EE 321 Chapter 10 Solutions

Solution of Homework problems 2 in Section 10.2 Chapter 10, Solution 1. Known quantities: Transistor diagrams, as shown in Figure P10.1: (a) pnp, VEB = 0.6 V and VEC = 4.0 V (b) npn, VCB = 0.7 V and VCE = 0.2 V (c) npn, VBE = 0.7 V and VCE = 0.3 V (d) pnp, VBC = 0.6 V and VEC = 5.4 V Find: For each transistor shown in Figure P10.1, determine whether the BE and BC junctions are forward or reverse biased, and determine the operating region. Analysis: (a) VBE = - 0.6 V for a pnp transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 3.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. (b) VBC = - 0.7 V for a npn transistor implies that the CB junction is reverse-biased. VBE = VBC - VEC = -0.5 V. The BE junction is reverse-biased. Therefore, the transistor is in the cutoff region. (c) VBE = 0.7 V for a npn transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 0.4 V. The CB junction is forward-biased. Therefore, the transistor is in the saturation region. (d) VBC = 0.6 V for a pnp transistor implies that the CB junction is reverse-biased. VBE = VBC – VEC = - 4.8 V. The BE junction is forward-biased. Therefore, the transistor is in the active region. Chapter 10, Solution 2. Known quantities: Transistor type and operating characteristics: a) npn, VBE = 0.8 V and VCE = 0.4 V b) npn, VCB = 1.4 V and VCE = 2.1 V c) pnp, VCB = 0.9 V and VCE = 0.4 V d) npn, VBE = - 1.2 V and VCB = 0.6 V Find: The region of operation for each transistor.

Analysis: a) Since VBE = 0.8 V, the BE junction is forward-biased. VCB = VCE + VEB = - 0.4 V. Thus, the CB junction is forward-biased. Therefore, the transistor is in the saturation region. b) VBE = VBC + VCE = 0.7 V. The BE junction is forward-biased. VCB = 1.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. c) VCB = 0.9 V for a pnp transistor implies that the CB junction is forward-biased. VBE = VBC – VCE = - 1.3 V. The BE junction is forward-biased. Therefore, the transistor is in the saturation region. d) With VBE = - 1.2 V, the BE junction is reverse-biased. VCB = - 0.6 V. The CB junction is reverse-biased. Therefore, the transistor is in the cutoff region. Chapter 10, Solution 3. Known quantities: The circuit of Figure P10.3:  

IC  100 . IB

Find: The operating point and the state of the transistor. Analysis: VBE  0.6 V and the BE junction is forward biased.

VCC  820  103 I B  VBE  910 I E & I E  I C  I B  101I B VCC  VBE 12  0.6   12.5A 3 820  10  910  101 911910 I C    I B  1.25mA IB 

Writing KVL around the right-hand side of the circuit: VCC  I C RC  VCE  I E RE  0

VCE  VCC  I C RC  I C  I B RE  12  (1.25)(2.2)  (1.25  0.0125)(0.910)  8.1V VBC  VBE  VCE  0.6  8.1   7.5 V : the BC junction is reverse biased VCE  VBE 

The transistor is in the active region.

Chapter 10, Solution 4. Known quantities: The magnitude of a pnp transistor's emitter and base current, and the magnitudes of the voltages across the emitter-base and collector-base junctions: IE = 6 mA, IB = 0.1 mA and VEB = 0.65 V, VBC = 7.3 V. Find: a) VCE. b) IC. c) The total power dissipated in the transistor, defined as P  VEC I C  VEB I B . Analysis: a) VEC = VBC + VEB = 7.3 + 0.65 = 7.95 V. b) IC = IE - IB = 6 - 0.1 = 5.9 mA. c) The total power dissipated in the transistor can be found to be:

P  VEC I C  VEB I B  7.95  5.9  103  0.65  0.1  103  46.97 mW

Chapter 10, Solution 5. Known quantities: The circuit of Figure P10.5, assuming the BJT has V = 0.6 V. Find: Change 15 V to 15 V The emitter current and the collector-base voltage. Analysis:

 15  VBE   15  0.6     480 A  30000   30000 

Applying KVL to the right-hand side of the circuit, I E   Then, on the left-hand side, assuming  >> 1:

 10  I C RC  VCB  0

VCB  10  I C RC  10  480  10 6   15  103  2.8 V

Chapter 10, Solution 6. Known quantities: The circuit of Figure P10.6, assuming the BJT has VBE  0.6 V and  =150. Find: The operating point and the region in which the transistor operates. Analysis: Define RC  3.3 k, RE  1.2 k, R1  62 k, R2  15 k, VCC  18 V By applying Thevenin’s theorem from base and mass, we have

R B  R1 || R2  12.078 kΩ



V BB  IB 

R2 VCC  3.5 V R1  R2

V BB  V BE  15 μA R B  R E (1   )

I C   I B  2.25 mA VCE  VCC  RC I C  R E I E  18  3300  2.25  10 3  1200  151  15  10  6  7.857 V From the value of VCE it is clear that the BJT is in the active region. Chapter 10, Solution 7. Known quantities: The circuit of Figure P10.7, assuming the BJT has V  0.6 V . VCC = 20V Find: The emitter current and the collector-base voltage. Analysis: Applying KVL to the right-hand side of the circuit,

 VBB  I E RE  VEB  0 V  VEB 20  0.6 I E  BB   497.4 μA . Since RE 39  103

  1 , I C  I E  497.4 μA

VBB = 20V

VCB  I C RC  VCC  0 VCB  I C RC  VCC  497.4  20  10  20  10.05V

Applying KVL to the left-hand side:

3

Chapter 10, Solution 9. Known quantities: The collector characteristics for a certain transistor, as shown in Figure P10.9. Find: a) The ratio IC/IB for VCE = 10 V and I B  100 A, 200A, and 600 A

b) VCE, assuming the maximum allowable collector power dissipation is 0.5 W for I B  500 A . Analysis: a) For IB = 100 µA and VCE = 10 V, from the characteristics, we have IC = 17 mA. The ratio IC / IB is 170. For IB = 200 µA and VCE = 10 V, from the characteristics, we have IC = 33 mA. The ratio IC / IB is 165. For IB = 600 µA and VCE = 10 V, from the characteristics, we have IC = 86 mA. The ratio IC / IB is 143. b) For IB = 500 A, and if we consider an average  from a., we have IC = 159·500 10-3= 79.5 mA. The power dissipated by the transistor is P  VCE I C  VBE I B  VCE I C , therefore: VCE 

P 0.5   6.29 V. I C 79.5103

Chapter 10, Solution 10. 

Known quantities: Figure P10.10, assuming both transistors are silicon-based with   100 . Find: a) IC1, VC1, VCE1. b) IC2, VC2, VCE2. Analysis: a) From KVL:

30  I B1RB1  VBE1  0 30  0.7 I B1   39.07 μA 750  103



I C1    I B1  3.907 mA



VC1  30  RC1I C1  30  3.907  6.2  5.779 V VCE1  VC1  5.779 V .

b) Again, from KVL: 5.779  VBE2  I E 2 RE 2  0 

I E2 

5.779  0.7 4.7  103

 1.081 mA

    100    1.081     1.07 mA .   1  101   

and I C 2  I E 2 

Also, 30  I C 2 ( RC 2  RE 2 )  VCE 2  0  VCE 2  30  (1.07)  (20  4.7)  3.574 V . Finally, I C 2 

30  VC 2 RC 2

 VC 2  30  (1.07)  (20)  8.603 V .

Chapter 10, Solution 11. Known quantities: Collector characteristics of the 2N3904 npn transistor, see data sheet pg. 560. Find: The operating point of the transistor in Figure P10.11, and the value of  at this point. Analysis: Construct a load line. Writing KVL, we have: 50  5000  IC  VCE  0 . Then, if I C  0 , VCE  50 V ; and if VCE  0 , I C  10 mA . The load line is shown superimposed on the collector characteristic below: The operating point is at the intersection of the load line and the I B  20 A line of the characteristic. Therefore, I CQ  5 mA and VCEQ  20 V . Under these conditions, an 5 A increase in I B yields an increase in I C of approximately 6  5  1 mA .



Therefore,

I C 1  103   200 I B 5  10 6

The same result can be obtained by checking the hFE gain from the data-sheets corresponding to 5 mA.

Load line

Chapter 10, Solution 14. Known quantities: The circuit of Figure P10.14, VCEsat=0.1V, VBEsat=0.6V, and β=50. Find: The base voltage required to saturate the transistor. Analysis: The collector current is 12  0.1 IC   11.9 mA 1

The base current is I 11.9 IB  C   0.238 mA  238A  50

And since 

V  VBEsat I B  BB mA 10

Therefore, V VBB  0.238 mA 10k  0.6  2.98 V

Chapter 10, Solution 16. Known quantities: Collector characteristics of 2N3904 npn transistor; Transistor circuits; Find: The operating point; Analysis: From KVL, 50  5kI C  VCE  5k ( IC  20A)  0

or VCE  10kI C  50  0.1  49.9

If VCE  0 , I C 

49.9  4.99mA , and if I C  0 , VCE  49.9V . The load 10k

line is shown superimposed on the collector characteristic below: The operating point is at the intersection of the load line and the I B  20A line of the characteristic. Therefore, I CQ  3mA and VCEQ  8V .

Under these conditions, a 10A increase in I B yields an increase in I C of approximately 5mA  3mA  2mA . Therefore, 

I C 2mA   200 I B 10A

Addition of the emitter resistor effectively increased the current gain by decreasing the magnitude of the slope of the load line.

Chapter 10, Solution 17. Known quantities: For the circuit shown in Figure 10.14 in the text:

Voff  0 V, Von  5 V, I B  5 mA, RB  1 kΩ, VCC  5 V, V  0.7V, VCEsat  0.2 V,   95, V

LED

 1.4 V, I LED  10 mA, Pmax  100 mW

Find: Range of RC. Analysis: RC 

VCC  V

LED

I LED

 VCEsat



5  1.4  0.2  340  0.01

From the maximum power I LED max  RC 

Pmax 0.1   71 mA V LED 1.4

VCC  V LED  VCEsat  47  I LED max

Therefore, RC [47, 340] 

Chapter 10, Solution 22. Known quantities: For the circuit shown in Figure 10.14 in the text: Voff  0 V, Von  5 V, I B max  1 mA, RB  1 kΩ, R  12 Ω, VCC  13 V, V  0.7V, VCEsat  1 V, IC  1A

Find:

Minimum value of  that will ensure the correct operation of the fuel injector. Analysis: V  VCEsat 13  1 I C  CC   1A R 12 I 1  min  C   1000 I B max 1  103

Chapter 10, Solution 25. Known quantities: The circuit of Figure P10.25: IC = 40 mA; Transistor large signal parameters. Find:

Design a constant-current battery charging circuit, that is, find the values of VCC, R1, R2 that will cause the transistor Q1 to act as a 40-mA constant current source. Assumptions: Assume that the transistor is forward biased. Use the large-signal model with  = 100. Analysis: The battery charging current is 40 mA, IC = 40 mA. Thus, the emitter current must be I E 

 1 I  40.4mA .  E

Since the base-emitter junction voltage is assumed to be 0.6 V, then resistor R2 has a voltage: V2  Vz  V  5.6  0.6  5 V , so the required value of R2 to be: R2 

V 5   123.8 I E 0.0404

Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply enough current fro the Zener to operate, for example R1 > 100 , so that there will be as little current flow through this resistance as possible. Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the sum of the battery voltage, the CE junction voltage and the voltage across R2. That is, VCC  9  VCE  5 . A collector supply of 24 V will be more than adequate for this task.

Chapter 10, Solution 26. Known quantities: The circuit of Figure of P10.26. Find:

Analyze the operation of the circuit and explain how I E is decreasing until the battery is full. Find the values of VCC, R1 that will result in a practical design.

Assumptions: Assume that the transistor is forward biased. Analysis: When the Zener Diode works in its reverse breakdown area, it provides a constant voltage: Vz  11 V . That means: VB  VZ  11 V . When the transistor is forward biased, according to KVL, VZ  I BE  RBE  V  Vbattery , where RBE is the base resistance. As the battery gets charged, the actual battery charging voltage Vbattery will increase from 9.6 V to 10.4 V. As Vbattery increases gradually, VZ and V stay unchanged, then we can see that I BE will decrease gradually. So I E    1I BE will also decrease at the same time. Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply enough current fro the Zener to operate, for example R1 > 100 , so that there will be as little current flow through this resistance as possible.

Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the sum of the battery voltage, the CE junction voltage. That is, VCC  11  VCE . A collector supply of 12 V should be adequate for this task.

Chapter 10, Solution 32. Known quantities: For the circuit shown in Figure P10.32: VCC  12 V

  130

R1  82 kΩ

R2  22 kΩ

RE  0.5 kΩ

RL  16 Ω .

Find: VCEQ at the DC operating point. Analysis: Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network (R1,, R2, VCC) in the base circuit: 12  22 V CC R 2 = = 2.538 V + 82  22 R1 R 2 R1 R 2 = 82  22 = 17.35 kΩ R B = R eq = 82  22 R1 + R 2

VD : V BB = V TH = V OC = Suppress V CC :

Redraw the circuit using the Thèvenin equivalent. The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages.

Assume the transistor is operating in its active region. Then, the base-emitter junction is forward biased. V BEQ  700 mV [Si]

I EQ = [  + 1 ] I BQ

KVL : - V BB + I BQ R B + V BEQ + I EQ R E = 0 - V BB + I BQ R B + V BEQ + [  + 1 ] I BQ R E = 0

I BQ =

V BB - V BEQ 2.538  0.7 = = 22.18 μA 17350  130  1  500 R B +  +1  R E

6 I EQ =  + 1  I BQ = 130 + 1   22.18  10 = 2.906 mA

KVL : - I EQ R E - V CEQ + V CC = 0 V CEQ = V CC - I EQ R E = 12  2.906  0.5 = 10.55 V

The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid.

Chapter 10, Solution 33. Known quantities: For the circuit shown in Figure P10.33: VCC  12 V   100 VEE  4 V RC  3 kΩ RE  3 kΩ RL  6 kΩ

RS  0.6 kΩ

RB  100 kΩ

vS  1 cos(6.28  103 t ) mV .

Find: VCEQ and the region of operation. Analysis: The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. Assume the transistor is operating in its active region; then, the baseemitter junction is forward biased and:

VBEQ  700 mV [ Si ] I CQ    I BQ

I EQ  (   1) I BQ

KVL :  VEE  I BQ RB  VBEQ  I EQ RE  0 I BQ 

VEE  VBEQ

RB    1RE





4  0.7  8.189A 100000  (100  1)(3000)

I CQ    I BQ  (100)  8.189  10  6  818.9A I EQ  (   1)  I BQ  (100  1)  8.189  10  6  827.0 A KVL :  VEE  I EQ RE  VCEQ  I CQ RC  VCC  0



VCEQ  VEE  VCC  I CQ RC  I EQ RE  4  12  818.9  10  6  3000  827.0  10  10  6  3000  11.06 V

The collector-emitter voltage is greater (more positive) than its saturation value (+ 0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. Notes: 1. DC power may be supplied to an npn BJT circuit by connecting the positive terminal of a DC source to the collector circuit, or, by connecting the negative terminal of a DC source to the emitter circuit, or, as was done here, both. 2. In a pnp BJT circuit the polarities of the sources must be reversed. Negative to collector and positive to emitter. Chapter 10, Solution 35. Known quantities: For the circuit shown in Figure P10.35: vS  3 V

  100

RB  60 kΩ

Find: a) The value of RE so that I E is 1 mA. b) RC so that VC is 5 V. c) The small-signal equivalent circuit of the amplifier for RL  5k d) The voltage gain. Analysis: (a) With RB  60 k and VB  3 V , applying KVL, we have 3  I B RB  0.6  (1   ) I B RE 2.4 IB  60k  101RE

I E  101

2.4  1mA 60k  101RE

Therefore, RE 

101  2.4  60  1.81 k 101

(b) VCE  15  IC RC  I E RE From (a), we have I C  I E Therefore, RC 

  0.99 mA  1

15  5  1.81  8.27 k 0.99

(c) The small signal equivalent circuit is shown below I B 

(d)

VS RB  hiw

 V 1   I C  out  h fe I B vout  I C  RL   1 hoe hoe   V 0.6 hie  BE I BQ   60.6k I B 0.0099  103

RB +

vS h ie -

B

C

RC

 IB h fe  I

 IC

B

1 RL h oe

E

Since hoe is not given, we can reasonably assume that 1/hoe is very large. Therefore, AV 

vout 100  RL   4.15 vs RB  hie

Chapter 10, Solution 36. Known quantities: For the circuit shown in Figure P10.36: RC  200 kΩ

Find: e) The operating point of the transistor. f) Voltage gain vout vin ; current gain iout iin g) Input resistance ri h) Output resistance ro Analysis: (a) VB  VCC

R2  6.1 V R1  R2

RB  R1 || R2  3749.87 

Assuming VBE  0.6 V , we have VEV  VB  VBE  5.5 V

+ v OUT -

V I E  E  22 mA RE I I B  E  0.088 mA b 1

and

VCE  VC  VE  V.CC  RC I C   5.5  15 - 200  21.912  10-3  5.5  5.12 V

(b) The AC equivalent circuit is shown on the right: hie 

VBE I B

I BQ



0.6 0.088  103

 6.82k

vout  RE ( I B  I C )  250(250  1) I B vin  I B hie  vout  I B hie  250  251 I B

Therefore, the voltage gain is AV 

vout  0.902 vin

and

iout  I B  I C  (  1)  I B v iin  I B  in  I B  ( I B hie  250  251  I B ) RB RB

and the current gain is iout (   1) I B   12.84 iin I B  ( I B hie  250  251  I B ) RB

(c) To find the input resistance we compute: vin  I B hie  250  251  I B iin  I B  ( I B hie  250  251  I B ) RB

Therefore. the input resistance is ri 

vin  3558 iin

(d) To find the output resistance we compute vout  RE ( I B  I C )  250(250  1) I B

iout  I B  I C  (  1)  I B

Therefore, the output resistance is ro 

vout  250 iout

Chapter 10, Solution 41. Known quantities: The circuit given in Figure P10.41. Find: Show that the given circuit functions as an OR gate if the output is taken at v01. Analysis: Construct a state table. This table clearly describes an AND gate when the output is taken at vo1 . v1

v2

Q1

Q2

Q3

vo1

vo2

0

0

off

off

on

0

5V

0

5V

off

on

off

5V

0

5V

0

on

off

off

5V

0

5V

5V

on

on

off

5V

0

Chapter 10, Solution 42. Known quantities: The circuit given in Figure P10.41. Find: Show that the given circuit functions as a NOR gate if the output is taken at v02. Analysis: See the state table constructed for Problem 10.41. This table clearly describes a NOR gate when the output is taken at vo 2 .

Chapter 10, Solution 45. Known quantities: In the circuit given in Figure P10.45 the minimum value of vin for a high input is 2.0 V.

Assume that the transistor Q1 has a  of at least 10. Find: The range for resistor RB that can guarantee that the transistor is on. Analysis: ic 

5  0.2  2.4 mA , therefore, iB = iC/ = 0.24 mA. 2000

(vin)min = 2.0 V and (vin)max = 5.0 V, therefore, applying KVL:

-vin +RB iB + 0.6 = 0

vin  0.6 . Substituting for (vin)min and (vin)max , we find the following range for RB: iB 5.833 kΩ  RB  18.333k Ω

or

RB 

Chapter 10, Solution 46. Known quantities: For the circuit given in Figure P10.46: R1C  R2C  10 kΩ , R1B  R2B  27 kΩ .

Find: a) vB, vout, and the state of the transistor Q1 when vin is low. b) vB, vout, and the state of the transistor Q1 when vin is high. Analysis: a)

vin is low  Q1 is cutoff vB = 5 V  Q2 is in saturation vout = low = 0.2 V.

b) vin is high  Q1 is in saturation vB = 0.2 V  Q2 is cutoff vout = high = 5 V.