱopen.

C) R2 isȱopen.

D) R1 ȱisȱopen.

Diff:ȱ2

15) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱanȱemitter-bypassȱcapacitorȱwereȱinstalled,ȱtheȱvalueȱofȱR in wouldȱbe A) 50ơ̇̄.

B) 175ơ̇̄.

C) 378ơ̇̄.

D) 500ơ̇̄.

Diff:ȱ2

16) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱanȱemitter-bypassȱcapacitorȱwereȱinstalled,ȱtheȱnewȱAVȱwouldȱbe A) 4.96.

B) 125.

C) 398.

Diff:ȱ2

317

D) 600.

17) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱAV1 = 75ȱandȱAV2 = 95,ȱAVT wouldȱbe A) 75.

B) 95.

C) 1275.

D) 7125.

Diff:ȱ2

18) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheȱfollowingȱfaultsȱwouldȱaccountȱforȱtheȱoutputȱvoltage beingȱlowerȱthanȱnormal,ȱbutȱnotȱaȱcompleteȱlossȱofȱoutputȱvoltage? A) AnȱopenȱC1

B) AnȱopenȱC3

C) AnȱopenȱC4

D) AnȱopenȱC5

Diff:ȱ3

19) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱV B2 wereȱhigherȱthanȱnormal,ȱtheȱproblem,ȱifȱany,ȱcouldȱbeȱthat A) C3 ȱisȱshorted.

B) R3 isȱopen.

C) C1 isȱopen.

D) C2 ȱisȱopen.

Diff:ȱ3

20) Referȱtoȱtheȱfigureȱabove.ȱȱInȱservicingȱthisȱamplifierȱV out hasȱaȱdcȱcomponentȱinȱadditionȱto theȱnormalȱacȱoutputȱvoltage.ȱThisȱwouldȱbeȱcausedȱbyȱa(n) A) openȱC3.

B) openȱC2.

C) openȱbase-emitterȱofȱC2 .

D) shortedȱC5.

Diff:ȱ3

21) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱoutputȱsignalȱatȱtheȱcollectorȱofȱ Q1 isȱhigherȱthanȱnormal.ȱThis couldȱbeȱcausedȱbyȱa(n) A) openȱC1.

B) openȱC3.

C) openȱbase-emitterȱofȱQ 1.

D) openȱC2.

Diff:ȱ3

22) Theȱbestȱselectionȱforȱaȱhighȱinputȱimpedanceȱamplifierȱisȱa A) lowȱgainȱcommon-emitter.

B) common-base.

C) common-collector.

D) highȱgainȱcommon-emitter.

Diff:ȱ2

318

23) Whichȱofȱtheȱfollowingȱisȱnotȱaȱcharacteristicȱofȱtheȱcommon-baseȱamplifier? A) Highȱinputȱimpedance

B) Currentȱgainȱofȱ1

C) Highȱvoltageȱgain

D) Highȱoutputȱimpedance

Diff:ȱ2

24) Whichȱofȱtheȱfollowingȱisȱnotȱaȱcharacteristicȱofȱtheȱemitter-follower? A) Voltageȱgainȱofȱ1

B) Lowȱinputȱimpedance

C) Lowȱoutputȱimpedance

D) Highȱcurrentȱgain

Diff:ȱ2

25) Theȱbestȱchoiceȱforȱaȱveryȱhighȱpowerȱamplifierȱisȱa(n) A) common-collector.

B) common-base.

C) common-emitter.

D) emitter-follower.

Diff:ȱ2

26) Forȱtransistors A) theȱdcȱandȱacȱcurrentȱgainsȱareȱtheȱsame. B) theȱdcȱcurrentȱgainȱisȱzero. C) theȱdcȱandȱacȱcurrentȱgainsȱareȱusuallyȱdifferent. D) amplificationȱofȱsignalȱvoltageȱisȱnotȱpossible. Diff:ȱ2

27) Theȱacȱresistanceȱofȱtheȱemitterȱdiodeȱrȇe equals A)

25ȱmV . IE

B) 25ȱmVȱ× IC.

C)

Diff:ȱ2

28) Inȱgeneral,ȱcouplingȱcapacitorsȱcanȱbeȱconsidered A) openȱforȱsignalȱvoltageȱandȱaȱshortȱforȱdc. B) shortȱforȱsignalȱvoltageȱandȱanȱopenȱforȱdc. C) lossy. D) shortȱforȱsignalȱvoltageȱandȱaȱshortȱforȱdc. Diff:ȱ2

319

25ȱmV . IB

D)

25ȱmV . IC

29) Theȱprimaryȱreasonȱanȱacȱloadȱlineȱdiffersȱfromȱaȱdcȱloadȱlineȱis A) theȱeffectiveȱacȱcollectorȱresistanceȱisȱgreaterȱthanȱtheȱdcȱcollectorȱresistance. B) theȱeffectiveȱacȱcollectorȱresistanceȱisȱlessȱthanȱtheȱdcȱcollectorȱresistance. C) changesȱinȱcurrentȱareȱnonlinearȱforȱsmall-signalȱamplifierȱoperation. D) theȱacȱloadȱlineȱisȱnotȱasȱsteepȱasȱtheȱdcȱloadȱline. Diff:ȱ3

30) Theȱh-parameter,ȱh fe ,ȱisȱtheȱsameȱasȱ________ȱofȱtheȱtransistor. A) dcȱBeta

B) acȱBeta

C) maximumȱcollectorȱcurrent

D) minimumȱholdȱcurrent

Diff:ȱ2

31) Theȱcapacitorȱthatȱproducesȱanȱacȱgroundȱatȱaȱpointȱinȱaȱcircuitȱisȱcalledȱa(n) A) bypassȱcapacitor.

B) couplingȱcapacitor.

C) dcȱshort.

D) acȱopen.

Diff:ȱ2

32) Reducingȱallȱdcȱsourcesȱtoȱzeroȱisȱdoneȱtoȱhelpȱobtainȱthe A) dcȱequivalentȱcircuit.

B) acȱequivalentȱcircuit.

C) completeȱamplifierȱcircuit.

D) voltage-dividerȱbiasedȱcircuit.

Diff:ȱ2

33) Theȱinputȱresistance,ȱRin(base) ,ȱofȱaȱcommon-emitterȱamplifier,ȱconsistsȱof A) rbȱ?ȱΆre.

C) re ? Άrȇe.

B) Ά ac rȇe.

D) RGȱ?ȱrc ? Άrȇe.

Diff:ȱ3

34) Theȱthreeȱfactorsȱthatȱmustȱbeȱtakenȱintoȱaccountȱwhenȱdeterminingȱtheȱactualȱsignalȱvoltage atȱtheȱbaseȱofȱaȱsmallȱsignalȱbipolarȱamplifierȱare A) sourceȱresistance,ȱemitterȱresistance,ȱandȱinputȱresistance. B) sourceȱresistance,ȱbiasȱresistance,ȱandȱinputȱresistance. C) sourceȱresistance,ȱcollectorȱresistance,ȱinternalȱresistance. D) sourceȱresistance,ȱbiasȱresistance,ȱandȱloadȱresistance. Diff:ȱ2

35) Theȱvalueȱofȱoutputȱresistanceȱinȱaȱcommon-emitterȱamplifier,ȱR out,ȱconsistsȱof A) RC.

B) RL + RC.

C) Ά ? RC.

Diff:ȱ2

320

D) RLȱ?ȱRC.

36) Theȱvoltageȱgainȱofȱanȱamplifierȱisȱdefinedȱasȱthe A) acȱinputȱvoltageȱdividedȱbyȱtheȱacȱoutputȱvoltage. B) acȱcollectorȱcurrentȱdividedȱbyȱtheȱacȱbaseȱcurrent. C) acȱoutputȱvoltageȱdividedȱbyȱtheȱacȱinputȱvoltage. D) acȱcollectorȱcurrentȱdividedȱbyȱtheȱacȱemitterȱcurrent. Diff:ȱ2

37) Removingȱaȱbypassȱcapacitorȱfromȱaȱfullyȱbypassed,ȱcommon -emitterȱamplifierȱcircuitȱwill ________ȱvoltageȱgainȱandȱ________ȱacȱinputȱresistance. A) increase,ȱdecrease

B) decrease,ȱincrease

C) decrease,ȱdecrease

D) increase,ȱincrease

Diff:ȱ3

38) Theȱvoltageȱgainȱofȱaȱcommon-emitterȱamplifier,ȱAV,ȱcanȱbeȱdefinedȱas IeRC Vb . B) AV = IC × RC. C) AVȱ= . A) AVȱ=ȱ Ierȇe Vc

D) AVȱ=ȱ

RC RCȱ+ȱ1

Diff:ȱ3

39) Forȱaȱbypassȱcapacitorȱtoȱworkȱproperly,ȱthe A) XCȱshouldȱbeȱtenȱtimesȱsmallerȱthanȱR E atȱtheȱminimumȱoperatingȱfrequency. B) XCȱshouldȱequalȱRE. C) XCȱshouldȱbeȱtenȱtimesȱgreaterȱthanȱR E atȱtheȱminimumȱoperatingȱfrequency. D) XCȱshouldȱbeȱtwiceȱtheȱvalueȱofȱtheȱRE. Diff:ȱ2

40) Aȱbypassȱcapacitorȱisȱplacedȱacrossȱtheȱemitterȱresistorȱinȱaȱvoltage-dividerȱbiased common-emitterȱamplifierȱcircuit.ȱThisȱwill A) placeȱtheȱemitterȱatȱacȱground.

B) shiftȱtheȱQ-pointȱonȱtheȱdcȱloadȱline.

C) reduceȱtheȱemitterȇsȱdcȱvoltageȱtoȱzero.

D) Allȱofȱtheȱabove.

Diff:ȱ2

41) Removingȱtheȱemitterȱbypassȱcapacitorȱfromȱaȱcommon-emitterȱamplifier A) increasesȱRinȱandȱdecreasesȱvoltageȱgain. B) decreasesȱRinȱandȱincreasesȱvoltageȱgain. C) doesȱnotȱaffectȱR in. D) increasesȱtheȱdistortion. Diff:ȱ3

321

.

42) IncreasingȱtheȱresistanceȱofȱtheȱloadȱresistorȱinȱanȱRCȱcoupledȱcommon -emitterȱamplifierȱwill haveȱwhatȱeffectȱonȱvoltageȱgain? A) Doesȱnotȱaffectȱtheȱvoltageȱgain

B) Decreasesȱtheȱvoltageȱgain

C) Increasesȱtheȱvoltageȱgain

D) Noneȱofȱtheȱabove.

Diff:ȱ3

43) Leavingȱsomeȱofȱtheȱemitterȱresistanceȱunbypassedȱinȱaȱcommon-emitterȱamplifierȱwill A) reduceȱdistortion.

B) stabilizeȱtheȱvoltageȱgain.

C) increaseȱtheȱinputȱimpedance.

D) Allȱofȱtheȱabove.

Diff:ȱ2

44) Inȱaȱswampedȱamplifier,ȱtheȱeffectsȱofȱtheȱemitterȱdiodeȱ(rȇe)ȱbecome A) importantȱtoȱvoltageȱgain.

B) criticalȱtoȱinputȱimpedance.

C) significantȱtoȱtheȱanalysis.

D) insignificant.

Diff:ȱ2

45) Toȱreduceȱtheȱdistortionȱofȱanȱamplifiedȱsignal,ȱyouȱcanȱincreaseȱthe A) collectorȱresistance.

B) emitterȱfeedbackȱresistance.

C) generatorȱresistance.

D) loadȱresistance.

Diff:ȱ3

46) Anȱemitterȱfollowerȱhasȱaȱvoltageȱgainȱthatȱis A) muchȱlessȱthanȱone.

B) approximatelyȱequalȱtoȱone.

C) greaterȱthanȱone.

D) zero.

Diff:ȱ1

47) Whereȱisȱtheȱoutputȱcouplingȱcapacitorȱconnectedȱonȱaȱcommon-collectorȱamplifier? A) Base

B) Emitter

C) Collector

D) dcȱpowerȱsupply

Diff:ȱ2

48) Theȱinputȱresistanceȱofȱtheȱbaseȱofȱanȱemitterȱfollowerȱisȱusually A) low.

B) high.

C) shortedȱtoȱground.

D) open.

Diff:ȱ2

322

49) Oftenȱaȱcommon-collectorȱwillȱbeȱtheȱlastȱstageȱbeforeȱtheȱload;ȱtheȱmainȱfunction(s)ȱofȱthis stageȱisȱto A) provideȱaȱlargeȱvoltageȱgain. B) bufferȱtheȱvoltageȱamplifiersȱfromȱtheȱlowȱresistanceȱloadȱandȱprovideȱimpedance matchingȱforȱmaximumȱpowerȱtransfer. C) provideȱphaseȱinversion. D) provideȱaȱhighȱfrequencyȱpathȱtoȱimproveȱtheȱfrequencyȱresponse. Diff:ȱ2

50) Outputȱresistanceȱinȱaȱcommon-collectorȱamplifierȱcircuitȱisȱsteppedȱdownȱbyȱaȱfactorȱof A) alphaȱ΅.

C) RE ? RL.

B) BetaȱΆ.

D) rȇeȱ+ȱre.

Diff:ȱ3

51) IfȱtwoȱtransistorsȱareȱconnectedȱasȱaȱDarlingtonȱpairȱandȱeachȱtransistorȱhasȱaȱ Betaȱofȱ175,ȱthe overallȱcurrentȱgainȱofȱtheȱpairȱequals A) 30,625.

B) 3,625.

C) 10,000.

D) 5,000.

Diff:ȱ1

52) Aȱthree-stageȱamplifierȱhasȱaȱgainȱofȱ20ȱforȱeachȱstage.ȱTheȱoverallȱdecibelȱvoltageȱgainȱis A) 60ȱdB.

B) 400ȱdB.

C) 78ȱdB.

D) 8,000ȱdB.

Diff:ȱ2

53) Inȱaȱtwo-stageȱamplifier,ȱtheȱinputȱresistanceȱofȱtheȱsecondȱstage A) doesȱnotȱaffectȱtheȱvoltageȱgainȱofȱtheȱfirstȱstage. B) affectsȱtheȱvoltageȱgainȱofȱtheȱfirstȱstage. C) isȱinȱparallelȱwithȱtheȱcollectorȱresistorȱofȱtheȱfirstȱstage. D) BothȱBȱandȱCȱabove. Diff:ȱ2

54) Inȱaȱtwo-stageȱamplifier,ȱtheȱvoltageȱgainȱofȱtheȱfirstȱstageȱisȱ80ȱandȱtheȱvoltageȱgainȱofȱthe secondȱstageȱisȱ50.ȱHowȱmuchȱisȱtheȱoverallȱvoltageȱgain? A) 72

B) 130

C) 4,000

D) 400

Diff:ȱ1

55) IfȱaȱCEȱstageȱisȱdirectȱcoupledȱtoȱanȱemitter-follower,ȱhowȱmanyȱcouplingȱcapacitorsȱareȱthere betweenȱtheȱtwoȱstages? A) 0

B) 1

C) 2

Diff:ȱ2

323

D) 3

56) Theȱquantityȱrƍ e A) isȱexternalȱtoȱtheȱtransistor.

B) isȱtemperatureȱdependent.

C) hasȱnoȱeffectȱonȱgainȱinȱtransistors.

D) isȱaȱdcȱvalue.

Diff:ȱ2

57) Aȱdifferentialȱamplifierȱprovidesȱanȱoutputȱthatȱisȱtheȱ________ȱofȱtheȱtwoȱinputȱquantities. A) square

B) difference

C) multiplication

D) sine

Diff:ȱ1

58) Aȱdifferentialȱamplifierȱshouldȱexhibitȱaȱ________ȱdifferentialȱgainȱandȱaȱ________ȱcommon modeȱgain. A) high,ȱlow

B) low,ȱhigh

C) high,ȱhigh

D) low,ȱlow

Diff:ȱ2

59) Assumeȱthatȱaȱcertainȱdifferentialȱamplifierȱhasȱaȱdifferentialȱgainȱofȱ3,000ȱandȱaȱcommon modeȱgainȱofȱ0.25.ȱWhatȱisȱtheȱCMRR? A) 66.89

B) 750

C) 3,025

D) 12,000

Diff:ȱ2

60) Assumeȱthatȱaȱcertainȱdifferentialȱamplifierȱhasȱaȱdifferentialȱgainȱofȱ5,000ȱandȱaȱcommon modeȱgainȱofȱ0.3.ȱWhatȱisȱtheȱCMRRȱinȱdB? A) 0.3ȱdB

B) 62.12ȱdB

C) 84.44ȱdB

Diff:ȱ2

324

D) 1,500ȱdB

Chapterȱ7 PowerȱAmplifiers 1) TheȱclassȱAȱamplifierȱisȱusuallyȱbiasedȱbelowȱcutoff. Diff:ȱ2

2) Darlingtonȱpairȱtransistorsȱareȱoftenȱusedȱinȱpowerȱamplifiersȱbecauseȱtheȱinputȱimpedanceȱis veryȱlow. Diff:ȱ2

3) Forȱcertainȱapplicationsȱwithȱlow-resistanceȱloads,ȱaȱpush-pullȱamplifierȱusingȱDarlington transistorsȱcanȱbeȱusedȱtoȱdecreaseȱtheȱinputȱresistanceȱpresentedȱtoȱtheȱdrivingȱamplifierȱand avoidȱgreatlyȱreducingȱvoltageȱgain. Diff:ȱ2

4) AȱclassȱBȱamplifierȱconductsȱforȱ________ȱofȱtheȱcycle. A) 45°

B) 90°

C) 180°

D) 360°

Diff:ȱ2

5) Theȱclassȱofȱamplifiersȱthatȱisȱtheȱmostȱefficientȱandȱhasȱtheȱmostȱdistortionȱisȱclassȱ________ amplifiers. A) A

B) B

C) C

D) AB

Diff:ȱ2

6) Push-pullȱamplifiersȱoftenȱuseȱclassȱ________ȱamplifiers. A) A

B) BȱorȱAB

C) C

Diff:ȱ2

7) IfȱaȱclassȱAȱamplifierȱhasȱaȱvoltageȱgainȱofȱ50ȱandȱaȱcurrentȱgainȱofȱ75,ȱtheȱpowerȱgainȱis A) 50.

B) 75.

C) 1500.

D) 3750.

Diff:ȱ2

8) IfȱaȱclassȱAȱamplifierȱhasȱR Cȱ= 4.7ȱk̛ andȱRE = 1.5ȱk̛ andȱVCC = 24ȱV,ȱIC(sat)ȱis A) 5.1ȱmA.

B) 16ȱmA.

C) 3.87ȱmA.

D) 0ȱmA.

Diff:ȱ2

9) Ifȱanȱapplicationȱforȱanȱamplifierȱrequiresȱoperationȱinȱaȱlinearȱmode,ȱtheȱmostȱlikelyȱchoice wouldȱbeȱa A) classȱA.

B) classȱB.

C) classȱC.

Diff:ȱ2

325

D) classȱAB.

10) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR 1 opened,ȱandȱV in atȱtheȱbaseȱwasȱlarge,ȱV outȱatȱtheȱcollector would A) increase.

B) decrease.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ3

11) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱdiodeȱopened,ȱthisȱamplifierȱwouldȱbeȱoperatingȱas A) classȱA.

B) classȱB.

C) classȱC.

D) classȱAB.

Diff:ȱ3

12) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱpurposeȱofȱtheȱdiodeȱisȱtoȱbiasȱtheȱamplifierȱas A) classȱA.

B) classȱB.

C) classȱC.

D) classȱAB.

Diff:ȱ2

13) TheȱmaximumȱefficiencyȱforȱaȱclassȱAȱamplifierȱisȱabout A) 25%.

B) 50%.

C) 75%.

D) 100%.

Diff:ȱ2

14) Theȱamplifierȱwithȱtheȱmostȱdistortionȱwouldȱbeȱaȱ________ȱamplifier. A) classȱA

B) classȱB

C) classȱC

Diff:ȱ2

326

D) classȱAB

15) Referȱtoȱtheȱfigureȱabove.ȱȱWithȱnoȱsignalȱinput,ȱtheȱdcȱemitterȱvoltageȱwithȱrespectȱtoȱground is A) 10.7ȱV.

B) 9.3ȱV.

C) 0ȱV.

D) 10ȱV.

Diff:ȱ2

16) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱamplifierȱonlyȱshowsȱaȱpositiveȱalternationȱatȱtheȱoutput.ȱȱThe possibleȱtroubleȱmightȱbeȱthat A) C3 ȱisȱshorted.

B) BE1 isȱopen.

C) BE2 isȱopen.

D) R1 ȱisȱopen.

Diff:ȱ3

17) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱdcȱvoltageȱacrossȱR L wasȱmeasuredȱatȱ10ȱV.ȱȱAȱpossible problem,ȱifȱany,ȱmightȱbeȱthat A) C1 ȱisȱopen.

B) C3 isȱshorted.

C) R1 isȱopen.

D) C3 ȱisȱopen.

Diff:ȱ3

18) Referȱtoȱtheȱfigureȱabove.ȱȱDuringȱtheȱpositiveȱinputȱalternation,ȱQ 1 isȱ________ȱandȱQ 2 is ________. A) on,ȱon

B) on,ȱoff

C) off,ȱoff

Diff:ȱ2

19) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱpurposeȱforȱtheȱdiodesȱD 1 andȱD 2 is A) toȱapplyȱequalȱsignalsȱtoȱeachȱtransistor. B) toȱallowȱtheȱcorrectȱbiasȱvoltagesȱonȱtheȱtwoȱbases. C) toȱmaintainȱconstantȱbiasȱwithȱtemperatureȱchanges. D) Allȱofȱtheȱabove. Diff:ȱ3

327

D) off,ȱon

20) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱcombinationȱofȱtheȱtwoȱtransistorsȱisȱcalled A) same.

B) complementary.

C) NPN.

D) PNP.

Diff:ȱ2

21) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱcircuitȱisȱoperatingȱasȱa A) classȱAȱpush-pull.

B) classȱABȱpush-pull.

C) classȱCȱpush-pull.

D) classȱBȱpush-pull.

Diff:ȱ2

22) Anȱapplicationȱforȱaȱpowerȱamplifierȱtoȱoperateȱatȱradioȱfrequenciesȱisȱneeded.ȱTheȱmostȱlikely choiceȱwouldȱbeȱaȱ_________ȱamplifier. A) classȱA

B) classȱB

C) classȱC

D) classȱAB

Diff:ȱ2

23) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱapproximateȱvoltagesȱonȱtheȱbase,ȱcollector,ȱandȱemitter, respectively,ȱare A) 0.7ȱV,ȱ6.8ȱV,ȱ0ȱV

B) 0ȱV,ȱ0ȱV,ȱ0ȱV

C) 0.7ȱV,ȱ15ȱV,ȱ0ȱV

D) 0.7ȱV,ȱ0ȱV,ȱ15ȱV

Diff:ȱ2

328

24) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR L showsȱaȱzeroȱsignalȱvoltageȱonȱanȱoscilloscope,ȱtheȱproblem mightȱbe that A) C3 ȱisȱopen.

B) BE1 isȱopen.

C) BE2 isȱopen.

D) R1 ȱisȱopen.

Diff:ȱ3

25) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱthereȱwereȱnoȱoutputȱsignal,ȱandȱtheȱmeasuredȱdcȱvoltageȱofȱ Q1 emitterȱwereȱ0ȱV,ȱtheȱtroubleȱmightȱbeȱthat A) D1 ȱisȱshorted.

B) D2 isȱshorted.

C) R1 ȱisȱopen.

D) Noȱtrouble;ȱeverythingȱisȱnormal.

Diff:ȱ3

26) AȱclassȱABȱamplifierȱisȱbiased A) atȱcutoff. B) slightlyȱaboveȱtheȱcenterȱofȱtheȱloadȱline. C) slightlyȱaboveȱcutoff. D) atȱtheȱcenterȱofȱtheȱloadȱline. Diff:ȱ3

27) Heatȱsinksȱreduceȱthe A) transistorȱpower.

B) ambientȱtemperature.

C) junctionȱtemperature.

D) collectorȱcurrent.

Diff:ȱ1

28) Anȱamplifierȱhasȱtwoȱloadȱlinesȱbecause A) itȱhasȱacȱandȱdcȱcollectorȱresistances.

B) itȱhasȱtwoȱequivalentȱcircuits.

C) theȱdcȱandȱacȱcircuitsȱareȱdifferent.

D) Allȱofȱtheȱabove.

Diff:ȱ3

329

29) WhenȱtheȱQ-pointȱisȱatȱtheȱcenterȱofȱtheȱacȱloadȱline,ȱaȱmaximumȱ________ȱsignalȱcanȱbe obtained. A) classȱA

B) classȱB

C) classȱC

D) Noneȱofȱtheȱabove.

Diff:ȱ3

30) Forȱmaximumȱpeak-to-peakȱoutputȱvoltage,ȱtheȱQ-pointȱshouldȱbe A) nearȱsaturation.

B) nearȱcutoff.

C) atȱtheȱcenterȱofȱtheȱdcȱloadȱline.

D) atȱtheȱcenterȱofȱtheȱacȱloadȱline.

Diff:ȱ3

31) Theȱacȱloadȱlineȱisȱtheȱsameȱasȱtheȱdcȱloadȱlineȱwhenȱtheȱacȱcollectorȱresistanceȱequalsȱthe A) dcȱemitterȱresistance. B) acȱemitterȱresistance. C) dcȱcollectorȱresistance. D) supplyȱvoltageȱdividedȱbyȱcollectorȱcurrent. Diff:ȱ3

32) ForȱaȱQ-pointȱnearȱtheȱcenterȱofȱtheȱdcȱloadȱline,ȱclippingȱisȱmoreȱlikelyȱtoȱoccurȱonȱthe A) positiveȱpeakȱofȱinputȱvoltage.

B) negativeȱpeakȱofȱoutputȱvoltage.

C) positiveȱpeakȱofȱoutputȱvoltage.

D) negativeȱpeakȱofȱemitterȱvoltage.

Diff:ȱ3

33) Theȱacȱloadȱlineȱusually A) equalsȱtheȱdcȱloadȱline.

B) hasȱlessȱslopeȱthanȱtheȱdcȱloadȱline.

C) isȱsteeperȱthanȱtheȱdcȱloadȱline.

D) isȱhorizontal.

Diff:ȱ2

34) ForȱaȱclassȱAȱCEȱamplifier,ȱtheȱpowerȱdissipation,ȱP DQ A) isȱmaximumȱwhenȱthereȱisȱnoȱinputȱsignal. B) increasesȱwhenȱtheȱpeak-to-peakȱloadȱvoltageȱincreases. C) isȱzeroȱwithȱnoȱinputȱsignal. D) isȱmaximumȱwhenȱtheȱtransistorȱisȱdrivenȱtoȱcutoff. Diff:ȱ2

35) AȱCEȱamplifierȱhasȱaȱloadȱpowerȱofȱ10ȱmWȱandȱtheȱdcȱpowerȱisȱ215ȱmW.ȱȱTheȱefficiencyȱis A) 46.5%.

B) 4.65%.

C) 25%.

Diff:ȱ2

330

D) 0%.

36) Toȱimproveȱtheȱefficiencyȱofȱanȱamplifier,ȱyouȱhaveȱto A) reduceȱloadȱpower.

B) decreaseȱunwantedȱpowerȱlosses.

C) reduceȱtheȱsupplyȱvoltage.

D) increaseȱtheȱdcȱcurrent.

Diff:ȱ2

37) Theȱquiescentȱcollectorȱcurrentȱisȱtheȱsameȱas A) acȱcollectorȱcurrent.

B) acȱloadȱresistorȱcurrent.

C) dcȱcollectorȱcurrent.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

38) ForȱeachȱtransistorȱinȱaȱclassȱBȱamplifier,ȱthereȱisȱcollectorȱcurrentȱfor A) 270°ȱofȱtheȱinputȱcycle.

B) 180° ofȱtheȱinputȱcycle.

C) 360°ȱofȱtheȱinputȱcycle.

D) 90° ofȱtheȱinputȱcycle.

Diff:ȱ2

39) InȱaȱclassȱABȱpush-pullȱamplifier,ȱtheȱtransistorsȱareȱbiasedȱslightlyȱaboveȱcutoffȱtoȱavoid A) unusuallyȱhighȱefficiency.

B) negativeȱfeedback.

C) crossoverȱdistortion.

D) aȱlowȱinputȱimpedance.

Diff:ȱ2

40) ForȱaȱclassȱABȱpush-pullȱamplifierȱtoȱworkȱproperly,ȱtheȱemitterȱdiodesȱmust A) matchȱtheȱcompensatingȱdiodes. B) beȱgermaniumȱandȱtheȱcompensatingȱdiodesȱmustȱbeȱsilicon. C) beȱsiliconȱandȱtheȱcompensatingȱdiodesȱmustȱbeȱgermanium. D) notȱmatchȱtheȱcompensatingȱdiodes. Diff:ȱ2

41) ForȱaȱclassȱABȱpush-pullȱamplifier,ȱdiodeȱbiasȱisȱusedȱto A) allowȱtheȱtransistorsȱtoȱconductȱforȱ360°. B) ensureȱthermalȱrunaway. C) avoidȱthermalȱrunaway. D) saturateȱtheȱoutputȱtransistors. Diff:ȱ2

42) TheȱmaximumȱefficiencyȱofȱaȱclassȱBȱpush-pullȱamplifierȱis A) 25ȱpercent.

B) 50ȱpercent.

C) 79ȱpercent.

Diff:ȱ1

331

D) 100ȱpercent.

43) Underȱno-signalȱorȱquiescentȱconditions,ȱtheȱtransistorsȱofȱaȱclassȱBȱpush-pullȱamplifier A) haveȱexcessivelyȱhighȱpowerȱdissipation. B) getȱquiteȱhot. C) areȱinȱsaturation. D) dissipateȱveryȱlittleȱpower. Diff:ȱ2

44) PowerȱMOSFETsȱhaveȱseveralȱadvantagesȱoverȱbipolarȱpowerȱtransistors.ȱWhichȱofȱthe followingȱstatementsȱisȱnotȱcorrect? A) LessȱproneȱtoȱESDȱdamageȱwhenȱcomparedȱtoȱBJTs B) Haveȱsimplerȱbiasingȱcircuits C) Canȱbeȱconnectedȱinȱparallelȱtoȱincreaseȱcurrentȱcapacity D) Haveȱaȱlowȱvoltageȱdropȱacrossȱtheȱdeviceȱunderȱhighȱvoltageȱandȱcurrentȱconditions Diff:ȱ3

45) ForȱaȱclassȱCȱamplifier,ȱthereȱisȱcollectorȱcurrentȱfor A) 0°ȱofȱtheȱinputȱcycle.

B) lessȱthanȱ180° ofȱtheȱinputȱcycle.

C) 210°ȱofȱtheȱinputȱcycle.

D) 360° ofȱtheȱinputȱcycle.

Diff:ȱ2

46) ClassȱCȱamplifiersȱareȱalmostȱalways A) transformer-coupledȱbetweenȱstages.

B) operatedȱatȱaudioȱfrequencies.

C) tunedȱRFȱamplifiers.

D) wideband.

Diff:ȱ2

47) TheȱinputȱsignalȱofȱaȱclassȱCȱamplifier A) isȱnegativelyȱclampedȱatȱtheȱbase. B) isȱamplifiedȱandȱinverted. C) producesȱbriefȱpulsesȱofȱcollectorȱcurrent. D) Allȱofȱtheȱabove. Diff:ȱ2

48) TheȱcollectorȱcurrentȱofȱaȱclassȱCȱamplifier A) isȱanȱamplifiedȱversionȱofȱtheȱinputȱvoltage. B) hasȱharmonics. C) isȱnegativelyȱclamped. D) flowsȱforȱhalfȱaȱcycle. Diff:ȱ2

332

Chapterȱ8 Field-EffectȱTransistors (FETs) 1) AȱFETȱhasȱthreeȱterminalsȱnamedȱtheȱsource,ȱdrain,ȱandȱgate. Diff:ȱ2

2) TheȱJFETȱoperatesȱwithȱaȱforward-biasedȱgate-sourceȱpnȱjunction. Diff:ȱ2

3) AnȱE-MOSFETȱcanȱbeȱusedȱasȱaȱswitch. Diff:ȱ2

4) AȱD-MOSFETȱcanȱoperateȱwithȱbothȱpositiveȱandȱnegativeȱvaluesȱofȱV GS. Diff:ȱ2

5) TheȱslopeȱofȱtheȱcharacteristicȱcurveȱinȱtheȱohmicȱregionȱofȱaȱJFETȱisȱdefinedȱbyȱtheȱequation: VDS . RDSȱ=ȱ ID Diff:ȱ2

6) SpecialȱcareȱisȱrequiredȱinȱhandlingȱaȱMOSFET. Diff:ȱ2

7) JFETȱdataȱsheetsȱspecifyȱinputȱresistanceȱbyȱgivingȱtheȱvaluesȱforȱV GS andȱIDSS. Diff:ȱ2

8) TheȱohmicȱregionȱofȱaȱJFETȱcharacteristicȱcurveȱisȱroughlyȱparabolicȱinȱshape. Diff:ȱ2

9) Underȱnoȱsignalȱconditions,ȱmidpointȱbiasȱallowsȱtheȱmaximumȱamountȱofȱdrainȱcurrent swingȱbetweenȱI DSSȱandȱzero. Diff:ȱ2

10) Aȱself-biasedȱȱnȬchannelȱJFETȱhasȱaȱV D = 8ȱV,ȱV GS = -5ȱV.ȱȱTheȱvalueȱofȱVDSȱis A) 3ȱV.

C) -5ȱV.

B) 8ȱV.

Diff:ȱ2

333

D) -3ȱV.

11) Fieldȱeffectȱtransistorsȱareȱalsoȱknownȱas A) unipolarȱdevices.

B) bipolarȱdevices.

C) three-chargeȱcarrierȱdevices.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

12) TheȱFETȱthatȱhasȱnoȱphysicalȱchannelȱis A) theȱDȱMOSFET.

B) theȱEȱMOSFET.

C) theȱJFET.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

13) AȱFETȱthatȱhasȱnoȱIDSSȱparameterȱisȱthe A) JFET.

B) DEȱMOSFET.

C) VȱMOSFET.

D) EȱMOSFET.

Diff:ȱ2

14) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱThisȱsymbolȱidentifies A) aȱpȬchannelȱEȱMOSFET.

B) anȱnȬchannelȱDȱMOSFET.

C) aȱpȬchannelȱDȱMOSFET.

D) anȱnȬchannelȱEȱMOSFET.

Diff:ȱ2

15) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱThisȱsymbolȱidentifies A) aȱpȬchannelȱEȱMOSFET.

B) anȱnȬchannelȱDȱMOSFET.

C) aȱpȬchannelȱDȱMOSFET.

D) anȱnȬchannelȱEȱMOSFET.

Diff:ȱ2

16) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱThisȱsymbolȱidentifies A) aȱpȬchannelȱEȱMOSFET.

B) anȱnȬchannelȱDȱMOSFET.

C) aȱpȬchannelȱDȱMOSFET.

D) anȱnȬchannelȱEȱMOSFET.

Diff:ȱ2

334

17) Referȱtoȱ(d)ȱinȱtheȱfigureȱabove.ȱȱThisȱsymbolȱidentifies A) aȱpȬchannelȱEȱMOSFET.

B) anȱnȬchannelȱDȱMOSFET.

C) aȱpȬchannelȱDȱMOSFET.

D) anȱnȬchannelȱEȱMOSFET.

Diff:ȱ2

18) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱIfȱI D = 4ȱmA,ȱtheȱvalueȱofȱV DS is A) 12ȱV.

B) 8ȱV.

C) 4ȱV.

D) 0ȱV.

Diff:ȱ2

19) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱIfȱID = 4ȱmA,ȱtheȱvalueȱofȱV GS is A) 20ȱV.

B) 11.2ȱV.

C) 8.8ȱV.

D) 0ȱV.

Diff:ȱ2

20) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱIfȱID = 2ȱmA,ȱtheȱvalueȱofȱV DS is A) 4.4ȱV.

B) 10ȱV.

C) 15.6ȱV.

D) 20ȱV.

Diff:ȱ2

21) AȱJFETȱmanufacturerȇsȱdataȱsheetȱspecifiesȱV GS(off) = -8ȱVȱandȱIDSS = 6ȱmA.ȱWhenȱV GS = -4 V,ȱtheȱvalueȱofȱI Dȱwouldȱbe A) 6ȱmA.

B) 1.25ȱmA.

C) 1.5ȱmA.

Diff:ȱ3

335

D) 4ȱmA.

22) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱvalueȱofȱtheȱvoltageȱdropȱacrossȱR D is A) 20ȱV.

B) 12ȱV.

C) 6ȱV.

D) 3ȱV.

Diff:ȱ2

23) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱamplifierȱisȱbiasedȱfor A) linearȱoperation.

B) pinch-offȱoperation.

C) saturation.

D) operationȱasȱaȱswitch.

Diff:ȱ2

24) Referȱtoȱtheȱfigureȱabove.ȱȱInȱthisȱcircuit,ȱV GS isȱbiasedȱcorrectlyȱforȱproperȱoperation.ȱȱThis meansȱthat VGSȱis A) positive.

B) negative.

C) eitherȱnegativeȱorȱpositive.

D) 0ȱV.

Diff:ȱ3

25) Referȱtoȱtheȱfigureȱabove.ȱȱCalculateȱtheȱvalueȱofȱV D. A) 20ȱV

B) 8ȱV

C) 6ȱV

D) 2ȱV

Diff:ȱ2

26) Referȱtoȱtheȱfigureȱabove.ȱȱCalculateȱtheȱvalueȱofȱV DS. A) 0ȱV

B) 2ȱV

C) 4ȱV

D) -2ȱV

Diff:ȱ2

27) Forȱproperȱoperation,ȱanȱnȬchannelȱE-MOSFETȱshouldȱbeȱbiasedȱsoȱthatȱV GSȱis A) eitherȱpositiveȱorȱnegative.

B) negative.

C) positive.

D) -4ȱV.

Diff:ȱ2

336

28) AȱgoodȱapplicationȱforȱaȱVMOSFETȱwouldȱbeȱasȱa A) powerȱamplifier.

B) lowȱpowerȱamplifier.

C) lowȱinputȱimpedanceȱdevice.

D) substituteȱforȱaȱdiode.

Diff:ȱ2

29) AȱVMOSFETȱdeviceȱoperatesȱin A) theȱdepletionȱmode. B) theȱenhancementȱmode. C) aȱJFETȱmode. D) inȱeitherȱenhancementȱorȱdepletionȱmode. Diff:ȱ2

30) Theȱgate-sourceȱjunctionȱofȱaȱJFETȱis A) normallyȱnotȱbiased. B) normallyȱforward-biased. C) normallyȱreverse-biased. D) aȱlowȱresistanceȱpathȱforȱdcȱcurrentȱwhenȱreverse-biased. Diff:ȱ2

31) TheȱchannelȱwidthȱinȱaȱJFETȱisȱcontrolledȱby A) varyingȱgateȱvoltage. B) varyingȱdrainȱvoltage. C) increasingȱforwardȱbiasȱonȱtheȱgate-sourceȱjunction. D) increasingȱreverseȱbiasȱonȱtheȱdrain-sourceȱjunction. Diff:ȱ2

32) Whenȱoperatedȱinȱtheȱohmicȱarea,ȱaȱJFETȱactsȱlikeȱa(n) A) smallȱresistor.

B) voltageȱsource.

C) currentȱsource.

D) insulator.

Diff:ȱ2

33) VDSȱequalsȱpinch-offȱvoltageȱdividedȱbyȱthe A) baseȱcurrent.

B) gateȱcurrent.

C) idealȱdrainȱcurrent.

D) drainȱcurrentȱforȱzeroȱgateȱvoltage.

Diff:ȱ3

337

34) IDSSȱcanȱbeȱdefinedȱasȱthe A) minimumȱpossibleȱdrainȱcurrent. B) maximumȱpossibleȱcurrentȱwithȱtheȱdrainȱshortedȱtoȱtheȱsource. C) maximumȱcurrentȱdrain-to-sourceȱwithȱaȱshortedȱgate. D) maximumȱdrainȱcurrentȱwithȱtheȱsourceȱshorted. Diff:ȱ2

35) Theȱpinch-offȱvoltageȱhasȱtheȱsameȱmagnitudeȱasȱthe A) gateȱvoltage.

B) drain-sourceȱvoltage.

C) gate-sourceȱvoltage.

D) gate-sourceȱcutoffȱvoltage.

Diff:ȱ3

36) JFETsȱareȱoftenȱcalled A) one-wayȱswitches.

B) two-wayȱswitches.

C) bipolarȱdevices.

D) unipolarȱdevices.

Diff:ȱ2

37) TheȱtransconductanceȱcurveȱofȱaȱJFETȱisȱaȱgraphȱof A) ISȱversusȱV DS.

B) IC versusȱV CE.

C) IDȱversusȱV GS.

D) ID Λ RDS.

Diff:ȱ2

38) ForȱaȱJFET,ȱthereȱisȱmaximumȱdrainȱcurrentȱwhen A) VGSȱequalsȱVGS(off).

B) VDS isȱzero.

C) theȱdrainȱandȱsourceȱareȱinterchanged.

D) VGS isȱzero.

Diff:ȱ2

39) TheȱtransconductanceȱcurveȱofȱaȱJFETȱis A) hyperbolic.

B) linear.

C) nonlinear.

D) symmetrical.

Diff:ȱ2

40) Aȱ________ȱchangeȱinȱV DSȱwillȱproduceȱaȱ________ȱchangeȱinȱI D. A) small,ȱlarge

B) large,ȱsmall

C) large,ȱlarge

Diff:ȱ3

338

D) small,ȱsmall

41) Toȱgetȱaȱnegativeȱgate-sourceȱvoltageȱinȱaȱself-biasedȱJFETȱcircuit,ȱyouȱmustȱuseȱa A) voltageȱdivider.

B) sourceȱresistor.

C) ground.

D) negativeȱgateȱsupplyȱvoltage.

Diff:ȱ3

42) TheȱeasiestȱwayȱtoȱbiasȱaȱJFETȱinȱtheȱohmicȱregionȱisȱwith A) voltage-dividerȱbias.

B) self-bias.

C) gateȱbias.

D) sourceȱbias.

Diff:ȱ3

43) Oneȱadvantageȱofȱvoltage-dividerȱbiasȱisȱthatȱtheȱdependencyȱofȱdrainȱcurrentȱI D,ȱonȱthe rangeȱofȱQ-pointsȱis A) increased.

B) reduced.

C) notȱaffected.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

44) Theȱdepletion-modeȱMOSFETȱcan A) operateȱwithȱonlyȱpositiveȱgateȱvoltages. B) operateȱwithȱonlyȱnegativeȱgateȱvoltages. C) notȱoperateȱinȱtheȱohmicȱregion. D) operateȱwithȱpositiveȱasȱwellȱasȱnegativeȱgateȱvoltages. Diff:ȱ2

45) AnȱnȬchannelȱE-MOSFETȱconductsȱwhenȱitȱhas A) VGSȱ>ȱV P. B) aȱthinȱlayerȱofȱpositiveȱchargesȱinȱtheȱsubstrateȱregionȱnearȱtheȱSiO 2 ȱlayer. C) VDSȱ>ȱ0. D) aȱthinȱlayerȱofȱnegativeȱchargesȱinȱtheȱsubstrateȱregionȱnearȱtheȱSiO 2 ȱlayer. Diff:ȱ3

46) Forȱanȱenhancement-modeȱMOSFET,ȱtheȱminimumȱV GS requiredȱtoȱproduceȱdrainȱcurrentȱis calledȱthe A) thresholdȱvoltage,ȱdesignatedȱV GS(th).

B) blockingȱvoltage,ȱdesignatedȱVB.

C) breakoverȱvoltage.

D) IDss.

Diff:ȱ2

339

47) SpecialȱhandlingȱprecautionsȱshouldȱbeȱtakenȱwhenȱworkingȱwithȱMOSFETs.ȱWhichȱofȱthe followingȱisȱnotȱoneȱofȱtheseȱprecautions? A) Allȱtestȱequipmentȱshouldȱbeȱgrounded. B) MOSFETȱdevicesȱshouldȱhaveȱtheirȱleadsȱshortedȱtogetherȱforȱshipmentȱandȱstorage. C) NeverȱremoveȱorȱinsertȱMOSFETȱdevicesȱwithȱtheȱpowerȱon. D) WorkersȱhandlingȱMOSFETȱdevicesȱshouldȱnotȱhaveȱgroundingȱstrapsȱattachedȱtoȱtheir wrists. Diff:ȱ2

48) TheȱsimplestȱmethodȱtoȱbiasȱaȱD-MOSFETȱisȱto A) setȱVGSȱ=ȱ+4.

B) setȱVGS = -4.

C) setȱVGSȱ=ȱ0.

D) selectȱtheȱcorrectȱvalueȱR D.

Diff:ȱ3

49) Theȱtype(s)ȱofȱbiasȱmostȱoftenȱusedȱwithȱE-MOSFETȱcircuitsȱis A) drain-feedback.

B) constantȱcurrent.

C) voltage-divider.

D) BothȱAȱandȱCȱabove.

Diff:ȱ2

340

Chapterȱ9 FETȱAmplifiers and Switching Circuits 1) Theȱformulaȱforȱtheȱvoltageȱgainȱofȱaȱcommon-sourceȱamplifierȱisȱRD/gm . Diff:ȱ2

2) Loadȱresistanceȱaddedȱtoȱtheȱoutputȱofȱanȱamplifierȱincreasesȱtheȱvoltageȱgain. Diff:ȱ2

3) Theȱadditionȱofȱaȱsourceȱbypassȱcapacitorȱwillȱincreaseȱtheȱvoltageȱgain. Diff:ȱ2

4) FETsȱareȱsuperiorȱtoȱBJTsȱinȱalmostȱallȱswitchingȱapplications. Diff:ȱ2

5) FETȱamplifiersȱgenerallyȱhaveȱlowerȱdistortionȱthanȱBJTȱamplifiers. Diff:ȱ2

6) Cascodeȱamplifiersȱareȱusedȱprimarilyȱinȱradioȱfrequencyȱapplications. Diff:ȱ2

7) MostȱofȱtheȱgainȱinȱaȱJFET-basedȱcascodeȱamplifierȱisȱprovidedȱbyȱtheȱCSȱamplifier. Diff:ȱ3

8) ClassȱDȱamplifiersȱusuallyȱemployȱJFETs. Diff:ȱ2

9) Theȱ3ȱstagesȱinȱaȱclassȱDȱamplifierȱareȱtheȱpulse-width-modulator,ȱtheȱswitchingȱamplifier, andȱtheȱlow-passȱfilter. Diff:ȱ3

10) Aȱpulse-width-modulatorȱemploysȱaȱsquare-waveȱgeneratorȱandȱaȱcomparator. Diff:ȱ3

11) AȱMOSFETȱswitchȱisȱturnedȱonȱandȱoffȱbyȱchangingȱtheȱsourceȱvoltage. Diff:ȱ2

12) AnȱanalogȱMOSFETȱswitchȱisȱturnedȱonȱwhenȱtheȱgateȱreceivesȱaȱpositiveȱvoltageȱpulse. Diff:ȱ2

341

13) CMOSȱcombinesȱn-channelȱandȱp-channelȱD-MOSFETS. Diff:ȱ3

14) PowerȱMOSFETsȱhaveȱaȱnegativeȱtemperatureȱcoefficientȱandȱthereforeȱareȱlessȱproneȱto thermalȱrunaway. Diff:ȱ3

15) InȱanȱamplifierȱusingȱaȱJFET,ȱtheȱgateȱcurrentȱisȱapproximatelyȱ0. Diff:ȱ2

16) Aȱcommon-sourceȱamplifierȱhasȱaȱ________ȱphaseȱshiftȱbetweenȱtheȱinputȱandȱtheȱoutput. A) 45°

B) 90°

C) 180°

D) 360°

Diff:ȱ2

17) Referȱtoȱtheȱfigureȱabove.ȱȱAssumingȱmidpointȱbiasing,ȱifȱV GS = -4ȱV,ȱtheȱvalueȱofȱR Sȱthatȱwill provideȱthisȱvalueȱis A) 600ơ̇̄.

B) 1.2ȱk̛.

C) 80ơ̇̄.

D) 800ơ̇̄.

Diff:ȱ2

18) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱV in = 50ȱmVp-p,ȱtheȱoutputȱvoltageȱis A) 50ȱmVp-p.

B) 4.4ȱVp-p.

C) 0.044ȱVp-p.

D) 440ȱmVp-p.

Diff:ȱ3

19) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱmeasuredȱvalueȱofȱV out wereȱbelowȱnormal,ȱtheȱproblem mightȱbeȱthat A) RDȱisȱopen.

B) C2 isȱshorted.

C) C2 ȱisȱopen.

D) Vin hasȱincreased.

Diff:ȱ3

342

20) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱV in = 1ȱVp-p,ȱtheȱoutputȱvoltageȱVout wouldȱbe A) undistorted.

B) clippedȱonȱtheȱnegativeȱpeaks.

C) clippedȱonȱtheȱpositiveȱpeaks.

D) 0ȱV.

Diff:ȱ2

21) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱID = 6ȱmA,ȱtheȱvalueȱofȱV GS is A) 9ȱV.

B) -9ȱV.

C) -19.8ȱV.

D) -10.2ȱV.

Diff:ȱ2

22) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱgm = 6500ȱΐSȱandȱanȱinputȱsignalȱofȱ125ȱmVp-pȱisȱappliedȱtoȱthe gate,ȱtheȱoutputȱvoltageȱV outȱis A) 2.68ȱVp-p.

B) 0.8125ȱVp-p.

C) 1.625ȱVp-p.

D) 6.25ȱVp-p.

Diff:ȱ2

23) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱC2 opened,ȱtheȱoutputȱsignalȱwould A) increaseȱinȱvalue.

B) decreaseȱinȱvalue.

C) notȱchange.

D) decreaseȱandȱthenȱincrease.

Diff:ȱ2

24) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱID = 4ȱmA,ȱI DSS = 16ȱmA,ȱandȱV GS(off)ȱ= -8ȱV,ȱVDSȱwouldȱbe A) 19.2ȱV.

B) -6ȱV.

C) 10.8ȱV.

D) 30ȱV.

Diff:ȱ2

25) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱgm = 4000ȱΐSȱandȱaȱsignalȱofȱ75ȱmVȱrmsȱisȱappliedȱtoȱtheȱgate,ȱthe p-pȱoutputȱvoltageȱis A) 990ȱmV.

B) 1.13ȱVp-p.

C) 2.8ȱVp-p.

Diff:ȱ3

343

D) 990ȱVp-p.

26) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR 7 wereȱtoȱincreaseȱinȱvalue,ȱVout would A) increase.

B) decrease.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ3

27) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱC2 opened,ȱVout would A) increase.

B) decrease.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ2

28) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR 1 opened,ȱVout would A) increase.

B) decrease.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ3

29) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱC3 opened,ȱVout would A) increase.

B) decrease.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ2

30) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR 3 opened,ȱVout would A) increase.

B) decrease.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ2

31) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR 5 opened,ȱVout would A) increase.

B) decrease.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ2

344

32) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱsource-drainȱofȱQ2 shorted,ȱtheȱoutputȱsignalȱfromȱQ 1 ȱwould A) increase.

B) decrease.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ3

33) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱAV1 = 18ȱandȱAVt = 288,ȱtheȱvalueȱofȱA V2 wouldȱbe A) 5184.

B) 18.

C) 49.18.

D) 16.

Diff:ȱ2

34) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱC4 opened,ȱtheȱsignalȱvoltageȱatȱtheȱdrainȱofȱQ 1 ȱwould A) increase.

B) decrease.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ3

35) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱC5 opened,ȱtheȱsignalȱvoltageȱacrossȱR7 would A) increase.

B) beȱzero.

C) remainȱtheȱsame.

D) distort.

Diff:ȱ3

36) WhatȱcomponentȱofȱanȱFETȱamplifierȱneedsȱtoȱbeȱatȱleastȱtenȱtimesȱgreaterȱthanȱR Dȱtoȱensure maximumȱvoltageȱgain? A) rȇds

B) RS

C) Rs

D) gm

Diff:ȱ2

37) Inȱaȱself-biasedȱcommon-sourceȱamplifier,ȱwhatȱpurposeȱdoesȱresistorȱRGȱserve? A) Keepsȱtheȱgateȱatȱapproximatelyȱzeroȱvolts B) Developsȱtheȱgate-sourceȱbiasȱcurrent C) Preventsȱloadingȱofȱtheȱacȱsignalȱsource D) BothȱAȱandȱCȱabove. Diff:ȱ3

38) AȱCSȱamplifierȱhasȱaȱvoltageȱgainȱof A) gm Rd.

B) gm Rs.

C)

Diff:ȱ3

345

gm Rs . (1ȱ+ȱg m Rd)

D)

gm Rd . (1ȱ+ȱg m Rd)

39) Theȱcapacitorȱconnectedȱtoȱtheȱsourceȱofȱaȱcommonȱsourceȱamplifier A) isȱusedȱforȱinputȱcoupling. B) isȱusedȱforȱoutputȱcoupling. C) makesȱtheȱsourceȱanȱacȱgroundȱpoint. D) isȱneededȱforȱextraȱpowerȱsupplyȱfiltering. Diff:ȱ2

40) Approximatelyȱhowȱmuchȱsignalȱvoltageȱshouldȱbeȱmeasuredȱatȱtheȱbypassedȱsourceȱofȱa commonȱsourceȱamplifier? Vin A) Vin B) 2 C) theȱsameȱamountȱasȱtheȱdrain

D) 0ȱV

Diff:ȱ2

41) Aȱsourceȱfollowerȱhasȱaȱvoltageȱgainȱof gm A) gm Rd. B) . RS

C)

gm Rs . (1ȱ+ȱg m RS)

D)

gm Rd . (1ȱ+ȱg m Rd)

Diff:ȱ2

42) Whenȱtheȱinputȱsignalȱisȱlarge,ȱaȱsourceȱfollowerȱhas A) aȱvoltageȱgainȱofȱlessȱthanȱone.

B) aȱsmallȱdistortion.

C) aȱhighȱinputȱresistance.

D) Allȱofȱtheȱabove.

Diff:ȱ2

43) Changingȱ________ȱcanȱcontrolȱtheȱvoltageȱgainȱofȱaȱcommon -sourceȱamplifier. A) theȱinputȱvoltage

B) gm

C) VDD

D) RS

Diff:ȱ2

44) Whereȱisȱtheȱinputȱsignalȱappliedȱinȱaȱcommon-gateȱamplifier? A) Source

B) Gate

C) Drain

D) Base

Diff:ȱ1

45) IfȱtheȱinterstageȱcouplingȱcapacitorȱopensȱinȱaȱtwoȱstageȱFETȱamplifier,ȱtheȱoutputȱsignalȱwill A) doubleȱinȱamplitude.

B) beȱslightlyȱdecreased.

C) beȱslightlyȱincreased.

D) beȱzero.

Diff:ȱ2

346

46) TwoȱFETȱamplifiersȱareȱoperatedȱinȱcascade.ȱIfȱtheȱgainȱofȱtheȱamplifiersȱisȱequalȱandȱtheȱtotal overallȱgainȱisȱ25,ȱtheȱgainȱofȱeachȱamplifierȱis A) 2.

B) 5.

C) 7.25.

D) 12.5.

Diff:ȱ2

47) TwoȱFETȱamplifiersȱoperatedȱinȱcascadeȱhaveȱgainsȱofȱ6ȱdBȱandȱ8ȱdB.ȱTheȱtotalȱoverallȱgainȱin dBȱis A) 2ȱdB.

B) 14ȱdB.

C) 48ȱdB.

D) 64ȱdB.

Diff:ȱ2

48) Howȱmuchȱdcȱvoltageȱshouldȱalwaysȱbeȱmeasuredȱatȱtheȱgateȱofȱaȱcommon -gateȱamplifier? +VDD C) 0.7ȱV D) 0ȱV A) +VDD B) 2 Diff:ȱ2

49) WhenȱtroubleshootingȱaȱmultipleȱstageȱCSȱamplifier,ȱsignalȱtracingȱrevealsȱaȱsignificant amountȱofȱsignalȱvoltageȱatȱtheȱsourceȱofȱtheȱfirstȱstage.ȱTheȱmostȱlikelyȱproblemȱis A) anȱopenȱcouplingȱcapacitorȱfromȱsignalȱsource. B) aȱshortedȱloadȱinȱthirdȱstage. C) anȱopenȱsourceȱbypassȱcapacitor. D) tooȱmuchȱdcȱsupplyȱvoltage. Diff:ȱ3

50) TheȱmainȱreasonȱthatȱtheȱinputȱresistanceȱofȱFETȱamplifiersȱisȱsoȱhighȱisȱthatȱthe A) FETȱjunctionsȱareȱforward-biasedȱandȱmadeȱofȱgermanium. B) biasingȱresistorsȱareȱextremelyȱsmall. C) powerȱsuppliesȱusedȱareȱhighȱvoltage/lowȱcurrent. D) gateȱjunctionȱisȱreverse-biased. Diff:ȱ2

51) TheȱacȱequivalentȱcircuitȱofȱaȱFETȱamplifierȱassumesȱthatȱtheȱcapacitors A) areȱtreatedȱlikeȱshorts. B) haveȱanȱXCȱofȱ600ơ̇̄ȱatȱtheȱsignalȱfrequency. C) areȱtreatedȱlikeȱopens. D) haveȱanȱXCȱofȱ1ȱM̛ȱatȱtheȱsignalȱfrequency. Diff:ȱ2

347

52) WhichȱofȱtheȱfollowingȱisȱnotȱaȱFETȱamplifier? A) Source-follower

B) Common-gate

C) Emitter-follower

D) Common-drain

Diff:ȱ1

53) IfȱRSȱisȱnotȱbypassed A) theȱdcȱpowerȱsupplyȱwillȱhaveȱexcessiveȱripple. B) theȱgainȱwillȱbeȱextremelyȱhigh. C) theȱgainȱwillȱbeȱreduced. D) noȱsignalȱwillȱbeȱcoupledȱtoȱtheȱoutput. Diff:ȱ2

54) Twoȱcommon-sourceȱamplifiersȱareȱcascaded.ȱTheȱoverallȱphaseȱshiftȱfromȱtheȱinputȱofȱthe firstȱamplifierȱtoȱtheȱoutputȱofȱtheȱsecondȱamplifierȱis A) 0°.

B) 45°.

C) 90°.

D) 180°.

Diff:ȱ2

55) WhichȱofȱtheȱfollowingȱisȱnotȱpartȱofȱaȱclassȱDȱamplifier? A) oscillator

B) pulse-widthȱmodulator

C) low-passȱfilter

D) switchingȱamplifier

Diff:ȱ3

56) WhichȱofȱtheȱfollowingȱisȱnotȱaȱcharacteristicȱofȱaȱJFET-basedȱcascodeȱamplifier? A) highȱinputȱresistance

B) highȱinputȱcapacitance

C) highȱfrequencyȱresponse

D) highȱgain

Diff:ȱ3

57) Whatȱisȱtheȱroleȱofȱtheȱlow-passȱfilterȱinȱaȱclassȱDȱamplifier? A) limitȱamplifierȱfrequencyȱresponse B) removeȱhum C) preventȱoscillation D) removeȱtheȱmodulatingȱfrequencyȱandȱharmonics Diff:ȱ3

58) WhichȱofȱtheȱfollowingȱisȱnotȱanȱapplicationȱofȱanȱanalogȱMOSFETȱswitch A) sample-and-holdȱcircuit

B) analogȱmultiplexer

C) inverter

D) switchedȱcapacitorȱcircuit

Diff:ȱ2

348

Chapterȱ10 AmplifierȱFrequencyȱResponse 1) Couplingȱandȱbypassȱcapacitorsȱlimitȱtheȱlow-frequencyȱresponseȱofȱanȱamplifier. Diff:ȱ2

2) High-frequencyȱresponseȱisȱlimitedȱbyȱtheȱinternalȱcapacitancesȱofȱaȱtransistor. Diff:ȱ2

3) Atȱtheȱcutoffȱfrequency,ȱtheȱoutputȱisȱdownȱbyȱ3ȱdB. Diff:ȱ2

4) Anȱoctaveȱofȱfrequencyȱisȱaȱten-timesȱchange. Diff:ȱ2

5) Theȱbandwidthȱisȱtheȱsumȱofȱtheȱtwoȱcutoffȱfrequencies. Diff:ȱ2

6) ToȱeffectivelyȱanalyzeȱanȱRCȱcoupledȱamplifierȇsȱhighȱfrequencyȱresponse,ȱyouȱonlyȱneedȱto considerȱtheȱcouplingȱandȱbypassȱcapacitances.ȱTheȱinternalȱcapacitanceȱcanȱbeȱignored. Diff:ȱ2

7) Ifȱanȱamplifierȱhasȱanȱoutputȱvoltageȱofȱ12.7ȱVp-pȱatȱtheȱmidpointȱofȱtheȱfrequencyȱrange,ȱthe outputȱvoltageȱatȱtheȱcutoffȱfrequencyȱwouldȱbe A) 12.7ȱVp-p.

B) 4.49ȱVp-p.

C) 5.89ȱVp-p.

D) 8.98ȱVp-p.

Diff:ȱ2

8) Ifȱanȱamplifierȱhasȱanȱinputȱsignalȱvoltageȱofȱ0.37ȱmVȱandȱanȱoutputȱvoltageȱofȱ16.8ȱV,ȱthe voltageȱgainȱinȱdBȱwouldȱbe A) 45.4ȱdB.

B) 33.1ȱdB.

C) 93.1ȱdB.

D) 46.6ȱdB.

Diff:ȱ2

9) Ifȱanȱamplifierȱhasȱaȱvoltageȱgainȱofȱ54ȱdB,ȱandȱanȱinputȱsignalȱofȱ22ȱmV,ȱtheȱoutputȱsignal voltageȱwouldȱbe A) 11ȱV.

B) 55.3ȱV.

C) 2.45ȱV.

D) 24.5ȱV.

Diff:ȱ3

10) Ifȱanȱamplifierȱhasȱaȱbandwidthȱofȱ47ȱkHzȱandȱaȱhigherȱcutoffȱfrequencyȱofȱ104ȱkHz,ȱtheȱlower cutoffȱfrequencyȱwouldȱbe A) 151ȱkHz.

B) 57ȱkHz.

C) 47ȱkHz.

Diff:ȱ2

349

D) 104ȱkHz.

11) IfȱanȱamplifierȱhasȱanȱR inȱ=ȱ950ơ̇̄,ȱandȱaȱcouplingȱcapacitorȱofȱvalueȱ3.3ȱ ΐF,ȱtheȱapproximate cutoffȱfrequencyȱwouldȱbe A) 508ȱHz.

B) 50.8ȱkHz.

C) 50.8ȱHz.

D) 5.08ȱHz.

Diff:ȱ2

12) TheȱfcȱofȱaȱcertainȱRCȱnetworkȱthatȱhasȱvaluesȱofȱRȱ = 470ơ̇̄ andȱCȱ= 0.005ȱΐFȱis A) 67.7ȱkHz.

B) 425ȱkHz.

C) 213ȱkHz.

D) 12ȱkHz.

Diff:ȱ2

13) Referȱtoȱtheȱfigureȱabove.ȱȱLowȱfrequencyȱresponseȱisȱaffectedȱby A) RC.

B) CBE.

C) C3.

D) Allȱofȱtheȱabove.

Diff:ȱ2

14) Referȱtoȱtheȱfigureȱabove.ȱȱHighȱfrequencyȱresponseȱisȱaffectedȱby A) RC.

B) CBE.

C) C3.

D) Allȱofȱtheȱabove.

Diff:ȱ2

15) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱoutputȱvoltageȱatȱtheȱupperȱcutoffȱfrequencyȱwasȱ7.19ȱVp -p, theȱoutputȱvoltageȱthatȱwouldȱbeȱexpectedȱatȱtheȱlowerȱcutoffȱfrequencyȱis A) 5.08ȱVp-p.

B) 7.19ȱVp-p.

C) 10.17ȱVp-p.

Diff:ȱ3

350

D) 2.11ȱVp-p.

16) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱoutputȱvoltageȱatȱf cl isȱ22ȱmV.ȱȱIdeally,ȱV outȱatȱtheȱmidpoint frequencyȱwouldȱbe A) 22ȱmV.

B) 17ȱmV.

C) 31.1ȱmV.

D) Notȱenoughȱinformationȱtoȱdetermine.

Diff:ȱ3

17) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱbandwidthȱofȱthisȱamplifierȱis A) theȱsumȱofȱtheȱupperȱandȱlowerȱfrequencies. B) theȱupperȱfrequencyȱdividedȱbyȱ0.707. C) theȱdifferenceȱbetweenȱtheȱupperȱandȱlowerȱfrequencies. D) theȱlowerȱfrequencyȱtimesȱ0.707. Diff:ȱ3

18) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱcapacitanceȱC bc affects A) high-frequencyȱresponse.

B) low-frequencyȱresponse.

C) mid-rangeȱresponse.

D) nothing.

Diff:ȱ2

19) Referȱtoȱtheȱfigureȱabove.ȱAȱdefiniteȱreductionȱinȱtheȱoutputȱvoltageȱisȱnoticed.ȱTheȱtroubleȱis that A) C3 ȱhasȱshorted.

B) C1 hasȱopened.

C) C2 ȱhasȱopened.

D) C3 hasȱopened.

Diff:ȱ3

20) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱreductionȱinȱtheȱoutputȱatȱveryȱhighȱfrequenciesȱisȱdueȱto A) theȱnegativeȱfeedbackȱeffectȱofȱR E.

B) theȱnegativeȱfeedbackȱeffectȱofȱCbc.

C) theȱpositiveȱfeedbackȱeffectȱofȱV BE.

D) RL decreasingȱinȱvalue.

Diff:ȱ2

21) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱoutputȱvoltageȱatȱf cl isȱ12ȱmV,ȱtheȱactualȱoutputȱvoltageȱatȱthe midpointȱfrequencyȱwouldȱbe A) 12ȱmV.

B) 12ȱmVp-p.

C) 16.97ȱmV.

D) 8.48ȱmV.

Diff:ȱ2

22) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR L increasesȱinȱvalue,ȱtheȱoutputȱvoltageȱwould A) increase.

B) decrease.

C) remainȱtheȱsame.

D) Cannotȱbeȱdetermined.

Diff:ȱ2

351

23) AȱBodeȱplot A) isȱaȱtestingȱmethodȱusedȱforȱdcȱamplifiers. B) providesȱaȱvisualȱpresentationȱofȱdecibelȱvoltageȱgainȱvs.ȱfrequency. C) indicatesȱvoltageȱgainȱwithȱnoȱreferenceȱtoȱfrequency. D) indicatesȱvoltageȱgainȱonlyȱatȱ0ȱHz. Diff:ȱ2

24) Theȱcutoffȱfrequencyȱofȱaȱlowȱpassȱfilterȱoccursȱat A) -5ȱdB.

B) -3ȱdB.

C) +3ȱdB.

D) -20ȱdB.

Diff:ȱ2

25) Aȱhighȱpassȱfilterȱmayȱbeȱusedȱto A) passȱlowȱfrequencies. B) passȱhighȱfrequencies. C) passȱfrequenciesȱbetweenȱlowȱandȱhigh. D) BothȱAȱandȱBȱabove. Diff:ȱ2

26) Aȱroll-offȱofȱ20ȱdBȱperȱdecadeȱisȱequivalentȱtoȱaȱroll-offȱofȱ________ȱperȱoctave. A) 3ȱdB

B) 13ȱdB

C) 12ȱdB

D) 6ȱdB

Diff:ȱ2

27) Whichȱofȱtheȱfollowingȱcapacitancesȱaffectsȱtheȱhighȱfrequencyȱresponseȱofȱanȱamplifier? A) Strayȱwiringȱcapacitance

B) Internalȱpnȱjunctionȱcapacitance

C) Couplingȱandȱbypassȱcapacitors

D) BothȱAȱandȱBȱabove.

Diff:ȱ1

28) Atȱlowȱfrequencies,ȱtheȱcouplingȱcapacitorsȱproduceȱaȱdecreaseȱin A) inputȱresistance.

B) voltageȱgain.

C) generatorȱresistance.

D) generatorȱvoltage.

Diff:ȱ2

29) Ifȱtheȱvalueȱofȱaȱfeedbackȱcapacitorȱisȱ50ȱpF,ȱwhatȱisȱtheȱinputȱMillerȱcapacitanceȱwhenȱA V = 200? A) 1ȱΐF

B) 10ȱΐF

C) 100ȱΐF

Diff:ȱ2

352

D) 10ȱnF

30) Theȱcriticalȱfrequenciesȱofȱanȱamplifierȱareȱtheȱfrequenciesȱwhereȱtheȱoutputȱvoltageȱis A) halfȱofȱV mid.

B) 0.707ȱofȱV mid.

C) zero.

D) 0.25ȱofȱVmid.

Diff:ȱ2

31) Theȱvoltageȱgainȱofȱanȱamplifierȱisȱ200.ȱȱTheȱdecibelȱvoltageȱgainȱis A) 23ȱdB.

B) 46ȱdB.

C) 200ȱdB.

D) 106ȱdB.

Diff:ȱ2

32) Ifȱtheȱvoltageȱgainȱdoubles,ȱtheȱdecibelȱvoltageȱgainȱincreasesȱby A) aȱfactorȱofȱ2.

B) 3ȱdB.

C) 6ȱdB.

D) 10ȱdB.

Diff:ȱ2

33) InȱtheȱmidbandȱofȱaȱCEȱamplifier A) theȱemitterȱisȱnotȱatȱacȱground. B) theȱMillerȱeffectȱhasȱmaximumȱinfluence. C) theȱvoltageȱgainȱisȱmaximum. D) theȱcouplingȱandȱbypassȱcapacitorsȱappearȱopen. Diff:ȱ2

34) Ifȱtheȱpowerȱgainȱdoubles,ȱtheȱdecibelȱpowerȱgainȱincreasesȱby A) aȱfactorȱofȱ2.

B) 3ȱdB.

C) 6ȱdB.

D) 10ȱdB.

Diff:ȱ2

35) Atȱlowȱfrequencies,ȱtheȱemitter-bypassȱcapacitor A) isȱnoȱlongerȱanȱacȱshort.

B) hasȱminimumȱXC.

C) increasesȱtheȱoutputȱvoltage.

D) increasesȱtheȱvoltageȱgain.

Diff:ȱ2

36) Raisingȱtheȱfrequencyȱofȱ1ȱkHzȱbyȱ2ȱoctavesȱresultsȱinȱaȱfrequencyȱof A) 4ȱkHz.

B) 500ȱHz.

C) 250ȱHz.

D) 2ȱMHz.

Diff:ȱ2

37) Theȱfrequencyȱresponseȱofȱanȱamplifierȱisȱaȱgraphȱof A) voltageȱversusȱcurrent.

B) voltageȱversusȱtime.

C) outputȱvoltageȱversusȱfrequency.

D) inputȱvoltageȱversusȱfrequency.

Diff:ȱ1

353

38) Theȱvoltageȱgainȱofȱanȱamplifierȱisȱ150.ȱȱIfȱtheȱoutputȱvoltageȱdoublesȱ(forȱtheȱsameȱamountȱof inputȱvoltage),ȱtheȱvoltageȱgainȱequals A) 21.7ȱdb.

B) 43.5ȱdb.

C) 49.5ȱdb.

D) 114ȱdb.

Diff:ȱ2

39) Atȱtheȱlowerȱorȱupperȱcutoffȱfrequency,ȱtheȱvoltageȱgainȱis A) 0.35ȱAv mid.

B) 0.5ȱAvmid.

C) 0.707ȱAv mid.

D) 0.995 ȱAv mid

.

Diff:ȱ1

40) PhaseȱshiftȱinȱtheȱinputȱofȱanȱRCȱcircuitȱwillȱapproachȱ90 ° whenȱfrequencyȱapproaches A) zero.

B) maximum.

C) mid-range.

Diff:ȱ2

41) WhatȱeffectȱdoesȱlowȱfrequencyȱhaveȱonȱtheȱemitterȱbypassȱRCȱcircuit? A) Decreasesȱimpedanceȱandȱincreasesȱvoltageȱgain B) Increasesȱimpedanceȱandȱdecreasesȱvoltageȱgain C) Increasesȱimpedanceȱandȱincreasesȱvoltageȱgain D) Decreasesȱimpedanceȱandȱdecreasesȱvoltageȱgain Diff:ȱ2

42) Forȱaȱlagȱnetworkȱaboveȱtheȱcutoffȱfrequency,ȱtheȱvoltageȱgain A) decreasesȱatȱtheȱrateȱofȱ20ȱdbȱperȱdecade. B) increasesȱatȱtheȱrateȱofȱ6ȱdbȱperȱoctave. C) decreasesȱatȱtheȱrateȱofȱ6ȱdbȱperȱoctave. D) AȱandȱCȱabove. Diff:ȱ3

43) Semilogȱgraphȱpaperȱhas A) twoȱ(2)ȱlinearȱaxis. B) oneȱverticalȱlinearȱaxisȱandȱoneȱlogarithmicȱhorizontalȱaxis. C) oneȱhorizontalȱlinearȱaxisȱandȱoneȱverticalȱlogarithmicȱaxis. D) twoȱ(2)ȱlogarithmicȱaxis. Diff:ȱ1

354

D) cutoff.

44) Theȱunity-gainȱfrequencyȱ(fT)ȱequalsȱtheȱproductȱofȱmid-rangeȱvoltageȱgainȱ(A V(mid))ȱand the A) compensatingȱcapacitance.

B) fcu.

C) BW.

D) loadȱresistance.

Diff:ȱ2

45) Atȱtheȱunity-gainȱfrequency,ȱtheȱopen-loopȱvoltageȱgainȱis A) 1.

B) Amid.

C) zero.

Diff:ȱ2

355

D) veryȱlarge.

Chapterȱ11 Thyristors 1) TheȱSCRȱisȱaȱdeviceȱthatȱcanȱbeȱtriggeredȱonȱbyȱaȱpulseȱappliedȱtoȱtheȱgate. Diff:ȱ2

2) Aȱdeviceȱthatȱconductsȱcurrentȱinȱonlyȱoneȱdirectionȱisȱcalledȱaȱdiac. Diff:ȱ2

3) AȱdeviceȱthatȱcanȱbeȱturnedȱonȱorȱoffȱbyȱaȱgateȱpulseȱisȱcalledȱanȱSCS. Diff:ȱ2

4) Aȱtriacȱcanȱbeȱturnedȱonȱbyȱaȱpulseȱappliedȱtoȱtheȱgate. Diff:ȱ2

5) AȱUJTȱisȱturnedȱonȱbyȱaȱnegativeȱpulseȱatȱtheȱbase. Diff:ȱ2

6) Theȱforcedȱcommutationȱmethodȱrequiresȱmomentarilyȱforcingȱcurrentȱinȱtheȱdirection oppositeȱtoȱtheȱreverseȱconductionȱsoȱthatȱtheȱnetȱreverseȱcurrentȱisȱreducedȱbelowȱtheȱholding value. Diff:ȱ3

7) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱTheȱsymbolȱisȱfor A) aȱtriac.

B) aȱUJT.

C) aȱdiac.

D) aȱPUT.

E) anȱSCR.

D) aȱPUT.

E) anȱSCR.

D) aȱPUT.

E) anȱSCR.

Diff:ȱ2

8) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱTheȱsymbolȱisȱfor A) aȱtriac.

B) aȱUJT.

C) aȱdiac.

Diff:ȱ2

9) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱTheȱsymbolȱisȱfor A) aȱtriac.

B) aȱUJT.

C) aȱdiac.

Diff:ȱ2

356

10) Referȱtoȱ(d)ȱinȱtheȱfigureȱabove.ȱȱTheȱsymbolȱisȱfor A) aȱtriac.

B) aȱUJT.

C) aȱdiac.

D) aȱPUT.

E) anȱSCR.

Diff:ȱ2

11) ToȱavoidȱconfusingȱtheȱschematicȱsymbolsȱforȱaȱJFETȱandȱaȱUJT,ȱkeepȱinȱmindȱthatȱtheȱUJT symbol A) hasȱtwoȱcircles. B) hasȱnoȱarrow. C) usesȱaȱdot. D) hasȱfourȱterminals. E) hasȱanȱarrowȱatȱanȱangle. Diff:ȱ2

12) Aȱgoodȱchoiceȱtoȱtriggerȱaȱtriacȱwouldȱbeȱa(n) A) 4-layerȱdiode.

B) SCR.

C) diac.

D) Schockleyȱdiode.

Diff:ȱ2

13) Theȱbestȱchoiceȱtoȱcontrolȱaȱsmallȱvariableȱspeedȱacȱmotorȱisȱa(n) A) triac.

B) diac.

C) SCR.

D) UJT.

E) PUT.

Diff:ȱ2

14) Theȱbestȱchoiceȱtoȱshutȱdownȱaȱdcȱpowerȱsupplyȱinȱcaseȱofȱanȱabnormallyȱhighȱvoltage conditionȱisȱa(n) A) triac.

B) diac.

C) SCR.

D) UJT.

E) PUT.

Diff:ȱ2

15) Eachȱofȱtheȱfollowingȱhasȱaȱgateȱexceptȱthe A) triac.

B) diac.

C) SCR.

D) PUT.

Diff:ȱ2

16) Theȱmostȱlikelyȱdeviceȱtoȱbeȱusedȱinȱanȱoscillatorȱisȱa(n) A) triac.

B) diac.

C) SCR.

D) UJT.

E) PUT.

Diff:ȱ2

17) AnȱSCSȱhasȱaȱuniqueȱabilityȱto A) beȱturnedȱonȱorȱoffȱwithȱaȱpulse.

B) beȱturnedȱoffȱonlyȱwithȱaȱpulse.

C) controlȱaȱPUT.

D) Allȱofȱtheȱabove.

Diff:ȱ2

357

18) IfȱanȱSCRȱstartsȱtoȱconductȱwhenȱaȱgateȱcurrentȱisȱestablished,ȱwhenȱtheȱgateȱcircuitȱis interruptedȱtheȱSCR A) willȱturnȱoff.

B) currentȱwillȱincrease.

C) willȱcontinueȱtoȱconduct.

D) gateȱcurrentȱwillȱincrease.

Diff:ȱ3

19) AnȱSCRȱwillȱturnȱoffȱwhenȱtheȱvoltageȱacrossȱitȱis A) increasedȱtoȱtheȱsupplyȱvoltage. B) decreasedȱtoȱnearȱ0ȱV. C) timedȱout. D) decreasedȱbyȱtheȱvalueȱofȱtheȱgateȱvoltage. Diff:ȱ2

20) Aȱtriacȱisȱusedȱforȱacȱbecause A) itȱconductsȱonȱbothȱalternations. B) itȱisȱturnedȱoffȱwhenȱtheȱacȱvoltageȱreachesȱzero. C) itȱcanȱdeliverȱmoreȱpowerȱtoȱtheȱloadȱinȱaȱcycle. D) Allȱofȱtheȱabove. Diff:ȱ2

21) AȱdeviceȱisȱneededȱtoȱtriggerȱanȱSCR.ȱȱOneȱpossibleȱdeviceȱtoȱuseȱmightȱbeȱa(n) A) SCR.

B) UJT.

C) Schockleyȱdiode.

D) PUT.

Diff:ȱ2

22) Theȱbestȱdeviceȱtoȱbeȱusedȱtoȱcontrolȱaȱdcȱmotorȱisȱthe A) triac.

B) PUT.

C) SCR.

D) diac.

Diff:ȱ2

23) AnȱSCRȱisȱusedȱtoȱcontrolȱtheȱspeedȱofȱaȱdcȱmotorȱbyȱ________ȱtheȱ________ȱofȱtheȱpulse deliveredȱtoȱtheȱmotor. A) varying,ȱwidth

B) increasing,ȱamplitude

C) decreasing,ȱgateȱwidth

D) Noneȱofȱtheȱabove.

Diff:ȱ3

358

24) Ofȱtheȱfollowingȱapplications,ȱtheȱlikelyȱoneȱtoȱuseȱaȱdiacȱisȱa(n) A) batteryȱcharger.

B) oscillator.

C) highȱfrequencyȱamplifier.

D) lampȱdimmer.

Diff:ȱ2

25) Ifȱtheȱgateȱcircuitȱisȱopenȱonȱaȱtriacȱusedȱforȱacȱmotorȱcontrol,ȱtheȱmotor A) canȱneverȱturnȱon.

B) willȱconstantlyȱrunȱatȱhalfȱspeed.

C) willȱconstantlyȱrunȱatȱfullȱspeed.

D) willȱrunȱatȱ60ȱRPM.

Diff:ȱ3

26) Aȱthyristorȱcanȱbeȱusedȱas A) aȱresistor.

B) anȱamplifier.

C) aȱswitch.

D) aȱpowerȱsource.

Diff:ȱ2

27) Theȱminimumȱinputȱcurrentȱthatȱcanȱturnȱonȱaȱthyristorȱisȱcalledȱthe A) holdingȱcurrent.

B) switchingȱcurrent.

C) breakdownȱcurrent.

D) low-currentȱdropoutȱcurrent.

Diff:ȱ2

28) Theȱminimumȱcurrentȱthatȱkeepsȱaȱlatchȱclosedȱisȱcalledȱthe A) pick-upȱcurrent.

B) criticalȱrateȱofȱcurrentȱrise.

C) triggerȱcurrent.

D) holdingȱcurrent.

Diff:ȱ2

29) Theȱonlyȱwayȱtoȱmakeȱaȱfour-layerȱdiodeȱconductȱȱisȱwith A) aȱtriggerȱinputȱappliedȱtoȱtheȱgate.

B) forwardȱbreakoverȱvoltage.

C) low-currentȱdropout.

D) Noneȱofȱtheȱabove.

Diff:ȱ3

30) TheȱonlyȱwayȱtoȱstopȱanȱSCRȱthatȱisȱconductingȱisȱby A) aȱpositiveȱtrigger.

B) low-currentȱdropȱout.

C) breakover.

D) reverse-biasȱtriggering.

Diff:ȱ2

359

31) Aȱsiliconȱcontrolledȱrectifierȱhas A) twoȱexternalȱleads.

B) threeȱexternalȱleads.

C) fourȱexternalȱleads.

D) threeȱdopedȱregions.

Diff:ȱ1

32) AnȱSCRȱisȱusuallyȱturnedȱonȱby A) breakover.

B) aȱgateȱtrigger.

C) breakdown.

D) holdingȱcurrent.

Diff:ȱ2

33) SCRsȱare A) low-powerȱdevices.

B) four-layerȱdiodes.

C) high-currentȱdevices.

D) bidirectional.

Diff:ȱ2

34) TheȱtriggerȱvoltageȱofȱanȱSCRȱisȱclosestȱto A) 0ȱV.

B) 0.7ȱV.

C) 4ȱV.

D) breakoverȱvoltage.

Diff:ȱ2

35) TheȱvoltageȱacrossȱaȱconductingȱSCR A) equalsȱtheȱbreakoverȱvoltage.

B) isȱapproximatelyȱzero.

C) decreasesȱwithȱmoreȱanodeȱcurrent.

D) equalsȱtheȱsourceȱvoltage.

Diff:ȱ2

36) DisconnectingȱtheȱgateȱleadȱonȱanȱSCR A) willȱresultȱinȱmaximumȱgateȱcurrent. B) resultsȱinȱtheȱSCRȱstayingȱonȱallȱtheȱtime. C) destroysȱtheȱdevice. D) meansȱthatȱtheȱSCRȱcannotȱbeȱtriggeredȱon. Diff:ȱ3

37) AȱconductingȱSCRȱcanȱbeȱturnedȱoffȱby A) reducingȱtheȱsourceȱvoltageȱtoȱzero. B) openingȱtheȱanodeȱcircuit. C) temporarilyȱshortingȱacrossȱtheȱSCRȇsȱanodeȱtoȱcathode. D) Allȱofȱtheȱabove. Diff:ȱ2

360

38) AnȱSCRȱcrowbarȱcircuitȱisȱusedȱto A) protectȱagainstȱundervoltageȱconditions. B) detectȱwhenȱaȱcircuitȱisȱoverheating. C) protectȱagainstȱovervoltageȱconditions. D) controlȱthreeȱphaseȱacȱmotors. Diff:ȱ2

39) Onceȱaȱdiacȱisȱconducting,ȱtheȱonlyȱwayȱtoȱturnȱitȱoffȱisȱwith A) low-currentȱdropout.

B) breakover.

C) aȱnegativeȱgateȱvoltage.

D) aȱpositiveȱgateȱvoltage.

Diff:ȱ2

40) Theȱdiacȱisȱequivalentȱto A) twoȱfour-layeredȱdiodesȱinȱparallel.

B) twoȱSCRsȱinȱparallel.

C) twoȱfour-layerȱdiodesȱinȱseries.

D) twoȱtriacsȱinȱparallel.

Diff:ȱ3

41) Theȱdiacȱisȱa A) transistor.

B) unidirectionalȱdevice.

C) three-layerȱdevice.

D) bidirectionalȱdevice.

Diff:ȱ2

42) Aȱtriacȱactsȱlike A) twoȱSCRsȱinȱparallel.

B) twoȱSCRsȱinȱseries.

C) twoȱdiacsȱinȱparallel.

D) aȱnormalȱtransistor.

Diff:ȱ1

43) Aȱtriac A) canȱtriggerȱonlyȱonȱpositiveȱgateȱvoltages. B) canȱtriggerȱonlyȱonȱnegativeȱgateȱvoltages. C) canȱbeȱtriggeredȱbyȱeitherȱaȱpositiveȱorȱaȱnegativeȱgateȱvoltage. D) cannotȱbeȱtriggeredȱwithȱgateȱvoltages. Diff:ȱ2

361

44) Aȱsilicon-controlledȱswitch A) hasȱonlyȱ1ȱgateȱlead. B) hasȱtwoȱgateȱleads. C) canȱonlyȱbeȱturnedȱoffȱwithȱlow-currentȱdropout. D) isȱbidirectional. Diff:ȱ2

45) AȱUJTȱhas A) twoȱbaseȱleads.

B) oneȱemitterȱlead.

C) twoȱemitterȱandȱoneȱbaseȱleads.

D) bothȱAȱandȱBȱabove.

Diff:ȱ2

46) ForȱaȱUJT,ȱtheȱintrinsicȱstandoffȱratioȱis A) equalȱtoȱ1.

B) alwaysȱgreaterȱthanȱ1.

C) alwaysȱlessȱthanȱ1.

D) usuallyȱaroundȱ1ȱk̛ȱorȱso.

Diff:ȱ2

47) TheȱPUTȱ(ProgrammableȱUnijunctionȱTransistor)ȱisȱactuallyȱaȱtypeȱof A) thyristor.

B) FETȱdevice.

C) triac.

Diff:ȱ2

362

D) SCR.

Chapterȱ12 TheȱOperationalȱAmplifier 1) Aȱgoodȱop-ampȱhasȱhighȱvoltageȱgain,ȱlowȱoutputȱimpedance,ȱandȱhighȱinputȱimpedance. Diff:ȱ2

2) Anȱinvertingȱamplifierȱhasȱanȱinputȱresistanceȱequalȱtoȱtheȱinputȱresistor. Diff:ȱ2

3) CMRRȱisȱtheȱmeasureȱofȱanȱop-ampȇsȱvoltageȱgainȱforȱanȱinvertingȱamplifier. Diff:ȱ2

4) Theȱdifferentialȱamplifierȱisȱusedȱasȱtheȱinputȱstageȱofȱanȱoperationalȱamplifier. Diff:ȱ2

5) Aȱvoltageȱfollowerȱhasȱaȱveryȱhighȱinputȱimpedance,ȱandȱisȱoftenȱusedȱasȱaȱhighȱvoltageȱgain amplifier. Diff:ȱ2

6) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱcircuitȱisȱtheȱinvertingȱamplifier? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

7) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱcircuitȱisȱaȱvoltageȱfollower? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

8) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱcircuitȱisȱtheȱnoninvertingȱamplifier? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

363

9) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱcircuitȱhasȱaȱvoltageȱgainȱofȱ1? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

10) Seeȱtheȱfigureȱabove.ȱȱWhichȱcircuitȱhasȱanȱinputȱimpedanceȱofȱaboutȱ5ȱk ̛? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ3

11) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱcircuitȱhasȱaȱvoltageȱgainȱofȱ10? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

12) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱcircuitȱhasȱaȱvoltageȱgainȱofȱ20? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

13) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱIfȱthisȱcircuitȱhasȱaȱV in = 12ȱVp-p,ȱtheȱvalueȱofȱV outȱwouldȱbe A) 20ȱV.

B) -20ȱV.

C) 8.48ȱVp-p.

D) 12ȱVp-p.

Diff:ȱ2

14) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱIfȱthisȱcircuitȱhasȱaȱV in = 0.7,ȱV out wouldȱbe A) 14.7ȱV.

B) -14.7ȱV.

C) 14ȱV.

D) 0ȱV.

Diff:ȱ2

15) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱV in = -60ȱmV.ȱȱTheȱvalueȱofȱV out is A) 600ȱmV.

B) -600ȱmV.

C) 660ȱmV.

D) aboutȱ16ȱV.

Diff:ȱ2

16) Seeȱtheȱfigureȱabove.ȱȱIfȱtheseȱthreeȱcircuitsȱwereȱconnectedȱasȱaȱmultiple-stageȱamplifier,ȱthe totalȱvoltageȱgainȱwouldȱbe A) 1.

B) 10.

C) 21.

Diff:ȱ2

364

D) 210.

17) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱIfȱR f isȱchangedȱtoȱ1ȱM̛,ȱtheȱnewȱAcl wouldȱbe A) 20.

B) -20.

C) 21.

D) -21.

Diff:ȱ2

18) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱIfȱanȱamplifierȱwithȱanȱinputȱimpedanceȱofȱ12ȱk ̛ȱandȱthe sameȱvoltageȱgainȱisȱneeded,ȱtheȱnewȱvalueȱofȱR 1 ȱwouldȱbeȱ________ȱandȱtheȱnewȱvalueȱofȱRf wouldȱbeȱ________. A) 10ȱk̛,ȱ100ȱk̛

B) 13.3ȱk̛,ȱ120ȱk̛

C) 12ȱk̛,ȱ108ȱk̛

D) 12ȱk̛,ȱ120ȱk̛

Diff:ȱ2

19) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱop-ampȱhasȱaȱslewȱrateȱofȱ1.33ȱV/ΐs.ȱȱHowȱlongȱwouldȱitȱtake theȱoutputȱvoltageȱtoȱchangeȱfromȱ-12ȱVȱtoȱ+12ȱV? A) 18ȱΐs

B) 16ȱΐs

C) 36ȱΐs

D) 48ȱΐs

Diff:ȱ2

20) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱcomponentsȱareȱusedȱtoȱsetȱinputȱimpedanceȱandȱvoltage gain? A) R4

B) R3

C) R1 andȱR2

D) R3 ȱandȱR4

Diff:ȱ2

21) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱcomponentsȱareȱusedȱforȱoffsetȱvoltageȱcompensation? A) R4

B) R3

C) R1ȱandȱR2

D) R2

Diff:ȱ3

22) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱcomponentsȱareȱusedȱforȱbiasȱcurrentȱcompensation? A) R4

B) R3

C) R1 andȱR2

Diff:ȱ2

365

D) R2

23) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱpurposeȱofȱR1 andȱR2 is A) forȱbiasȱcurrentȱcompensation. B) forȱinputȱoffsetȱvoltageȱcompensation. C) toȱsetȱinputȱimpedanceȱonly. D) toȱsetȱinputȱimpedanceȱandȱvoltageȱgain. Diff:ȱ2

24) Itȱtakesȱanȱop-ampȱ22ȱΐsȱtoȱchangeȱitsȱoutputȱfromȱ-15ȱVȱtoȱ+15ȱV.ȱȱTheȱslewȱrateȱforȱthis amplifierȱis A) 1.36ȱV/ΐs.

B) 0.68ȱV/ΐs.

C) 0.73ȱV/ΐs.

D) 660ȱV/ΐs.

Diff:ȱ2

25) Aȱvoltageȱfollowerȱamplifierȱcomesȱtoȱyouȱforȱservice.ȱȱWithȱaȱsineȱwaveȱinputȱofȱ1ȱV,ȱthe outputȱisȱsquaredȱoffȱatȱapproximatelyȱ30ȱVȱpeak-to-peak.ȱTheȱmostȱlikelyȱproblemȱis A) noȱdcȱsupplyȱvoltage.

B) theȱinputȱisȱshortedȱtoȱground.

C) openȱfeedbackȱloop.

D) theȱoutputȱisȱshortedȱtoȱground.

Diff:ȱ3

26) Theȱop-ampȱcanȱamplify A) acȱsignalsȱonly.

B) dcȱsignalsȱonly.

C) bothȱacȱandȱdcȱsignals.

D) neitherȱacȱnorȱdcȱsignals.

Diff:ȱ1

27) Theȱtypicalȱinputȱstageȱofȱanȱop-ampȱhasȱa(n) A) emitterȱfollower. B) single-endedȱinputȱandȱdifferentialȱoutput. C) push-pullȱcircuit. D) two-inputȱdifferentialȱamplifier. Diff:ȱ2

28) Theȱcommon-modeȱsignalȱisȱappliedȱto A) theȱnoninvertingȱinput.

B) theȱinvertingȱinput.

C) bothȱinputs.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

366

29) Ifȱanȱop-ampȱwereȱperfect,ȱtheȱCMRRȱwould A) beȱzero.

B) approachȱinfinity.

C) beȱveryȱsmall.

D) beȱlessȱthanȱone.

Diff:ȱ2

30) Inȱanȱop-amp,ȱtheȱCMRRȱisȱlimitedȱmostlyȱbyȱthe A) theȱcommon-modeȱgainȱofȱtheȱop-amp. B) gain-bandwidthȱproduct. C) supplyȱvoltages. D) toleranceȱofȱtheȱresistors. Diff:ȱ2

31) Theȱopen-loopȱvoltageȱgainȱ(A ol)ȱofȱanȱop-ampȱisȱthe A) externalȱvoltageȱgainȱtheȱdeviceȱisȱcapableȱof. B) internalȱvoltageȱgainȱtheȱdeviceȱisȱcapableȱof. C) mostȱcontrolledȱparameter. D) sameȱasȱA cm. Diff:ȱ2

32) Theȱinputȱoffsetȱcurrentȱis A) theȱdifferenceȱbetweenȱtheȱinputȱbiasȱcurrents. B) notȱrelatedȱtoȱtheȱinputȱbiasȱcurrents. C) lessȱthanȱtheȱinputȱoffsetȱvoltage. D) unimportantȱwhenȱaȱbaseȱresistorȱisȱused. Diff:ȱ2

33) Withȱbothȱinputsȱgrounded,ȱtheȱonlyȱoffsetȱthatȱproducesȱanȱerrorȱisȱthe A) inputȱoffsetȱcurrent.

B) inputȱbiasȱcurrent.

C) inputȱoffsetȱvoltage.

D) inputȱshortȱcircuitȱcurrent.

Diff:ȱ2

34) Theȱinputȱoffsetȱcurrentȱequalsȱthe A) differenceȱbetweenȱtwoȱbaseȱcurrents. B) averageȱofȱtwoȱbaseȱcurrents. C) collectorȱcurrentȱdividedȱbyȱcurrentȱgain. D) differenceȱbetweenȱtwoȱbase-emitterȱvoltages. Diff:ȱ2

367

35) Theȱidealȱop-ampȱhas A) infiniteȱinputȱimpedanceȱandȱzeroȱoutputȱimpedance. B) infiniteȱoutputȱimpedanceȱandȱzeroȱinputȱimpedance. C) infiniteȱvoltageȱgainȱandȱinfiniteȱbandwidth. D) BothȱAȱandȱC. Diff:ȱ2

36) Theȱtwoȱbasicȱwaysȱofȱspecifyingȱinputȱimpedanceȱofȱanȱop-ampȱare A) differentialȱandȱextremelyȱhigh.

B) differentialȱandȱcommon-loop.

C) differentialȱandȱcommon-mode.

D) closed-loopȱandȱcommon-mode.

Diff:ȱ2

37) Whatȱtypeȱofȱinputȱisȱprovidedȱtoȱanȱop-ampȱforȱslew-rateȱmeasurement? A) Constantȱdcȱvoltage

B) Sinusoidalȱvoltage

C) Stepȱinputȱvoltage

D) Triangleȱwaveform

Diff:ȱ2

38) Whenȱtheȱinitialȱslopeȱofȱaȱsineȱwaveȱisȱgreaterȱthanȱtheȱslewȱrate A) distortionȱoccurs.

B) theȱoutputȱlooksȱexactlyȱlikeȱtheȱinput.

C) voltageȱgainȱisȱmaximum.

D) theȱop-ampȱworksȱbest.

Diff:ȱ2

39) Whichȱop-ampȱparameterȱisȱdependentȱuponȱtheȱhighȱfrequencyȱresponseȱofȱtheȱamplifier stagesȱinsideȱtheȱop-amp? A) Inputȱbiasȱcurrent

B) Short-circuitȱoutputȱcurrent

C) Slewȱrate

D) Inputȱoffsetȱvoltage

Diff:ȱ2

40) Withȱnegativeȱfeedback,ȱtheȱreturningȱsignal A) aidsȱtheȱinputȱsignal. B) opposesȱtheȱinputȱsignal. C) isȱproportionalȱtoȱoutputȱcurrent. D) isȱproportionalȱtoȱdifferentialȱvoltageȱgain. Diff:ȱ2

368

41) Theȱclosed-loopȱvoltageȱgainȱofȱanȱinvertingȱamplifierȱequals A) theȱratioȱofȱtheȱinputȱresistanceȱtoȱtheȱfeedbackȱresistance. B) theȱopen-loopȱvoltageȱgain. C) theȱfeedbackȱresistanceȱdividedȱbyȱtheȱinputȱresistance. D) theȱinputȱresistance. Diff:ȱ2

42) Anȱinvertingȱamplifierȱwithȱaȱgainȱofȱ-8ȱmeansȱthat A) theȱsignalȱisȱattenuatedȱbyȱaȱfactorȱofȱ8. B) acȱandȱdcȱsignalsȱareȱincreasedȱbyȱaȱfactorȱofȱ8ȱwithȱnoȱphaseȱshift. C) theȱsignalȱamplitudeȱisȱincreasedȱbyȱaȱfactorȱofȱ8ȱwithȱaȱ180 ° phaseȱshift. D) theȱstageȱreducesȱtheȱsignalȱbyȱaȱfactorȱofȱ8ȱfromȱinputȱtoȱoutput. Diff:ȱ2

43) Theȱvoltageȱfollowerȱhasȱa A) closed-loopȱvoltageȱgainȱofȱunity.

B) smallȱopen-loopȱvoltageȱgain.

C) closed-loopȱbandwidthȱofȱzero.

D) largeȱclosed-loopȱoutputȱimpedance.

Diff:ȱ2

44) TheȱfeedbackȱfractionȱȈBȈ A) isȱalwaysȱlessȱthanȱ1.

B) isȱusuallyȱgreaterȱthanȱ1.

C) mayȱequalȱ1.

D) mayȱnotȱequalȱ1.

Diff:ȱ2

45) TheȱloopȱgainȱA OLB A) isȱusuallyȱmuchȱsmallerȱthanȱ1.

B) isȱusuallyȱmuchȱgreaterȱthanȱ1.

C) mayȱnotȱequalȱ1.

D) isȱbetweenȱ0ȱandȱ1.

Diff:ȱ2

46) Thereȱcanȱbeȱnoȱcurrentȱtoȱgroundȱthrough A) aȱmechanicalȱground.

B) anȱacȱground.

C) aȱvirtualȱground.

D) anȱordinaryȱground.

Diff:ȱ2

369

47) Theȱclosed-loopȱinputȱimpedanceȱinȱaȱnoninvertingȱamplifierȱis A) muchȱgreaterȱthanȱtheȱopen-loopȱinputȱimpedance. B) equalȱtoȱtheȱopen-loopȱinputȱimpedance. C) sometimesȱlessȱthanȱtheȱopen-loopȱinputȱimpedance. D) ideallyȱzero. Diff:ȱ2

48) Whatȱisȱtheȱvalueȱofȱcompensatingȱresistorȱneededȱforȱaȱnoninvertingȱop-ampȱcircuitȱusingȱ10 k̛ȱresistorsȱforȱbothȱRIȱandȱRF? A) 5ȱk̛

B) 10ȱk̛

C) 20ȱk̛

D) 100ȱk̛

Diff:ȱ2

49) Theȱvoltageȱgainȱofȱanȱop-ampȱisȱunityȱatȱthe A) cutoffȱfrequency.

B) unity-gainȱfrequency.

C) generatorȱfrequency.

D) powerȱbandwidth.

Diff:ȱ2

50) Forȱaȱgivenȱop-amp,ȱwhichȱofȱtheseȱisȱconstant? A) fC(CL)

B) feedbackȱvoltage

C) AOL

D) AOL f c(OL)

Diff:ȱ3

51) Ifȱtheȱunit-gainȱfrequencyȱisȱ10ȱMHzȱandȱtheȱmid -bandȱopen-loopȱvoltageȱgainȱisȱ200,000,ȱthe cutoffȱfrequencyȱis A) 5ȱkHz.

B) 50ȱHz.

C) 5ȱHz.

Diff:ȱ1

370

D) 25ȱMHz.

Chapterȱ13 BasicȱOp-AmpȱCircuits 1) Anȱop-ampȱcomparatorȱhasȱanȱoutputȱdependentȱuponȱtheȱpolaritiesȱofȱtheȱtwoȱinputs. Diff:ȱ2

2) Theȱoutputȱvoltageȱofȱaȱsummingȱamplifierȱisȱproportionalȱtoȱtheȱproductȱofȱtheȱinput voltages. Diff:ȱ2

3) Integrationȱisȱaȱmathematicalȱprocessȱforȱdeterminingȱtheȱareaȱunderȱaȱcurve. Diff:ȱ2

4) Aȱsquareȱwaveȱinputȱtoȱanȱop-ampȱintegratorȱwillȱproduceȱaȱsineȱwaveȱoutput. Diff:ȱ2

5) Boundingȱallowsȱtheȱop-ampȱtoȱhaveȱunlimitedȱoutputȱvoltageȱupȱtoȱtheȱvalueȱofȱtheȱsupply voltages. Diff:ȱ2

6) Anȱop-ampȱhasȱanȱopen-loopȱgainȱofȱ100,000.ȱȱV sat = +/-12ȱV.ȱȱAȱdifferentialȱsignalȱvoltageȱof 150ȱΐVp-pȱisȱappliedȱbetweenȱtheȱinputs.ȱȱTheȱoutputȱvoltageȱis A) 12ȱV.

B) -12ȱV.

C) 12ȱVp-p.

D) 24ȱVp-p.

Diff:ȱ2

7) Aȱsummingȱamplifierȱcanȱadd A) dcȱvoltages.

B) acȱvoltages.

C) dcȱtoȱacȱvoltages.

D) Allȱofȱtheȱabove.

Diff:ȱ2

8) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱIfȱV CC = 15ȱV,ȱtheȱapproximateȱoutputȱvoltageȱis A) 1ȱV.

B) -1ȱV.

C) 13ȱV.

Diff:ȱ2

371

D) -13ȱV.

9) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱIfȱVsat = +/-12ȱV,ȱtheȱapproximateȱoutputȱvoltageȱis A) 12ȱV.

B) -12ȱV.

C) 2ȱV.

D) -2ȱV.

Diff:ȱ2

10) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱWithȱtheȱinputsȱshown,ȱtheȱoutputȱvoltageȱwouldȱbe A) 7ȱV.

B) -7ȱV.

C) +Vsat.

D) -Vsat.

Diff:ȱ2

11) Eachȱofȱtheȱfollowingȱisȱusedȱtoȱestablishȱaȱreferenceȱvoltageȱonȱaȱcomparatorȱinputȱexceptȱa A) voltageȱdivider.

B) zenerȱdiode.

C) LED.

D) battery.

Diff:ȱ2

12) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱIfȱaȱsolderȱsplashȱshortedȱtheȱtwoȱendsȱofȱtheȱfeedbackȱresistor toȱeachȱother,ȱtheȱoutputȱvoltageȱwouldȱbe A) 0.5ȱV.

B) -0.5ȱV.

C) 0ȱV.

D) -Vsat.

Diff:ȱ3

13) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱAȱvoltmeterȱplacedȱfromȱtheȱinvertingȱinputȱtoȱgroundȱwould read A) -0.925ȱV.

B) -2.775ȱV.

C) 2.775ȱV.

Diff:ȱ3

372

D) Yȱ0ȱV.

14) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱThisȱcircuitȱisȱknownȱasȱa A) multivibrator.

B) zeroȱlevelȱdetector.

C) comparatorȱwithȱhysteresis.

D) noninvertingȱamplifier.

Diff:ȱ2

15) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱThisȱtypeȱofȱcircuitȱwillȱusuallyȱhave A) aȱsquareȱwaveȱoutputȱifȱtheȱinputȱisȱaȱsineȱwave. B) aȱtriangleȱwaveȱoutput. C) aȱrampȱoutputȱforȱaȱsquareȱwaveȱinput. D) Noneȱofȱtheȱabove. Diff:ȱ2

16) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱTheȱoutputȱvoltageȱwithȱtheȱinputsȱasȱshownȱis A) +Vsat.

B) -Vsat.

C) 26ȱVp-p.

Diff:ȱ3

17) Seeȱ(a)ȱinȱtheȱfigureȱabove.ȱȱThisȱcircuitȱisȱknownȱas A) aȱnoninvertingȱamplifier.

B) anȱintegrator.

C) aȱdifferentiator.

D) aȱsummingȱamplifier.

Diff:ȱ2

373

D) 17.06ȱVp-p.

18) Referȱtoȱtheȱfigureȱabove.ȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱanȱintegrator? A) (a)

B) (b)

C) Neitherȱofȱtheȱabove.

Diff:ȱ2

19) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱAȱsquareȱwaveȱinputȱisȱappliedȱtoȱthisȱcircuit.ȱȱTheȱoutput voltageȱisȱmostȱlikelyȱtoȱbe A) aȱsquareȱwave.

B) aȱtriangleȱwave.

C) aȱsineȱwave.

D) noȱoutput.

Diff:ȱ2

20) AȱSchmittȱtriggerȱisȱaȱcomparatorȱwith A) hysteresis.

B) oneȱtriggerȱpoint.

C) twoȱtriggerȱpoints.

D) AȱandȱCȱabove.

Diff:ȱ2

21) Anȱop-ampȱhasȱanȱopen-loopȱgainȱofȱ90,000.ȱȱV sat = +/-13ȱV.ȱȱAȱdifferentialȱvoltageȱofȱ0.1 Vp-pȱisȱappliedȱbetweenȱtheȱinputs.ȱȱTheȱoutputȱvoltageȱis A) 13ȱV.

B) -13ȱV.

C) 13ȱVp-p.

D) 26ȱVp-p.

C) sawtooth.

D) squareȱwave.

Diff:ȱ3

22) TheȱoutputȱofȱaȱSchmittȱtriggerȱisȱa A) triangleȱwave.

B) sineȱwave.

Diff:ȱ2

23) Anȱintegratorȱcircuit A) usesȱaȱcapacitorȱinȱitsȱfeedbackȱcircuit. B) producesȱaȱrampȱvoltageȱatȱitsȱoutputȱforȱaȱstepȱinputȱvoltage. C) usesȱanȱinductorȱinȱitsȱfeedbackȱcircuit. D) AȱandȱBȱabove. Diff:ȱ2

24) Aȱdifferentiatorȱcircuit A) usesȱaȱresistorȱinȱitsȱfeedbackȱcircuit. B) usesȱaȱcapacitorȱinȱitsȱfeedbackȱcircuit. C) producesȱaȱsquareȱwaveȱatȱitsȱoutputȱforȱaȱtriangleȱwaveȱinput. D) AȱandȱCȱabove. Diff:ȱ2

374

25) Hysteresisȱvoltageȱisȱdefinedȱas A) theȱvoltageȱofȱtheȱlowerȱtriggerȱpoint. B) theȱvoltageȱofȱtheȱupperȱtriggerȱpoint. C) theȱdifferenceȱinȱvoltageȱbetweenȱtheȱupperȱandȱtheȱlowerȱtriggerȱpoints. D) theȱsumȱofȱvoltagesȱofȱtheȱupperȱandȱtheȱlowerȱtriggerȱpoints. Diff:ȱ2

26) Aȱcomparatorȱisȱanȱexampleȱofȱa(n) A) activeȱfilter.

B) currentȱsource.

C) linearȱcircuit.

D) nonlinearȱcircuit.

Diff:ȱ2

27) Ifȱtheȱinputȱtoȱaȱcomparatorȱisȱaȱsineȱwave,ȱtheȱoutputȱisȱa A) rampȱvoltage.

B) sineȱwave.

C) rectangularȱwave.

D) sawtoothȱwave.

Diff:ȱ2

28) Aȱzeroȱcrossingȱdetectorȱisȱa A) comparatorȱwithȱnoȱoutput. B) comparatorȱwithȱaȱtripȱpointȱreferencedȱtoȱzero. C) peakȱdetector. D) limiter. Diff:ȱ2

29) Aȱwindowȱcomparator A) hasȱonlyȱoneȱusableȱthreshold. B) usesȱhysteresisȱtoȱspeedȱupȱresponse. C) clampsȱtheȱinputȱpositively. D) detectsȱanȱinputȱvoltageȱbetweenȱtwoȱlimits. Diff:ȱ2

30) AȱSchmittȱtriggerȱhas A) onlyȱoneȱtripȱpoint.

B) onlyȱnegativeȱfeedback.

C) twoȱslightlyȱdifferentȱtripȱpoints.

D) aȱtriangularȱoutput.

Diff:ȱ2

375

31) AȱSchmittȱtriggerȱisȱaȱcomparatorȱwith A) negativeȱfeedback.

B) positiveȱfeedback.

C) NeitherȱAȱnorȱB.

D) BothȱAȱandȱB.

Diff:ȱ2

32) Theȱ________ȱcircuitȱovercomesȱtheȱproblemȱofȱfalseȱswitchingȱcausedȱbyȱnoiseȱonȱtheȱinput(s). A) inputȱbuffer

B) Schmittȱtrigger

C) inputȱnoiseȱeliminator

D) differentiator

Diff:ȱ2

33) TheȱamountȱofȱhysteresisȱinȱaȱSchmittȱtriggerȱisȱdefinedȱby VUTP . A) VUTPȱ-ȱVLTP. B) VUTP xȱVLTP. C) VLTP

D) VUTPȱ+ VLTP.

Diff:ȱ3

34) Outputȱboundingȱisȱtheȱprocessȱofȱ________ȱtheȱoutputȱvoltageȱrangeȱofȱaȱcomparator. A) extending

B) limiting

C) comparing

D) filtering

Diff:ȱ2

35) Ifȱallȱtheȱresistorsȱinȱaȱsummingȱamplifierȱareȱequal,ȱtheȱoutputȱwillȱbeȱequalȱtoȱthe A) averageȱofȱtheȱindividualȱinputs. B) invertedȱaverageȱofȱtheȱindividualȱinputs. C) sumȱofȱtheȱindividualȱinputs. D) invertedȱsumȱofȱtheȱindividualȱinputs. Diff:ȱ2

36) IfȱtheȱvalueȱofȱresistorȱR fȱinȱaȱaveragingȱamplifierȱcircuitȱisȱequalȱtoȱtheȱvalueȱofȱoneȱinput resistorȱdividedȱbyȱtheȱnumberȱofȱinputs,ȱtheȱoutputȱwillȱbeȱequalȱto A) averageȱofȱtheȱindividualȱinputs. B) invertedȱaverageȱofȱtheȱindividualȱinputs. C) sumȱofȱtheȱindividualȱinputs. D) invertedȱsumȱofȱtheȱindividualȱinputs. Diff:ȱ3

37) Theȱ________ȱinputȱmakesȱtheȱsummingȱamplifierȱcircuitȱpossible. A) virtualȱgroundȱatȱtheȱnoninverting

B) virtualȱgroundȱatȱtheȱinverting

C) lowȱvoltage

D) highȱvoltage

Diff:ȱ3

376

38) AȱD/Aȱconverterȱisȱanȱapplicationȱofȱthe A) adjustableȱbandwidthȱcircuit.

B) noninvertingȱamplifier.

C) voltage-to-currentȱconverter.

D) scalingȱadder.

Diff:ȱ2

39) Inȱanȱop-ampȱintegrator,ȱtheȱcurrentȱthroughȱtheȱinputȱresistorȱisȱintoȱthe A) invertingȱinput.

B) noninvertingȱinput.

C) bypassȱcapacitor.

D) feedbackȱcapacitor.

Diff:ȱ2

40) Aȱmathematicalȱoperationȱthatȱdeterminesȱtheȱrateȱofȱchangeȱofȱaȱcurveȱisȱcalled A) differentiation.

B) integration.

C) curveȱaveraging.

D) linearȱregression.

Diff:ȱ2

41) Aȱmathematicalȱoperationȱforȱfindingȱtheȱareaȱunderȱtheȱcurveȱofȱaȱgraphȱisȱcalled A) differentiation.

B) integration.

C) curveȱaveraging.

D) linearȱregression.

Diff:ȱ2

42) TheȱformulaȱICȱ=ȱȱ

VC t

ȱCȱshowsȱthatȱforȱaȱgivenȱcapacitor,ȱifȱtheȱvoltageȱchangesȱatȱaȱconstant

rateȱwithȱrespectȱtoȱtime,ȱthenȱcurrentȱwill A) increase.

B) decrease.

C) beȱconstant.

D) decreaseȱlogarithmically.

Diff:ȱ3

43) Theȱoutputȱofȱanȱop-ampȱdifferentiatorȱwithȱaȱrectangularȱinputȱisȱa A) seriesȱofȱpositiveȱandȱnegativeȱspikes. B) sineȱwave. C) rampȱvoltage. Diff:ȱ2

44) Aȱboundedȱcomparatorȱusesȱ________ȱinȱtheȱfeedbackȱcircuit. A) equal-valuedȱresistors

B) zenerȱdiodes

C) anȱinductor

D) aȱresonantȱfilter

Diff:ȱ2

377

45) Aȱtwo-inputȱsummingȱamplifierȱhasȱinputȱvoltagesȱofȱ3ȱVȱandȱ4ȱV.ȱIfȱRf isȱopen,ȱtheȱoutput voltageȱwillȱbeȱapproximately A) +7ȱV.

B) -7ȱV.

C) +Vsat.

D) -Vsat.

Diff:ȱ3

46) AnȱR/2Rȱladderȱcircuitȱwouldȱbeȱfoundȱinȱa A) commonȱemitterȱamplifier.

B) differentialȱamplifier.

C) Schmittȱtriggerȱcircuit.

D) D/Aȱconverterȱcircuit.

Diff:ȱ2

378

Chapterȱ14 Special-PurposeȱOp-AmpȱCircuits 1) Aȱbasicȱinstrumentationȱamplifierȱhasȱthreeȱop-amps. Diff:ȱ2

2) Oneȱofȱtheȱkeyȱcharacteristicsȱofȱanȱinstrumentationȱamplifierȱisȱitsȱlowȱinputȱimpedance. Diff:ȱ2

3) Theȱvoltageȱgainȱofȱanȱinstrumentationȱamplifierȱisȱsetȱwithȱanȱexternalȱresistor. Diff:ȱ2

4) Aȱbasicȱisolationȱamplifierȱhasȱtwoȱelectricallyȱisolatedȱsections. Diff:ȱ2

5) Mostȱmodernȱisolationȱamplifiersȱuseȱtransformerȱcouplingȱforȱisolation. Diff:ȱ2

6) OTAȱstandsȱforȱoperationalȱtransistorȱamplifier. Diff:ȱ2

7) AȱlogȱamplifierȱhasȱaȱJFETȱinȱtheȱfeedbackȱloop. Diff:ȱ2

8) AnȱantilogȱamplifierȱhasȱaȱBJTȱinȱseriesȱwithȱtheȱinput. Diff:ȱ2

9) Theȱmainȱpurposeȱofȱanȱinstrumentationȱamplifierȱisȱtoȱamplifyȱcommonȱmodeȱvoltage. Diff:ȱ2

10) TheȱOTAȱisȱaȱvoltage-to-currentȱamplifier. Diff:ȱ2

11) Linearȱsignalȱcompressionȱscalesȱdownȱallȱsignalȱlevelsȱbyȱtheȱsameȱamount. Diff:ȱ2

12) TheȱOTAȱhasȱaȱ________ȱinputȱimpedanceȱandȱaȱ________ȱCMRR. A) high,ȱlow

B) low,ȱhigh

Diff:ȱ2

379

C) high,ȱhigh

13) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱanȱantilogȱamplifier? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

14) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱconstant-currentȱsource? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

15) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱanȱisolationȱamplifier? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

16) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱanȱinstrumentationȱamplifier? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

17) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱIfȱR 1 = R2ȱ= 30ȱk̛ andȱtheȱclosedȱloopȱgainȱisȱ450,ȱtheȱvalueȱof theȱexternalȱgain-settingȱresistorȱR Gȱis A) 133.63ȱk̛.

B) 133.63ơ̇̄.

C) 13.36ơ̇̄.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

380

18) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱIfȱR 1 = R2 = 28ȱk̛ andȱRG = 100ơ̇̄,ȱtheȱAclȱwouldȱbe A) 5.60.

B) 56.1.

C) 561.

D) 560.

Diff:ȱ2

19) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱIfȱVin = 5ȱVȱandȱR in = 22ȱk̛,ȱtheȱcurrentȱthruȱtheȱloadȱR L wouldȱbe A) 227.27ȱmA.

B) 2.27ȱmA.

C) 22.72ȱmA.

D) 227.27ȱΐA.

Diff:ȱ2

20) Referȱtoȱ(d)ȱinȱtheȱfigureȱabove.ȱȱIfȱV in = 200ȱmV,ȱRF = 52ȱk,ȱandȱIEBO = 50ȱnA,ȱtheȱV outȱwould be A) 7.75ȱmV.

B) -5.41ȱmV.

C) -7.75ȱV.

D) -775ȱmV.

Diff:ȱ2

21) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱvoltage-to-currentȱconverter? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

22) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱcurrent-to-voltageȱconverter? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

23) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱcontainsȱanȱOTA? A) (a)

B) (b)

C) (c)

Diff:ȱ2

381

D) (d)

24) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱpeakȱdetector? A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

25) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱIfȱR L = 20ȱk,ȱR1 = 1.2ȱk,ȱandȱV in = 2.5ȱV,ȱtheȱloadȱcurrentȱI L wouldȱbe A) 20.83ȱmA.

B) 2.083ȱmA.

C) 2.083ȱA.

D) 208.3ȱΐA.

Diff:ȱ2

26) Referȱtoȱ(d)ȱinȱtheȱfigureȱabove.ȱȱIfȱg m = 25ȱmSȱandȱR L = 25ȱk̛,ȱtheȱvoltageȱgainȱwouldȱbe A) 625.

B) 62.5.

C) 6.25.

D) Notȱenoughȱinformationȱisȱgiven.

Diff:ȱ2

27) Theȱinputȱsignalȱforȱanȱinstrumentationȱamplifierȱusuallyȱcomesȱfrom A) anȱinvertingȱamplifier.

B) aȱfilteredȱdcȱpowerȱsupply.

C) aȱdifferentialȱamplifier.

D) aȱWheatstoneȱbridge.

Diff:ȱ2

28) Instrumentationȱamplifiersȱareȱwellȱsuitedȱforȱenvironmentsȱwith A) noȱsignalȱandȱnoȱnoise. B) highȱsignalȱlevelsȱthatȱneedȱtoȱbeȱattenuated. C) noȱpowerȱsupplyȱvoltages. D) smallȱsignalsȱsuperimposedȱonȱlargerȱcommon-modeȱvoltages. Diff:ȱ3

29) Anȱinstrumentationȱamplifierȱhasȱaȱhigh A) outputȱimpedance.

B) powerȱgain.

C) CMRR.

D) supplyȱvoltage.

Diff:ȱ2

30) Anȱinputȱtransducerȱconverts A) voltageȱtoȱcurrent. B) currentȱtoȱvoltage. C) anȱelectricalȱquantityȱtoȱaȱnonelectricalȱquantity. D) aȱnonelectricalȱquantityȱtoȱanȱelectricalȱquantity. Diff:ȱ2

382

31) Inȱsomeȱrespectsȱanȱisolationȱamplifierȱisȱnothingȱmoreȱthanȱanȱelaborate A) op-amp.

B) instrumentationȱamplifier.

C) rectifierȱandȱfilter.

D) BothȱAȱandȱBȱabove.

Diff:ȱ2

32) Theȱprimaryȱfunctionȱofȱtheȱoscillatorȱinȱanȱisolationȱamplifierȱisȱto A) convertȱdcȱtoȱhighȱfrequencyȱac. B) convertȱdcȱtoȱlowȱfrequencyȱac. C) rectifyȱhighȱfrequencyȱacȱtoȱdc. D) produceȱdualȱpolarityȱdcȱvoltagesȱforȱtheȱinputȱtoȱtheȱdemodulator. Diff:ȱ3

33) TheȱvoltageȱgainȱofȱanȱOTAȱcanȱbeȱcalculatedȱusingȱtheȱformula A) AVȱ=ȱȱ

Rf . ȱRi

B) AV = g m RL.

Rf C) AVȱ=ȱ Ri ȱ+ȱ1.

D) AVȱ=ȱ

2Rf ȱRi

.

Diff:ȱ2

34) Ifȱanȱoperationalȱtransconductanceȱamplifierȱ(OTA)ȱisȱusedȱasȱaȱnonlinearȱmixerȱandȱanȱaudio signalȱisȱmixedȱwithȱanȱRFȱsignal,ȱtheȱoutputȱwillȱbeȱa(n)ȱ________ȱsignal. A) squareȱwave

B) triangularȱwave

C) frequencyȱmodulatedȱ(FM)

D) amplitudeȱmodulatedȱ(AM)

Diff:ȱ3

35) WhenȱusingȱanȱOTAȱinȱaȱSchmitt-triggerȱconfiguration,ȱtheȱtriggerȱpointsȱareȱcontrolledȱby A) theȱIOUT.

B) theȱIBIAS.

C) theȱVOUT.

D) BothȱAȱandȱBȱabove.

Diff:ȱ3

36) Anȱantilogȱamplifierȱisȱformedȱbyȱconnectingȱaȱpnȱjunctionȱ(diodeȱorȱBJT)ȱtoȱthe A) input.

B) output.

C) feedbackȱloop.

D) invertingȱandȱnoninvertingȱinputs.

Diff:ȱ3

383

37) Toȱscaleȱdownȱlargeȱsignalȱvoltagesȱwithoutȱobscuringȱlowerȱsignalȱvoltages,ȱ________ȱshould beȱused. A) signalȱcompression

B) logarithmicȱsignalȱcompression

C) naturalȱlogarithmicȱsignalȱcompression

D) antilogarithmicȱsignalȱcompression

Diff:ȱ3

38) Aȱvoltage-to-currentȱconverterȱisȱusedȱinȱapplicationsȱwhereȱitȱisȱnecessaryȱtoȱhaveȱanȱoutput loadȱcurrentȱthatȱisȱcontrolledȱby A) inputȱvoltage.

B) inputȱresistance.

C) outputȱresistance.

D) inputȱfrequency.

Diff:ȱ3

39) Theȱoutputȱofȱaȱpeakȱdetectorȱisȱalways A) 70.7%ȱofȱinput. B) equalȱtoȱtheȱmaxȱvalueȱofȱtheȱpeakȱlevelȱreceivedȱsinceȱtheȱlastȱresetȱpulse. C) equalȱtoȱtheȱminȱvalueȱofȱtheȱpeakȱlevelȱreceivedȱsinceȱtheȱlastȱresetȱpulse. D) Noneȱofȱtheȱabove. Diff:ȱ2

40) Whichȱofȱtheȱfollowingȱcorrectlyȱdescribesȱtheȱrelationshipȱbetweenȱbandwidthȱandȱgainȱforȱan instrumentationȱamplifier? A) Instrumentationȱamplifiersȱhaveȱnoȱbandwidthȱsinceȱtheyȱcanȱonlyȱamplifyȱdc. B) Bandwidthȱandȱgainȱareȱbothȱzero. C) Bandwidthȱincreasesȱasȱgainȱincreases. D) Bandwidthȱdecreasesȱasȱgainȱincreases. Diff:ȱ3

41) Isolationȱamplifiersȱcanȱaccomplishȱisolationȱusing A) transformerȱcoupling.

B) opticalȱcoupling.

C) capacitiveȱcoupling.

D) Allȱofȱtheȱabove.

Diff:ȱ2

42) Theȱvoltageȱgainȱofȱtheȱinputȱstageȱofȱanȱisolationȱamplifierȱisȱ12.ȱIfȱtheȱoutputȱstageȱhasȱaȱgain ofȱ6,ȱtheȱtotalȱvoltageȱgainȱis A) 6.

B) 18.

C) 72.

Diff:ȱ2

384

D) 144.

43) Theȱtotalȱvoltageȱgainȱofȱanȱisolationȱamplifierȱisȱ49.ȱIfȱtheȱvoltageȱgainȱofȱtheȱinputȱandȱoutput stagesȱareȱequal,ȱtheȱvoltageȱgainȱforȱeachȱstageȱis A) 4.9.

B) 7.

C) 24.5.

D) 98.

Diff:ȱ3

44) Logȱamplifiersȱareȱusedȱto A) compressȱanȱanalogȱsignal. B) linearizeȱanȱexponentialȱoutputȱfromȱaȱtransducer. C) performȱanalogȱmultiplicationȱandȱdivision. D) Allȱofȱtheȱabove. Diff:ȱ2

45) Whatȱisȱtheȱrequiredȱvalueȱofȱfeedbackȱresistanceȱneededȱinȱaȱcurrent-to-voltageȱconverter whereȱeachȱ100ȱΐAȱofȱinputȱcurrentȱneedsȱtoȱresultȱinȱanȱoutputȱofȱ1ȱV? A) 1ȱk̛

B) 10ȱk̛

C) 1ȱM̛

D) 10ȱM̛

Diff:ȱ2

46) Theȱcurrentȱappliedȱtoȱaȱcurrent-to-voltageȱconverterȱcircuitȱchangesȱfromȱ4ȱtoȱ20ȱmA.ȱIfȱRf is 625ơ̇̄,ȱwhatȱisȱtheȱchangeȱinȱoutputȱvoltage? A) 1ȱV

B) 5ȱV

C) 10ȱV

Diff:ȱ2

385

D) 20ȱV

Chapterȱ15 ActiveȱFilters 1) Theȱbandwidthȱofȱaȱband-passȱfilterȱisȱtheȱdifferenceȱbetweenȱtheȱtwoȱcutoffȱfrequencies. Diff:ȱ2

2) Butterworthȱfiltersȱhaveȱaȱroll-offȱofȱ40ȱdB/decadeȱandȱaȱwidelyȱvaryingȱoutputȱinȱthe passband. Diff:ȱ2

3) Aȱhigh-passȱfilterȱpassesȱhighȱfrequenciesȱeasilyȱandȱattenuatesȱallȱothers. Diff:ȱ2

4) Aȱlow-passȱfilterȱattenuatesȱlowȱfrequencies. Diff:ȱ2

5) Aȱband-passȱfilterȱpassesȱallȱfrequenciesȱwithinȱaȱspecifiedȱbandȱandȱblocksȱallȱother frequencies. Diff:ȱ2

6) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱThisȱisȱtheȱfrequencyȱresponseȱcurveȱforȱa A) low-passȱfilter.

B) high-passȱfilter.

C) band-passȱfilter.

D) band-stopȱfilter.

Diff:ȱ2

386

7) Referȱtoȱ(b)ȱinȱtheȱfigureȱabove.ȱȱThisȱisȱtheȱfrequencyȱresponseȱcurveȱforȱa A) low-passȱfilter.

B) high-passȱfilter.

C) band-passȱfilter.

D) band-stopȱfilter.

Diff:ȱ2

8) Referȱtoȱ(c)ȱinȱtheȱfigureȱabove.ȱȱThisȱisȱtheȱfrequencyȱresponseȱcurveȱforȱa A) low-passȱfilter.

B) high-passȱfilter.

C) band-passȱfilter.

D) band-stopȱfilter.

Diff:ȱ2

9) Referȱtoȱ(d)ȱinȱtheȱfigureȱabove.ȱȱThisȱisȱtheȱfrequencyȱresponseȱcurveȱforȱa A) low-passȱfilter.

B) high-passȱfilter.

C) band-passȱfilter.

D) band-stopȱfilter.

Diff:ȱ2

10) Referȱtoȱtheȱfigureȱabove.ȱȱIdentifyȱtheȱactiveȱsingle-poleȱhigh-passȱfilter. A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

11) Referȱtoȱtheȱfigureȱabove.ȱȱIdentifyȱtheȱhigh-passȱfilterȱwithȱaȱ40ȱdB/decadeȱroll-off. A) (a)

B) (b)

C) (c)

Diff:ȱ2

12) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱlow-passȱfilterȱwithȱaȱ20ȱdB/decadeȱroll-offȱis A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

387

D) (d)

13) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱband-passȱfilterȱis A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

14) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱlow-passȱfilterȱwithȱaȱroll-offȱofȱ40ȱdB/decadeȱis A) (a)

B) (b)

C) (c)

D) (d)

Diff:ȱ2

15) Referȱtoȱ(a)ȱinȱtheȱfigureȱabove.ȱȱThisȱcircuitȱwasȱcheckedȱforȱproperȱoperationȱandȱf Cȱwas correctȱbutȱtheȱvoltageȱgainȱisȱ1.ȱTheȱcauseȱofȱthisȱproblemȱmightȱbeȱthat A) theȱ1.2ȱk̛ȱresistorȱisȱopen.

B) theȱcapacitorȱisȱshorted.

C) RAȱisȱopen.

D) RBȱisȱopen.

Diff:ȱ3

16) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱcircuitȱisȱknownȱasȱaȱ________ȱandȱtheȱroll -offȱrateȱisȱ________. A) low-passȱfilter,ȱ60ȱdB/decade

B) high-passȱfilter,ȱ20ȱdB/decade

C) high-passȱfilter,ȱ80ȱdB/decade

D) band-passȱfilter,ȱ80ȱdB/decade

Diff:ȱ3

17) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱcutoffȱfrequencyȱforȱtheȱfirstȱfilterȱsectionȱisȱ________ȱtheȱcutoff frequencyȱforȱtheȱsecondȱsection. A) equalȱto

B) higherȱthan

C) lowerȱthan

D) Noneȱofȱtheȱabove.

Diff:ȱ2

18) Anotherȱnameȱforȱaȱband-stopȱfilterȱisȱa(n)ȱ________ȱfilter. A) low-cut

B) all-stop.

C) notch

Diff:ȱ2

388

D) all-pass

19) Aȱhigh-passȱactiveȱfilterȱhasȱaȱcutoffȱfrequencyȱofȱ1.23ȱkHz.ȱȱTheȱbandwidthȱofȱthisȱfilterȱis A) 2.46ȱkHz.

B) 1.23ȱkHz.

C) 644ȱHz.

D) Cannotȱdetermine.

Diff:ȱ2

20) Aȱlow-passȱfilterȱwithȱaȱroll-offȱrateȱofȱ60ȱdB/decadeȱisȱneeded.ȱȱTheȱbestȱcombinationȱtoȱuseȱis A) aȱ2-poleȱfilterȱfollowedȱbyȱanotherȱ2-pole. B) twoȱsingle-poleȱfiltersȱinȱseries. C) aȱ2-poleȱfilterȱfollowedȱbyȱaȱ1-pole. D) Noneȱofȱtheȱabove. Diff:ȱ2

21) Aȱhigh-passȱfilterȱhasȱRȱ=ȱ47ȱk̛ andȱCȱ= 0.002ȱΐF.ȱȱTheȱcutoffȱfrequencyȱis A) 1.69ȱkHz.

B) 10.6ȱkHz.

C) 3.39ȱHz.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

22) Aȱpoleȱisȱaȱnetworkȱthatȱcontains A) aȱresistorȱandȱaȱcapacitor.

B) aȱresistorȱandȱanȱinductor.

C) aȱcapacitorȱandȱanȱinductor.

D) twoȱresistorsȱandȱoneȱinductor.

Diff:ȱ2

23) Aȱmaximallyȱflatȱfrequencyȱresponseȱisȱaȱcommonȱnameȱfor A) Chebyshev.

B) Bessel.

C) Butterworth.

D) Colpitts.

Diff:ȱ2

24) AȱsingleȱRCȱcircuitȱproducesȱaȱroll-offȱrateȱof A) -20ȱdB/decade.

B) -6ȱdB/octave.

C) -40ȱdB/decade.

D) AȱandȱBȱabove.

Diff:ȱ3

25) Aȱlow-passȱfilterȱhasȱaȱcutoffȱfrequencyȱofȱ1.23ȱkHz.ȱȱDetermineȱtheȱbandwidthȱofȱtheȱfilter. A) 2.46ȱkHz

B) 1.23ȱkHz

C) 644ȱHz

D) Notȱenoughȱinformationȱisȱgiven.

Diff:ȱ3

389

26) Aboveȱtheȱcutoffȱfrequencyȱofȱaȱlow-passȱfilter,ȱtheȱoutputȱvoltage A) doesȱnotȱchange. B) doublesȱforȱeveryȱ1ȱkHzȱincreaseȱinȱfrequency. C) increases. D) decreases. Diff:ȱ3

27) Theȱcenterȱfrequencyȱofȱaȱband-passȱfilterȱisȱalwaysȱequalȱtoȱthe A) bandwidth. B) geometricȱaverageȱofȱtheȱcutoffȱfrequencies. C) bandwidthȱdividedȱbyȱQ. D) 3-dBȱfrequency. Diff:ȱ1

28) Band-passȱfiltersȱareȱdesignedȱtoȱpassȱaȱbandȱofȱfrequenciesȱbetween A) fc1ȱandȱfc2.

B) aȱband-startȱandȱband-stop.

C) 1ȱkHzȱandȱ10ȱkHz.

D) 1ȱkHzȱandȱ10ȱMHz.

Diff:ȱ2

29) Low-Qȱfiltersȱareȱ________ȱcircuits,ȱandȱhigh-Qȱfiltersȱareȱ________ȱcircuits. A) band-pass,ȱband-stop

B) wideȱband-pass,ȱnarrowȱband-pass

C) lowȱpass,ȱhighȱpass

D) lowȱorder,ȱhighȱorder

Diff:ȱ2

30) Aȱnotchȱfilterȱisȱa(n) A) all-passȱfilter.

B) band-passȱcircuit.

C) band-stopȱcircuit.

D) time-delayȱcircuit.

Diff:ȱ2

31) Theȱtypeȱofȱfilterȱresponseȱwithȱaȱrippledȱpassbandȱisȱthe A) Butterworth.

B) Chebyshev.

C) InverseȱChebyshev.

D) Bessel.

Diff:ȱ2

32) Theȱfilterȱresponseȱcharacteristicȱthatȱdistortsȱpulsesȱtheȱleastȱisȱthe A) Butterworth.

B) Chebyshev.

C) Elliptic.

Diff:ȱ3

390

D) Bessel.

33) Theȱfilterȱwithȱtheȱslowestȱroll-offȱrateȱisȱthe A) Butterworth.

B) Chebyshev.

C) Elliptic.

D) Bessel.

Diff:ȱ2

34) Theȱdampingȱfactorȱ(DF)ȱofȱanȱactiveȱfilterȱdeterminedȱby A) theȱpositiveȱfeedbackȱofȱtheȱcircuit.

B) theȱnegativeȱfeedbackȱofȱtheȱcircuit.

C) theȱnumberȱofȱpoles.

D) theȱQȱofȱtheȱcircuit.

Diff:ȱ2

35) IfȱaȱButterworthȱfilterȱhasȱ9ȱsecond-orderȱstages,ȱitsȱrolloffȱrateȱis A) 20ȱdBȱperȱdecade.

B) 40ȱdBȱperȱdecade.

C) 180ȱdBȱperȱdecade.

D) 360ȱdBȱperȱdecade.

Diff:ȱ2

36) Sallen-Keyȱfiltersȱareȱalsoȱcalled A) VCVSȱfilters.

B) multipleȱfeedbackȱfilters.

C) biquadraticȱfilters.

D) state-variableȱfilters.

Diff:ȱ2

37) Byȱcascadingȱlow-passȱfilters,ȱ________ȱcanȱbeȱimproved. A) bandwidth

B) roll-offȱrate

C) Q-rating

Diff:ȱ2

38) Aȱmultiple-feedbackȱband-passȱfilter A) usesȱaȱminimumȱofȱtwoȱop-amps. B) isȱusedȱforȱaȱnarrowȱbandȱ(highȱQ)ȱfilter. C) isȱusedȱforȱaȱwideȱbandȱ(lowȱQ)ȱfilter. D) isȱalsoȱknownȱasȱaȱSallen-Keyȱfilter. Diff:ȱ3

39) Theȱstate-variableȱfilter A) isȱdifficultȱtoȱtune. B) usesȱfewerȱthanȱthreeȱop-amps. C) hasȱhighȱcomponentȱsensitivity. D) hasȱaȱlow-pass,ȱhigh-passȱandȱband-passȱoutput. Diff:ȱ2

391

D) phaseȱshift

40) TheȱQȱofȱaȱstate-variableȱfilterȱisȱcontrolledȱbyȱthe R5 . A) ratioȱofȱ R6 C) ratioȱofȱtheȱfeedbackȱresistors.

B) productȱofȱR5ȱ× R6. D) BothȱAȱandȱCȱabove.

Diff:ȱ3

41) Theȱlowestȱfrequencyȱthatȱanȱactiveȱlow-passȱfilterȱcanȱpassȱis A) determinedȱbyȱtheȱunityȱgainȱbandwidthȱofȱtheȱop-amp. B) determinedȱbyȱtheȱfeedbackȱcapacitorȱvalue. C) determinedȱbyȱtheȱdampingȱfactor. D) 0ȱHz. Diff:ȱ2

42) TheȱQȱofȱaȱband-passȱfilterȱis A) theȱratioȱofȱtheȱcenterȱfrequencyȱtoȱtheȱbandwidth. B) notȱrelatedȱtoȱtheȱdampingȱfactor. C) theȱdifferenceȱbetweenȱtheȱtwoȱop-ampȱpowerȱsupplyȱvoltages. D) theȱquiescentȱcurrentȱspecificationȱofȱtheȱop-ampȱinputȱstage. Diff:ȱ2

43) Aȱband-stopȱfilterȱisȱalsoȱknownȱas A) aȱband-rejectȱfilter.

B) aȱnotchȱfilter.

C) aȱband-eliminationȱfilter.

D) Allȱofȱtheȱabove.

Diff:ȱ2

44) Whichȱofȱtheȱfollowingȱisȱnotȱanȱadvantageȱofȱactiveȱfiltersȱoverȱpassiveȱfilters? A) Activeȱfiltersȱprovideȱfilteringȱwithȱgain. B) Activeȱfiltersȱhaveȱminimalȱloadingȱdueȱtoȱaȱhighȱinputȱimpedance. C) Activeȱfiltersȱrequireȱnoȱpowerȱtoȱoperate. D) Activeȱfiltersȱhaveȱaȱlowȱoutputȱimpedance. Diff:ȱ3

45) Whatȱisȱtheȱvalueȱofȱresistanceȱneededȱwithȱaȱ0.01ȱ ΐFȱcapacitorȱforȱaȱcriticalȱfrequencyȱofȱ23.4 kHz? A) 680ơ̇̄

B) 2.34ȱk̛

C) 18ȱk̛

Diff:ȱ2

392

D) 132.9ȱk̛

Chapterȱ16 Oscillators 1) Toȱoperateȱproperly,ȱanȱoscillatorȱrequiresȱanȱexternalȱacȱinputȱsignal. Diff:ȱ2

2) Anȱoscillatorȱcanȱproduceȱmanyȱtypesȱofȱoutputsȱsuchȱasȱsine,ȱtriangle,ȱorȱsquareȱwaves. Diff:ȱ2

3) Positiveȱfeedbackȱisȱrequiredȱforȱanȱoscillatorȱtoȱoperateȱproperly. Diff:ȱ2

4) Crystalȱoscillatorsȱareȱveryȱstable. Diff:ȱ2

5) AnȱRCȱphase-shiftȱoscillatorȱusesȱfeedbackȱfromȱaȱtankȱcircuit. Diff:ȱ2

6) TheȱTwin-TȱOscillatorȱisȱaȱpopularȱchoiceȱbecauseȱitȱworksȱwellȱoverȱaȱwideȱrangeȱof frequencies. Diff:ȱ2

393

7) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱColpittsȱoscillator? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

8) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱClappȱoscillator? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

9) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱHartleyȱoscillator? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

394

10) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱC5 wereȱtoȱopen A) theȱresonantȱfrequencyȱwouldȱbeȱhigherȱthanȱnormal. B) theȱresonantȱfrequencyȱwouldȱbeȱlowerȱthanȱnormal. C) thereȱwouldȱbeȱnoȱsignalȱpresentȱatȱVout. D) theȱcircuitȱwouldȱdrawȱnoȱcurrentȱfromȱ+Vcc. Diff:ȱ2

11) Ofȱtheȱfollowingȱoscillatorȱcircuitȱcomponents,ȱtheȱoneȱthatȱexhibitsȱtheȱpiezoelectricȱeffectȱis the A) inductor.

B) capacitor.

C) crystal.

D) resistor.

Diff:ȱ2

12) Aȱveryȱstableȱoscillatorȱisȱneededȱtoȱoperateȱonȱaȱsingleȱfrequency.ȱȱAȱgoodȱchoiceȱmightȱbeȱa A) Hartley.

B) VCO.

C) crystal.

Diff:ȱ2

395

D) Clapp.

13) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱvalueȱofȱV CC = 5ȱV,ȱtheȱoutputȱvoltageȱwouldȱbe A) aȱsquareȱwaveȱofȱ10ȱV p-p. B) aȱsquareȱwaveȱthatȱvariesȱbetweenȱ0ȱVȱandȱ5ȱV. C) aȱsineȱwave. D) 5ȱVȱdc. Diff:ȱ2

14) Nonsinusoidalȱoscillatorsȱproduce A) sineȱwavesȱonly.

B) triangleȱwaves.

C) squareȱwaves.

D) EitherȱBȱorȱCȱabove.

Diff:ȱ2

15) Theȱ555ȱtimerȱcontains A) 2ȱcomparators.

B) 3ȱcomparators.

C) 4ȱcomparators.

D) 5ȱcomparators.

Diff:ȱ2

16) Aȱstableȱtypeȱofȱoscillatorȱis A) theȱClappȱoscillator.

B) theȱHartleyȱoscillator.

C) theȱcrystalȱoscillator.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

396

17) Referȱtoȱtheȱfigureȱabove.ȱȱToȱreduceȱtheȱdutyȱcycleȱtoȱlessȱthanȱ50%,ȱwhichȱofȱtheȱfollowing circuitȱchangesȱwouldȱbeȱnecessary? A) ReduceȱtheȱsizeȱofȱR1.

B) ReduceȱtheȱsizeȱofȱR2.

C) IncreaseȱtheȱsizeȱofȱR 1.

D) ConnectȱaȱdiodeȱinȱparallelȱwithȱR2 .

Diff:ȱ2

18) Aȱcircuitȱthatȱcanȱchangeȱtheȱfrequencyȱofȱoscillationȱwithȱanȱapplicationȱofȱaȱdcȱvoltageȱis sometimesȱcalledȱa(n) A) voltage-controlledȱoscillator.

B) crystalȱoscillator.

C) Hartleyȱoscillator.

D) astableȱmultivibrator.

Diff:ȱ2

19) Whichȱofȱtheȱfollowingȱisȱnotȱanȱessentialȱrequirementȱofȱaȱfeedbackȱoscillator? A) Positiveȱfeedbackȱnetwork B) Negativeȱfeedbackȱnetwork C) Phaseȱshiftȱaroundȱtheȱfeedbackȱloopȱofȱ0° D) Amplifierȱcircuit Diff:ȱ2

20) Inȱorderȱtoȱsustainȱoscillationsȱinȱaȱfeedbackȱoscillator,ȱtheȱgainȱshouldȱbeȱ________ȱsoȱthe productȱofȱAVȱ×ȱBȱequalsȱ________. A) reduced,ȱone

B) reduced,ȱlessȱthanȱone

C) increased,ȱmoreȱthanȱone

D) increased,ȱmuchȱgreaterȱthanȱone

Diff:ȱ2

21) Theȱvoltageȱthatȱstartsȱaȱfeedbackȱoscillatorȱisȱcausedȱby A) rippleȱfromȱtheȱpowerȱsupply.

B) thermalȱnoiseȱinȱresistors.

C) theȱinputȱsignalȱfromȱaȱgenerator.

D) positiveȱfeedback.

Diff:ȱ2

397

22) AȱWien-bridgeȱoscillatorȱuses A) positiveȱfeedback.

B) negativeȱfeedback.

C) bothȱtypesȱofȱfeedback.

D) anȱLCȱtankȱcircuit.

Diff:ȱ2

23) TheȱRCȱfeedbackȱnetworkȱusedȱinȱtheȱWien -bridgeȱoscillatorȱhasȱaȱmaximumȱoutputȱvoltage when A) XCȱ =ȱXL.

B) XL = R.

C) R1ȱ=ȱR2 ȱandȱXC1ȱ=ȱXC2.

D) Rȱ= 0ơ̇̄.

Diff:ȱ3

24) Theȱclosed-loopȱvoltageȱgain,ȱA CL,ȱforȱaȱWien-bridgeȱoscillatorȱis A) 3,ȱafterȱtheȱoscillationsȱhaveȱbuiltȱup.

B) slightlyȱgreaterȱthanȱ1.

C) lessȱthanȱ1.

D) exactlyȱ1.

Diff:ȱ2

25) Inȱorderȱforȱfeedbackȱoscillatorsȱtoȱoperateȱproperly,ȱtheȱgainȱhasȱtoȱbe A) 1/4.

B) self-adjusting.

C) stabilized.

D) nonlinear.

Diff:ȱ2

26) Theȱphase-shiftȱoscillatorȱusuallyȱhas A) twoȱleadȱorȱlagȱcircuits.

B) threeȱleadȱorȱlagȱcircuits.

C) noȱRCȱsections.

D) noȱfeedbackȱloop.

Diff:ȱ2

27) OneȱwayȱtoȱrecognizeȱaȱColpittsȱoscillatorȱisȱbyȱthe A) tappedȱinductorsȱinȱtheȱtankȱcircuit. B) tappedȱcapacitorsȱinȱtheȱtankȱcircuit. C) threeȱlagȱnetworksȱinȱtheȱfeedbackȱpath. D) lead/lagȱnetworkȱinȱtheȱfeedbackȱpath. Diff:ȱ2

28) OneȱwayȱtoȱrecognizeȱaȱHartleyȱoscillatorȱisȱbyȱthe A) transformerȱusedȱforȱfeedback. B) threeȱleadȱnetworksȱinȱtheȱfeedbackȱpath. C) tappedȱcapacitorsȱinȱtheȱtankȱcircuit. D) tappedȱinductorsȱinȱtheȱtankȱcircuit. Diff:ȱ2

398

29) WhenȱQȱdecreasesȱinȱaȱColpittsȱoscillator,ȱtheȱfrequencyȱofȱoscillation A) decreases.

B) remainsȱtheȱsame.

C) increases.

D) isȱunpredictable.

Diff:ȱ2

30) TheȱHartleyȱoscillatorȱuses A) onlyȱresistorsȱandȱcapacitors.

B) twoȱinductors.

C) aȱtungstenȱlamp.

D) aȱticklerȱcoil.

Diff:ȱ2

31) Whichȱofȱtheȱfollowingȱcannotȱbeȱusedȱasȱtheȱamplifierȱelementȱwithinȱanȱoscillator? A) BJT

B) thermistor

C) FET

D) op-amp

Diff:ȱ2

32) WhichȱtypeȱofȱLCȱoscillatorȱusesȱaȱticklerȱcoilȱinȱtheȱfeedbackȱpath? A) Colpitts

B) Hartley

C) Armstrong

D) Clapp

Diff:ȱ2

33) Whichȱofȱtheȱfollowingȱoscillatorȱtypesȱdoesȱnotȱuseȱanyȱinductors? A) Wien-bridge

B) Hartleyȱoscillator

C) Colpittsȱoscillator

D) Clappȱoscillator

Diff:ȱ2

34) InȱaȱColpittsȱoscillator,ȱwhichȱcomponent(s)ȱdetermineȱtheȱfeedbackȱfraction,ȱB? A) TheȱresistorsȱR1 ȱandȱR2 ȱinȱtheȱbaseȱcircuit B) TheȱRFȱchokeȱinȱtheȱcollectorȱcircuit C) TheȱcapacitorsȱC 1 ȱandȱC2ȱinȱtheȱtankȱcircuit D) Theȱinductorȱinȱtheȱtankȱcircuit Diff:ȱ3

35) TheȱQȱofȱaȱcrystal A) isȱextremelyȱlow.

B) isȱaboutȱ10ȱorȱsoȱinȱmostȱcases.

C) isȱextremelyȱhigh.

D) Noneȱofȱtheȱabove.

Diff:ȱ2

399

36) Aȱcrystalȇsȱfundamentalȱfrequencyȱdependsȱupon A) crystalȱthickness.

B) typeȱofȱcut.

C) mechanicalȱdimensions.

D) Allȱofȱtheȱabove.

Diff:ȱ2

37) Theȱhigherȱresonantȱfrequenciesȱofȱaȱcrystalȱareȱcalled A) undertones.

B) overtones.

C) octaves.

Diff:ȱ2

400

D) decades.

Chapterȱ17 VoltageȱRegulators 1) Switchingȱregulatorsȱareȱveryȱefficient. Diff:ȱ2

2) Aȱzenerȱdiodeȱisȱsometimesȱusedȱasȱaȱvoltageȱregulator. Diff:ȱ2

3) Lineȱregulationȱisȱtheȱpercentageȱchangeȱinȱinputȱvoltageȱforȱaȱgivenȱchangeȱinȱoutputȱvoltage. Diff:ȱ2

4) Inȱaȱshuntȱregulator,ȱtheȱcontrolȱelementȱisȱinȱseriesȱwithȱtheȱload. Diff:ȱ2

5) Mostȱvoltageȱregulatorsȱincludeȱsomeȱkindȱofȱprotectionȱcircuitry. Diff:ȱ2

6) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱzenerȱhadȱaȱvoltageȱratingȱofȱ3.7ȱV,ȱV outȱwouldȱbe A) 25ȱV.

B) 20.2ȱV.

C) 18.2ȱV.

D) 7.1ȱV.

Diff:ȱ2

7) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱaȱwireȱclippingȱwereȱtoȱshortȱQ 1 emitterȱtoȱcollector,ȱtheȱproblem thatȱmightȱresultȱis A) R2 ȱwouldȱopen.

B) VOUT wouldȱincreaseȱtoȱ25ȱV.

C) Q1ȱwouldȱfail.

D) theȱzenerȱwouldȱopen.

Diff:ȱ3

401

8) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱshuntȱregulator? A) (a)

B) (b)

C) (c)

D) (a)ȱorȱ(b)

Diff:ȱ2

9) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱstep-upȱswitchingȱregulator? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

10) Referȱtoȱtheȱfigureȱabove.ȱȱWhichȱofȱtheseȱcircuitsȱisȱknownȱasȱaȱseriesȱregulator? A) (a)

B) (b)

C) (c)

D) Noneȱofȱtheȱabove.

Diff:ȱ2

11) Referȱtoȱtheȱfigureȱabove.ȱȱTheȱpurposeȱforȱtheȱop-ampȱisȱto A) supplyȱaȱreferenceȱvoltage.

B) senseȱtheȱerrorȱsignal.

C) limitȱtheȱinputȱvoltageȱtoȱtheȱcircuit.

D) amplifyȱtheȱerrorȱsignal.

Diff:ȱ2

402

12) Referȱtoȱtheȱfigureȱabove.ȱȱAnȱincreaseȱinȱV OUT willȱcauseȱQ1 to A) conductȱless.

B) conductȱtheȱsame.

C) conductȱmore.

D) open.

Diff:ȱ3

13) Referȱtoȱtheȱfigureȱabove.ȱȱToȱincreaseȱtheȱcurrentȱhandlingȱcapabilityȱofȱthisȱregulator,ȱbeyond theȱ5ȱAȱratingȱofȱtheȱtransistor,ȱtheȱreasonableȱthingȱtoȱdoȱwouldȱbeȱto A) placeȱanotherȱtransistorȱinȱseriesȱwithȱQ 1. B) increaseȱtheȱvalueȱofȱtheȱzenerȱdiode. C) placeȱanotherȱtransistorȱinȱparallelȱwithȱQ 1. D) changeȱtheȱvaluesȱofȱR 2 ȱandȱR3. Diff:ȱ2

14) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱtheȱoutputȱvoltageȱtendsȱtoȱincreaseȱdueȱtoȱaȱdecreaseȱinȱload current,ȱtheȱtransistorȱwillȱconductȱforȱ________ȱtimeȱeachȱcycle. A) aȱlonger

B) aȱshorter

C) theȱsame

D) exactlyȱhalfȱthe

Diff:ȱ2

403

15) Referȱtoȱtheȱfigureȱabove.ȱȱThisȱcircuitȱisȱbroughtȱinȱforȱrepair.ȱȱTheȱmeasuredȱoutputȱvoltage wasȱ25ȱVȱunderȱallȱloadȱconditions.ȱȱAȱpossibleȱcauseȱofȱthisȱsymptomȱmightȱbeȱthat A) R2 ȱhasȱopened.

B) Q1 base-emitterȱhasȱopened.

C) R3 ȱhasȱopened.

D) Vin hasȱdecreased.

Diff:ȱ3

16) Referȱtoȱtheȱfigureȱabove.ȱȱIfȱR 1 opened,ȱVOUT would A) increase.

B) decreaseȱtoȱzero.

C) remainȱtheȱsame.

D) Cannotȱbeȱdetermined.

Diff:ȱ2

404

17) Referȱtoȱtheȱfigureȱabove.ȱInȱallȱofȱtheseȱcircuits,ȱtheȱzenerȱisȱused A) toȱsenseȱtheȱchangeȱinȱoutputȱvoltage.

B) asȱaȱreferenceȱvoltage.

C) toȱsupplyȱtheȱop-ampȱwithȱVCC.

D) toȱregulateȱtheȱoutputȱvoltageȱdirectly.

Diff:ȱ2

18) Referȱtoȱtheȱfigureȱabove.ȱTheȱcircuitȱthatȱwillȱalsoȱregulateȱtheȱoutputȱvoltageȱwhenȱV in varies is A) (a).

B) (b).

C) (c).

D) Allȱofȱtheȱabove.

Diff:ȱ2

19) Referȱtoȱtheȱfigureȱabove.ȱThisȱcircuitȱoperatesȱatȱaȱrelativelyȱ________ȱfrequencyȱandȱits efficiencyȱisȱ________. A) low,ȱlow

B) low,ȱhigh

C) high,ȱhigh

Diff:ȱ2

405

D) high,ȱlow

20) Referȱtoȱtheȱfigureȱabove.ȱTheȱpurposeȱforȱtheȱdiodeȱD 1 isȱto A) supplyȱaȱreferenceȱvoltage.

B) amplifyȱtheȱerrorȱsignal.

C) senseȱtheȱerrorȱsignal.

D) limitȱtheȱinputȱvoltageȱtoȱtheȱcircuit.

Diff:ȱ2

21) Referȱtoȱtheȱfigureȱabove.ȱIfȱaȱsolderȱsplashȱshortedȱtheȱendsȱofȱR 1 toȱeachȱother,ȱtheȱresult wouldȱbeȱthat A) theȱop-ampȱwouldȱfail.

B) Q1ȱwouldȱopen.

C) theȱoutputȱvoltageȱwouldȱnotȱchange.

D) theȱzenerȱwouldȱfail.

Diff:ȱ3

22) Aȱvoltageȱregulatorȱwithȱaȱno-loadȱdcȱoutputȱofȱ15ȱVȱisȱconnectedȱtoȱaȱloadȱwithȱaȱresistanceȱof 12ơ̇̄.ȱȱIfȱtheȱloadȱvoltageȱdecreasesȱtoȱ14.5ȱV,ȱtheȱpercentȱregulationȱwouldȱbe A) 96.7%.

B) 3.33%.

C) 3.45%.

Diff:ȱ2

23) Anȱadvantageȱofȱaȱȱswitchingȱregulatorȱis A) lessȱheatȱandȱwastedȱpower. B) theȱcircuitȱisȱveryȱefficient. C) voltagesȱcanȱbeȱstepped-upȱorȱstepped-down. D) Allȱofȱtheȱabove. E) Noneȱofȱtheȱabove. Diff:ȱ2

406

D) 100%.

24) Aȱvoltageȱregulatorȱhasȱaȱno-loadȱoutputȱofȱ18ȱVȱandȱaȱfullȱloadȱoutputȱofȱ17.3ȱV.ȱTheȱpercent loadȱregulationȱis A) 0.25%.

B) 96.1%.

C) 4.05%.

D) 1.04%.

Diff:ȱ2

25) Aȱvoltageȱregulatorȱwithȱaȱno-loadȱoutputȱdcȱvoltageȱofȱ12ȱVȱisȱconnectȱtoȱaȱloadȱwithȱa resistanceȱofȱ10ơ̇̄.ȱȱIfȱtheȱloadȱresistanceȱdecreasesȱtoȱ7.5ơ̇̄,ȱtheȱloadȱvoltageȱwillȱdecreaseȱto 10.9ȱV.ȱTheȱloadȱcurrentȱwillȱbeȱ________ȱandȱtheȱpercentȱloadȱregulationȱisȱ________. A) 1.45ȱA,ȱ90.8%

B) 1.45ȱA,ȱ10.09%

C) 1.6ȱA,ȱ90.8%

D) 1.6ȱA,ȱ9.17%

Diff:ȱ2

26) Anȱincreaseȱofȱlineȱvoltageȱintoȱaȱpowerȱsupplyȱusuallyȱproduces A) aȱdecreaseȱinȱloadȱresistance. B) anȱincreaseȱinȱloadȱvoltage. C) aȱdecreaseȱinȱefficiency. D) lessȱpowerȱdissipationȱinȱtheȱrectifierȱdiodes. Diff:ȱ2

27) ________ȱisȱaȱmeasurementȱofȱhowȱwellȱtheȱpowerȱsupplyȱmaintainsȱaȱconstantȱoutputȱvoltage withȱchangesȱinȱinputȱvoltage. A) Voltageȱcontrol

B) Loadȱvoltageȱcontrol

C) Loadȱregulation

D) Lineȱregulation

Diff:ȱ2

28) Ifȱtheȱoutputȱofȱaȱvoltageȱregulatorȱvariesȱfromȱ15ȱtoȱ14.7ȱVȱbetweenȱtheȱminimumȱand maximumȱloadȱcurrent,ȱtheȱloadȱregulationȱis A) 0.

B) 1%.

C) 2%.

D) 5%.

Diff:ȱ2

29) Ifȱtheȱoutputȱofȱaȱvoltageȱregulatorȱvariesȱfromȱ20ȱtoȱ19.8ȱVȱwhenȱtheȱlineȱvoltageȱvariesȱover itsȱspecifiedȱrange,ȱtheȱloadȱregulationȱis A) 0.

B) 1%.

C) 2%.

Diff:ȱ2

30) Aȱseriesȱregulatorȱisȱanȱexampleȱofȱa A) linearȱregulator.

B) switchingȱregulator.

C) shuntȱregulator.

D) ac-to-dcȱconverter.

Diff:ȱ2

407

D) 5%.

31) Withoutȱcurrentȱlimiting,ȱaȱshortedȱloadȱwillȱprobably A) produceȱzeroȱloadȱcurrent. B) destroyȱdiodesȱandȱtransistors. C) haveȱaȱloadȱvoltageȱequalȱtoȱtheȱzenerȱvoltage. D) haveȱtooȱlittleȱloadȱcurrent. Diff:ȱ2

32) Simpleȱcurrentȱlimitingȱmayȱproduceȱtooȱmuchȱheatȱinȱthe A) zenerȱdiode.

B) loadȱresistor.

C) passȱtransistor.

D) ambientȱair.

Diff:ȱ2

33) Withȱfoldbackȱcurrentȱlimiting,ȱtheȱloadȱvoltageȱapproachesȱzero,ȱandȱtheȱloadȱcurrent approaches A) aȱsmallȱvalue.

B) infinity.

C) theȱzenerȱcurrent.

D) aȱdestructiveȱlevel.

Diff:ȱ3

34) Ifȱtheȱloadȱisȱshorted,ȱtheȱpassȱtransistorȱhasȱtheȱleastȱpowerȱdissipationȱwhenȱtheȱregulator has A) foldbackȱlimiting.

B) lowȱefficiency.

C) buckȱtopology.

D) aȱhighȱzenerȱvoltage.

Diff:ȱ2

35) Switchingȱregulatorȱconfigurationsȱinclude A) step-down.

B) inverting.

C) step-up.

D) Allȱofȱtheȱabove.

Diff:ȱ2

36) Anȱadvantageȱofȱshuntȱregulationȱis A) built-inȱshort-circuitȱprotection. B) lowȱpowerȱdissipationȱinȱtheȱpassȱtransistor. C) highȱefficiency. D) littleȱwastedȱpower. Diff:ȱ2

408

37) Toȱgetȱmoreȱoutputȱvoltageȱfromȱaȱstep-downȱswitchingȱregulator,ȱyouȱhaveȱto A) decreaseȱtheȱdutyȱcycle.

B) decreaseȱtheȱinputȱvoltage.

C) increaseȱtheȱdutyȱcycle.

D) increaseȱtheȱswitchingȱfrequency.

Diff:ȱ3

38) Aȱ________ȱmaintainsȱaȱconstantȱoutputȱvoltageȱbyȱcontrollingȱtheȱdutyȱcycleȱofȱaȱswitchȱin seriesȱwithȱtheȱload. A) shuntȱregulator

B) linearȱregulator

C) seriesȱregulator

D) switchingȱregulator

Diff:ȱ2

39) Switchingȱregulatorsȱhaveȱ________ȱthanȱlinearȱregulators. A) moreȱheat-sinkingȱrequirements

B) simplerȱcircuitry

C) lowerȱefficiency

D) greaterȱefficiency

Diff:ȱ2

40) Inȱaȱstep-upȱregulator,ȱtheȱoutputȱvoltageȱisȱfilteredȱwithȱa A) choke-inputȱfilter.

B) capacitor-inputȱfilter.

C) diode.

D) voltageȱdivider.

Diff:ȱ2

41) Theȱ7912ȱproducesȱaȱregulatedȱoutputȱvoltageȱof A) +5ȱV.

B) +9ȱV.

C) -12ȱV.

D) +12ȱV.

Diff:ȱ2

42) Theȱ7800ȱseriesȱofȱvoltageȱregulatorsȱproducesȱanȱoutputȱvoltageȱthatȱis A) positive.

B) negative.

C) eitherȱpositiveȱorȱnegative.

D) unregulated.

Diff:ȱ2

43) TheȱLM317ȱregulatorȱprovidesȱa(n) A) fixedȱpositiveȱoutputȱvoltage.

B) adjustableȱpositiveȱoutputȱvoltage.

C) fixedȱnegativeȱoutputȱvoltage.

D) adjustableȱnegativeȱoutputȱvoltage.

Diff:ȱ2

409

44) Theȱmainȱdifferenceȱbetweenȱtheȱ78XXȱandȱ79XXȱseriesȱregulatorsȱisȱthat A) theȱ78XXȱseriesȱisȱadjustableȱandȱtheȱ79XXȱseriesȱisȱfixed. B) theȱ78XXȱseriesȱisȱfixedȱandȱtheȱ79XXȱseriesȱisȱadjustable. C) theȱ78XXȱseriesȱprovidesȱaȱpositiveȱfixedȱoutputȱvoltageȱandȱtheȱ79XXȱseriesȱprovidesȱa negativeȱfixedȱoutputȱvoltage. D) theȱ78XXȱseriesȱprovidesȱaȱnegativeȱfixedȱoutputȱvoltageȱandȱtheȱ79XXȱseriesȱprovidesȱa positiveȱfixedȱoutputȱvoltage. Diff:ȱ2

45) Aȱ7805ȱregulatorȱhasȱaȱ+20ȱVȱinput.ȱIfȱtheȱinputȱandȱoutputȱpinsȱofȱtheȱregulatorȱareȱshorted togetherȱbyȱaȱsolderȱbridge,ȱtheȱoutputȱvoltageȱwillȱbe A) 0ȱV.

C) +5ȱV.

B) 0.7ȱV.

Diff:ȱ3

410

D) +20ȱV.

Chapterȱ18 Communications 1) BalancedȱmodulationȱisȱusedȱinȱcertainȱtypesȱofȱcommunicationsȱsuchȱasȱAMȱbroadcast systems,ȱbutȱisȱnotȱusedȱinȱsingleȱside-bandȱsystems. Diff:ȱ2

2) Fiber-opticȱcablesȱhaveȱwiderȱbandwidth,ȱbutȱareȱmoreȱsusceptibleȱtoȱinterference. Diff:ȱ2

3) Scatteringȱisȱtheȱlossȱofȱopticalȱsignalȱintoȱtheȱcladdingȱofȱtheȱfiber-opticȱcable. Diff:ȱ2

4) CombiningȱanȱaudioȱsignalȱwithȱanȱRFȱcarrierȱinȱaȱnonlinearȱdeviceȱisȱcalled A) neutralization.

B) demodulation.

C) filtering.

D) modulation.

Diff:ȱ2

5) Theȱprocessȱofȱmodifyingȱaȱhighȱfrequencyȱcarrierȱbyȱtheȱinformationȱtoȱbeȱtransmittedȱis called A) modulation.

B) multiplexing.

C) detection.

D) discrimination.

Diff:ȱ2

6) Mixingȱtwoȱsignalsȱbyȱaȱnonlinearȱprocessȱisȱcalled A) sumȱandȱproductȱfrequencies.

B) heterodyning.

C) spectrumȱmodulation.

D) bandwidthȱdiscrimination.

Diff:ȱ3

7) Theȱoutlineȱofȱtheȱpeaksȱofȱtheȱmodulatedȱcarrierȱhasȱtheȱshapeȱofȱtheȱinformationȱsignal,ȱand isȱcalledȱthe A) envelope.

B) lower-sideȱfrequency.

C) RFȱindex.

D) dutyȱcycle.

Diff:ȱ2

8) Higherȱmodulatingȱfrequenciesȱareȱamplifiedȱmoreȱthanȱtheȱlowerȱfrequenciesȱatȱthe transmittingȱendȱofȱanȱFMȱsystemȱbyȱaȱprocessȱcalled A) pre-emphasis.

B) fullȱduplex.

C) de-emphasis.

Diff:ȱ3

411

D) filtering.

9) Theȱquadrantȱclassificationȱofȱaȱlinearȱmultiplierȱindicatesȱtheȱnumberȱofȱ________ȱthatȱthe multiplierȱcanȱhandle. A) inputȱpolarityȱcombinations

B) outputȱpolarityȱcombinations

C) transferȱcharacteristics

D) scaleȱfactors

Diff:ȱ3

10) Theȱpurposeȱofȱaȱbalancedȱmodulatorȱisȱtoȱeliminate A) theȱupperȱside-band.

B) theȱlowerȱside-band.

C) bothȱside-bands.

D) theȱcarrier.

Diff:ȱ2

11) Theȱproductȱofȱtwoȱsinusoidalȱsignalsȱisȱcalled A) balancedȱmodulation.

B) lowerȱsideȱfrequency.

C) suppressed-carrierȱmodulation.

D) BothȱAȱandȱCȱabove.

Diff:ȱ2

12) IfȱyouȱreceiveȱanȱAMȱsignalȱmodulatedȱbyȱaȱpureȱsinusoidalȱsignalȱinȱtheȱaudioȱfrequency range,ȱyouȱwillȱhear A) aȱsingleȱtone.

B) static.

C) distortion.

D) nothing.

C) demodulator.

D) modulator.

Diff:ȱ2

13) Aȱmixerȱisȱbasicallyȱaȱfrequency A) doubler.

B) converter.

Diff:ȱ2

14) AȱbasicȱIFȱAmplifierȱalwaysȱhas A) aȱtunedȱ(resonant)ȱcircuitȱonȱtheȱinput. B) aȱtunedȱ(resonant)ȱcircuitȱonȱtheȱoutput. C) aȱtunedȱ(resonant)ȱcircuitȱonȱbothȱtheȱinputȱandȱoutput. D) Noneȱofȱtheȱabove. Diff:ȱ2

15) TheȱmainȱdifferenceȱbetweenȱanȱFMȱreceiverȱandȱanȱAMȱreceiverȱisȱtheȱmethodȱusedȱto recoverȱtheȱaudioȱsignalȱfromȱthe A) modulatedȱIF.

B) carrier.

C) mixer.

Diff:ȱ2

412

D) detector.

16) TheȱphaseȱdetectorȱinȱaȱPLLȱisȱfollowedȱbyȱaȱlow-passȱfilter.ȱTheȱlow-passȱfilterȱpassesȱthe ________ȱandȱrejectsȱallȱotherȱfrequencies. A) inputȱsignal B) feedbackȱsignal C) sumȱofȱtheȱinputȱandȱfeedbackȱsignals D) differenceȱofȱtheȱinputȱandȱfeedbackȱsignals Diff:ȱ2

17) InȱaȱPLL,ȱtoȱobtainȱlock,ȱtheȱsignalȱfrequencyȱmustȱcomeȱwithinȱthe A) lockȱrange.

B) closedȱrange.

C) captureȱrange.

D) deviationȱrange.

Diff:ȱ2

18) ForȱaȱPLL,ȱtheȱcaptureȱrangeȱis A) alwaysȱgreaterȱthanȱtheȱlockȱrange.

B) alwaysȱtheȱsameȱasȱtheȱlockȱrange.

C) usuallyȱlessȱthanȱtheȱlockȱrange.

D) alwaysȱtwoȱtimesȱtheȱlockȱrange.

Diff:ȱ2

19) AȱPLLȱcanȱbeȱusedȱasȱa(n) A) seriesȱvoltageȱregulator.

B) FMȱdemodulator.

C) audioȱamplifier.

D) BothȱBȱandȱCȱabove.

Diff:ȱ2

20) Aȱphaseȱdetectorȱhas A) oneȱinputȱsignalȱandȱtwoȱoutputȱsignals. B) twoȱinputȱsignalsȱandȱoneȱoutputȱsignal. C) noȱinputȱsignals. D) threeȱoutputȱsignals. Diff:ȱ2

21) Theȱbandwidthȱofȱtheȱlow-passȱfilterȱinȱaȱPLLȱdeterminesȱthe A) captureȱrange.

B) lockȱrange.

C) free-runningȱfrequency.

D) phaseȱdifference.

Diff:ȱ2

413

22) MostȱVCOȇsȱusedȱinȱPLLȇsȱoperateȱonȱtheȱprincipleȱof A) variableȱpower.

B) variableȱinductance.

C) variableȱreactance.

D) phaseȱdetection.

Diff:ȱ2

23) TheȱIFȱamplifiersȱwithinȱanȱFMȱradioȱreceiverȱareȱtunedȱto A) 455ȱkHz.

B) 4.55ȱMHz.

C) 10.7ȱMHz.

D) 100.7ȱMHz.

Diff:ȱ2

24) Theȱsignalȱlossȱthatȱoccursȱwhenȱphotonsȱinteractȱwithȱtheȱmoleculesȱofȱtheȱcoreȱinȱa fiber-opticȱcableȱisȱcalled A) refraction.

B) absorption.

C) scattering.

Diff:ȱ3

414

D) scattering.

Answer Key CHAPTER 1

CHAPTER 2

CHAPTER 3

1) FALSE 2) FALSE 3) TRUE 4) FALSE 5) TRUE 6) C 7) B 8) A 9) A 10) D 11) D 12) C 13) B 14) D 15) A 16) C 17) B 18) A 19) C 20) A 21) C 22) B 23) D 24) B 25) B 26) A 27) B 28) D 29) A 30) A 31) D 32) A 33) B 34) D 35) D 36) A 37) C 38) C 39) A

1) TRUE 2) FALSE 3) TRUE 4) FALSE 5) FALSE 6) D 7) D 8) A 9) C 10) D 11) B 12) B 13) C 14) D 15) B 16) D 17) B 18) A 19) C 20) B 21) C 22) D 23) E 24) B 25) D 26) B 27) A 28) C 29) B 30) A 31) B 32) C 33) D 34) C 35) C 36) B 37) A 38) C 39) B 40) D 41) B 42) D 43) B 44) A

1) TRUE 2) FALSE 3) C 4) B 5) C 6) A 7) A 8) A 9) D 10) D 11) D 12) D 13) B 14) B 15) B 16) A 17) D 18) E 19) C 20) B 21) D 22) E 23) C 24) A 25) B 26) A 27) D 28) D 29) A 30) C 31) D 32) C 33) A 34) B 35) B 36) B 37) A 38) B 39) B 40) D 41) C 42) C 43) B 44) A 45) A 46) B

415

CHAPTER 4

CHAPTER 5

CHAPTER 6

1) TRUE 2) FALSE 3) FALSE 4) TRUE 5) TRUE 6) TRUE 7) TRUE 8) D 9) C 10) A 11) B 12) D 13) C 14) B 15) B 16) D 17) C 18) A 19) C 20) A 21) A 22) D 23) C 24) A 25) D 26) B 27) C 28) B 29) B 30) C 31) B 32) B 33) B 34) C 35) C 36) C 37) A 38) C 39) B 40) D 41) A 42) C 43) B 44) B 45) B 46) B 47) D

1) TRUE 2) FALSE 3) FALSE 4) TRUE 5) TRUE 6) C 7) C 8) D 9) B 10) E 11) D 12) C 13) B 14) D 15) B 16) A 17) C 18) B 19) C 20) B 21) D 22) B 23) D 24) C 25) A 26) C 27) B 28) C 29) C 30) D 31) A 32) B 33) C 34) B 35) B 36) A 37) D 38) D 39) A 40) A 41) B 42) B 43) D

1) FALSE 2) TRUE 3) TRUE 4) FALSE 5) FALSE 6) D 7) C 8) C 9) A 10) B 11) B 12) D 13) A 14) D 15) C 16) D 17) D 18) C 19) A 20) D 21) B 22) C 23) A 24) B 25) C 26) C 27) A 28) B 29) B 30) B 31) A 32) B 33) B 34) B 35) A 36) C 37) B 38) C 39) A 40) A 41) A 42) C 43) D 44) D 45) B 46) B 47) B 48) B 49) B 50) B 51) A 52) C 53) D

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54) C 55) A 56) B 57) B 58) A 59) D 60) C

CHAPTER 7

CHAPTER 8

1) FALSE 2) FALSE 3) FALSE 4) C 5) C 6) B 7) D 8) C 9) A 10) D 11) A 12) D 13) A 14) C 15) D 16) C 17) B 18) B 19) D 20) B 21) B 22) C 23) C 24) A 25) C 26) C 27) C 28) D 29) A 30) D 31) C 32) C 33) C 34) A 35) B 36) B 37) C 38) B 39) C 40) A 41) C 42) C 43) D 44) A 45) B 46) C 47) D 48) B

1) TRUE 2) FALSE 3) TRUE 4) TRUE 5) FALSE 6) TRUE 7) FALSE 8) TRUE 9) FALSE 10) A 11) A 12) B 13) D 14) B 15) C 16) D 17) A 18) B 19) D 20) C 21) C 22) B 23) A 24) B 25) B 26) B 27) C 28) A 29) B 30) C 31) A 32) A 33) D 34) C 35) D 36) D 37) C 38) D 39) C 40) B 41) B 42) A 43) B 44) D 45) D 46) A 47) D 48) C 49) D

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CHAPTER 9 1) FALSE 2) FALSE 3) TRUE 4) TRUE 5) FALSE 6) TRUE 7) FALSE 8) FALSE 9) TRUE 10) FALSE 11) FALSE 12) TRUE 13) FALSE 14) FALSE 15) TRUE 16) C 17) D 18) D 19) C 20) A 21) B 22) A 23) B 24) C 25) C 26) A 27) B 28) C 29) B 30) B 31) D 32) B 33) D 34) C 35) B 36) A 37) D 38) A 39) C 40) D 41) C 42) D 43) B 44) A 45) D 46) B 47) B 48) D 49) C 50) D 51) A 52) C 53) C

CHAPTER 11

54) A 55) A 56) B 57) D 58) C

CHAPTER 10 1) TRUE 2) TRUE 3) TRUE 4) FALSE 5) FALSE 6) FALSE 7) D 8) C 9) A 10) B 11) C 12) A 13) C 14) B 15) B 16) C 17) C 18) A 19) D 20) B 21) C 22) A 23) B 24) B 25) B 26) D 27) D 28) B 29) D 30) B 31) B 32) B 33) C 34) B 35) A 36) A 37) C 38) C 39) C 40) A 41) B 42) D 43) B 44) C 45) A

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1) TRUE 2) FALSE 3) TRUE 4) TRUE 5) FALSE 6) FALSE 7) E 8) B 9) D 10) C 11) E 12) C 13) C 14) C 15) B 16) D 17) A 18) C 19) B 20) D 21) B 22) C 23) A 24) D 25) A 26) C 27) B 28) D 29) B 30) B 31) B 32) B 33) C 34) B 35) B 36) D 37) D 38) C 39) A 40) A 41) D 42) A 43) C 44) B 45) D 46) C 47) A

CHAPTER 12

CHAPTER 13

CHAPTER 14

1) TRUE 2) TRUE 3) FALSE 4) TRUE 5) FALSE 6) C 7) A 8) B 9) A 10) D 11) C 12) D 13) D 14) A 15) A 16) D 17) B 18) D 19) A 20) C 21) A 22) B 23) D 24) A 25) C 26) C 27) D 28) C 29) B 30) A 31) B 32) A 33) C 34) A 35) D 36) C 37) C 38) A 39) C 40) B 41) C 42) C 43) A 44) A 45) B 46) C 47) A 48) A 49) B 50) D 51) B

1) TRUE 2) FALSE 3) TRUE 4) FALSE 5) FALSE 6) D 7) D 8) C 9) B 10) C 11) C 12) C 13) D 14) C 15) A 16) C 17) C 18) B 19) B 20) D 21) D 22) D 23) D 24) D 25) C 26) D 27) C 28) B 29) D 30) C 31) B 32) B 33) A 34) B 35) D 36) B 37) B 38) D 39) D 40) A 41) B 42) C 43) A 44) B 45) D 46) D

1) TRUE 2) FALSE 3) TRUE 4) TRUE 5) FALSE 6) FALSE 7) FALSE 8) TRUE 9) FALSE 10) TRUE 11) TRUE 12) C 13) D 14) B 15) C 16) A 17) B 18) C 19) D 20) C 21) B 22) A 23) D 24) C 25) B 26) A 27) D 28) D 29) C 30) D 31) D 32) A 33) B 34) D 35) D 36) A 37) B 38) A 39) B 40) D 41) D 42) C 43) B 44) D 45) B 46) C

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CHAPTER 15

CHAPTER 16

CHAPTER 17

1) TRUE 2) FALSE 3) TRUE 4) FALSE 5) TRUE 6) C 7) B 8) A 9) D 10) A 11) D 12) D 13) C 14) B 15) D 16) C 17) A 18) C 19) D 20) C 21) A 22) A 23) C 24) D 25) B 26) D 27) B 28) A 29) B 30) C 31) B 32) D 33) D 34) B 35) D 36) A 37) B 38) C 39) D 40) D 41) D 42) A 43) D 44) C 45) A

1) FALSE 2) TRUE 3) TRUE 4) TRUE 5) FALSE 6) FALSE 7) A 8) B 9) C 10) C 11) C 12) C 13) C 14) D 15) A 16) C 17) D 18) A 19) B 20) A 21) B 22) C 23) C 24) A 25) B 26) B 27) B 28) D 29) A 30) B 31) B 32) C 33) A 34) C 35) C 36) D 37) B

1) TRUE 2) TRUE 3) FALSE 4) FALSE 5) TRUE 6) B 7) B 8) B 9) D 10) A 11) D 12) A 13) C 14) B 15) A 16) B 17) B 18) D 19) C 20) A 21) D 22) C 23) D 24) C 25) B 26) B 27) D 28) C 29) B 30) A 31) B 32) C 33) A 34) A 35) D 36) A 37) C 38) D 39) D 40) B 41) C 42) A 43) B 44) C 45) D

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CHAPTER 18 1) FALSE 2) FALSE 3) TRUE 4) D 5) A 6) B 7) A 8) A 9) A 10) D 11) D 12) A 13) B 14) C 15) A 16) D 17) C 18) C 19) B 20) B 21) A 22) C 23) C 24) B

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