BITS-PILANI GOA CAMPUS
EEE Analog Electronics Lab Manual
Common Emitter Amplifier
Aim: To design and study the performance of a common emitter transistor amplifier.
Apparatus and components required:
D.C. power supply, Cathode Ray Oscilloscope, function generator, decade resistance box,
resistors, capacitors, and transistor BC 107B.
Theory:
When the transistor is used as an amplifier, the first step is to choose a proper configuration in
which device is to used. Then the transistor is biased to get the desired Q-point, at the middle
of the DC load line, in its output characteristics. The signal is applied to the amplifier input and
gain is achieved. The most convenient way to analyse the circuit analytically is by replacing
he device by its model. The behaviour of an amplifier depends upon the magnitude and
frequency of the input signal. Accordingly different models of the transistor are used, e.g. small
signal model, high frequency model, and large signal model. In this experiment we shall study
the small performance of a common-emitter amplifier.
For low frequency and small signal operation of the transistor, the transistor is replaced
by its small signal model. Figure 1 shows the equivalent circuit.
The relationship of the parameters of the model is given below:
r = Base emitter resistance =
VT
kT
where VT =
= 26 mV at room temperature and IB is the
IB
q
dc Q-point base current.
IC
, where IC is the dc Q-point collector current.
VT
= ac current gain = g m r
v = vbe = ac base-emitter voltage
ro= output resistance
gm= tranconductance=
Consider the CE amplifier circuit with voltage divider biasing arrangement as shown in figure
2. Here C1 and C2 are called coupling capacitors and CE the bypass capacitor.
DC Analysis:
Let us first design the voltage divider bias circuit. Assume the following data for the design.
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EEE Analog Electronics Lab Manual
vo
Figure 2. CE Amplifier circuit
R1= 39 k 
R2 = 8.2 k
RL= 
Rs= 50 
Vcc= 10 V
Ic= 1 mA
Vce= 5 V Vbe= 0.6 V
 =325
CC= 4 F
CE= 100 F and let Vs = 40 mV
fT=100 MHz and C= 2.5 pF
From the DC equivalent circuit the Thevenin voltage is given by
V R
............
Vth = cc 2 =
= ........V
R1 + R2 ...........
and Thevenin resistance is given by
RR
............
Rth = 1 2 =
= .........k
R1 + R2 .............
Considering the base-emitter loop and applying KVL, we get
Vth − VBE
I
....
. Since IC is given I B = C =
IB =
 ....
RB + (1 +  ) RE
 Vth − VBE 

 − Rth
IB
.....................


 RE is given by R E =
=
1+ 
.......................
The value of VCE is given by VCE = VCC − I C RC − ( I B + I C ) RE
V − I B + I C ) RE
This equation gives RC as RC = CC
IC
The values of IC and VCE specify the dc operating point or Q-point of the transistor.
AC Analysis:
Assume all capacitors as short at the given signal frequency. To determine the performance
parameters like voltage gain, current gain, input resistance, output resistance of the amplifier,
consider the ac equivalent circuit as shown in figure 3. Since dc power supply voltage does not
change with the signal, it is assumed short for ac.
Let us first find the value of r and g m .
r =
VT
........
=
= .........
I B ............
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gm =

r
EEE Analog Electronics Lab Manual
= ..........S
Voltage Gain Avs:
Vo = − g m v RC where v =
Rth || r
= ........V
RS + Rth || r
Vo = ………V
VO
.....
=
= ........
VS ........
The voltage gain is negative indicating 180  phase reversal between the input and output
Therefore Avs =
signals.
Current Gain Ais:
VS
is =
= ..........A
RS + RB || r
V
io = O = ........... A
RC
i
Ais = o = ..........
is
Input Resistance Ri:
Ri = Rth || r = .............
Output Resistance Ro:
Since the value of RL is infinite, the output resistance Ro=Rc.
Frequency Response:
Due to the coupling and bypass capacitors, at low frequencies, the gain of the amplifier is
reduced. The internal transistor capacitors lower the voltage gain and input impedance at high
frequencies. At high frequencies the coupling, bypass and transistor parasitic capacitors
provide very low impedances. The low impedance due to parasitic capacitors reduces the gain
at high frequencies. Thus the gain of the amplifier reduces both at low and high frequencies.
For a single stage amplifier as shown in fig, the lower cutoff frequency fL is due to the input
coupling capacitor C1 is given by
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fL =
EEE Analog Electronics Lab Manual
1
where R=Rs + Ri
2RC1
The high frequency -model is as shown in figure 4. here rx is the bulk base resistance also
called base spreading resistance. C is the base-emitter capacitance; C is the collector-base
junction capacitance. The value of C is either directly given in the data sheets or can be
calculated using C =
gm
where fT is unity common-emitter short-circuit current gain
2f T
frequency.
Using the high-frequency model, the complete circuit of CE configuration at high frequency is
shown in fig below.
Here C= C+(Avs C) .
The output cutoff frequency fH is given by
fH =
1
2RC
The bandwidth of the amplifier is defined as BW= fH - fL.
Procedure:
Voltage Gain: From the designed component values construct the circuit as shown in
figure
1. Measure the voltages at collector, base, and emitter of the transistor. From these values
determine the Q-point values. Then, feed 40 mV sine wave signal at 10 kHz and measure the
output voltage vo. Hence, calculate the small signal voltage gain Avs =
vo
.
vi
Frequency response: To study the frequency response, vary the input signal frequency from
100 Hz to 1MHz and measure the output voltage in each step. Then calculate the voltage gain
and plot a semi-log graph with frequency along x-axis and voltage gain along y-axis. Find the
cut-off frequencies fH and fL corresponding
Av (max)
2
and then bandwidth = fH – fL.
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EEE Analog Electronics Lab Manual
Input Resistance Ri: To find the input resistance connect a resistance box (with R=0 ) in
series with the input signal source keeping the input voltage at 40 mV and frequency at
10
kHz (mid-band frequency) as shown in figure 5. Now increase the resistance in the resistance
box until the input voltage reduces to half and note down the resistance. This is the input
resistance of the amplifier.
Output Resistance: To find the output
resistance connect a resistance box (with R=100 k) between the output and ground. Set the
input voltage at 40 mV and frequency at 10 kHz (mid-band frequency) and measure the output
voltage. Now decrease the resistance in the resistance box until the output voltage reduces to
half and note down the resistance. This is the output resistance (Ro) of the amplifier.
Current Gain Ai: The input current is =
VS
v
and output current, io = O . Now the current
Ri
Ro
io
is calculated.
is
Result: A Common Emitter transistor amplifier circuit is designed as per the specifications and
gain Ai =
the results are tabulated as shown below.
Parameter
Designed Measured
Base Voltage VB (V)
Collector Voltage VC (V)
Emitter Voltage VE (V)
Collector current IC (mA)
Base Current IB (A)
DC current gain 
Voltage gain Avs
Input Resistance Ri ()
Output Resistance Ro ()
Current gain Ai
Lower Cutoff frequency f1 (Hz)
Upper Cutoff frequency f2 (Hz)
Bandwidth= f2-f1 (Hz)
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