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Tutorial Problems: Bipolar Junction Transistor (Basic BJT Amplifiers)
Part A. Common-Emitter Amplifier
1. For the circuit shown in Figure 1, the transistor parameters are ! = 100 and V A = !. Design the
circuit such that I CQ = 0.25 mA and V CEQ = 3 V. Find the small-signal voltage gain Av = vo / v s. Find
the input resistance seen by the signal source v s.
Figure 1
Solution:
For dc analysis, the capacitors C C and C E both act as open circuit .
Given the desired operating point I CQ = 0.25 mA and V CEQ = 3 V, we have:
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The small-signal parameters are:
For small-signal ac analysis, all dc voltages and capacitors act as short circuit . The following expressions
are obtained:
The input resistance Ri seen by the signal source v s is:
2. Consider the circuit shown in Figure 2. The transistor parameters are ! = 100 and V A = 100 V.
Determine Ri, Av = vo / v s and Ai = io / i s.
Figure 2
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Solution:
A dc analysis is performed to determine the dc operating point by treating all capacitors as open circuit .
The small-signal parameters are:
For small-signal ac analysis, all dc voltages and capacitors act as short circuit . The following small-signalac equivalent circuit is obtained:
Small-signal model of transistor circuit (*g mV " = ! ib )
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The input resistance Ri is:
3. The parameters of the transistor in Figure 3 are ! = 100 and V A = 100 V.
(a) Find the dc voltages at the base and emitter terminals.
(b)
Find RC such that V CEQ = 3.5 V.(c) Assuming C C and C E act as short circuits, determine the small-signal voltage gain Av = vo /
v s.
(d) Repeat part (c) if a 500 " source resistor is in series with the v s signal source.
Figure 3
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Solution:
(a)
A dc analysis is performed to determine the dc operating point by treating all capacitors as open circuit .
(b)
Given V CEQ is desired to be 3.5 V, hence:
(c)
The small-signal parameters are:
Using the small-signal ac equivalent circuit, the following expressions are obtained:
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(d)
If the source resistor is changed to 500 ", the new value of Av is:
Therefore the voltage gain Av decreases as the source resistance RS increases due to a larger voltage drop
across the source resistor.
4. The transistor in the circuit in Figure 4 has a dc current gain of ! = 100.
(a) Determine the small-signal voltage gain Av = vo / v s.
(b) Find the input and output resistances Ri and Ro.
Figure 4
Solution:
(a)
A dc analysis is performed to determine the dc operating point by treating all capacitors as open circuit .
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The small-signal parameters are:
Using the small-signal ac equivalent circuit, the following expressions are obtained:
(b)
The input resistance Ri is:
To calculate the output resistance Ro, the signal source v s is short-circuited and this gives ib = 0. Thefollowing equation can be written by KCL at node vo:
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Part B. Common-Collector Amplifier (Emitter Follower)
5. The transistor parameters for the circuit in Figure 5 are ! = 180 and V A = !.
(a) Find I CQ and V CEQ.(b) Plot the dc and ac load lines.
(c) Calculate the small-signal voltage gain.
(d) Determine the input and output resistances Rib
and Ro.
Figure 5
Solution:
(a)
For dc analysis, the capacitors C C 1 and C C 2 act as open circuit .
(b)
The dc load line is given by:
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The ac load line is given by:
(c)
The small-signal parameters are:
The small-signal ac equivalent circuit becomes:
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(d)
The input resistance Rib is:
To calculate the output resistance Ro, the signal source v s is short-circuited and the following equationscan be written by KCL at node vo and node vb:
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6. For the circuit shown in Figure 6, let V CC = 5 V, R L = 4 k ", R E = 3 k ", R1 = 60 k ", and R2 = 40 k ".
The transistor parameters are ! = 50 and V A = 80 V.
(a) Determine I CQ and V ECQ.
(b) Plot the dc and ac load lines.
(c) Determine Av = vo / v s and Ai = io / i s.(d) Determine Rib and Ro.
Figure 6
Solution:
(a)
For dc analysis, the capacitors C C 1 and C C 2 act as open circuit .
(b)
The dc load line is given by:
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The ac load line is given by:
(c)
The small-signal parameters are:
The small-signal ac equivalent circuit becomes:
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(d)
The input resistance Rib is:
To calculate the output resistance Ro, the signal source v s is short-circuited and the following equationscan be written by KCL at node vo:
7. For the transistor in Figure 7, the parameters are ! = 100 and V A = !.
(a)
Design the circuit such that I EQ = 1 mA and the Q-point is in the center of the dc load line.(b) If the peak-to-peak sinusoidal output voltage is 4 V, determine the peak-to-peak sinusoidal
signals at the base of the transistor and the peak-to-peak value of v s.
(c) If the load resistor R L = 1 k " is connected to the output through a coupling capacitor,
determine the peak-to-peak value in the output voltage, assuming v s is equal to the value
determined in part (b).
Figure 7
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Solution:
(a)
For dc analysis, the capacitor C C acts as open circuit .
(b)
The small-signal ac equivalent circuit is given by:
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If the peak-to-peak output voltage vo(peak-peak) is 4 V,
(c)
If the load resistor R L = 1 k " is added in parallel to R E , Eq. (4) must be modified accordingly:
Therefore vo(peak-peak) becomes smaller due to the loading effect by R L.
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8. An emitter-follower amplifier, with the configuration shown in Figure 8, is to be designed such that
an audio signal given by v s = 5 sin(3000t ) V but with a source resistance of RS = 10 " can drive a
small speaker. Assume the supply voltages are V + = + 12 V and V # = # 12 V and ! = 50. The load,
representing the speaker, is R L = 12 ". The amplifier should be capable of delivering approximately
1 W of average power to the load. What is the signal power gain of your amplifier?
Figure 8
Solution:
To deliver 1 W of average power to the load, the peak-to-peak output voltage should be:
The required voltage gain Av is:
Choose I EQ = 0.8 A and V CEQ = 12 V,
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The small-signal ac equivalent circuit is given by:
Choosing I EQ = 0.5 A gives:
The small-signal voltage gain is taken from Q.7 with some modifications:
Due to the presence of the source resistance RS (loading effect) the required voltage gain of Av = 0.9798
cannot be achieved. Note that Av = 0.9951 if RS = 0.
Therefore the maximum achievable peak output voltage is:
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Hence the output power delivered to the load R L is:
The input power delivered by the signal source v s is:
Hence the signal power gain of the amplifier is:
Part C. AC Load Line Analysis / Maximum Symmetrical Swing
9. For the circuit in Figure 9, the transistor parameters are ! = 100 and V A = 100 V. The values of RC ,
R E and R L are as shown in the figure. Design a bias-stable circuit to achieve the maximum
undistorted swing in the output voltage if the total instantaneous C-E voltage is to remain in the
range 1 $ vCE $ 8 V and the minimum collector current is to be iC (min) = 0.1 mA.
Figure 9
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Solution:
To obtain a bias-stable circuit, let:
The dc load line of the circuit is given by:
The ac load line of the circuit is given by:
Given vCE (min) = 1 V and iC (min) = 0.1 mA, the maximum swing of vCE and iC from the Q-point ( I CQ, V CEQ)
would be:
Since and are related by the ac load line,
Solving (1) and (3) at the Q-point ( I CQ, V CEQ):
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To decide the value for V TH :
10. In the circuit in Figure 10 with transistor parameters ! = 180 and V A = !, design the bias resistors R1
and R2 to achieve maximum symmetrical swing in the output voltage and to maintain a bias-stable
circuit. The total instantaneous C-E voltage is to remain in the range 0.5 $ vCE $ 4.5 V and the total
instantaneous collector current is to be iC % 0.25 mA.
Figure 10
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Solution:
To obtain a bias-stable circuit, let:
The dc load line of the circuit is given by:
The ac load line of the circuit is given by:
Given vCE (min) = 0.5 V and iC (min) = 0.25 mA, the maximum swing of vCE and iC from the Q-point ( I CQ,
V CEQ) would be:
Since and are related by the ac load line,
Solving (1) and (3) at the Q-point ( I CQ, V CEQ):
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To decide the value for V TH :