# Tutorial Problems: Bipolar Junction Transistor (Basic BJT Amplifiers)

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1 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 β = 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 A v = v o / 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: 1

2 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 R i seen by the signal source v s is: 2. Consider the circuit shown in Figure 2. The transistor parameters are β = 100 and V A = 100 V. Determine R i, A v = v o / v s and A i = i o / i s. Figure 2 2

3 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-signal ac equivalent circuit is obtained: Small-signal model of transistor circuit (*g m V π = βi b ) 3

4 The input resistance R i is: 3. The parameters of the transistor in Figure 3 are β = 100 and V A = 100 V. (a) Find the dc voltages at the base and emitter terminals. (b) Find R C such that V CEQ = 3.5 V. (c) Assuming C C and C E act as short circuits, determine the small-signal voltage gain A v = v o / v s. (d) Repeat part (c) if a 500 Ω source resistor is in series with the v s signal source. Figure 3 4

5 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: 5

6 (d) If the source resistor is changed to 500 Ω, the new value of A v is: Therefore the voltage gain A v decreases as the source resistance R S 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 β = 100. (a) Determine the small-signal voltage gain A v = v o / v s. (b) Find the input and output resistances R i and R o. Figure 4 Solution: (a) A dc analysis is performed to determine the dc operating point by treating all capacitors as open circuit. 6

7 The small-signal parameters are: Using the small-signal ac equivalent circuit, the following expressions are obtained: (b) The input resistance R i is: To calculate the output resistance R o, the signal source v s is short-circuited and this gives i b = 0. The following equation can be written by KCL at node v o : 7

8 Part B. Common-Collector Amplifier (Emitter Follower) 5. The transistor parameters for the circuit in Figure 5 are β = 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 R ib and R o. Figure 5 Solution: (a) For dc analysis, the capacitors C C1 and C C2 act as open circuit. (b) The dc load line is given by: 8

9 The ac load line is given by: (c) The small-signal parameters are: The small-signal ac equivalent circuit becomes: 9

10 (d) The input resistance R ib is: To calculate the output resistance R o, the signal source v s is short-circuited and the following equations can be written by KCL at node v o and node v b : 10

11 6. For the circuit shown in Figure 6, let V CC = 5 V, R L = 4 kω, R E = 3 kω, R 1 = 60 kω, and R 2 = 40 kω. The transistor parameters are β = 50 and V A = 80 V. (a) Determine I CQ and V ECQ. (b) Plot the dc and ac load lines. (c) Determine A v = v o / v s and A i = i o / i s. (d) Determine R ib and R o. Figure 6 Solution: (a) For dc analysis, the capacitors C C1 and C C2 act as open circuit. (b) The dc load line is given by: 11

12 The ac load line is given by: (c) The small-signal parameters are: The small-signal ac equivalent circuit becomes: 12

13 13

14 (d) The input resistance R ib is: To calculate the output resistance R o, the signal source v s is short-circuited and the following equations can be written by KCL at node v o : 7. For the transistor in Figure 7, the parameters are β = 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 kω 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 14

15 Solution: (a) For dc analysis, the capacitor C C acts as open circuit. (b) The small-signal ac equivalent circuit is given by: 15

16 If the peak-to-peak output voltage v o(peak-peak) is 4 V, (c) If the load resistor R L = 1 kω is added in parallel to R E, Eq. (4) must be modified accordingly: Therefore v o(peak-peak) becomes smaller due to the loading effect by R L. 16

17 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 R S = 10 Ω can drive a small speaker. Assume the supply voltages are V + = + 12 V and V = 12 V and β = 50. The load, representing the speaker, is R L = 12 Ω. 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 A v is: Choose I EQ = 0.8 A and V CEQ = 12 V, 17

18 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 R S (loading effect) the required voltage gain of A v = cannot be achieved. Note that A v = if R S = 0. Therefore the maximum achievable peak output voltage is: 18

19 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 β = 100 and V A = 100 V. The values of R C, 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 v CE 8 V and the minimum collector current is to be i C (min) = 0.1 ma. Figure 9 19

20 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 v CE(min) = 1 V and i C(min) = 0.1 ma, the maximum swing of v CE and i C 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 ): 20

21 To decide the value for V TH : 10. In the circuit in Figure 10 with transistor parameters β = 180 and V A =, design the bias resistors R 1 and R 2 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 v CE 4.5 V and the total instantaneous collector current is to be i C 0.25 ma. Figure 10 21

22 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 v CE(min) = 0.5 V and i C(min) = 0.25 ma, the maximum swing of v CE and i C 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 ): 22

23 To decide the value for V TH : 23

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