semiconductor devices and circuit lab manual

1 M.M.UNIVERSITY, SADOPUR (AMBALA)

LABORATORY MANUAL

PRACTICAL EXPERIMENT INSTRUCTION SHEET AIM: To measure and study P-N junction diode V- I characteristics. EXPERIMENT NO. : ECE-209-01

ISSUE DATE  :

DEPTT.: ELECTRONICS AND COMMUNICATION ENGINEERING LABORATORY :

SEMESTER : III

NO. OF PAGES : 04

Aim: - To measure and study P-N junction diode V- I characteristics. Apparatus : - Diode Characteristics Kit and connecting leads. Theory: -

A P-N junction is known as Semiconductor diode or Crystal diode. It is the combination of P-type & N-type Semiconductor which offers nearly zero resistance to current on forward biasing & nearly infinite Resistance to the flow of current when in reverse b iased. Forward biasing

When P-type semiconductor is connected to the +ve terminal and N-type to  – ve ve terminal of voltage source nearly zero resistance is offered to the flow of current. Reverse biasing

When P-type semiconductor is connected to the  – ve ve terminal and N-type to +ve terminal nearly zero current flow in this condition.

Fig. 1.1 Forward biased P-N junction

PREPARED BY: Deepak Dhadwal

Fig. 1.2 Reverse biased P-N junction

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2

Procedure : -

(1) Connect the circuit as shown in figure. (2) Switch on the power supply. (3) Vary the value of input dc supply in steps. (4) Note down the ammeter & voltmeter readings for ea ch step. (5) Plot the graph of Voltage Vs Current.

Observations: Table 1.1 Observation Table.

Fig. 1.3 Forward and Reverse Diode I-V characteristics

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

2

Procedure : -

(1) Connect the circuit as shown in figure. (2) Switch on the power supply. (3) Vary the value of input dc supply in steps. (4) Note down the ammeter & voltmeter readings for ea ch step. (5) Plot the graph of Voltage Vs Current.

Observations: Table 1.1 Observation Table.

Fig. 1.3 Forward and Reverse Diode I-V characteristics

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

3 Precautions :

(1)Always connect the voltmeter in parallel & ammeter in series as shown in fig. (2)Connection should be proper & tight.

(3)Switch „ON‟ the supply after completing the circuit. (4)DC supply should be increased slowly in steps (5)Reading of voltmeter & Ammeter should be accurate.

Result: - The graph has been plotted between voltage and current.

QUESTION-ANSWER:

Q1. What is extrinsic semiconductor? ANS. Intrinsic semiconductor conducts a small amount of current. To make it more conductive a small amount of impurity is added in it. Then this semicondu ctor is called extrinsic semiconductor.

Q2. What is an intrinsic semiconductor? ANS. Semiconductor in its purest form is called intrinsic semiconductor.

Q3. Which semiconductor is mostly used germanium or silicon? ANS. Silicon is mostly used because germanium has some free electrons at room temperature where as silicon have no free electrons at room temperature.

Q4. What is n- type semiconductor? ANS. When a small amount of Pentavalent impurity is added to pure semiconductor which results a large number of free electrons, this semiconductor is known as n- type semiconductor.

Q5. What is p- type semiconductor? ANS. When a small amount of trivalent impurity is added to pure semiconductor which results a large number of holes, this semiconductor is known as p- type semiconductor.

PREPARED BY: Deepak Dhadwal

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4 Q6.What is P-N Junction? ANS. The P-N junction is control element for semiconductor devices, when a P-type semiconductor is suitably  joined to an N-type semiconductor this formation is called P-N junction.

Q7. What do you understand by potential barrier? ANS.The electric field between donor and acceptor ions is called potential barriers.

Q8. What do you mean by knee voltage? ANS. The forward voltage at which the current through the diode starts rising abruptly is known as knee voltage. Its value for germanium is 0.3v and for silicon it is 0.7v.

Q9.What is break down voltage? ANS. The reverse voltage at which p-n junction breaks is known as breakdown voltage.

Q10. Explain peak inverse voltage (PIV)? ANS. The maximum value of reverse voltage that p-n junction can withstand without getting damaged is called  peck inverse voltage.

Q11. What are the application areas of diode? ANS. Diode is mainly used in rectifier circuits, detector and switching circuit.

Q12. What do you mean by forward biased? ANS. When +ve terminal of battery is connected to P side & -ve terminal to N side of diode.

Q13. What do you mean by reverse biased? ANS. When +ve terminal of battery is connected to N side & -ve terminal to P side of diode.

Q14. Define max. Power rating? ANS. Max. Power that can be dissipated at junction without damage to it. PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

5 M.M.UNIVERSITY, SADOPUR (AMBALA)

LABORATORY MANUAL

PRACTICAL EXPERIMENT INSTRUCTION SHEET AIM: To Study Half wave and Full wave Rectifier. EXPERIMENT NO. : ECE-209-02

ISSUE DATE  :

DEPTT.: ELECTRONICS AND COMMUNICATION ENGINEERING LABORATORY :

SEMESTER : III

NO. OF PAGES: 06

Aim: -To Study Half wave and Full wave Rectifier. Apparatus : - Rectifier kit, CRO and connecting leads. Theory: Half wave rectifier: Rectification is a process of conversion of AC to DC. In half-wave rectifier, only one diode is used . During +ve half Cycle the diode is forward biased &, it conducts current through the load resistor R .During  – ve half cycle diode is reverse  biased Hence, no current flow through the circuit. Only +ve half cycle appears across the load, whereas, the

 – ve half

Cycle is suppressed.

Full wave rectifier:

In full-wave rectification, When A.C supplied at the input , both the half cycles current flows through the load in the same direction. The following two circuits are commonly employed.

Centre-tap full-wave Rectifier :

In this rectifier, two diodes & a center-tap transformer is used. During +ve half cycle the diode D1 is forward  biased & D2 is reverse biased .Output will be obtained across load resistor R .During – ve half cycle diode D1 is reverse biased &D2 is forward biased. Output will be obtained across load resistor R again & the direction of output is same i.e, DC output is obtained.

Bridge Rectifier :

The ckt. Contains four diodes connected to form a bridge. In this an ordinary Transformer is used. During +ve half cycle of secondary voltage, diodes D1 & D3 are forward biased & diodes PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

6 D2& D4 are reverse biased & vice versa. Diagrams :

Fig. 2.1 Circuit Diagram of Half Wave Rectifier

Fig. 2.2 Circuit Diagram of Centre-tap full-wave Rectifier

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

7

Fig. 2.3 Circuit Diagram of full-wave Bridge Rectifier Procedure: 1.

Connect the circuit as shown in figure.

2.

Supply the input AC signal to the circuit.

3.

Output signal is obtained on CRO which shows the DC (pulsating output).

4.

Measure voltage on Meter.

5.

Draw the wave form.

Wave forms:

Fig. 2.4 Wave form For Half wave Rectifier

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

8

Fig. 2.5 Wave form For Full wave Rectifier Observation Table: -

S.NO.

Measurements Parameter

1.

Vdc

2.

RMS Value of Voltage

Half Wave Rectifier

Full wave Rectifier

Vrms 3.

Ripple factor=

√( Vrms / Vdc)2--1 4.

Input Frequency

5.

Output Frequency

6.

Peak inverse voltage

Precautions: -

(a)Connection should be proper & tight.

(b)Switch „ON‟ the supply after completing the circuit. (c)Note down the input & output wave accurately.

Result: - The input and output waveforms of full wave rectifier has been drawn.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

9 QUESTION-ANSWER:

Q1. What is rectifier? ANS. Rectifier is basically a p-n junction diode w hich converts alternating current into unidirectional current.

Q2. What is need of rectifier? ANS. Most of the electronics components are operated with dc voltage. Therefore it is necessary to convert alternating voltage into dc voltage.

Q3.What is the different type of rectifier? ANS. There are mainly two types of rectifier, 1.Half wave rectifier

2. Full wave rectifier.

Q4.What is half wave rectifier? ANS. When ac supply is applied at the input of rectifier it will suppress its one half and will give only one half at output.

Q5. What do you mean by rectifier efficiency? ANS. The ratio of DC power output to the AC power input is called rectifier efficiency.

Q6. What do you understand by ripple factor? ANS. Ripple factor is defined as the ratio of the effective value of the AC component of voltage or current to the average value.

Q7. What is the value of form factor in half wave rectifier? ANS.It is defind as the ratio of rms value to the average value. The value of form factoe in half wave rectifier is 1.57.

Q8. What do you mean by full wave rectifier?

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

10

ANS. In this type of circuit‟s more than one diode is used. It enables the circuit to process both cycles of the AC supply. During both the cycles, current flow through the load in the same direction.

Q9. What is Centre- tap rectifier? ANS. In this type of circuit a transformer with the centre tapped secondary is used. Two diodes work together in such a way that during the positive half cycle one diode is forward biased and the other is reverse biased. For st

negative half cycle the second diode will be forward biased and 1  diode is reverse biased. Hence we can get output in the both the helf cycle.

Q10.What is step- down transformers? ANS. In step down transformer the number of turns in primary winding is greater than the number of turns in secondary winding.

Q11. What is the value of PIV in case of half wave rectifier? ANS. PIV= Vm

Q12. What is the value of P IV in case of Centre- tap wave rectifier? ANS. PIV= 2Vm

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

11 M.M.UNIVERSITY, SADOPUR (AMBALA)

LABORATORY MANUAL

PRACTICAL EXPERIMENT INSTRUCTION SHEET AIM: To study diode as clipper and clamper. EXPERIMENT NO. : ECE-209-03

ISSUE DATE  :

DEPTT.: ELECTRONICS AND COMMUNICATION ENGINEERING LABORATORY :

SEMESTER : III

NO. OF PAGES: 04

Aim: - To study diode as clipper and clamper. Apparatus: - Function Generator, Oscilloscope, DC Power Supply, Breadboard, Diodes, Capacitors and

Resistor. Theory: - This experiment studies the applications of the diode in the clipping & clamping operations. 1. Clipping Circuits: -

Figure (l) shows a biased positive clipper, for the diode to turn in the input voltage must be greater +V, when Vm is greater than +V , the diode acts like a closed switch (ideally) & the voltage across the output equals +V , this output stays at +V as long as the input voltage exceeds +V. When the input voltage is less than +V , the diode opens and the circuit acts as a voltage divider, as usual , R L should be much greater than R, in this way , most of input voltage appears across the output. The output waveforms of Figure (1) summarize the circuit action. The biased clipper removes all signals above the (+V) level. It is required to remove a portion of negative half cycle of the input voltage, the circuit is improved as shown in fig (2). Such a circuit as known as  biased negative clipper.

2. Clamping Circuits: -

A circuit that shifts either positive or negative peak or the signal at the desired DC level is known as clamping circuit. (1) Positive clamper: - A circuit that shifts the signal the positive side in such a way that the negative peak of the signal falls on the zero level is a called a positive clamper. A positive clamper as shown in fig (3). It contains a diode D and a capacitor C. the output is taken across the load resistor R L

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

12

Fig. 3.1 Positive Clipping

Fig. 3.2 Negative Clipping

Fig. 3.3 Positive Clamping

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

13

Fig. 3.4 Negative Clamping

Procedure: Clipping Circuit :

1. Connect the circuit shown in Figure (3). 2. Ensure that the variable DC is at minimum and the source is at 10VP.P. 3. Observe and Sketch the input and output waveforms. 4. Increase the variable DC voltage to 4V, and notice to what voltage are the positive peaks chopped off, sketch the waveforms.

Clamping Circuit :

1. Connect the circuit shown in Figure (4). 2. Ensure the variable DC is at minimum. 3. Set the sine wave generator frequency to 1 KHz and its output amplitude to 10VP.P 4. Observe and sketch the input waveform with the variable DC at minimum, Sketch the output waveform.

Precautions: -

(a)Connection should be proper & tight.

(b)Switch „ON‟ the supply after completing the ckt. (c)Note down the input & output wave accurately.

Result: - Output voltage V0 =__________ during positive half cycle

Output voltage V0 =__________ during negative half cycle PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

14 QUESTION-ANSWER:

Q1.What is non-linear wave shaping? ANS. Non linear wave shaping is the process, on applying any wave at input of a non-linear device the shape of the output wave varies non-linearly with the input wave. Q2. Which are the circuits for nonlinear wave shaping? ANS. Clipping circuit & Clamping circuits are the circuits for non-linear wave shaping. Q3. According to non-linear devices how clippers can be classified? ANS. According to non-linear devices clippers can be classified as diodes clippers & transistor clippers. Q4. According to configuration used classify clippers? ANS. According to configuration used classify clippers can b e classified as a)Series diode clipper. b) Parallel or shunt diode c lipper. c) Combination clippers. Q5. Classify clippers according to level of clippers? ANS. According to level of clipping the clippers may be a) Positive clippers. b) Negative clippers c) Biased clippers d) Combinational clippers. Q6. What is positive clipper circuit? ANS. Positive clipper is one which removes the positive half cycles of the input voltage Q7. What is negative clipper circuit? ANS. Negative clipper is one which removes the negative half cycles of the input voltage Q8. What is clamping? ANS. A circuit that places either the positive or negative peak of a signal at a desired level is known as Clamping circuit. Q9. How many types of clampers are there? ANS. There are 2 types of clampers a) Positive clamper. b)Negative clamper Q10. What is clipping circuit ? ANS. A wave-shaping circuit which controls the output waveform by removing or clipping a portion of the applied wave is known as clipping circuit.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

15 M.M.UNIVERSITY, SADOPUR (AMBALA)

LABORATORY MANUAL

PRACTICAL EXPERIMENT INSTRUCTION SHEET AIM: To study common emitter (CE) amplifier  –  it’s current & power gains and input, output impedances. EXPERIMENT NO. : ECE-209-04

ISSUE DATE  :

DEPTT.: ELECTRONICS AND COMMUNICATION ENGINEERING LABORATORY :

SEMESTER : III

NO. OF PAGES: 05

Aim: -To study common emitter (CE) amplifier  –  it’s  current & power gains and input, output impedances. Apparatus required: - NV6542 trainer kit, Digital Multimeter (DMM), Multimeter probes, 2mm Patch chords,

mains cord. Theory: -

The common-emitter (CE) amplifier provides an output voltage that is 180° out of phase with the input voltage. This voltage phase shift can be explained as follows: 1.

The input voltage and current are in phase.

2. The input and output currents are in phase. Therefore, output current is in phase with the input voltage. 3. An increase in output current results in a decrease in output voltage, and vice versa (as given by). Therefore, output voltage is 180° out of phase with output current. Since the output current is in phase with the input voltage, the input and output voltages are 180° out of phase. Various parameters associated with BJT CE amplifier are:1. Voltage Gain (Av):Voltage gain is the factor by which ac signal voltage increases from the amplifier input to the amplifier output. Stated mathematically,

Since IE

≈ IC, the voltage gain of a CE amplifier also equals the ratio of ac collector resistance to ac emitter

resistance.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

16 Theoretical → Av = -R C/ R e Practical → Av = Vout / Vin 0

Note: Negative sign reveals an 180  phase shift between Vout and Vin

2. Input impedance, Zin:The input impedance of a CE amplifier equals the parallel combination of the base biasing resistor(s) and the input impedance to the transistor base. Theoretical → Zin = R 1 ║ R eq

Practical →Zin =Vin / Iin 3. Output Impedance, Zout:Theoretical → Zout = R c

Practical →Zout = Vout / Iout 4. Current gain, Ai:Ai = Iout/ Iin Iin = Vin / Zin And Iout = - Vout / R L

Procedure:To measure Voltage gain, A v

1. Connect circuit as shown in fig. 2.

Connect CRO probe from sockets „Vin1‟ and „G1‟ to channel CH1 of CRO.

3. Switch on the power supply. 4.

Using the „Frequency control‟ and „Amplitude control‟ knobs of the „Sine Wave generator‟ section, set

the input signal at 2Vp-p, 1 KHz sine wave signal and observe the same on CRO channel CH1. 5. Switch off the power supply. Remove CRO probe. 6. Now set the multimeter dial to AC voltage mode. 7. Connect multimeter probes to test points „Vin1‟ and „G1‟. PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

17 8. Switch on the power supply.

 Note down the value observed on millimeters‟ display. It is Vin. 10. Connect multimeter probes to test points „ Vout1 ‟ and „G2‟. 9.

11. Switch on the power supply. 12. Note down the value observed on multimeter display. It is Vout 13. Switch off the power supply. 14. Now calculate the Voltage Gain (Av) using the formula Av = Vout /Vin 15. Now to verify this value of voltage gain we will calculate its theoretical value such that Av = -R C /R  Note: The value of Rc is 94 Ω and Re is 51.1 Ω.

To calculate Input impedance, Z in:-

1. Note down the value of Vin obtained. 2. Now remove the patch cord between „Vin1‟ and „1‟. 3. Set the multimeter dial to AC current mode. 4. Connect multimeter probes between sockets „Vin1‟ and „1‟ 5. Note down the value observed on multimeter‟s display. It is Iin. 6. Switch off the power supply. 7. Now calculate the input impedance (Zin) using the formula Zin =VIN / Iin 8. Now to verify this value of input impedance we will calculate its theoretical value such that Zin = R1 ║ R eq

The value of R1 is 14.7 KΩ, R2 is 6.6 KΩ, take hFE = 50 (as per datasheet) and RE or Re is equal to 51.1Ω.

 Note:

To calculate Output impedance, Z out:-

1. Note down the value of Vout obtained. PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

18 2. Set the multimeter dial to AC current mode. 3. Remove patch cords between sockets „B‟ and „C‟. 4. Now connect multimeter probes between sockets „B‟ and „C‟ . 5. Switch on the power supply & note down the value observed on multimeters display. It is Iout. 6. Switch off the power supply. 7. Now calculate the output impedance (Zout) using the formula Zout = Vout / Iout 8. Now to verify this value of input impedance we will calculate its theoretical value such that Zo = R c Note: The value of R c is 94 Ω.

To measure Current Gain, Ai:1. Note down the value of Iin obtained. 2. Note down the value of Iout obtained. 3. Now calculate the Current Gain (Ai) using the formula Ai = Iout/ Iin 4. Now to verify this value of voltage gain we will calculate its theoretical value such that Ai = β  Note: The value of β is between 25- 50 as per the specification sheet. Observations:

Percentage Error (%) = {(Calculated value –  Theoretical value)/Theoretical value }x 100 Table 4.1 Observation Table

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

19 Result:-

The calculated values of various parameters are: 1. Voltage gain, Av = …………. 2. Current gain, AI = ……… 3. Output Impedance Zo =…….ohms 4. Input impedance, Zin =……..ohms

QUESTION-ANSWER:

Q1. What do you mean by biasing of transistor? ANS. When dc voltages are applied across the different terminals of transistor, it is called biasing. Q2. What is d.c. current gain in common base configuration? ANS. It is ratio of collector current(Ic) to emitter current (Ie). Q3. What is typical value for dc current gain? ANS. 0.99 Q4. What is a.c. current gain in CB configuration? ANS. It is ratio of change in collector current to change in emitter current. Q5. What are input characteristics? ANS. These curves relate i/p current & i/p voltage for a given value of o/p voltage. Q6. What are output characteristics? ANS. The curves relate o/p voltage & o/p current for a given value of input current. Q7. Which configuration has highest voltage gain? ANS. Common Emitter. Q8. Which configuration is most widely used? ANS. Common Emitter. Q9. What is operating point? ANS. The zero signal values of Ic & Vce. Q10. Which region is heavily doped in Transistor? ANS. Emitter.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

20 M.M.UNIVERSITY, SADOPUR (AMBALA)

LABORATORY MANUAL

PRACTICAL EXPERIMENT INSTRUCTION SHEET AIM: To study the frequency response of RC-Coupled Amplifier. EXPERIMENT NO. : ECE-209-05

ISSUE DATE  :

DEPTT.: ELECTRONICS AND COMMUNICATION ENGINEERING SEMESTER : III

LABORATORY :

NO. OF PAGES: 05

Aim: - To study the frequency response of RC-Coupled Amplifier. Apparatus required:- NV6542 trainer, CRO, CRO probes, Function generator, Function generator probes, 2

mm patch cords, mains cord. Theory: RC- Coupled Amplifier

One of the coupling methods to couple two stages of an amplifier is RC-coupling. An RC Coupled network is shown in figure5.1.

Fig. 5.1 RC Coupled Amplifier Stages

The network of R2 and Cl enclosed in the dashed lines is the coupling network. C1 is the coupling capacitor which connects the output of Ql to the input of Q2. R2 will develop the signal to be applied to the base of Q2. Cl acts as a limiting factor at low frequencies because its reactance increases with a decrease in frequency and some point will be reached when a voltage drop will appear across it. This will reduce the size of the signal PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

21  being applied to Q2. At medium frequencies the reactance of Cl is so small that it can be considered a short to the signal. C1 will also isolate any DC voltage developed at the collector of Q1 from the DC bias developed at the base of Q2. A frequency-response curve is a graphical representation of the relationship between amplifier gain and operating frequency. A generic frequency response curve is shown in Figure. This particular curve illustrates the relationship between power gain and frequency.

Fig. 5.2 Frequency-response curve

As shown: 1. The circuit power gain remains relatively constant across the mid band range of frequencies. 2. As operating frequency decreases from the mid band area of the curve, a point is reached where the power gain begins to drop off. The frequency at which power gain equals 50% of its mid band value is called the lower cutoff frequency ( f c1 ).

3. As operating frequency increases from the mid band area of the curve, a point is reached where the power gain begins to drop off again. The frequency at which power gain equals 50% of its mid band value is called the upper cutoff frequency (f c2 ).  Note that the bandwidth of the circuit is found as the difference between the cutoffs frequencies. By formula, BW = f C2 –  f C1

Procedure:1. Connect „1‟ and „2‟ to channel CH1 of CRO.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

22 2. Switch on the power supply. 3.

Using the „Frequency Control ‟ and „Amplitude Control ‟ knobs of the Sine Wave Generator

section, set

the Input signal at 1Vp-p, 100 Hz sine wave signal and observe the same on CRO channel CH1. 4. Switch off the power supply. 5. Remove CRO probes. 6. Connect a patch cord across sockets ‘+12V’ of DC power su pply and ‘+12V’ of RC coupled Amplifier

section.

Connect a patch cord between points „Vin1‟ and „1‟ and another patch cord  between sockets „G1‟ and „2‟without disturbing the „Frequency Control’ and ‘Amplitude Control’ knobs of the Sine wave generator 7.

section. 8. Connect a patch cord between „B‟ and „C‟. 9. Connect CRO probes from Vin1 and G1 to CRO channel CH1. 10. Connect point Vout1 with point Vin2 and point G2 with G3 (This will cascade the two stages through RC

coupling. Where Cc is the coupling capacitor and the

Combination of R‟1 and R‟2 will act as coupling

resistance) 11. To observe the output waveform at Stage 2 between points Vout2 and G4, use a CRO probe from Vout2 and G4 to CRO channel CH2. 12. Increase the amplitude of Input signal until you get maximum undistorted output signal. Note down the

value of Input signal (Vin). 13. Keep on rotating the Frequency Control  pot while recording the values of both the input frequency (Hz)

and output voltage amplitude (Vout). 14. Now we will calculate voltage gain Av given by Vout /Vin. 15. Now calculate Voltage gain Av (dB) = 20 log A v. 16. Plot the graph between Input Frequency (Hz) and Voltage Gain (dB).

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

23 Observation Table: S.NO.

Frequency

1

100Hz

2

500Hz

3

1KHz

4

3 KHz

5

5 KHz

6

10 KHz

7

20 KHz

Vin

Vout

Av = Vout / Vin

Av (dB) = 20 log Av

Result:- Frequency responses of RC-Coupled Amplifier have been plotted. QUESTION-ANSWER:

Q1. In RC coupled amplifier which component is responsible for reduction in voltage gain in the high frequency range? ANS. Shunt capacitance in the input circuit.

Q2. In RC coupled amplifier which component‟s value is responsible for low 3-Db frequency? ANS. Increasing the value of coupling capacitor Cb.

Q3. In RC coupled amplifier which component‟s value is responsible for high 3-dB frequency? ANS. By reducing the total effective shunt capacitance in the input circuit of hybrid pie model. Q4. In a single stage RC coupled amplifier, what is the phase shift introduced in the true middle frequency? ANS. 180° Q5. Which type of coupling capacitor is used in RC coupled amplifier?

ANS. 0.05 μf paper capacitor.

Q6. What is the application of RC coupled amplifier? ANS. It is widely used as a voltage amplifier. Q7. In single stage RC coupled amplifier, what is the phase shift at low 3-dB frequenc y? ANS. 225° Q8. In single stage RC coupled amplifier, what is the phase shift at high 3- dB frequency? ANS. 135° PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

24

Q9. In RC coupled amplifier what is the effect of low 3-dB frequency by increasing the value of coupling capacitor C b? ANS. Decreasing. Q10. In RC coupled amplifier what is the effect of low 3-dB frequency by increasing the value of total effective shunt capacitor? ANS. Decreasing.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

25 M.M.UNIVERSITY, SADOPUR (AMBALA)

LABORATORY MANUAL

PRACTICAL EXPERIMENT INSTRUCTION SHEET AIM: Study of Unipolar Junction Transistor (UJT) Characteristics. EXPERIMENT NO. : ECE-209-06

ISSUE DATE  :

DEPTT.: ELECTRONICS AND COMMUNICATION ENGINEERING LABORATORY :

SEMESTER : III

NO. OF PAGES: 05

Aim: -Study of Unipolar Junction Transistor (UJT) Characteristics. Apparatus required: - UJT trainer kit, connecting lead. Theory: -

An unipolar   junction transistor consists a bar of n - type silicone which is lightly doped, having high ohmic contacts at both side of the bar. These are called as base1 and base 2. A small piece of p - type material is diffused at the bar and an ohmic contact is brought out called the emitter. The whole arrangement is treated as a series combination of two base resistances as RB1 and RB2 called as inter base resistance is equal to RBB and a diode. Under operating conditions the emitter is forward biased with respect to base1asVEB1. Without applying VBB the junction exhibit a p n junction diode forward biased characteristics with slightly higher resistance than diode. When VBB is also applied such that the base2 is +ve than base1, the friction of voltage appears across RB1 as;

η=R B1/R BB=VRB1/R B

--------1

Generally η At small value of VEB1 when it is less than η VBB, the p - n junction is reverse biased and a small current flows ( in μA ). When VEB1 is greater than ( > V BB + VD ) the diode conducts and holes are injected into R B1. They are attracted towards the -ve of VBB. This causes an equal number of electrons to flow from R B1to the emitter, which leads to decrease R B1. Thus the emitter current increases and VEB1 drops. This negative resistance effect extends from peak point voltage to the valley point voltage as VP to VV. Increase in VEB1Further results in increase of emitter current gradually. The peak po int voltage is related to η as;

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

26 η = RB1/RBB = (VP - VD)/VBB

---------------2

Where the VD is the drop across the forward biased silicone diode ( = 0.6V) The negative resistance region of the uni junction transistor is used as relaxation oscillator in power control. Circuit Diagram: -

Fig. 6.1 Circuit arrangement for V e - Ie characteristics.

Fig. 6.2 Ve - Ie characteristics for unipolar  junction transistor

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

27 Procedure: -

1. Connect the circuit as shown in fig 1. Keep both supplies to minimum. Switch ON the power. Now V EB and VBB = 0.

2. Connect the milliammeter as shown in fig 1. Slowly increase the VEB supply and note the voltage value at which the mA meter starts to deflect. This will give inherent emitter  –   base Forward voltage drop VD. Increase voltage further and note voltage and current readings. Bring VEB supply back to zero.

3. Adjust the VBB voltage about +10V. 4. Increase VBE until it approaches the VP value. At VP the current rise suddenly and the emitter voltage drops to a value equal to valley voltage VV. Increase VBE Further and note more readings of VE and IE. Bring VBE supply back to minimum.

5. Repeat step 4 at different values of VBB. Note VE, IE, VP and VV for each setting of VBB. 6. Connect mA meter at + VBB and B2 of UJT as shown in fig 2. 7. Fix the VBB voltage at +15V. Note the deflection in mA meter since R BB is presented there. Increase VBE to fire the UJT. Note the sudden rise in current at B2 terminal, which show the decrement of R BB at trigger  point = VPeak .

8. Plot graphs between VE and IE from the observations, Find out the negative resistance region and saturation region from the graph. The declined graph shows the conductivity modulation. Calculate the resistance at VP and VV. The result shows fall in resistance, which indicates the negative resistance region of the device.

Δ V / Δi, where Δ V is equal to (Vp - VV.) R BB = VBB / IBB, where VEB = 0.

Observations: Table 6.1 Observation Table

S.No.

VBB = 0V VE

VBB = 10V IE1

VE

IE2

1. 2. 3. PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

28 Precautions :

(1)Always connect the voltmeter in parallel & ammeter in series as shown in figure. (2)Connection should be proper & tight.

(3)Switch „ON‟ the supply after completing the c ircuit. (4)DC supply should be increased slowly in steps. (5)Reading of voltmeter & Ammeter should be accurate.

Result: -- The graph has been plotted between voltage and current.

QUESTION-ANSWER: Q1. What are h parameters of a transistor? ANS. The h parameter or the hybrid parameters of a transistor helps us to analyze the amplifying action of transistor for small signal .it is necessary for practical purposes. In generally there are four type of h parameter. These are : h11 h12 h21 h22

Q2. What are benefits of H-parameter? ANS. Following are the advantages of h-parameters:I. h-parameters are Real Numbers up to radio freq uency II. They are easy to measure III. They can be determined from transistor static characteristic curves IV. They are convenient to use in circuit analysis and design V. Easily convertible from one configuration to other VI. Readily supplied by manufacturers Q3. H parameter operates in which frequency? ANS. up to radio frequencies i.e 20-20khz Q4. Which of the h-parameters corresponds to re in a common-base configuration? ANS. hib Q5. Which of the following is referred to as the reverse transfer voltage ratio? ANS.hr Q6.What is admittance? ANS. In electrical engineering, the admittance (Y ) is a measure of how easily a circuit or device will allow a current to flow. It is defined as the inverse of the impedance ( Z ). The SI unit of admittance is the siemens (symbol S). Q7. What is conductance? PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

29 ANS. The electrical resistance of an electrical element is the opposition to the passage of an electric current through that element; the inverse quantity is electrical conductance , the ease at which an electric current  passes. Electrical resistance shares some conceptual parallels with the mechanical no tion of friction. The SI unit

of electrical resistance is the ohm (Ω), while electrical conductance is measured in siemens (S). Q8. The lowest output impedance is obtained in case of BJT amplifiers for ANS. CC Configuration

Q9. Which of the following parameters is used for distinguishing between a small signal and a large-signal amplifier? ANS. (A) Voltage gain (B) Frequency response (C) Harmonic Distortion (D) Input/output impedances Q10. Removing bypass capacitor across the emitter-leg resistor in a CE amplifier causes (A) increase in current gain. (B) decrease in current gain. (C) increase in voltage gain. (D ) decrease in voltage gain.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

30 M.M.UNIVERSITY, SADOPUR (AMBALA)

LABORATORY MANUAL

PRACTICAL EXPERIMENT INSTRUCTION SHEET AIM: To study Push - Pull amplifier EXPERIMENT NO. : ECE-209-07

ISSUE DATE  :

DEPTT.: ELECTRONICS AND COMMUNICATION ENGINEERING LABORATORY :

SEMESTER : III

NO. OF PAGES: 04

Aim: -To study Push - Pull amplifier.

Apparatus required: -  Push - Pull amplifier trainer kit and connecting lead.

Theory: -

In a push - pull power amplifier, class B is used to obtain high percentage of efficiency (the relation between taken and given). But these amplifiers suffer from the crossover distortion. It is due to keep the bias nearly cut off point which cause to flow current in transistor in later cycle.Such class B push - pull output stage is shown in fig. where the voltages at emitters of Q3 and Q2, are pulled to ground through RE while the bias is set for 0.25Vdc just below half of required cut  –   off potential (in these transistor it is 0.6Vdc). Due to it the after zero crossing, part of input wave is not able to perform amplification and cause to produce the crossover distortion, where the cut - off bias cause to flow no current when there is no signal and quiescent current flows according to input  –   output signal strength. This enables this mode of configuration to obtain low power dissipation in transistors but high power conversion called the efficiency.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

31 Circuit Diagram: -

Fig. 7.1 Circuit Diagram of push-pull amplifier Procedure: -

1. Switch on power. Observe there is no quiescent current flow through the output transistors. 2.  Now connect the function generator across the input. Connect CRO as shown in fig. Apply sine wave 1Khz signal of such amplitude to Obtain 3Vpp signal across RL. 3. Observe the current in transistors and the crossover distortion at the zero crossing points. 4. Adjust input amplitude to obtain 6Vpp signal across RL. Note the quiescent current from meter. Measure voltage at collector of Q2 as Vc. Now from the observed values: The dc power, Pdc = (Vc) x dc current 2

The ac power, Pac = (Vo rms) ÷ RL Where Vo rms = (Vo pp / 2) †√2

... watt ... watt

... V rms

The percent efficiency of amplifier, η % = (Pac / Pdc) x 100

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

32

Fig. 7.2 Crossover distortion in a push-pull amplifier

Precautions: -

1. Connect the circuit properly as shown in fig7.1. 2. Set the input waveform of correct amplitude and frequency. 3. Connect the CRO to the output terminal. 4. All connection should be tight. 5. Take proper care while taking reading.

Result: -- The class B amplifier has high efficiency (about 70%) but has Crossover distortion which leads to

increase in distortion particularly at low input signals.

QUESTION-ANSWER:

Q1. What is feedback in amplifiers? Ans. The process of combining a fraction of output energy back to the input is called feedback. Q2. What is the application of negative feedback amplifier? Ans. Negative feedback amplifier makes the circuit stable. Q3. What is voltage series feedback amplifier? Ans. It is that amplifier in which output voltage feedback in voltage series with input Signal, resulting in an overall gain reduction. Q4. By Which factor reduces the input noise & non-linear distortions of the amplifier?

Ans. (1+Aβ)

Q5. what is the effect of frequency on phase shift of an amplifier? PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

33 Ans. Shift of an amplifier will change with frequency. Q6. How does negative feedback increase bandwidth of an amplifier? Ans. The bandwidth of an amplifier without feedback is equal to separation between 3 db frequencies f1 and f2. If A is the gain, then gain bandwidth product is A* BW. With the negative feedback the amplifier gain is reduced and since gain bandwidth product has to remain constant in both cases, so the bandwidth will increase to compensate for the reduction in gain. Q7.How do series and shunt feedback differ from each other? Ans. Series means feedback connecting in series with input signal while shunt means feedback connecting in shunt with input signal. Q8. Distortion in an amplifier with negative feedback increases or decreases? Ans. Decreases Q9. Feedback in an amplifier always helps to Ans. Control its output Q10.When negative feedback is applied to an amplifier, its bandwidth: Ans. Increased.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

34 M.M.UNIVERSITY, SADOPUR (AMBALA)

LABORATORY MANUAL

PRACTICAL EXPERIMENT INSTRUCTION SHEET AIM: To study the characteristics of Silicon Controlled Rectifier (SCR) / Thyristor. EXPERIMENT NO. : ECE-209-08

ISSUE DATE  :

DEPTT.: ELECTRONICS AND COMMUNICATION ENGINEERING LABORATORY :

SEMESTER : III

NO. OF PAGES: 05

Aim: -To study the characteristics of Silicon Controlled Rectifier (SCR) / Thyristor. Apparatus:-  NV6530 SCR Characteristic Trainer, 2mm Patch cords Theory: -

The Silicon Controlled Rectifier (SCR) is a semiconductor device that is a member of a family of control devices known as Thyristors. The SCR has become the work house of the industrial control industry. Its evolution over the year has yielded a device that is less expensive, more reliable and smaller in size than ever  before. Typical applications include: DC motor control, generator field regulation, variable frequency drive (VFD) DC bus voltage control solid state Relays and lighting system control. The SCR is a three  –   lead device with an anode and a cathode (us with a Standard diode) plus a third control lead or gate. As the name implies, it is a Rectifier which can be controlled  –   or more correctly one that can be triggered

to the “ON” state by applying a small positive voltage (VTM) to the gate lead. Once gated ON, the

trigger signal may be removed and the SCR will remain conducting as long as current flows through the device. The load to be controlled by the SCR is normally placed in the anode circuit.

Commutation:-

For the SCR to turn OFF current flow through the device must be interrupted, or drop below the maximum holding current(IH) , for a short period of time (typically 10-20 microseconds) which is known as the commutated turn –  off time (tq). When applied to alternating current circuits or pulsating DC system, the device will self  –  commutate at the end of every half –  cycle when the current goes through zero. When applied to pure DC circuits, in applications such as alarm or trip circuit latching, the SCR can be reset manually by interrupting the current with a push button.

PREPARED BY: Deepak Dhadwal

APPROVED BY: HOD(ECE)

35 When used in VFD's or inverters, SCRs are electronically forced OFF using additional Commutating circuitry, Such as smaller SCRs and Capacitors, which momentarily apply an opposing reverse bias voltage across the SCR(This is complicated everything has to be exactly right).

Basic three modes of operation of SCR:

1. Reverse blocking mode

Cathode is positive with respect to anode with gate open.SCR is in reverse bias i.e., junction J1 & J3 in reverse  bias J2 is in forward bias. The device act as two PN diode connected in series with reverse voltage applied across it. Small leakage current of the order of a few mill ampere or microampere flows; this is off state of SCR. If reverse voltage increases, then at critical breakdown level or reverse breakdown voltage (VBR) an avalanche occurs at J1 & J3 & reverse current increase rapidly, so more loss in SCR. This may lead to SCR damage  because Junction temperature is increasing. Maximum working reverse voltage across SCR does not exceed VBR. If applied reverse voltage across SCR < VBR, then the device offers high impedance in reverse direction. SCR is treated as open switch.

2. Forward blocking mode (off state mode)

Anode is positive with respect to cathode with gate open. SCR is forward bias, junction J1 & J3 is forward bias and J2 is reverse bias. Here small forward leakage current flow. If forward voltage increases then J2 junction (rev. bias) will have avalanche breakdown called forward break over voltage (VBO) Maximum working forward voltage across SCR does not exceed VBO. If forward voltage