ec - 1 - labmanual

EINSTEIN COLLEGE OF ENGINEERING Sir.C.V.Raman Nagar, Tirunelveli-12

Department of Electronics and communication Engg.

Subject Code: EC 38

Electronic Circuits - I Lab

Name :

Reg No :

Branch :

Year & Semester :

www.Vidyarthiplus.com


Sub Code: EC38 EC-I Lab

Einstein College of Engineering

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INDEX

Exp. No.

DATE TITLE OF EXPERIMENT PAGE NO. MARKS

STAFF INITIAL

1 FIXED BIAS AMPLIFIER

2 COMMON EMITTER AMPLIFIER WITH AND WITHOUT FEEDBACK

3 COMMON COLLECTOR AMPLIFIER

4 BOOTSTRAPPED SOURCE FOLLOWER

5 DIFFERENTIAL AMPLIFIER

6 CLASS A POWER AMPLIFIER

7 COMPLEMENTARY SYMMETRY CLASS B PUSH PULL AMPLIFIER

8 HALF WAVE RECTIFIER

9 FULL WAVE RECTIFIER

10 DARLINGTON EMITTER FOLLOWER





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FIXED BIAS AMPLIFIER CIRCUIT DIAGRAM: LOAD LINE:





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Ex.No : 1 Date :

FIXED BIAS AMPLIFIER AIM: To design a fixed bias amplifier using BJT and to determine the gain bandwidth product from its frequency response curve. APPARATUS REQUIRED:

1. Resistor (2) 2. Transistor (1) BC 107 3. Capacitor (2) 4. RPS (1) (0-30)V 5. AFO(1) 6. CRO 7. Bread board 8. Connecting wires

DESIGN:





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MODEL GRAPH: TABULATION: Vin = Frequency (HZ) Output voltage ( Volt) Gain=Vo/Vin Gain in

dB=20log(Vo/Vi)





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THEORY: To use a transistor in any application, it is necessary to provide external DC voltages to operate it in the active region. This process is known as Biasing. The biasing circuit should be designed to fix the operating point at the center of the active region. The circuit design should provide a degree of temperature stability.

In this fixed bias amplifier a high resistance RB is connected between the base and positive end of supply. The value of RB can be directly found by applying KVL to the input and output loop. For this reason, this method is called fixed bias method.

This fixed biasing circuit is very simple and biasing conditions can easily be set and also the calculations are very simple. But it provides very poor stabilization. PROCEDURE:

1. Connect the circuit as per the circuit diagram. 2. Set the input and apply the input signal from the FG and observe the

output. 3. Vary the input frequency from 10Hz to 1MHz and the output amplitude is

noted. 4. Calculate the gain and draw the frequency response curve. 5. Note the gain bandwidth product.

RESULT: Thus the fixed bias amplifier was designed and the gain bandwidth product was determined from the frequency response curve. Gain =

Bandwidth= Gain Bandwidth product =





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COMMON EMITTER AMPLIFIER WITH AND WITHOUT FEEDBACK CIRCUIT DIAGRAM:

CE Amplifier with feedback CE Amplifier without feedback





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Ex.No : 2 Date :

COMMON EMITTER AMPLIFIER WITH AND WITHOUT FEEDBACK AIM: To design a CE amplifier under voltage divider bias with and without feedback and to determine the gain bandwidth product from its frequency response curve. APPARATUS REQUIRED:

Sl.No Apparatus Range Quantity 1.

2. 3. 4. 5. 6. 7. 8.

Resistor

Transistor capacitor

RPS AFO CRO

Bread board Connecting wires

BC 547

(0-30)V

1 1 1 1 1 2 1 1 1 1

few

DESIGN:





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LOAD LINE: MODEL GRAPH:





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THEORY:

A CE amplifier is a circuit it can be used to increase the magnitude of the

input voltage at the output by means of energy drawn from an external source.

The CE amplifier construct with voltage divider biasing circuit. Biasing circuit

sets the proper operating point for the CE amplifier. The input coupling capacitor

couples the input signal to the base of the transistor. It blocks any dc component

present in the signal and passes only ac signal. Because of this, biasing conditions are

maintained constant. The output coupling capacitor couples the output of the amplifier

to the load. It blocks dc and passes only ac part of the amplified signal. An emitter

bypass capacitor is connected in parallel with the emitter resistance, to provide a low

reactance path to the amplified ac signal.

The input signal is applied to the base of the transistor through the coupling

capacitor .During the positive half cycle of the input signal, the transistor is forward

biased, and hence the collector current increases and collector voltage goes in

negative direction. During the negative half cycle, transistor becomes reverse biased;

it reduces the collector current and increases the collector voltage. Therefore, we can

say that there is an 180o phase shift between input and output voltages for a CE

amplifier. A CE amplifier with unbypassed resistance RE (with feedback) gives higher

bandwidth than the CE amplifier without feedback.





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TABULATION(With feedback): Vi=

Vin = Frequency (HZ)

Output voltage ( Volt)

Gain=Vo/Vin Gain in dB=20log(Vo/Vi)

TABULATION(Without feedback):








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Frequency Response of CE amplifier:

The voltage gain of an amplifier varies with signal frequency. It is because

reactance of the capacitors in the circuit changes with signal frequency and hence

affects the output voltage.

The curve between voltage gain and signal frequency of an amplifier is known

as frequency response. CE amplifier gives a wideband frequency response without

peak at any frequency and hence used to cover a complete AF amplifier bands (20 HZ

-20 KHZ).

Bandwidth:

The range of frequency over which the gain is equal to or greater than

70.7% of the maximum gain is known as bandwidth. So, for any frequency lying

between FL and FH, the gain is equal to or greater than 70.7% of the maximum gain.

PROCEDURE: 1. Connect the circuit as per the circuit diagram. 2. Set the input and apply the input signal from the FG and observe the output. 3. Vary the input frequency from 10Hz to 1MHz and the output amplitude is

noted. 4. Removing the bypass capacitor, the process is repeated for CE amplifier with

feedback. 5. Calculate the gain and draw the frequency response curve. 6. Note the gain bandwidth product.

RESULT: Thus the CE amplifier with voltage divider biasing circuit with and without feedback was designed and Bandwidth was determined from the frequency response curve. With feedback Without feedback Gain = Gain = Bandwidth= Bandwidth= Gain Bandwidth product = Gain Bandwidth product =





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COMMON COLLECTOR AMPLIFIER

CIRCUIT DIAGRAM: MODEL GRAPH:





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Ex.No : 3 Date :

COMMON COLLECTOR AMPLIFIER AIM: To design a CC amplifier and to determine the gain bandwidth product from its frequency response curve. APPARATUS REQUIRED:

Sl.No Apparatus Range Quantity

1.

2. 3. 4. 5. 6. 7. 8.

Resistor

Transistor capacitor

RPS AFO CRO

Bread board Connecting wires

BC 547

(0-30)V

1 1 1 1 1 2 1 1 1 1

few

DESIGN:





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TABULATION: Vi=








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THEORY: In common collector amplifier, input is given to the base terminal and output is taken from emitter, collector terminal is common to both the input and output. During the positive half cycle of the input, the transistor is forward biased and hence the emitter current (o/p current) increases (1+) times the base current. [IE=IB (1+)] .Hence the output voltage VE is also increases. VE=VB-VBE From this equation we understand that increase in VB , increases the output voltage as VBE constant. Thus the output of a common collector amplifier is as its input voltage. The voltage gain of CC amplifier is unity, thus it is used as a buffer amplifier and also it is used for impedance matching. PROCEDURE: 1. Connect the circuit as per the circuit diagram. 2. Set the input and apply the input signal from the FG and observe the output.

3. Vary the input frequency from 10Hz to 1MHz and the output amplitude is noted.

4. Calculate the gain and draw the frequency response curve.

5. Note the gain bandwidth product.

RESULT: Thus the CC amplifier with voltage divider biasing circuit was constructed and the frequency response curve was plotted. Gain = Bandwidth= Gain Bandwidth product =





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BOOTSTRAPPED SOURCE FOLLOWER

CIRCUIT DIAGRAM: WITH BOOTSTRAPPED: WITHOUT BOOTSTRAPPED:





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Ex.No : 4 Date :

BOOTSTRAPPED SOURCE FOLLOWER AIM: To construct a source follower and to find its input & output resistance with and without bootstrapping gate resistance. APPARATUS REQUIRED:

S.No Apparatus Range Quantity

1.

2.

3.

4. 5.

6. 7. 8.

9.

Resistor Transistor Capacitor RPS AFO Multimeter CRO Bread board Connecting wires

10K 100K 2.2M 150K

BFW 10

10 nf 10 pf

(0-30)V

1 1 1 1 1 1 1 1 1 1 1 1

few THEORY: It is similar to Common emitter amplifier. Here Rg is used to bias the FET .The Capacitors C1and C2 are used to couple the ac input voltage source and the output voltage respectively, these are known as coupling capacitors. The Capacitor Cs keeps the source of the FET effectively at a.c ground and is known as by-pass capacitors.

The operation of the circuit may be understood from the assumption that when a small ac signal is applied to the gate, it produces variations in the drain current. As the gate to source voltage increases, the drain current also increases .As a result of this the voltage across the resistor (RD) also increases. This causes the drain voltage to decreases.





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CALCULATION:





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It means that the positive half cycle of the input voltages produces the

negative half cycle of the output voltages. In other words, the output voltages are 180 degree out of phase with the input voltage. This phenomenon of phase inversion is similar to common emitter amplifier. PROCEDURE: 1. Connect the circuit as per the circuit diagram. 2. Set the input and apply the input signal from the FG and observe the output.

3. Connect the ammeter in the circuit as shown and note the current value. 4. Calculate the gain and draw the frequency response curve.

5. Calculate the input & output impedance.

6. Note the gain bandwidth product.

RESULT: Thus the source follower with Bootstrapped gate resistance was constructed.

Input & Output impedance with bootstrapped = ---------- Input & Output impedance without bootstrapped = ----------





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DIFFERENTIAL AMPLIFIER CIRCUIT DIAGRAM: Common mode: Differential mode:





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Ex.No : 5 Date :

DIFFERENTIAL AMPLIFIER AIM: To construct a differential amplifier using BJT and to measure CMRR. APPARATUS REQUIRED:


1.

2. 3. 4. 5. 6.

Resistor Transistor RPS Bread board Multimeter Connecting wires

4.7K 7.7K 5.6K

SL 100 (0-30)V

2 2 1 2 2 1 1

few

FORMULA USED:

1. CMRR = Ad/Ac 2. Ad = Vo / (V1-V2) 3. Ac = 2V0 / (V1+V2)

Where, Ad ----- Differential mode gain. Ac ------ Common mode gain. THEORY: The differential amplifier amplifies the difference between two input voltage signals. It uses the emitter biased circuit. The two transistors have exactly matched characteristics. The two collector resistances RC1 and RC2 are equal and the magnitude of +VCC and VEE are also same. There are two modes of operation in differential amplifier. In common mode operation, same voltage with same polarity is applied to the base of the transistors by using RPS. The output is taken across the collector terminals of the two transistors which is zero. In difference mode operation, same voltage with opposite polarity is applied to the base of the transistors by using





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TABULATION: COMMON MODE

V1(V) V2(V) V0(V) AC=2V0/(V1+V2)

DIFFERENTIAL MODE

V1(V) V2(V) V0(V) AD= V0/(V1-V2)

CALCULATION:





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RPS. The output is taken across the collector terminals of the two transistors. The difference output is twice as large as the input signal voltage. APPLICATION:

1. Used in operational amplifier. 2. Used in integrated circuits. 3. Used in instrumentation system waveform generators & A/D converters

PROCEDURE: 1. Connect the circuit as per the circuit diagram.

2. For differential mode operation, apply two different voltages at the inputs

V1 & V2 and the measure the output voltage. 3. For common mode operation, apply same voltage at the inputs V1 & V2 and the measure the output voltage.

4 . Calculate the difference mode gain (Ad), common mode gain (Ac) using their formula.

5. Determine the CMRR. RESULT:

Thus the differential amplifier using BJT was constructed and CMRR was calculated.

CMRR in dB =





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CLASS A POWER AMPLIFIER CIRCUIT DIAGRAM: MODEL GRAPH:





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Ex.No : 6 Date : CLASS A POWER AMPLIFIER AIM: To construct a class A power amplifier and to determine the efficiency from its output wave form. APPARATUS REQUIRED:


1.

2.

3. 4. 5. 6. 7. 8. 9.

Resistor Capacitor Ammeter Transistor RPS Multimeter CRO FG Bread board Connecting wires

100 10

560

2.2f 5 f

(0-50)mA SL 100

(0-30)V

2 1 1 1 1 1 1 2 1 1 1 1

few





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TABULATION: Voltage(V) Time(ms) Current(mA)

Input waveform

Output waveform

CALCULATION:





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THEORY: Transistor power amplifiers handle large signals. If the current flows at all times during the full cycle of the signal, the power amplifier is known as class A amplifier. Obviously, for this to happen the power amplifier must be biased in such a way that no part of the signal is cut off. The efficiency of class A amplifier is only 50%. But it provides less power dissipation. Also there is no distortion in class A power amplifier. PROCEDURE:

1. Connect the circuit as per the circuit diagram. 2. Set the input and apply the input signal from the FG and observe the

output. 3. Note down the collector current using ammeter. 4. Draw the input and output waveform. 5. By using specified formula, AC output power, DC input power and

efficiency are calculated. RESULT: Thus the class A amplifier was designed and constructed also its efficiency was calculated. Efficiency =





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COMPLEMENTARY SYMMETRY CLASS B PUSH PULL AMPLIFIER CIRCUIT DIAGRAM: MODEL GRAPH:





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Ex.No : 7 Date :

COMPLEMENTARY SYMMETRY CLASS B PUSH PULL AMPLIFIER

AIM: To construct a complementary symmetry push pull power amplifier to observe the output waveform with crossover distortion and to determine its maximum power output & efficiency. APPARATUS REQUIRED:


1.

2.

3. 4. 5. 6. 7. 8.

Resistor Capacitor Transistor RPS CRO FG Bread board Connecting wires

100

1k

5 f

SL 100 (0-30V)

1 1 2 2 1 1 1

few

THEORY:

In a complementary symmetry class B amplifier, a pnp-npn transistor pair is used. When common emitter configuration is used in push pull amplifier, it becomes difficult to match the output impedance for maximum power transfer without an output transformer. Hence a matched pair of complementary transistors is used in common collector configuration in this arrangement. The common collector configuration has the lowest output impedance and hence impedance matching is possible.





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CALCULATION:





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PROCEDURE:

1. Connect the circuit as per the circuit diagram. 1. Sinusoidal input is given to the amplifier and output of an amplifier with

and without crossover distortion is noted 2. By using specified formula, AC output power, DC input power and

efficiency are calculated.

RESULT:

Thus the complementary symmetry push pull amplifier was constructed and the output waveforms are drawn. efficiency =





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HALF WAVE RECTIFIER

CIRCUIT DIAGRAM: Half wave rectifier with filter: MODEL GRAPH:

Vin (Volts)

t (ms)

Vo (Volts)

t (ms)

t (ms)

Vo (Volts)

Without Filter

With Filter





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Ex.No : 8 Date :

HALF WAVE RECTIFIER

AIM: To construct a half wave rectifier with simple capacitor filter and to measure its ripple factor from the output waveforms. APPARATUS REQUIRED:


1. 2. 3. 4.

5. 6. 7.

Resistor Capacitor Diode Stepdown Transformer Bread board Connecting wires CRO

1K

470 f IN 4001

2 1 1 1 1

Few 1

THEORY: A Half wave rectifier is a device which converts ac voltage to pulsating dc voltage using one PN junction diode. The ac voltage (230 V, 50 HZ) is connected to the primary of the transformer. The transformer steps down the ac voltage. Thus, with suitable turns ratio we get desired ac secondary voltage. The rectifier circuit converts this ac voltage in to a pulsating dc voltage. Half wave rectifier conducts during positive half cycle and gives output in the form of positive sinusoidal pulses. Hence the output is called pulsating dc. A pulsating dc voltage containing large varying component called ripple in it. The capacitor filter is used after rectifier circuit, which reduces the ripple content in the pulsating dc. Thus filter converts pulsating dc in to pure dc. Ripple Factor: The output of the rectifier is of pulsating dc type. The amount of ac content in the output can be mathematically expressed by a factor called ripple factor.





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TABULATION:

Parameter

Output with simple C filter

Vm (V)

Vrms(V)

CALCULATION:





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Ripple Factor = R.M.S Value of ac component Average dc component

= Vrms/Vdc) 2 Where, Vrms = Vm/2 ; Vdc = Vm/ PROCEDURE:

1. The circuit connections are made as per the circuit diagram. 2. First without connecting the capacitive filter, note down the amplitude and time period of the rectified waveform. 3. Now connect the capacitive filter and note down the amplitude and time period of the rectified waveform. 4. Connect the CRO across the load and measure the full load voltage then remove the load and measure the no load voltage. 5. Plot the graph and calculate the efficiency.

RESULT: Thus the half wave rectifier was constructed and input, output waveforms were drawn. Ripple Factor r =





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FULL WAVE RECTIFIER CIRCUIT DIAGRAM: Full wave rectifier with filter: MODEL GRAPH:

Vin (Volts)

t (ms)

Vo (Volts)

t (ms)

t (ms)

Vo (Volts)

Without Filter

With Filter





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Ex.No : 9 Date :

FULL WAVE RECTIFIER AIM: To construct a full wave rectifier with simple capacitor filter and to measure its ripple factor from the output waveforms. APPARATUS REQUIRED:


1. 2. 3. 4.

5. 6. 7.

Resistor Capacitor Diode Stepdown Transformer Bread board Connecting wires CRO

1K

100 f IN 4001

2 1 2 1 1

Few 1

THEORY: A full wave rectifier is a device which converts ac voltage to pulsating dc voltage using two PN junction diode. The ac voltage (230 V, 50 HZ) is connected to the primary of the transformer. The transformer steps down the ac voltage. Thus, with suitable turns ratio we get desired ac secondary voltage. The rectifier circuit converts this ac voltage in to a pulsating dc voltage. Full wave rectifier conducts during both positive and negative half cycle of input ac supply. Because two diodes are used in this cir circuit .It gives output in the form of positive sinusoidal pulses. Hence the output is called pulsating dc. A pulsating dc voltage containing large varying component called ripple in it. The capacitor filter is used after rectifier circuit, which reduces the ripple content in the pulsating dc. Thus filter converts pulsating dc in to pure dc.





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TABULATION:

Parameter Output with simple C filter

Vm (V)

Vrms(V)

CALCULATION:





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Ripple Factor: The output of the rectifier is of pulsating dc type. The amount of ac content in the output can be mathematically expressed by a factor called ripple factor.

Ripple Factor = R.M.S Value of ac component Average dc component = Vrms/Vdc) 2 Where, Vrms = Vm/2 ; Vdc = 2Vm/

PROCEDURE:

1. The circuit connections are made as per the circuit diagram. 2. First without connecting the capacitive filter, note down the amplitude and time period of the rectified waveform. 3. Now connect the capacitive filter and note down the amplitude and time period of the rectified waveform. 4. Connect the CRO across the load and measure the full load voltage then remove the load and measure the no load voltage. 5. Plot the graph and calculate the efficiency.

RESULT: Thus the full wave rectifier was constructed and input, output waveforms were drawn. Ripple Factor with filter =





Page 41 of 51

DARLINGTON EMITTER FOLLOWER

CIRCUIT DIAGRAM: MODEL GRAPH:





Page 42 of 51

Ex.No : 10 Date :

DARLINGTON EMITTER FOLLOWER AIM: To construct a BJT Darlington Emitter follower and plot the frequency response curve and also to determine the gain, input & output impedances. APPARATUS REQUIRED:

Apparatus Range Quantity

Resistor Capacitor Transistor RPS CRO FG Mulimeter Bread board Connecting wires

100

1k

5 f

BC 107 (0-30V)

1 1

2 2 1 1 1 1

Few





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TABULATION: Frequency(Hz) Voltage(mV) Gain=V0/Vin Gain in

dB=20log(gain)





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THEORY:

Darlington connection is used to achieve larger input impedances. This circuit consists of two cascaded emitter followers. The main feature of this connection is that the composite transistor acts a single unit with a current gain that is the product of the current gains of the individual transistors. The voltage gain of Darlington circuit is close to unity, but its deviation from unity is slightly greater than that of the emitter follower. The first draw back of the Darlington transistor pair is that the quiescent current of the first stage is much smaller than that of the second. The second draw back is that the leakage current of the first transistor is amplified buy the second, and hence the overall leakage current may be high.

For these two reasons, a Darlington connection of three or more transistor is usually impractical.

PROCEDURE: 1. Connect the circuit as per the circuit diagram. 2. Set Vi= 1V, using the signal generator. 3. Keeping the input voltage constant, vary the frequency from 0Hz to 1Mhzin regular steps of 10 and note down corresponding output voltage. 4. Plot the frequency response: Gain(dB) Vs Frequency(Hz). 5. Find the input and output impedances. 6. Calculate the Bandwidth from the graph.

RESULT: Thus the Darlington emitter follower was constructed and the frequency response curve was plotted. Gain = Input impedance =

Output impedance =





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ELECTRONIC CIRCUITS1 LAB Viva Questions

1.Why do we choose q point at the center of the loadline? 2. Name the two techniques used in the stability of the q point.

3. Define current amplification factor? 4. Why is the transistor called a current controlled device? 5. What do you meant by thermal runway?





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6. When does a transistor act as a switch? 7. What is biasing? 8. What are the requirements for biasing circuits? 9. Define input bias current? 10. What are the different types of biasing circuits? 11. What is operating point? 12. What is stability factor? 13. What is d.c load line?





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14. What are the advantages of fixed bias circuit? 15. Explain about the various regions in a transistor? 16. What is the necessary of the coupling capacitor? 17. What is reverse saturation current? 18.What is an amplifier? 19. What are the basic rules of an operating amplifier? 20. What are the two techniques used to increase the input impedance?





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21. What is a darlington transistor? 22. How are amplifiers classified according to the input? 23. How are amplifiers classified according to the transistor configuration? 24. What is the different analysis available to analyze a transistor? 25. What is feed back? 26. What are the types of feed back? 27. What is positive feedback?





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28. What is negative feed back? 29. which feedback increases the gain of the amplifier? 30. Which feedback decreases the gain of the amplifier? 31. What is the disadvantage of negative feed back? 32. What is an op-amp? 33. What are the DC characteristics of an op-amp? 34. What are the AC characteristics of an op-amp?





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35. Define Common Mode Rejection Ratio. 36. What is Q factor? 37. What is the maximum efficiency of Class A & Class B amplifier? 38. What is cross over distortion? 39. What is a rectifier? 40. Define ripple factor.





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