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

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

1. The transistor, in the following circuit, has a very high current gain with negligible early effect. What is the voltage gain of the circuit?

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

Explanation: The small signal model of the transistor is used to derive the voltage gain of the follower shown above.

We note that the output voltage is a result of the total current flowing into the emitter * R1. Hence a KCL at X reveals that would give us a relation between the output voltage and v. Now we note that a KVL: VIN=v + VOUT and find that for a very high current gain, the voltage gain is approximated to R1 / R1 + 1/gm since R1 / gm * r = β.

2. The transistor, in the following circuit, has a very high current gain with negligible early effect. What is the input resistance of the circuit?

a) rπ+(β+1)*R1
b) (β+1)*R1
c) rπ+R1
d) rπ+(β-1)*R1

Explanation: The input resistance of the following circuit is similar to the input resistance of a degenerated C.E. stage. This makes the input impedance as rπ+(β+1)*R1. This leads to a direct observation that the input resistance is unaffected by where the voltage node is selected.

3. The transistor, in the following circuit, has a very high current gain with negligible early effect. What is the input resistance of the circuit?

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

Explanation: The output impedance of the above follower is equal to R1 in parallel to the resistance looking into the emitter of the follower. We note that in absence of early effect, the input impedance is simply 1/gm. The overall resistance becomes 1/gm || R1.

4. The transistor, in the following circuit, has a very high current gain with considerable early effect. What is the input resistance of the circuit?

a) 1/gm || R1 || rO
b) R1 || rO
c) 1/gm || rO
d) 1/gm || R1

Explanation: The output impedance of the above follower is equal to R1 in parallel to the resistance looking into the emitter of the follower. We note that in presence of early effect, the input impedance is 1/gm || rO . The overall resistance becomes 1/gm || R1.

5. What happens to the gain of a follower if the emitter of the transistor is connected to a current source? Assume that the transistor has a high current gain but negligible early effect.

a) Nearer to 1
b) Infinity
c) 1/gm
d) Much less than 1

Explanation: The gain of a follower stage is given by RE/RE+1/gm. Note that RE tends to infinity since the output impedance of the current source is infinity. This makes the gain reach unity and the follower stage is a perfect buffer.

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

a) R3 + rπ
b) rπ + (β + 1) * R1
c) R3 + (β + 1) * R1
d) R3 + rπ + (β + 1) * R1

Explanation: The input resistance of the above circuit is R3 in series with the input impedance looking into the base of the transistor. This resistance is equal to the input impedance of a degenerated C.E. stage which is rπ + (β + 1) * R1. The overall resistance is R3 + rπ + (β+1) * R1.

7. Assume that the transistor has a high current gain but negligible early effect. The following circuit behaves as a buffer stage. What device can be used to realize this current source?

a) A pnp transistor
b) A npn transistor
c) A photodiode
d) An oscillator

Explanation: A npn transistor behaves as a current source from a node to ground while a pnp transistor behaves as a current source from one node to another node. This can be observed from the small signal model of each transistor. A diode cannot behave as a current source and hence, the correct answer is a npn transistor.

8. Assume that the transistor has a high current gain but negligible early effect. What is the input impedance of the following circuit?

a) Infinity
b) 0
c) rπ + (β + 1) * RI
d) rπ

Explanation: While calculating the input impedance of a circuit, all current sources are opened while the voltage sources are shorted to ground. The input impedance is simply rπ which can be seen directly from the small signal model.

9. Assume that the transistor has a very high current gain with negligible early effect. What is the output impedance of the following circuit?

a) (R3 / (β + 1) + 1/gm) || R1
b) (R3 / (β + 1) + 1/gm)
c) 1/gm || R1
d) (R3 / (β + 1) || R1

Explanation: The output impedance of the follower is R1 in parallel to the impedance looking into the collector of the transistor. This impedance is (R3 / (β + 1) + 1/gm) and can be calculated with the help of the small signal model.

We note that v=-Vout*r/(r+R3) and a KCL at the emitter note readily gives us the impedance as (R3 / (β + 1) and the overall impedance is (R3 / (β + 1) + 1/gm) || R1.

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

a) R1 / R1 + (R3 / (β + 1) + 1/gm)
b) R1 / R1 + (R3 / (β + 1))
c) R1 / R1 + 1/gm
d) 1/1 + (R3 / (β + 1) + 1/gm)

Explanation: The voltage gain of follower is R1 / R1 + (the total resistance looking into the emitter of the transistor). Since the resistance looking into the emitter is (R3 / (β + 1) + 1/gm), the overall voltage gain becomes R1 / R1 + (R3 / (β+1) + 1/gm).

11. Assume that the transistor has a very high current gain with negligible early effect. For a collector current of 1.3mA, β = 99 and R1=1KΩ, what is R3 such that the voltage gain is 0.8?

a) 4515 Ω
b) 1450 Ω
c) 1150 Ω
d) 5501 Ω

Explanation: The voltage gain of a follower stage is R1/R1 + (R3 / (β + 1) + 1/gm) and gm is I/VT which makes 1/gm equal to 20. Putting the value, we get R3 as 1150 Ω.

12. Assume that the transistor has a very high current gain with negligible early effect. For a collector current of 1.3mA, β = 99 and R1=50Ω, what is R3 such that the input impedance is 8KΩ?

a) 102Ω
b) 1.5KΩ
c) .5KΩ
d) 1.02KΩ

Explanation: The input impedance is R3 + rπ + (β+1) * R1 and we calculate rπ = β/gm which is equal to 1980Ω. We put the values gives and R3 comes out to be 1.02K. Note that this circuit provides a high input impedance and the voltage drop across R1 is greater than the voltage drop across R3 which makes the circuit less sensitive to β.

13. What happens to the input resistance of a follower if the emitter area increases by a factor of 2? Assume that the transistor has a very high current gain and negligible early effect.
a) It increases by a factor of 2
b) It decreases by a factor of 2
c) It increases heavily
d) It decreases slightly

Explanation: The transconductance of the device increases which increases rπ. The current gain is assumed to stay very high since the collector current is proportional to the emitter area. The input impedance increases quite drastically and more than a factor of 2.

14. What happens to the output resistance of a follower if the emitter area decreases by a factor of 2? Assume that the transistor has a very high current gain and negligible early effect.
a) It increases by a factor of 2
b) It decreases by a factor of 4
c) It increases heavily
d) It decreases

Explanation: In absence of early effect, the output impedance of a follower is 1/gm||Re. Increasing the emitter area by a factor of 2 increases the transconductance by factor of 2. Thereby, the output impedance decreases in absence of early effect.

15. What happens to the voltage gain of a follower if the emitter area decreases by a factor of 2? Assume that the transistor has a very high current gain and negligible early effect.
a) It increases by a factor of 2
b) It decreases by a factor of 4
c) It increases heavily
d) It decreases heavily

Explanation: The increase in the emitter area increases the impedance looking into the emitter of the transistor. This leads to a decrease in the voltage gain. If we look at the expression of voltage gain, this impedance is present in the denominator.

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