Frequency Response of A Single Stage RC Coupled Amplifier

Experiment No.
-1
Aim: To Plot the Frequency Response of a single stage RC Coupled Amplifier with feed back
and without feedback and find the following:
1. Voltage Gain
2. Lower cut off Frequency
3. Upper cut off Frequency
4. Bandwidth
5. Gain Bandwidth Product
Apparatus: 1. Digital Storage Oscilloscope (DSO)

2. Power Supply
3. Bread Board
4. Signal Generator
Component: Resistances: 100k Ω = 01, 4.7k Ω = 01, 22k Ω = 01, 220k Ω = 01, 220 Ω = 01
Capacitors: 10µf = 02, 47 µf = 01
Transistor BC107 = 01
Connecting wire and DSO Probe
Introduction
A practical amplifier circuit is meant to raise the voltage level of the input signal. This signal
may be obtained from anywhere e.g. radio or TV receiver circuit. Such a signal is not of a single
frequency. But it consists of a band of frequencies, e.g. from 20 Hz to 20 KHz. If the
loudspeakers are to reproduce the sound faithfully, the amplifier used must amplify all the
frequency components of signal by same amount. If it does not do so, the output of the
loudspeaker will not be the exact replica of the original sound. When this happen then it means
distortion has been introduced by the amplifier. Consider an RC coupled amplifier circuit shown
fig 1 shows frequency response curve of a RC coupled amplifier. The curve is usually plotted on
a semi log graph paper with frequency range on logarithmic scale so that large frequency range
can be accommodated. The gain is constant for a limited band of frequencies. This range is
called mid-frequency band and gain is called mid band gain. AVM. On both sides of the mid
frequency range, the gain decreases. For very low and very high frequencies the gain is almost
zero.
In mid band frequency range, the coupling capacitors and bypass capacitors are as good as short
circuits. But when the frequency is low. These capacitors can no longer be replaced by the short
circuit approximation.
At low frequency, output capacitor reactance increases. The voltage across RL reduces because
some voltage drop takes place across XC. Thus output voltage reduces.
The XC reactance not only reduces the gain but also change the phase between input and output.
It would not be exactly 180o but decided by the reactance. At zero frequency, the capacitors are
open circuited therefore output voltage reduces to zero.
Page 1 of 26
The gain is constant over a frequency range. The frequencies at which the gain reduces to 70.7%
of the maximum gain are known as cut off frequencies, upper cut off and lower cut off frequency
fig. 2, shows these two frequencies. The difference of these two frequencies is called Band width
(BW) of an amplifier.
BW = f2 – f1.
Fig. 2
At f1 and f2, the voltage gain becomes 0.707 Am(1 / 2). The output voltage reduces to 1 / 2
of maximum output voltage. Since the power is proportional to voltage square, the output power
at these frequencies becomes half of maximum power. The gain on dB scale is given by
20 log10(V2 / V1) = 10 log 10 (V2 / V1)2 = 3 dB.
20 log10(V2 / V1) = 20 log10(0.707) =10 log10 (1 / 2)2 = 10 log10(1 / 2) = -3 dB.
If the difference in gain is more than 3 dB, then it can be detected by human. If it is less than 3
dB it cannot be detected.
Procedure:
1. Connect the circuit as shown in the diagram.
2. Apply a sinusoidal input signal of = from a signal generator.
3. Connect the output to the DSO
4. Measure output voltage and calculate gain
5. Keeping the input voltage constant vary the input frequency and note the output
voltage of the Amplifier till the output decrease upto a -3dB point.
6. Draw a graph between input frequency Vs output voltage
7. For without feedback short terminal A and B by a connecting wire and repeat
procedure 1 to 6.
Page 2 of 26
Circuit:
VDC = + 12V
R1 R3
100K 4.7K
C2
To Digital Storage Oscilloscope (DSO)
Q1 1n
C1
Signal Generator 10UFD

BC107A
A
VOFF = 0V
VAMPL = 20MV
FREQ = 1KHZ C2
R2 R4 10UFD
22K 220
B
0
Observation table
Vin = 20mv
S. No Frequency Output voltage Voltage Gain Voltage gain in

dB
10 Hz
100 Hz
1k Hz
10 kHz
100 kHz
1MHz
Result:
1. Voltage Gain ………………………….
2. Lower cut off Frequency ……………………………
3. Upper cut off Frequency …………………………….
4. Bandwidth ……………………………
5. Gain Bandwidth Product …………………………..
Page 3 of 26
Experiment No. -2
Aim: To Plot the Frequency Response of a Double stage RC Coupled Amplifier find the
following:
1. Voltage Gain
4. Bandwidth
5. Gain Bandwidth Product and loading effect

2. Power Supply
3. Bread Board
4. Signal Generator
Component: Resistances: 100k Ω = 02, 4.7k Ω = 02, 22k Ω = 02, 220k Ω = 02, 220 Ω = 02
Capacitors: 10µf = 04, 47 µf = 02
Introduction
To increases the voltage gain of the amplifier, multiple amplifier are connects in cascade. The
output of one amplifier is the input to another stage. In this way the overall voltage gain can be
increased, when number of amplifier stages are used in succession it is called a multistage
amplifier or cascade amplifier. The load on the first amplifier is the input resistance of the
second amplifier. The various stages need not have the same voltage and current gain. In
practice, the earlier stages are often voltage amplifiers and the last one or two stages are current
amplifiers. The voltage amplifier stages assure that the current stages have the proper input
swing. The amount of gain in a stage is determined by the load on the amplifier stage, which is
governed by the input resistance to the next stage. Therefore, in designing or analyzing
multistage amplifier, we start at the output and proceed toward the input.
A n-stage amplifier can be represented by the block diagram as shown in fig. 1
Fig. 1
Page 4 of 26
In fig. 1, the overall voltage gain is the product of the voltage gain of each stage. That is, the
overall voltage gain is ABC.
To represent the gain of the cascade amplifier, the voltage gains are represents in dB. The two
power levels of input and output of an amplifier are compared on a logarithmic scale rather than
linear scale. The number of bels by which the output power P2 exceeds the input power P1 is
defined as
Because of dB scale the gain can be directly added when a number of stages are cascaded.
Types of Coupling:
In a multistage amplifier the output of one stage makes the input of the next stage. Normally a
network is used between two stages so that a minimum loss of voltage occurs when the signal
passes through this network to the next stage. Also the dc voltage at the output of one stage
should not be permitted to go to the input of the next. Otherwise, the biasing of the next stage are
disturbed.
The three couplings generally used are.
1. RC coupling
2. Impedance coupling
3. Transformer coupling.
Page 5 of 26
RC coupling:
Fig. 2 shows RC coupling the most commonly used method of coupling from one stage to the
next. An ac source with a source resistance R S drives the input of an amplifier. The grounded
emitter stage amplifies the signal, which is then coupled to next CE stage the signal is further
amplified to get larger output.
In this case the signal developed across the collector resistor of each stage is coupled into the
base of the next stage. The cascaded stages amplify the signal and the overall gain equals the
product of the individual gains.
The coupling capacitors pass ac but block dc Because of this the stages are isolated as for as dc is
concerned. This is necessary to avoid shifting of Q-points. The drawback of this approach is the
lower frequency limit imposed by the coupling capacitor.
The bypass capacitors are needed because they bypass the emitters to ground. Without them, the
voltage gain of each stage would be lost. These bypass capacitors also place a lower limit on the
frequency response. As the frequency keeps decreasing, a point is reached at which capacitors no
longer look like a.c. shorts. At this frequency the voltage gain starts to decrease because of the
local feedback and the overall gain of the amplifier drops significantly. These amplifiers are
suitable for frequencies above 10 Hz.
Circuit:
VDC = + 12V
R1 R3 R6 R5
100K 4.7K 100K 4.7K
C2 C4
Q1 10UFD Q2 1n
C1
Signal Generator 10UFD

BC107A BC107A
A A
VOFF = 0V
VAMPL = 20MV
FREQ = 1KHZ C3 C5
R2 R4 10UFD R7 R8 10UFD
22K 220 22K 220
B B
0 0
Fig. 2
Procedure:
Page 6 of 26
2. Apply a sinusoidal input signal of 20mv from a signal generator.
Observation table
Vin = 20mv

dB
10 Hz
100 Hz
1k Hz
10 kHz
100 kHz
1MHz
Result:
4. Bandwidth ……………………………
5. Gain Bandwidth Product …………………………..
Page 7 of 26
Experiment No. -3
Aim: To Plot the Frequency Response of a FET Common Source Amplifier find the
following:
1. Voltage Gain
4. Bandwidth

2. Power Supply
3. Bread Board
4. Signal Generator
Component: Resistances: 8.2kΩ = 01, 4.7kΩ = 01, 470Ω = 01, 100Ω = 01

Capacitors: 10µf = 02, 100µf = 01
FET BFW 11 = 01
Introduction
When a small ac signal is coupled into the gate it produces variations in gate source voltage.
This produces a sinusoidal drain current. Since an ac current flows through the drain resistor. An
amplified ac voltage is obtained at the output. An increase in gate source voltage produces more
drain current, which means that the drain voltage is decreasing. Since the positive half cycle of
input voltage produces the negative half cycle of output voltage, we get phase inversion in a CS
amplifier.
The ac equivalent circuit is shown in fig. 1
Fig. 1
The ac output voltage is
vout = - gm v gS RD
Page 8 of 26
Negative sign means phase inversion. Because the ac source is directly connected between the
gate source terminals therefore ac input voltage equals
Vin = Vgs
The voltage gain is given by
Circuit:
VDC = + 12V
R1 R3
8.2K 470
C2
10UFD
D
C1 FET
BFW11
Signal Generator 10UFD G
S
A
VOFF = 0V
VAMPL = 20MV
FREQ = 1KHZ C3
R2 R4 100UFD
4.7K 100
B
0
Procedure:
2. Apply a sinusoidal input signal of 20mv from a signal generator.
Page 9 of 26
Observation table
Vin = 20mv

dB
10 Hz
100 Hz
1k Hz
10 kHz
100 kHz
1MHz
Result:
4. Bandwidth ……………………………
5. Gain Bandwidth Product ……………………………
Page 10 of 26
Experiment No. -4
Aim: To study and calculate the input and output impedance of Darlington Amplifier and also
calculate the amplification factor.
Apparatus:
1. Power Supply
2. Bread Board
3. DMM
Component: Resistances: 100Ω = 01

Transistor SL100 = 01
Introduction
Theory: consists of two emitter followers in cascaded mode as shown in fig. 1. The overall gain
is close to unity. The main advantage of Darlington amplifier is very large increase in input
impedance and an equal decrease in output impedance .
Fig. 1
Page 11 of 26
Output impedance:
The Thevenin impedance at the input is given by
RTH = RS || R1 || R2
Similar to single stage common collector amplifier, the output impedance of the two stages
zout(1) and zout(2) are given by.
Therefore, t he output impedance of the amplifier is very small.
Circuit:
Darlington Trasistor Amplifier

VDC= +12V
+ C
B
VDC BC107A
Driver
- B SL100
E
Output
Vin +
-
E
R1
Vout
1k
Page 12 of 26
Procedure:
1. Connect the circuit as shown in diagram.
2. Connect the DC voltage source to the input of BC 107 transistor through a DMM
connected in a ammeter mode.
3. Connect 100Ω resistance to the emitter , through another DMM connected as an ammeter
4. Adjust the input voltage to 5v and note down the reading of input current and output
current.
5. Also measure the output voltage (voltage gain is less then 1)
6. With the value of ammeter , calculate the amplification factor of a transistor β1
7. Repeat same procedure for SL 100 transistor and find out the amplification factor of
SL100, is β2
8. Now, connect dc voltage to the base of transistor BC107 , connect BC 107 emitter to base
of SL100 and 100Ω resistance to SL 100 emitter , which form the Darlington circuit.
9. Measure the input current and voltage , output current and voltage.
10. Total current gain = β1× β2
11. Ration of input voltage and input current gives impedance and ration of output voltage
and output current gives output impedance
Result:
1. Amplification factor β1…………………………………
2. Amplification factor β2 …………………………………..
3. Total current gain β1× β2 ………………………………
4. Input impedance………………………………..
5. output impedance……………………………..
Page 13 of 26
Experiment No. -5
Aim: To Study BJT Differential Amplifier and measure its CMRR

2. Power Supply
3. Bread Board
4. Signal Generator
5. Multi meter
Component: Resistances: 1kΩ = 02, 470Ω = 02, 5.6KΩ = 01,

Introduction
A long tailed pair is a common design in electronics for implement a differential

amplifier. Its amplifies the current with very little voltage gain. It consist of two bipolar junction
transistor (BJT) , FET connected so that the BJT emitters or FET sources are connected together.
The common electrode s are then connected to a large voltage source through a large
resistor forming the long tail of the name the long tail providing an approximate constant current
source. In more sophisticated designs a true (active ) constant current source may be substitute
for a long tail.
Connect in fashion , this gives the circuit two input which are differentially amplifies
(Subtracted and multiplied) by the pair. The output may be a single ended or differential
depended on the need of the subsequent circuitry.
Circuit:
Emitter Coupled Diffential Amplifier
+12v
R1
1k R2
1k
VC2 out
VC1 out
+
R4 BC107 R5
VDC 470
BC107
470
- V1 in V2 in
R3
5.6k
-12v
Fig. 1
Page 14 of 26
Procedure: for DC characteristics
1. Connected the circuit as shown in figure.

2. Adjust the dc voltage of the voltage source to 0.15v with multi meter .
3. Connect DC voltage source output to one of the input terminals of the differential
amplifier and short the other input to ground.
4. Measure the DC out put voltage between two collector of the differential amplifier with
DMM
5. Calculate differential Gain AV output/ input
6. Now connect both the inputs together to the DC voltage source , and adjust the voltage to
2V
7. Again measure the DC output voltage between two collector out put of differential
amplifier with DMM
8. Calculate the common mode gain AC output/input
9. With the above value calculate the CMMR = Ad/AC
Procedure: for AC Characteristic
1. Adjust the AC voltage of the signal generator to 10mv RMS voltage (IKHz)
2. Connect signal generator out put to one of the input terminal of a differential amplifier
and short the other input to ground.
3. Measure the AC output voltage between two collector output of a differential amplifier
with DMM
4. Calculate the differential gain Ad = output/input
5. Now connect both input together to the signal generator , and adjust the voltage to 200mv
6. Again measure the AC output voltage between two collector outputs of the differential
amplification with DMM
7. Calculate the common mode gain Ac output / input
8. With the above values calculate the CMRR = Differential gain / common mode gain
Result:
Differential gain :………………………….
Common mode gain……………………….
CMRR :…………………………………..
Page 15 of 26
Experiment No. -6
Aim: To Construct RC phase shift oscillator and to measure the output frequency

2. Power Supply
3. Bread Board
4. Signal Generator
5. Multi meter
Component: Resistances: 100kΩ = 01, 10kΩ = 01, 5.1KΩ = 01, 1kΩ = 04

Variable resistance = 10kΩ = 01
Capacitor 0.1ufd = 06
Introduction
Transistor Phase Shift Oscillator:
At low frequencies (around 100 kHz or less), resistors and capacitors are usually employed to
determine the frequency of oscillation. Fig shows transistorized phase shift oscillator circuit
employing RC network. If the phase shift through the common emitter amplifier is 180°, then the
oscillation may occur at the frequency where the RC network produces an additional 180° phase
shift.
Since a transistor is used as the active element, the output across R of the feedback network is
shunted by the relatively low input resistance of the transistor, because input diode is a forward
biased diode
Hence, instead of employing voltage series feedback, voltage shunt feedback is used for a transistor
phase shift oscillator.
Circuit:
+12V
R2 R1
100K 5.1k
C2
DSO
Q1 0.1uf d
B C1
A OUTPUT
0.1uf d
signal generator V1
BC107A
VAMPL = 20mv
R3
FREQ = 1kHz 10K
R6 C3
1k 0.1uf d
R5
10K
Page 16 of 26
Fig.1
Feedback
C4 C5 C6
0.1uf d 0.1uf d
0.1uf d
R7 R10 R11
1k 1k 1k
+12V
R2 R1
100K 5.1k
C2
DSO
Q1 0.1uf d
B C1
OUTPUT
0.1uf d BC107A
R3
10K
R6 C3
1k 0.1uf d
R5
10K
Fig.2
Procedure :
1. Connect the circuit diagram shown in fig.1

2. Apply a sine wave from a signal generator to the input terminal of the network and observe
the out put of the circuit on DSO
3. Vary the frequency of the signal generator and find out the amplitude of output at each
frequency.
4. Find the frequency of signal generator at which the output of circuit becomes 1/29of the
input.
5. Also make sure that at this frequency the output and inputs of RC network are exactly 180o
out of phase.
6. Now connect the circuit shown in fig.2
7. Observe the output of the RC phase shift oscillator
8. Adjust the variable resistance until we get undistorted output
9. Find frequency of the out put
10. Calculate the theoretical frequency by fo = 1/2Π√6RC
Result:
Compare both theoretical and practical frequencies
Page 17 of 26
Page 18 of 26
Experiment No. -7
Aim: To study Wein Bridge Oscillator and to measure the output frequency

2. Power Supply
3. Bread Board
4. Multi meter
Component: Resistances: 10kΩ = 03, 2.7kΩ = 02, 100Ω = 01, 1kΩ = 03

Capacitor 0.1ufd = 02, 47ufd = 03
Theory Wien Bridge Oscillator:
The Wien Bridge oscillator is a standard oscillator circuit for low to moderate frequencies, in the
range 5Hz to about 1MHz. It is mainly used in audio frequency generators.
The Wien Bridge oscillator uses a feedback circuit called a lead lag network as shown in fig 1
Fig.1
At very low frequencies, the series capacitor looks open to the input signal and there is no
output signal. At very high frequencies the shunt capacitor looks shorted, and there is no
output. In between these extremes, the output voltage reaches a maximum value. The
frequency at which the output is maximized is called the resonant frequency. At this frequency,
the feedback fraction reaches a maximum value of 1/3.
At very low frequencies, the phase angle is positive, and the circuit acts like a lead network. On
the other hand, at very high frequencies, the phase angle is negative, and the circuit acts like a
lag network. In between, there is a resonant frequency fr at which the phase angle equals 0°.
Page 19 of 26
Circuit:
A
R9 R10
+12V
1k 1k R1 R3 R5 R7
Feedback
10k 100 15k 1k

C3
0.1uf d
C1
B C2
B
Q1 47uf d Q2 0.1uf d TO DSO
C D
E SL100 SL100
OUTPUT
R11 R12
0.1uf d 1k 1k R2 R6
2.7k 2.7k
R4
F 1k R8
1K C3
47uf d
Fig.1
Procedure :

2. Observe the output of the Wein Bridge Oscillator
5. Calculate the theoretical frequency by fo = 1/2ΠRC
Result:
Page 20 of 26
Experiment No. -8
Aim: To study Hartley and colpitts oscillator and to measure the output frequency

2. Power Supply
3. Bread Board
4. Signal Generator
5. Multi meter
Component: Resistances: 100kΩ = 01, 10kΩ = 01, 5.1KΩ = 01, 1kΩ = 04

Capacitor 0.1ufd = 06
Introduction
Wein bridge and RC phase shift oscillator is not suited to high frequencies (above 1MHz). The
main problem is the phase shift through the amplifier.
The alternative is an LC oscillator, a circuit that can be used for frequencies between 1MHz and
500MHz.. With an amplifier and LC tank circuit, we can feedback a signal with the right amplitude
and phase is feedback to sustain oscillations. shows the circuit of LC oscillator. The voltage divider
bias sets up a quiescent operating point. The circuit then has a low frequency voltage gain of rc /
r'e where rc is the ac resistance seen by the selector. Because of the base and collector lag networks,
the high frequency voltage gain is less then rc / r'e.
Circuit: Colpitt Oscillator
+12V
R3
100K 5.1k
C2
B
Q2 0.1uf d
C1
TO DSO
C
D SL100
0.1uf d
OUTPUT
0.1uf d 2 R2
10K R4
1k 0.1uf d
10mh
R5
0
.01uf d 1
1K
Fig.1
Page 21 of 26
Circuit: Hartley Oscillator
+12V
R3
100K 5.1k
C C2
B
D
Q2 0.1uf d
2 C1
TO DSO
10mh SL100
0.1uf d
OUTPUT
1
2 R2
0.1uf d
10K R4
1k 0.1uf d
0 10mh
R5
1
A 1K
Fig.2
Procedure: for colpitts oscillator

2. Observe the output of the colpitts Oscillator
5. Calculate the theoretical frequency by fo = 1/√2ΠLC
Result:
Procedure: Hartley oscillator

2. Observe the output of the Hartleys Oscillator
5. Calculate the theoretical frequency by fo = 1/√2ΠLC
Result:
Page 22 of 26
DELHI TECHNOLOGICAL UNIVERSITY
ELECTRONICS & COMMUNICATION ENGG. DEPTT.
ELECTRONICS DEVICES AND CIRCUITS- I LAB
List of Experiments for Odd Semester
1. To become familiar with the operation of basic laboratory instruments

1. Power supply
2. Signal generator
3. Digital Storage Oscilloscope (DSO)
4. Multi meter
2. To study the active and passive Electronic components
3. To plot forward and reverse bias characteristics for

PN Junction diode and
Zener diode
4. To study and plot the input and out put characteristics of the given transistor in CB
(common base) configuration.
5. To study and plot the input and out put characteristics of the given transistor in
CE(common emitter) configuration.
6. To study and plot drain characteristics and transfer characteristics of JFET
7. To study and plot drain characteristics and transfer characteristics of N-Channel

metal oxide semiconductor junction field effect transistor (MOSFET)
8. To study of rectifier circuits

1. Half wave rectifier
2. Full wave rectifier
3. Bridge rectifier
and find ripple factor with different filter circuits
9. To study different clipping and clamping circuits
Page 23 of 26
ELECTRONICS DEVICES AND CIRCUITS- II LAB
List of Experiments for Even Semester
1. To Plot the Frequency Response of a single stage RC Coupled Amplifier with feed back
and without feedback and find the following:
1. Voltage Gain
4. Bandwidth
2. To Plot the Frequency Response of a Double stage RC Coupled Amplifier find the
Following:
1. Voltage Gain
4. Bandwidth
5. Gain Bandwidth Product and loading effect
3. To Plot the Frequency Response of a FET Common Source Amplifier find the
following:
1. Voltage Gain
4. Bandwidth
4. To study and calculate the input and output impedance of Darlington Amplifier and also
calculate the amplification factor
5. To Study BJT Differential Amplifier and measure its CMRR
6. To Construct RC phase shift oscillator and to measure the output frequency
7. To study Wein Bridge Oscillator and to measure the output frequency
8. To study Hartley and colpitts oscillator and to measure the output frequency
Page 24 of 26
FOR
ELECTRONICS DEVICES AND CIRCUITS- I LAB
Mr. Alok Kumar Singh

Lab In-charge
Delhi Technological university
Page 25 of 26
FOR
ELECTRONICS DEVICES AND CIRCUITS- II LAB
Dr. Mrs. S.Indu

Lab In-charge
Delhi Technological University
Page 26 of 26

Frequency Response of A Single Stage RC Coupled Amplifier

Uploaded by

Copyright:

Available Formats

Frequency Response of A Single Stage RC Coupled Amplifier

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Frequency Response of A Single Stage RC Coupled Amplifier

Uploaded by

Copyright:

Available Formats

Experiment No.

Apparatus: 1. Digital Storage Oscilloscope (DSO)

20 log10(V2 / V1) = 10 log 10 (V2 / V1)2 = 3 dB.

20 log10(V2 / V1) = 20 log10(0.707) =10 log10 (1 / 2)2 = 10 log10(1 / 2) = -3 dB.

Signal Generator 10UFD

S. No Frequency Output voltage Voltage Gain Voltage gain in

2. Lower cut off Frequency ……………………………

3. Upper cut off Frequency …………………………….

5. Gain Bandwidth Product …………………………..

Apparatus: 1. Digital Storage Oscilloscope (DSO)

A n-stage amplifier can be represented by the block diagram as shown in fig. 1

The three couplings generally used are.

Signal Generator 10UFD

2. Apply a sinusoidal input signal of 20mv from a signal generator.

3. Connect the output to the DSO

4. Measure output voltage and calculate gain

6. Draw a graph between input frequency Vs output voltage

S. No Frequency Output voltage Voltage Gain Voltage gain in

2. Lower cut off Frequency ……………………………

3. Upper cut off Frequency …………………………….

5. Gain Bandwidth Product …………………………..

Apparatus: 1. Digital Storage Oscilloscope (DSO)

Component: Resistances: 8.2kΩ = 01, 4.7kΩ = 01, 470Ω = 01, 100Ω = 01

The ac equivalent circuit is shown in fig. 1

The ac output voltage is

The voltage gain is given by

1. Connect the circuit as shown in the diagram.

2. Apply a sinusoidal input signal of 20mv from a signal generator.

3. Connect the output to the DSO

4. Measure output voltage and calculate gain

6. Draw a graph between input frequency Vs output voltage

S. No Frequency Output voltage Voltage Gain Voltage gain in

2. Lower cut off Frequency ……………………………

3. Upper cut off Frequency …………………………….

5. Gain Bandwidth Product ……………………………

Component: Resistances: 100Ω = 01

The Thevenin impedance at the input is given by

Therefore, t he output impedance of the amplifier is very small.

Darlington Trasistor Amplifier

1. Connect the circuit as shown in diagram.

connected in a ammeter mode.

5. Also measure the output voltage (voltage gain is less then 1)

6. With the value of ammeter , calculate the amplification factor of a transistor β1

10. Total current gain = β1× β2

and output current gives output impedance

2. Amplification factor β2 …………………………………..

3. Total current gain β1× β2 ………………………………

Aim: To Study BJT Differential Amplifier and measure its CMRR

Apparatus: 1. Digital Storage Oscilloscope (DSO)

Component: Resistances: 1kΩ = 02, 470Ω = 02, 5.6KΩ = 01,

A long tailed pair is a common design in electronics for implement a differential

1. Connected the circuit as shown in figure.

Procedure: for AC Characteristic

Differential gain :………………………….

Common mode gain……………………….

Apparatus: 1. Digital Storage Oscilloscope (DSO)

Component: Resistances: 100kΩ = 01, 10kΩ = 01, 5.1KΩ = 01, 1kΩ = 04

Transistor Phase Shift Oscillator: