# Analog Circuits Questions and Answers – Effect of Various Capacitors on Frequency Response – 2

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This set of Analog Circuits Mcqs focuses on “Effect of Various Capacitors on Frequency Response – 2”.

1. Ignoring early effect, if R1 is the total resistance connected to the base and R2 is the total resistance connected at the collector, what could be the approximate input pole of a simple C.E. stage?
a) 1 / [R1 * (Cµ(2+gm*R2) + Cπ)]
b) 1 / [R1 * (Cµ(1+2*gm*R2) + Cπ)]
c) 1 / [R1 * (Cµ(1+gm*R2) + Cπ)]
d) 1 / [R1 * (Cµ(1-gm*R2) + Cπ)]

Explanation: The input pole can be approximately calculated by observing the input node. The input node is the node where the base of the B.J.T. is connected to the input voltage. The product of total resistance and capacitance connected at that particular node is R1 * Cin and Cin is Cµ(1+gm*R2) + Cπ- the inverse of this product gives us the input pole. Thus the correct option is 1 / [R1 * (Cµ(1+gm*R2) + Cπ)].

2. Ignoring early effect, if R2 is the total resistance at the collector, what could be the approximate output pole of a simple C.E. stage?
a) 1 / [R2 * (Ccs + Cµ*(1 + 2/gm*R2))]
b) 1 / [R2 * (Ccs – Cµ*(1 + 1/gm*R2))]
c) 1 / [R2 * (Ccs + Cµ*(1 – 1/gm*R2))]
d) 1 / [R2 * (Ccs + Cµ*(1 + 1/gm*R2))]

Explanation: The output pole can be approximately calculated by observing the output node. For a C.E. stage, the output node is the node where the Collector of the B.J.T. is connected to the output measuring device. The product of total resistance and capacitance connected at that particular node is R2 * Cout and Cout is (Ccs + Cµ*(1 + 1/gm*R2). The inverse of this product gives us the output pole. Thus the correct option is 1 / [R2 * (Ccs + Cµ*(1 + 1/gm*R2))].

3. If the load resistance of a C.E. stage increases by a factor of 2, what happens to the high frequency response?
a) The 3 db roll off occurs faster
b) The 3 db roll off occurs later
c) The input pole shifts towards origin
d) The input pole becomes infinite

Explanation: If the load resistance increases by a factor of 2, the output pole decreases since it’s inversely proportional to the load resistance. Hence the C.E. stage experiences a faster roll off due to the pole.

4. During high frequency applications of a B.J.T., which of the following three stages do not get affected by Miller’s approximation?
a) C.E.
b) C.B.
c) C.C.
d) Follower

Explanation: During the C.B. stage, the capacitance between the base and the collector doesn’t suffer from Miller approximation since the input is applied to the emitter of the B.J.T. There are no capacitors connected between two nodes having a constant gain. Hence the C.B. stage doesn’t get affected by miller approximation.

5. Ignoring early effect, if C1 is the total capacitance tied to the emitter, what is the input pole of a simple C.B. stage?
a) 1/gm * C1
b) 2/gm * C1
c) gm * C1
d) gm * 2C1

Explanation: The resistance looking into the emitter of the B.J.T. is 1/gm. The capacitance connected to the input node is C1 (as mentioned). The inverse product of these two provides us the input pole of the C.B. stage.

6. Ignoring early effect, if R1 is the total resistance connected to the collector; what is the output pole of a simple C.B. stage?
a) 1/[R1 * (Ccs + Cµ)]
b) 1/[R1* (Ccs + 2*Cµ)]
c) 1/[R1 * (2*Ccs + Cµ)]
d) 1/[R1 * 2*(Ccs + Cµ)]

Explanation: The output pole is calculated, approximately, by the inverse product of the total resistance and the capacitance connected at the output node. We find that the total resistance connected to the output node is R1 while the total capacitance is Ccs + Cµ. In absence of early effect, 1/[R1 * (Ccs + Cµ)] becomes the output pole.

7. If early effect is included, and R1 is the total resistance connected at the collector. What is the output pole of a simple C.B. stage?
a) 1/[(R1 || ro) * 2(Ccs + Cµ)]
b) 1/[(R1 || ro) * (Ccs + Cµ)]
c) 1/[(R1 || ro) * (2*Ccs + Cµ)]
d) 1/[(R1 || ro) * 2*(Ccs + 2*Cµ)]

Explanation: The output pole is calculated, approximately, by the inverse product of the total resistance and the capacitance connected at the output node. We find that the total resistance connected to the output node is R1 in parallel with ro, due to early effect, while the total capacitance is C2 ie Ccs + Cµ. Thus, the correct option is 1/[(R1 || ro) * (Ccs + Cµ)].

8. In a simple follower stage, C2 is a parasitic capacitance arising due to the depletion region between the collector and the substrate. What is the value of C2?
a) 0
b) Infinite
c) Ccs
d) 2*Ccs

Explanation: During the high frequency response, the capacitor between the collector and the substrate gets shorted to A.C. ground at both of its terminals. Hence, C2=0. The answer would have been Ccs for any other stage of B.J.T.

9. For a cascode stage, with input applied to the C.B. stage, the input capacitance gets multiplied by a factor of ____
a) 0
b) 1
c) 3
d) 2

Explanation: The small signal gain, of the C.B. stage, in a cascode stage is approximately equal to the ratio of the transconductances of the two B.J.T.’s. Since they are roughly same, the gain is 1. Miller multiplication leads to multiplying the capacitance, between base and collector, by a factor of (1 + small signal gain) which is 2. Hence, the correct option is 2.

10. If the B.J.T. is used as a follower, which capacitor experiences Miller multiplication?
a) Cπ
b) Cµ
c) Ccs
d) Cb

Explanation: We find that the input is given to the base of the B.J.T. while the output is sensed at the collector of the B.J.T. We observe that the only capacitance connected between two nodes- where there is an amplification unit between the nodes, is Cπ. Hence, the correct option is Cπ.

11. If 1/h12 = 10 for a C.E. stage, what is the value of the base to collector capacitance, after Miller multiplication, at the output side?
a) 1.1Cµ
b) 1.2Cµ
c) 2.1Cµ
d) 2.2Cµ

Explanation: At the output side of a C.E. stage, Cµ gets multiplied by a factor of (1+1/Av) where Av is the voltage gain. 1/h12 is nothing but Av. Hence, the value changes to 1.1Cµ.

12. If 1/h12 = 4, for a C.E. stage- what is the value of the base to collector capacitance, after Miller multiplication, at the input side?
a) 4Cµ
b) 5Cµ
c) 6Cµ
d) 1.1Cµ

Explanation: The capacitor, Cµ, gets multiplied by a factor of (1 + Av), at the input side of a C.E. stage. 1/h12 is equal to Av since h12 is the reverse voltage amplification factor. Hence, the final value becomes 5Cµ.

13. The transconductance of a B.J.T.is 5mS (gm) while a 2KΩ (Rl) load resistance is connected to the C.E. stage. Neglecting the Early effect, what is the Miller multiplication factor for the input side?
a) 21
b) 11
c) 20
d) 0

Explanation: The Miller multiplication factor for the input side of a C.E. stage is (1+Av). Now, Av is the small signal low frequency gain of the C.E. stage which is gm*RL=10. Hence, the Miller multiplication factor is 11. 