# Microelectronics Questions and Answers – Bipolar Amplifiers – Common Base Stage – Set 2

This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Bipolar Amplifiers – Common Base Stage – Set 2”.

1. Assume that the transistor has a very high current gain with negligible early effect. What is the output resistance of Q1?

a) R1 || 1/gm
b) (R1 || 1/gm) + R2
c) R1 || (1/gm + R2)
d) R1

Explanation: The output resistance is only R1. Observe that Q2 is unbiased so it is in the cut-off state. The emitter doesn’t contribute to any impedance at the collector of Q1.

2. Assume that the transistor has a very high current gain with negligible early effect. What is the output resistance of Q1?

a) R1 || 1/gm
b) (R1 + 1/gm) + R2
c) R1 + (1/gm || R2)
d) 2*R1

Explanation: Q2 is behaving as a common base stage. It offers an input impedance of 1/gm at the emitter of Q1 and this make the output impedance of Q1, R1 || 1/gm.

3. Assume that the transistor has a very high current gain with only Q2 has early effect. What is the output resistance of Q2?

a) {rO2+(1+gm2rO2)(R1||rπ2)}
b) (1+gm2rO2)(R1||rπ2)
c) {rO2+(1+gm2rO2)(R1||rπ2)}||R2
d) R2

Explanation: Q2 is degenerated by R1 and Q1 doesn’t contribute due to negligible early effect. This leads us to the output resistance as {rO2+(1+gmrO)(R1||rπ2)}||R2. Observe that even though Q2 is behaving as a common base stage, the output impedance is quite similar to a degenerated common emitter stage.

4. Assume that the transistor has a very high current gain while both has early effect. What is the output resistance of Q2?

a) R2
b) {rO2+(1+gm2rO2)(R1||rO1||rπ2)}||R2
c) rO2
d) R1||rO1||rπ2

Explanation: Q2 is degenerated by R1 and Q1 which contributes rO1 due to early effect. These two resistances come in parallel and are connected to the emitter of Q2. This leads us to the output resistance as b) {rO2+(1+gm2rO2)(R1||rO1||rπ2)}||R2.

5. Assume that the transistor has a very high current gain while both has early effect. What is the input resistance of Q2?

a) 1/gm2 || ro2
b) 1/gm2 || ro1
c) 1/gm1 || ro2
d) 1/gm2

Explanation: The input resistance of Q2, behaving as a common base stage, is 1/gm2 || ro2. This input impedance is independent of the previous stage. Note that in absence of early effect, the answer would’ve been 1/gm2.

6. Assume that the transistor has a very high current gain while both has early effect. What is the voltage gain till Q1?

a) -gm1 * (R1 || rO1 || 1/gm2 || rO2)
b) gm1 * (R1 || rO1)
c) -gm1 * (R1 || rO1 + 1/gm2 || rO2)
d) -gm1 * (R1 || rO1)

Explanation: The voltage gain till Q1 is similar to that of a C.E. stage and that is gm1 * (The total resistance connected to the collector of Q1). Note that three resistances are connected to the collector of Q1. First is R1, second is Q1 itself which offers rO1 due to early effect and third is the emitter of Q3 which offers a resistance of 1/gm2 || rO2. All these resistances are connected from the collector to ground and the total resistance connected to the collector is (R1 || rO1 || 1/gm2 || rO2).

7. Assume that the transistor has a very high current gain while both has early effect. What is the voltage gain till Q2?

a) -(R1 || rO1 || 1/gm2 || rO2) * {rO2 + (1 + gm2rO2)(R1 || rO1 || rπ2)} || R2
b) -gm1 * (R1 || rO1 || 1/gm2 || rO2) * {rO2 + (1 + gm2rO2)(R1 || rO1 || rπ2)} || R2
c) gm1 * (R1 || rO1 || 1/gm2 || rO2)
d) -gm1 * (R1 || rO1 || 1/gm2 || rO2) * gm2 * {rO2 + (1 + gm2rO2)(R1 || rO1 || rπ2)} || R2

Explanation: The voltage gain till Q2 is the product of the voltage gain from the input to Q1 and from Q1 to Q2. The first voltage gain is gm1 * (R1 || rO1 || 1/gm2 || rO2). The second voltage gain is gm2 * {rO2 + (1 + gm2rO2)(R1 || rO1 || rπ2)} || R2 since the output impedance of common base stage is degenerated by Q1 and R1 . The total voltage gain thus becomes
gm1 * (R1 || rO1 || 1/gm2 || rO2) * gm2 * {rO2 + (1 + gm2rO2)(R1 || rO1 || rπ2)} || R2.

8. Assume that the transistor has a very high current gain while only Q2 has early effect. What is the voltage gain till Q1?

a) gm1 * (R1 || 1/gm2)
b) gm1 * (R1 || rO2)
c) gm1 * (R1/gm2 || rO2)
d) gm1 * (R1 || 1/gm2 || rO2)

Explanation: The voltage gain till Q1 is gm1 * the total gain connected to the collector of Q1. The total resistance connected to it is R1 || 1/gm2 || rO2. The overall voltage gain is gm1 * (R1 || 1/gm2 || rO2).

9. Assume that the transistor has a very high current gain while only Q2 has early effect. What is the voltage gain till Q1?

a) gm1*(R1 || -1/gm2)
b) -gm1*(R1 || 1/gm2)
c) gm1*(R1 || 1/gm2)
d) -gm1*(R1 || 1/gm2)

Explanation: The voltage gain till Q1 is gm1 * the total resistance connected at the collector of Q1. The total resistance is R1 || 1/gm2 || rO2 and the overall voltage gain is gm1*(R1 || 1/gm2 || rO2).

10. Assume that the transistor has a very high current gain while only Q2 has early effect. What is the voltage gain till Q2?

a) gm2*{rO2+(1+gm2rO2)(R1||rO1||rπ2)}||R2*gm1*(R1||1/gm2||rO2)
b) gm1*(R1||1/gm2||rO2)
c) gm2*{rO2+(1+gm2rO2)(R1||rO1||rπ2)}||R2
d) -gm2*{rO2+(1+gm2rO2)(R1||rO1||rπ2)}||R2

Explanation: The voltage gain till Q2 will be the product of voltage gains from input to Q1 and from Q1 to ground. The first gain is gm1*(R1||1/gm2||rO2) while the second gain is gm2*{rO2+(1+gm2rO2)(R1||rO1||rπ2)}||R2. Multiplying both the gains yield the overall gain as
gm2*{rO2+(1+gm2rO2)(R1||rO1||rπ2)}||R2*gm1*(R1||1/gm2||rO2).

11. Assume that the transistor has a very high current gain with negligible early effect. What is the input resistance of the following circuit?

a) $$\frac {(R_4 || R_5)}{(\beta+1)}$$
b) 1/gm+R5/(β+1)
c) 1/gm+(R4)/(β+1)
d) 1/gm+(R4||R5)/(β+1)

Explanation: The input impedance looking into the base of the emitter is 1/gm + (The base resistance)/(β+1). The base resistance is approximated to R4 || R5 since the current gain is very high and the base current is nearly 0. The overall input impedance is 1/gm+(R4||R5)/(β+1).

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