Microelectronics Questions and Answers – Bipolar Amplifiers – Common Emitter Stage

This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Bipolar Amplifiers – Common Emitter Stage”.

1. In absence of early effect, what is the input resistance of the circuit?

a) rπ
b) rπ+(β+1)*R2
c) (β+1)*R2
d) Infinite
View Answer

Answer: b
Explanation: The above circuit is a degenerated common emitter stage and we invoke the small signal model to find the input resistance.

We use nodal analysis at the emitter node and then we find the emitter voltage. Rin=Vin-VE/(rπ) and Rin turns out to be rπ+(β+1)*R2.

2. In absence of early effect, what is the output resistance of the circuit?

a) ro+(rπ||R2)(1+gm*ro)
b) ro
c) rπ||R2
d) 1+gm*ro
View Answer

Answer: a
Explanation: We have to invoke the small signal model

Firstly, we need to use KCL at E and find the emitter voltage. Then we use KVL from Vin via ro and R2 and we derive Rout as ro+(rπ||R2)(1+gm*ro).

3. If the circuit has a very high current gain, find the gain of the following circuit.

a) -gm*RC
b) 0
c) Infinite
d) -RC/(1/gm + RE)
View Answer

Answer: d
Explanation: We will have to invoke the small signal model.

We use KCL at the emitter node and find the emitter voltage. Thereafter, we use KVL in the base-emitter loop and the gain is approximately -RC/(1/gm + RE) since the gain is very high.
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4. In absence of early effect, what is the gain of the following circuit?

a) -RC/(1/gm + RE)
b) RC/(1/gm + RE)
c) -RC*gm
d) RC*gm
View Answer

Answer: a
Explanation: In absence of early effect, the gain of the above circuit is similar to that of a degenerated common emitter stage which is -RC/(1/gm + RE). The biasing method only places and impact on gm but the expression of voltage gain is the same.

5. Find the input resistance of the following circuit.

a) R2 || R3
b) R2 || R3 || Rπ1
c) Rπ1
d) Infinite
View Answer

Answer: b
Explanation: The input resistance is literally independent of the cascaded structure. We can invoke the small signal model and the result will be the same as that of a simple C.E. stage. The input resistance, Rin, is R2||R3||Rπ1 which concerns the biasing resistors and the resistance in the base-emitter junction.

6. If all the transistors are identical and early effect is neglected, find the gain of the following circuit till Q1 . Q4 can be assumed to be properly biased.

a) 0
b) gm1*(R || rπ2)
c) gm1 * rπ2
d) Infinite
View Answer

Answer: b
Explanation: We draw the small signal model of Q1 & observe that the gain is simply gm1 * Rout. We only need to find Rout which is (R || rπ2). Rout is derived by turning all independent voltage sources to 0. The small signal model of both the transistors reveals that the collector of Q1 is connected to the base-emitter resistance and R. This makes the resistors connected between the same pair of terminals.

7. All the transistors have a high current gain and early effect is absent in each of them. What is the total gain till the collector of Q1?

a) gm5*(4/gm4||rπ1)*gm1*(rπ2 + (β+6)*1/gm3)
b) gm5*(3/gm4||rπ1)*gm1*(rπ2 + (β+1)*1/gm3)
c) gm5*(1/gm4||rπ1)*gm1*(rπ2 + (β+1)*1/gm3)
d) gm5*(7/gm4*rπ1)*gm1*(rπ2 + (β+5)*1/gm3)
View Answer

Answer: c
Explanation: We approach this circuit by first finding the gain till the collector of Q5 which is -gm5*(1/gm4||rπ1). Observe that Q1 is being used as a simple CE stage hence the gain for Q1 is gm1 times the total resistance connected to the collector. This resistance is the input resistance of Q2 which is further degenerated by Q3. Q3 offers a resistance of 1/gm3+rπ3 which is approximately 1/gm3 and the overall input resistance at Q2 is (rπ2 + (β+1)*1/gm3). The overall gain is the product of both the gains, becoming a positive quantity. and it turns out to be gm5*(1/gm4||rπ1)*gm1*(rπ2 + (β+1)*1/gm3).
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8. If the transistors are identical with high current gain and only Q2 and Q3 has early effect, what is the output resistance of Q2 if a resistor of resistance R is kept in between it’s collector and the supply voltage?

a) RC || ro2 + (rπ2 || ro3 || 1/gm3)(1+gm2*ro2)
b) RC || ro2
c) rπ2 || ro3 || 1/gm3
d) Infinite
View Answer

Answer: a
Explanation: The output resistance of Q5 is given by the total resistance looking its collector in parallel with RC. We note that Q2 has been degenerated by Q3, which offers a resistance of ro3 || 1/gm3 to the emitter of Q2. Therefore, the resistance looking into the collector is ro2 + (rπ2 || ro3 || 1/gm3)(1+gm2*ro2) and the total resistance becomes RC || ro2 + (rπ2 || ro3 || 1/gm3)(1+gm2*ro2).

9. Assume that all the transistors are identical with negligible early effect and very high current gain. What is the input resistance at the base of Q4?

a) rπ4
b) 0
c) Infinite
d) rπ4 + (β + 1) * rπ1
View Answer

Answer: d
Explanation: We note that Q4 is degenerated by Q5 and Q1. Q5 offers infinite output impedance while Q1 offers a resistance of rπ1. Note that they are in parallel and the overall resistance becomes rπ1. The overall resistance is rπ4 + (β+1)* rπ1 due to degeneration.
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10. Assume that all the transistors are identical with negligible early effect and very high current gain. In the following circuit, Q4 is behaving as a _____

a) Switch
b) Amplifier
c) Current Source
d) Cut-off
View Answer

Answer: c
Explanation: The input is applied to the base of Q5 while Q4 is simply biased to behave as a current source. It offers a resistance of ro4||1/gm4 at its emitter which is connected to the collector of Q5. But it doesn’t behave as an amplifier since the input is connected to the base of Q5.

11. Assume that all the transistors are identical with negligible early effect and very high current gain. What is the voltage gain till the collector of Q2?

a) gm1*[rπ2+(β+1)*(1/gm3||ro3)]*gm2*[(rπ2||ro3||1/gm3)(1+gm2*ro2)]||[rπ4+(β+1)*ro5]||R1
b) Infinite
c) gm1*[rπ2+(β+1)*(1/gm3||ro3)]*gm2*[(rπ2||ro3||1/gm3)(1+gm2*ro2)]
d) [rπ4+(β+1)*(ro5||1/gm5)]||R1
View Answer

Answer: a
Explanation: Firstly, we will calculate the voltage gain to Q1. This is given by gm1 times the total resistance connected to the collector. This resistance is given by the internal resistance of the degenerated Q2 which is rπ2+(β+1)*(1/gm3||ro3). Finally, the gain for Q2 is gm2 times the resistance connected to the collector of Q2. Observe that the total resistance connected at the collector are R1, the resistance looking into the base of Q4 which rπ4+(β+1)*ro5 and the output resistance of Q2 which is ro2+(rπ2||1/gm3||ro3)(1+gm3*ro2). All the resistance come in parallel to each other and the total resistance is ro2+[(rπ2||ro3||1/gm3)(1+gm2*ro2)]||[rπ4+(β+1)*(ro3||1/gm3)]||R1. The overall voltage gain is gm1*[rπ2+(β+1)*(1/gm3||ro3)]*gm2*[(rπ2||ro3||1/gm3)(1+gm2*ro2)]||[rπ4+(β+1)*ro5]||R1.

12. The supply voltage is 5V while the input voltage is 700mV. For a reverse saturation current of 10-14A and R1=1K, calculate the voltage gain without early effect.

a) 19
b) -188
c) 6.5
d) 0
View Answer

Answer: b
Explanation: The voltage gain is simply gm*R1 and we can calculate gm by finding the collector current first. The collector current is IS*exp(\(\frac {V_{BE}}{V_T}\)) and that turns out to be 4.9mA. Dividing this by the thermal voltage gives us the transconductance as 188mS and the overall voltage gain is 188.

13. The supply voltage is 5V while the input voltage is 700mV. For a reverse saturation current of 10-16A and R1=1K, calculate the conductance offered due to early voltage of 5V.

a) 750 S
b) 800 S
c) 816 S
d) Infinite
View Answer

Answer: c
Explanation: Firstly, we need to find the collector current which comes out to be 4.9mA. We simply divide this by the early voltage to ro and its inverse turn out to be 816 S.

Sanfoundry Global Education & Learning Series – Microelectronics.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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