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
View Answer

Answer: a
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.
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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
View Answer

Answer: a
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
View Answer

Answer: d
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
View Answer

Answer: b
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
View Answer

Answer: d
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.
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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
View Answer

Answer: b
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
View Answer

Answer: a
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
View Answer

Answer: a
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
View Answer

Answer: a
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.
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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
View Answer

Answer: a
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
View Answer

Answer: a
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
View Answer

Answer: a
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)
View Answer

Answer: c
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).
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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
View Answer

Answer: b
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)
View Answer

Answer: d
Explanation: We have a capacitor from the collector to substrate, Ccs, which comes in parallel to the miller approximation of the capacitance from base to collector. The miller approximation defines the latter as Cµ*(1 + 1/gm*R2). Since capacitors gets added, when in parallel, the correct option is Ccs + Cµ*(1+ 1/gm*R2).

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn