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

1. Ignoring early effect, if R_{1} is the total resistance connected to the base and R_{2} is the total resistance connected at the collector, what could be the approximate input pole of a simple C.E. stage?

a) 1 / [R_{1} * (C_{µ}(2+g_{m}*R_{2}) + C_{π})]

b) 1 / [R_{1} * (C_{µ}(1+2*g_{m}*R_{2}) + C_{π})]

c) 1 / [R_{1} * (C_{µ}(1+g_{m}*R_{2}) + C_{π})]

d) 1 / [R_{1} * (C_{µ}(1-g_{m}*R_{2}) + C_{π})]

View Answer

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 R

_{1}* C

_{in}and C

_{in}is C

_{µ}(1+g

_{m}*R

_{2}) + C

_{π-}the inverse of this product gives us the input pole. Thus the correct option is 1 / [R

_{1}* (C

_{µ}(1+g

_{m}*R

_{2}) + C

_{π})].

2. Ignoring early effect, if R_{2} is the total resistance at the collector, what could be the approximate output pole of a simple C.E. stage?

a) 1 / [R_{2} * (C_{cs} + C_{µ}*(1 + 2/g_{m}*R_{2}))]

b) 1 / [R_{2} * (C_{cs} – C_{µ}*(1 + 1/g_{m}*R_{2}))]

c) 1 / [R_{2} * (C_{cs} + C_{µ}*(1 – 1/g_{m}*R_{2}))]

d) 1 / [R_{2} * (C_{cs} + C_{µ}*(1 + 1/g_{m}*R_{2}))]

View Answer

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 R

_{2}* C

_{out}and C

_{out}is (C

_{cs}+ C

_{µ}*(1 + 1/g

_{m}*R

_{2}). The inverse of this product gives us the output pole. Thus the correct option is 1 / [R

_{2}* (C

_{cs}+ C

_{µ}*(1 + 1/g

_{m}*R

_{2}))].

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

View Answer

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

View Answer

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 C_{1} is the total capacitance tied to the emitter, what is the input pole of a simple C.B. stage?

a) 1/g_{m} * C_{1}

b) 2/g_{m} * C_{1}

c) g_{m} * C_{1}

d) g_{m} * 2C_{1}

View Answer

Explanation: The resistance looking into the emitter of the B.J.T. is 1/g

_{m}. The capacitance connected to the input node is C

_{1}(as mentioned). The inverse product of these two provides us the input pole of the C.B. stage.

6. Ignoring early effect, if R_{1} is the total resistance connected to the collector; what is the output pole of a simple C.B. stage?

a) 1/[R_{1} * (C_{cs} + C_{µ})]

b) 1/[R_{1}* (C_{cs} + 2*C_{µ})]

c) 1/[R_{1} * (2*C_{cs} + C_{µ})]

d) 1/[R_{1} * 2*(C_{cs} + C_{µ})]

View Answer

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 R

_{1}while the total capacitance is C

_{cs}+ C

_{µ}. In absence of early effect, 1/[R

_{1}* (C

_{cs}+ C

_{µ})] becomes the output pole.

7. If early effect is included, and R_{1} is the total resistance connected at the collector. What is the output pole of a simple C.B. stage?

a) 1/[(R_{1} || ro) * 2(C_{cs} + C_{µ})]

b) 1/[(R_{1} || ro) * (C_{cs} + C_{µ})]

c) 1/[(R_{1} || ro) * (2*C_{cs} + C_{µ})]

d) 1/[(R_{1} || ro) * 2*(C_{cs} + 2*C_{µ})]

View Answer

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 R

_{1}in parallel with ro, due to early effect, while the total capacitance is C

_{2}ie C

_{cs}+ C

_{µ}. Thus, the correct option is 1/[(R

_{1}|| ro) * (C

_{cs}+ C

_{µ})].

8. In a simple follower stage, C_{2} is a parasitic capacitance arising due to the depletion region between the collector and the substrate. What is the value of C_{2}?

a) 0

b) Infinite

c) C_{cs}

d) 2*C_{cs}

View Answer

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, C

_{2}=0. The answer would have been C

_{cs}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

View Answer

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) C_{π}

b) C_{µ}

c) C_{cs}

d) C_{b}

View Answer

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 C

_{π}. Hence, the correct option is C

_{π}.

11. If 1/h_{12} = 10 for a C.E. stage, what is the value of the base to collector capacitance, after Miller multiplication, at the output side?

a) 1.1C_{µ}

b) 1.2C_{µ}

c) 2.1C_{µ}

d) 2.2C_{µ}

View Answer

Explanation: At the output side of a C.E. stage, C

_{µ}gets multiplied by a factor of (1+1/A

_{v}) where A

_{v}is the voltage gain. 1/h

_{12}is nothing but A

_{v}. Hence, the value changes to 1.1C

_{µ}.

12. If 1/h_{12} = 4, for a C.E. stage- what is the value of the base to collector capacitance, after Miller multiplication, at the input side?

a) 4C_{µ}

b) 5C_{µ}

c) 6C_{µ}

d) 1.1C_{µ}

View Answer

Explanation: The capacitor, C

_{µ}, gets multiplied by a factor of (1 + A

_{v}), at the input side of a C.E. stage. 1/h

_{12}is equal to A

_{v}since h

_{12}is the reverse voltage amplification factor. Hence, the final value becomes 5C

_{µ}.

13. The transconductance of a B.J.T.is 5mS (g_{m}) while a 2KΩ (R_{l}) 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

View Answer

Explanation: The Miller multiplication factor for the input side of a C.E. stage is (1+A

_{v}). Now, Av is the small signal low frequency gain of the C.E. stage which is g

_{m}*R

_{L}=10. Hence, the Miller multiplication factor is 11.

**Sanfoundry Global Education & Learning Series – Analog Circuits.**

To practice MCQs on all areas of Analog Circuits, __here is complete set of 1000+ Multiple Choice Questions and Answers__.

**If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]**

**Related Posts:**

- Apply for Electrical & Electronics Engineering Internship
- Apply for Electrical Engineering Internship
- Practice Electrical & Electronics Engineering MCQs
- Practice Electrical Engineering MCQs
- Check Electrical & Electronics Engineering Books