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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 .


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)


(b)

Given V CEQ is desired to be 3.5 V, hence:

(c)


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)



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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 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 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 :

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