# Analog Circuits Questions and Answers – Effect of Various Capacitors on Frequency Response – 1

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This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Effect of Various Capacitors on Frequency Response – 1”.

1. During high frequency applications of a B.J.T., which parasitic capacitors arise between the base and the emitter?
a) Cje and Cb
b) Ccs
c) Cb
d) Ccs and Cb

Explanation: There are two capacitors which arise between bases and emitter. One is Cje due to depletion region associated between base and emitter. Cb is another capacitor which arises due to the accumulation of electrons in the base which further results into the concentration gradient within the base of the transistor.

2. During high frequency applications of a B.J.T., which parasitic capacitors arise between the collector and the emitter?
a) No capacitor arises
b) Ccs
c) Cb
d) Ccs and Cb

Explanation: The emitter and the collector are far away from each other when the B.J.T. is being constructed. Hence, we find that they don’t share a common junction where charges can accumulate. Thus, no such parasitic capacitors appear.

3. During high frequency applications of a B.J.T, which parasitic capacitors arise between the collector and the base?
a) Cje and Cb
b) Ccs
c) Cπ
d) Cµ

Explanation: Only one capacitor up between the base and the collector. This is due to the depletion region present between the base and the collector region.

4. Which parasitic capacitors are present at the collector terminal of the B.J.T.?
a) Cje and Cb
b) Ccs and Cµ
c) Cb
d) Ccs and Cb

Explanation: There are two capacitors attached to the collector terminal. The collector-base junction provides a depletion capacitance (Cµ) while the collector substrate junction provides a certain capacitance (Ccs).

5. Which parasitic capacitors do not affect the frequency response of the C.E. stage, of the B.J.T.?
a) Cje and Cb
b) Ccs and Cµ
c) Cb and Cµ
d) No parasitic capacitor gets deactivated

Explanation: While observing the frequency response of a C.E. stage, we find that all the parasitic capacitances of the B.J.T. end up slowing the speed of the B.J.T. The frequency response of this stage is affected by all the parasitic capacitors.

6. Which parasitic capacitors don’t affect the frequency response of the C.B. stage of the B.J.T.?
a) None of the parasitic capacitances
b) All the parasitic capacitances
c) Some of the coupling capacitors
d) Ccs and Cb

Explanation: All the parasitic capacitors of a B.J.T. affect the C.B. stage. None of the parasitic capacitors gets deactivated and they end up behaving as a pole during the frequency response of the C.B. stage.

7. Which parasitic capacitors don’t affect the frequency response of the C.C. stage of the B.J.T.?
a) Ccs
b) Ccs and Cb
c) Cb
d) Ccs and Cµ

Explanation: In the follower stage, the load is present at the emitter. The parasitic capacitors present between the collector and the substrate i.e. Cµ gets deactivated. This is observed from the small signal analysis where both the terminals of this capacitor get shorted to A.C. ground.

8. If the transconductance of the B.J.T increases, the transit frequency ______
a) Increases
b) Decreases
c) Doesn’t get affected
d) Doubles

Explanation: The transit frequency is directly proportional to the transconductance of the B.J.T. Hence, the correct option is increases. Since it hasn’t been mentioned that whether the transconductance has been doubled or not, we cannot conclude the option “doubles” as an answer.

9. If the total capacitance between the base and the emitter increases by a factor of 2, the transit frequency __________
a) reduces by 2
b) increases by 2
c) reduces by 4
d) increases by 4

Explanation: The transit frequency is almost inversely proportional to the total capacitance between the base and the emitter of the B.J.T. Hence, the transit frequency will approximately reduce by 2 and the correct option becomes reduces by 2.

10. Which effect plays a critical role in producing changes in the frequency response of the B.J.T.?
a) Thevenin’s effect
b) Miller effect
c) Tellegen’s effect
d) Norton’s effect

Explanation: The miller effect results in a change in the capacitance seen between the base and the collector. This is why, it affects the frequency response of the B.J.T. deeply by changing the poles and affecting the high frequency voltage gain stage.

11. If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the input side gets multiplied?
a) 1 + gmRl
b) 1 – gmRl
c) 1 + 2*gmRl
d) 1 – 2*gmRl

Explanation: The low frequency gain of the C.E. stage is gmRl. By the application of miller effect, we find that the capacitor between the base and the collector, looking into the input of the C.E. stage, will be increased by a factor of 1 + gmRl.

12. If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the output side gets multiplied?
a) 1 + 1/gmRl
b) 1 – 1/gmRl
c) 1 + 2/gmRl
d) 1 – 2/gmRl

Explanation: The low frequency response of the C.E. stage is gmRl. By the application of miller effect, we find that the capacitance between the base and the collector, looking from the output side, will be increased by a factor of 1 + 1/gmRl. Hence, the correct option is 1 + 1/ gmRl.

13. If a C.E. stage with early effect has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the output side, gets multiplied?
a) 1 + 2/gm*(Rl || ro)
b) 1 – 1/gm*(Rl || ro)
c) 1 + 1/gm*(Rl || ro)
d) 1 – 2/gm*(Rl || ro)

Explanation: If the early effect is considered, the low frequency response of the C.E. stage becomes gm*(Rl || ro). Thereby, miller approximation shows that the capacitance between the base and the collector, looking from the output side, will be increased by a factor of 1 + 1/gm*(Rl || ro). Hence the correct option is 1 + 1/ gm*(Rl || ro).

14. For a high frequency response of a simple C.E. stage with a transconductance of gm, what is Cin?
a) Cµ(1 + gm*R2) – Cπ
b) Cµ(1 + gm*R2) + Cπ
c) Cµ(1 – 2*gm*R2) + Cπ
d) Cµ(1 + 2*gm*R2) – Cπ

Explanation: The input capacitance is an equivalent of the base to emitter capacitance in parallel to the miller approximation of the base to collector capacitance. Due to miller approximation, the base to collector capacitance becomes Cµ(1+gm*R2) while the base to emitter capacitance is Cπ. Capacitors get added, when in parallel and thus Cµ(1+gm*R2) + Cπ is correct.

15. For a high frequency response of a simple C.E. stage with a transconductance of gm, what is Cout?
a) Ccs – Cµ*(2 + 1/gm*R2)
b) Ccs + Cµ*(1 + 2/gm*R2)
c) Ccs – Cµ*(1 + 1/gm*R2)
d) Ccs + Cµ*(1 + 1/gm*R2) 