This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Bipolar Cascode”.

1. What is the input impedance of the following circuit?

a) r_{π1}

b) r_{π2} +r_{π1}

c) r_{π2}

d) Infinite

View Answer

Explanation: The input impedance of the circuit is simply r

_{π1}. Q

_{2}doesn’t impact on the impedance and only the impedance looking into the base of Q

_{1}is the input impedance.

2. What is the input impedance of Q_{2}?

a) r_{π1} + (β+1) * r_{o2}

b) r_{π2} + (β+1) * r_{o1}

c) Infinite

d) 0

View Answer

Explanation: Q

_{2}is degenerated by the output impedance of Q

_{1}, looking into its collector, i.e. r

_{o1}. Henceforth, the input impedance becomes r

_{π2}+ (β+1) * r

_{o1}. This is directly derivable from the expression of input impedance of a degenerated B.J.T.

3. What is the impedance looking into the collector of Q_{1} in the following circuit?

a) r_{o2}

b) r_{o1}+r_{o2}

c) r_{o1}

d) r_{o1}+r_{π1}

View Answer

Explanation: The impedance looking into the collector of Q

_{1}is independent of the rest of the circuit. This can be verified by a small signal analysis and hence, the impedance is r

_{o1}.

4. What is the impedance looking into the collector of Q_{2} in the following circuit?

a) (r_{o2}+R_{1})*r_{π2}/(r_{π1}+(β+1)*r_{O2}+R_{1})

b) (r_{o1}+R_{1})*r_{π1}/(r_{π2}+(β+1)*r_{O1}+R_{1})

c) (r_{o1}+R_{1})*r_{π2}/(r_{π1}+(β+1)*r_{O2}+R_{1})

d) (r_{o2}+R_{1})*r_{π2}/(r_{π2}+(β+1)*r_{O2}+R_{1})

View Answer

Explanation: We draw the small signal model of Q

_{2}and readily note that V

_{X}=-V

_{π}and apply the Thevenin’s procedure of finding the resistance. Now, we apply a KCL at the collector node and find the current through r

_{o}. This current is g

_{m}*v

_{π}+(I

_{X}-V

_{X}/r

_{π}). Now, the current through R

_{C}is I

_{X}-V

_{X}/r

_{π}. Finally, we apply a KVL from V

_{IN}, though r

_{o}and R

_{1}to ground. After replacing the current and rearranging carefully, we reach the expression of impedance as (r

_{o2}+R

_{1})*r

_{π2}/(r

_{π2}+(β+1)*r

_{o2}+R

_{1}). Note that we have calculated the input impedance of a CG stage with early effect.

5. What is the impedance at node A if all the transistors experience early effect?

a) (r_{o2}+R_{1})*r_{π2}/(r_{π1}+(β+1)*r_{O1}+R_{1})||r_{o1}

b) (r_{o2}+R_{1})*r_{π2}/(r_{π2}+(β+1)*r_{O2}+R_{1})||r_{o1}

c) (r_{o1}+R_{1})*r_{π1}/(r_{π2}+(β+1)*r_{O1}+R_{1})||r_{o2}

d) (r_{o1}+R_{1})*r_{π1}/(r_{π2}+(β+1)*r_{O2}+R_{1})||r_{o2}

View Answer

Explanation: At node A, we have two branches. One branch is connected to the emitter of Q

_{2}which gives an impedance as (r

_{o2}+R

_{1})*r

_{π2}/(r

_{π2}+(β+1)*r

_{O2}+R

_{1}). The other branch is connected to the collector of Q

_{1}which gives a collector impedance of r

_{o2}. When calculating the impedance, we turn off all the independent voltage sources and apply a voltage at node A. Henceforth, we find that the resistances offered by these two branches appear in parallel and the overall impedance becomes (r

_{o2}+R

_{1})*r

_{π2}/(r

_{π2}+(β+1)*r

_{O2}+R

_{1})||r

_{o1}.

6. What is the output impedance of the following circuit?

a) (1+g_{m1}*r_{o1})(r_{o2}||r_{π2})+r_{o1}||R_{1}

b) (1+g_{m12}*r_{o1})(r_{o2}||r_{π1})+r_{o1}||R_{1}

c) (1+g_{m1}*r_{o2})(r_{o1}||r_{π2})+r_{o2}||R_{1}

d) (1+g_{m2}*r_{o2})(r_{o2}||r_{π1})+r_{o2}||R_{1}

View Answer

Explanation: The output impedance of this circuit can be calculated by using the small signal analysis. We find that Q

_{2}is degenerated by the output impedance of Q

_{1}– the impedance looking into the collector of Q

_{1}becomes (1+g

_{m2}*r

_{o2})(r

_{o2}||r

_{π1})+r

_{o2}. This impedance is parallel to R

_{1}and the overall impedance becomes (1+g

_{m2}*r

_{o2})(r

_{o2}||r

_{π1})+r

_{o2}||R

_{1}.

7. What is the transconductance of Q_{1}?

a) g_{m1}

b) g_{m2}

c) g_{m1}+g_{m2}

d) 0

View Answer

Explanation: The transconductance of Q

_{1}is only g

_{m1}. Given the transistor is not degenerated, the transconductance remains g

_{m1}.

8. What is the transconductance of Q_{2}?

a) g_{m1}

b) g_{m2}/1+g_{m2}*(r_{o1})

c) g_{m1}+g_{m2}

d) g_{m1}/1+(1/r_{π}+g_{m1}*(r_{o1}))

View Answer

Explanation: The transconductance of a transistor reduces after it gets degenerated. The new transistor is g

_{m1}/1+(1/r

_{π}+g

_{m1}*(degenerating resistance)). Therefore, for Q

_{2}, the new transconductance is g

_{m1}/1+(1/r

_{π}+g

_{m1}*(r

_{o1})). Note that option b is valid if the current gain is very high. Since it hasn’t been mentioned, we cannot take it as the correct answer.

9. What is the voltage gain from V_{IN} to node A in presence of Early Effect in all the transistors?

a) -g_{m1}*(r_{o2}+R_{1})*r_{π2}/(r_{π1}+(β+1)*r_{O1}+R_{1})||r_{o2}

b) -g_{m2}*(r_{o1}+R_{1})*r_{π1}/(r_{π2}+(β+1)*r_{O1}+R_{1})||r_{o2}

c) -g_{m1}*(r_{o2}+R_{1})*r_{π2}/(r_{π2}+(β+1)*r_{O2}+R_{1})||r_{o1}

d) -g_{m2}*(r_{o1}+R_{1})*r_{π1}/(r_{π2}+(β+1)*r_{O2}+R_{1})||r_{o1}

View Answer

Explanation: The voltage gain from V

_{IN}to node A is due to Q

_{1}behaving as a CS stage. The voltage gain of a CS stage is given by –(transconductance * (output impedance)). This impedance is given by the output impedance at node A. The output impedance is calculated as (r

_{o2}+R

_{1})*r

_{π2}/(r

_{π2}+(β+1)*r

_{O2}+R

_{1})||r

_{o1}. The overall gain is -g

_{m1}*(r

_{o2}+R

_{1})*r

_{π2}/(r

_{π2}+(β+1)*r

_{O2}+R

_{1})||r

_{o1}.

10. What is the voltage gain node A to V_{OUT} in presence of early effect?

a) (1/r_{o1}+g_{m1})(R_{1}||r_{o1})

b) (1/r_{o2}+g_{m2})(R_{1}||r_{o2})

c) (1/r_{o1}+g_{m2})(R_{1}||r_{o1})

d) (1/r_{o2}+g_{m1})(R_{1}||r_{o2})

View Answer

Explanation: We draw the small signal model of Q

_{2}& neglect Q

_{1}. Now, we calculate the current through r

_{o2}as V

_{IN}-V

_{OUT}/r

_{o2}. Now, we write a KCL at the collector node and carefully rearrange the terms to get the voltage gain as (1/r

_{o2}+g

_{m2})(R

_{1}||r

_{o2}).

11. What is the voltage gain for the following circuit in presence of early effect?

a) -g_{m1}*(r_{o1}+R_{1})*r_{π1}/(r_{π1}+(β+1)*r_{O1}+R_{1})||r_{o1}*(1/r_{o1}+g_{m2})(R_{1}||r_{o1})

b) -g_{m1}*(r_{o2}+R_{1})*r_{π1}/(r_{π2}+(β+1)*r_{O2}+R_{1})||r_{o2}*(1/r_{o2}+g_{m1})(R_{1}||r_{o1})

c) -g_{m1}*(r_{o1}+R_{1})*r_{π2}/(r_{π1}+(β+1)*r_{O1}+R_{1})||r_{o2}*(1/r_{o1}+g_{m1})(R_{1}||r_{o2})

d) -g_{m1}*(r_{o2}+R_{1})*r_{π2}/(r_{π2}+(β+1)*r_{O2}+R_{1})||r_{o1}*(1/r_{o2}+g_{m2})(R_{1}||r_{o2})

View Answer

Explanation: To calculate the voltage gain for the following circuit, we observe that V

_{OUT}/V

_{IN}=V

_{OUT}/V

_{A}*V

_{A}/V

_{IN}. Hence, we find that the overall voltage gain is the product of two voltage gain – one for Q

_{1}behaving as a CS stage and one for Q

_{2}behaving as a CG stage. For the CS stage, the gain is g

_{m1}*(r

_{o2}+R

_{1})*r

_{π2}/(r

_{π2}+(β+1)*r

_{O2}+R

_{1})||r

_{o1}while that of the CG stage is r

_{o1}*(1/r

_{o2}+g

_{m2})(R

_{1}||r

_{o2}). The overall voltage gain is the product of these two gains i.e. -g

_{m1}*(r

_{o2}+R

_{1})*r

_{π2}/(r

_{π2}+(β+1)*r

_{O2}+R

_{1})||r

_{o1}*(1/r

_{o2}+g

_{m2})(R

_{1}||r

_{o2}).

12. What is the role of Q_{2} in the following circuit?

a) Cascode Device

b) Degenerating Device

c) Follower

d) Driver

View Answer

Explanation: Q

_{2}is regarded as a Cascode Device. Q

_{2}is regarded as the degenerating device. None of the devices behave as a follower or driver since Q

_{2}behaves as a CG stage while Q

_{1}behaves as a CS stage.

13. Suppose that a parasitic resistance appears in the emitter of Q_{2}. What happens to the voltage gain of the following circuit?

a) 1/(1/((r_{o1}+1/g_{m1})(R_{1}||r_{o1}))+r_{p}/R_{1}

b) 1/(1/((r_{o2}+1/g_{m1})(R_{1}||r_{o1}))+r_{p}/R_{1}

c) 1/(1/((r_{o2}+1/g_{m2})(R_{1}||r_{o2}))+r_{p}/R_{1}

d) 1/(1/((r_{o2}+1/g_{m1})(R_{1}||r_{o2}))+r_{p}/R_{1}

View Answer

Explanation: We draw the small signal model of Q

_{2}& neglect Q

_{1}. Firstly, we perform a KVL from V

_{X}to ground via r

_{p}& r

_{π}. Next, we note that the current leaving the emitter node and goes toward the collector is V

_{O}/R

_{1}. Hence, we can now define V

_{π}in terms of V

_{in}& V

_{out}. Next, we perform a nodal analysis at the emitter node. Now, we note that the voltage gain from the emitter to V

_{OUT}is (r

_{o2}+1/g

_{m2})(R

_{1}||r

_{o2}). Henceforth, we replace the nodal voltage in terms of V

_{OUT}& we can also replace V

_{π}. After carefully rearranging the terms, we will get the voltage gain as 1/ (1/((r

_{o2}+1/g

_{m2})(R

_{1}||r

_{o2}))+r

_{p}/R

_{1}.

**Sanfoundry Global Education & Learning Series – Microelectronics**.

To practice all areas of Microelectronics, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.