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

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

1. Assuming a very high current gain and negligible early effect, what happens to the voltage gain of the follower stage if the base width is reduced by a factor of 2?
a) It doubles
b) It decreases by a factor of 4
c) It increases
d) It decreases

Explanation: If the base width reduces by a factor of 2, the transconductance increases by a factor of 2. Hence 1/gm decreases by a factor of 2 and not that this term is present in the denominator in the expression of voltage gain and the voltage gain increases slightly.

2. Assuming a very high current gain and negligible early effect, what happens to the output resistance of the follower stage if the base width is increased by a factor of 5?
a) It increases by a factor of 5
b) It decreases by a factor of 5
c) It increases by a factor of 3
d) It decreases by a factor of 3

Explanation: We note that as base width increases by a factor of 5, the transconductance decreases by a factor of 5. We have seen that the output impedance of the follower is 1/gm || Re and a decrease in the transconductance eventually increases the output resistance.

3. The transistor in the following circuit has a very high current gain with an inbuilt early voltage. What is the voltage gain of the circuit?

a) (Re||rO)/{(Re||rO)+1
b) (Re||rO)/1+R4/(β+1)+1/gm}
c) 1/{(Re||rO)+R4/(β+1)+1/gm}
d) (Re||rO)/{(Re||rO)+R4/(β+1)+1/gm}

Explanation: Firstly, we note that the voltage gain of the circuit is quite similar to a follower circuit with a base resistance. But in presence of early effect, we note that rO gets connected from the emitter to ground-this is observed from the small signal model. Hence, the only change in the expression of voltage gain is rO appears in parallel-the gain becomes (Re||rO)/{(Re||rO)+R4/(β+1)+1/gm}.

4. The transistor in the following circuit has a very high current gain with an inbuilt early voltage. What is the input impedance of the circuit?

a) rπ+(β+1)*(RE||rO)
b) rπ+(β+1)*RE
c) (β+1)*(RE||rO)
d) rπ

Explanation: We note that while calculating the input impedance, rO appears in parallel to RE. The early effected can be incorporated into the small signal model by placing rO in parallel to RE. The input impedance becomes rπ+(β+1)*(RE||rO).

5. The transistor in the following circuit has a very high current gain with an inbuilt early voltage. What is the output impedance of the circuit?

a) R4/(β+1)+1/gm||RE||rO
b) R4/(β+1)+1/gm||rO
c) R4/(β+1)||RE
d) RE||rO

Explanation: While calculating the output impedance, we turn off all independent voltage sources and note that rO comes in parallel to RE. Hence the output impedance can be modified to include early effect as R4/(β+1)+1/gm||RE||rO.

6. In the following circuit, VCC=5V, IS=10-14, R4=8KΩ, the current gain is 100 and RE=1KΩ. Calculate the operating point.

a) (2.3mA, 2.7V)
b) (2.8mA, 3.5V)
c) (2.3mA, 3.1V)
d) (2.3mA, 3.7V)

Explanation: To calculate the operating point, we use two sets of equations. The first equation would be KVL from the supply voltage via R4 to the ground via Re. The second equation is the exponential relationship between IC and VBE. We will have to solve both the equations by assuming a base-voltage and finding the collector current and perform iterations until the values converge. For 2 iterations, we find the collector current stays at 2.3mA. VCE becomes 3.7V and the operating point is (2.3mA, 3.7V). Note that the transistor is deep into saturation.

7. Assume that the transistor has a very high current gain and negligible early effect. What is the voltage gain from input to Q2?

a) gm2 * (R2 || 1/gm1)
b) -gm2 * (R2)
c) gm2 * (1/gm1)
d) -gm2 * (R2 || 1/gm1)

Explanation: While calculating the voltage gain from input to Q2, we only consider that connection of the emitter and collector and base of Q2. This directly leads us to recognizing that Q2 behaves as a C.E. stage and the voltage gain becomes -gm2 * the total resistance connected to the collector. This resistance is R2 in parallel to the resistance offered by the emitter of Q1 – the total output impedance is (R2 || 1/gm1). The overall voltage gain is -gm2 * (R2 || 1/gm1).

8. Assume that the transistor has a very high current gain and negligible early effect. what is the voltage gain from Q2 to Q1?

a) gm1*(rπ3 – (β+1)*RE) || R3
b) gm1*(rπ3 + (β+1)*RE) || R3
c) gm1*(rπ3 || R3)
d) gm1*(β+1)*(RE || R3)

Explanation: While considering the voltage gain from Q2 to Q1, we note that Q1 is behaving as a common gate stage. We readily calculate the output impedance of Q1 – that turns out to be R3 in parallel the resistance looking into the base of Q3. This resistance is (rπ3 + (β+1)*RE) and the overall output impedance is (rπ3 – (β+1)*RE) || R3. The overall voltage gain is gm1*(rπ3 + (β+1)*RE) || R3.

9. Assume that the transistor has a very high current gain and negligible early effect. what is the voltage gain from Q1 to Q3?

a) RE/RE + (RE/β+1) || 4/gm3
b) 1/RE + (R2/β+1) || 9/gm3
c) RE/RE + (R3/β+1) || 1/gm3
d) RE/1 + (R3/β+1) || 1/gm3

Explanation: Firstly, we identify that Q3 is behaving as a follower with a base resistance. The calculation of this base resistance is a bit tricky – noting that early effect is absent, we need to calculate the impedance offered by the collector of Q1 which is further degenerated by the output impedance of Q2 ie R2. So, we draw the small signal model of Q1.

Now, we note that r and R2 is in parallel to each other but it doesn’t really matter for the current flowing into R2 is the sum of currents coming from r and I. This current only exists if a base voltage is tied to the base of the transistor or the transistor is literally not biased. Hence, the output resistance at the collector is infinite and it appears in parallel to R3 (while calculating the impedance, we turn of all voltage sources) and the output resistance is only R3 and the base resistance is also R3. Now we can readily write the voltage gain as RE/RE + (R3/β+1) || 1/gm3.

10. Assume that the transistor has a very high current gain and negligible early effect. what is output impedance of Q2 if it gives rise to an early voltage along with Q2?

a) rO1||(1/gm1||rO1)
b) R2||(1/gm1||rO1)
c) rO1||R2||(1/gm1||rO1)
d) rO1||(1/gm1||rO1)

Explanation: As Q1 gives rise to an early voltage, we find that the impedance looking into the emitter is 1/gm1||rO1. This is connected to the collector of Q2, where the output is being sensed, which is connected to R2 & rO1 (early effect). All these resistances come in parallel while calculating the output impedance and the overall resistance is rO1||R2||(1/gm1||rO1).

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