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

1. The Cascode Stage is a Cascade of ______________

a) CS & CD stage

b) CG & CD stage

c) CS & CG stage

d) Driver and follower stage

View Answer

Explanation: The Cascode stage is a cascade of a CS & CG stage. The input is applied to the CS stage while the CG stage behaves more like a current source. The other cascaded stages aren’t referred to as Cascode stages.

2. Which MOSFET is behaving as the degenerating device?

a) M_{1}

b) M_{2}

c) Neither of the MOSFETs

d) Both the MOSFETs

View Answer

Explanation: M

_{2}degenerates the output impedance of M

_{2}. This output impedance is seen from the drain terminal.

3. Which MOSFET is behaving as the Cascode device?

a) M_{1}

b) M_{2}

c) Neither of the MOSFETs

d) Both the MOSFETs

View Answer

Explanation: M

_{2}is behaving as the Cascode device. It is being used as a current source to increase the output impedance of M

_{1}.

4. In absence of M_{2}, what is the minimum output voltage swing in the following circuit?

a) One overdrive voltage

b) Twice the overdrive voltage

c) Thrice the overdrive voltage

d) Four times the overdrive voltage

View Answer

Explanation: In absence of M

_{1}, the minimum voltage swing can be one overdrive voltage. This is because the input voltage has to be less than the output voltage by the threshold voltage. Hence, the output voltage faces a restriction of one overdrive voltage.

5. In absence of M_{2}, what is the minimum output voltage swing in the following circuit?

a) One overdrive voltage

b) Twice the overdrive voltage

c) Thrice the overdrive voltage

d) Four times the overdrive voltage

View Answer

Explanation: Firstly, the output voltage provides an amplified version of the input voltage at saturation. If the MOSFET has to operate in the saturation region, the drain voltage i.e. the output voltage has to more than the gate voltage by the threshold voltage. We have two MOSFETs and hence, the drain voltage of M

_{2}has to be greater than its V

_{GS2}by one threshold voltage. Now, V

_{S2}=V

_{D1}and V

_{D1}itself has to be more than its V

_{GS1}i.e. the input voltage by one threshold voltage. Hence, the minimum output voltage swing becomes two threshold voltage.

6. What is the output resistance at node B if both the MOSFETs suffer from channel length modulation?

a) (r_{o1}||1/g_{m2}||r_{o2})

b) (R_{3}+r_{o2})/(1+(g_{m2}*r_{o2})

c) (R_{3}+r_{o2})/(1+(g_{m2}*r_{o2})||r_{o1}

d) (r_{o1}||1/g_{m2}||r_{o3})

View Answer

Explanation: Two branches are present at node B. One connect to the source of M

_{2}which offers an impedance of (R

_{3}+r

_{o2})/(1+(g

_{m2}*r

_{o2})). The other connects to the drain of M

_{1}which offers an impedance of r

_{o1}. Hence, if we measure the output impedance at B, we follow the procedure of Thevenin and find that these two impedances will be in parallel to each other. The overall impedance becomes (R

_{3}+r

_{o2})/(1+(g

_{m2}*r

_{o2})||r

_{o1}.

7. What is the output resistance at node B if both the MOSFETs suffer from channel length modulation and M_{2} suffers from Body Effect?

a) (R_{3}+r_{o1})/(1+(g_{m2}-g_{mb2})*r_{o1})||r_{o2}

b) (R_{1}+r_{o2})/(1+(g_{mb2}*r_{o2}))||r_{o1}

c) (R_{2}+r_{o2})/(1+(g_{m2}+g_{mb2})*r_{o2})||r_{o1}

d) (R_{3}+r_{o2})/(1+(g_{m2}+g_{mb2})*r_{o2})||r_{o1}

View Answer

Explanation: Two branches are present at node B. One connect to the source of M

_{2}which offers an impedance of (R

_{3}+r

_{o2})/(1+(g

_{m2}+g

_{mb2})*r

_{o2}). The other connects to the drain of M

_{1}which offers an impedance of r

_{o1}. Hence, if we measure the output impedance at B, we follow the procedure of Thevenin and find that these two impedances will be in parallel to each other. The overall impedance becomes (R

_{3}+r

_{o2})/(1+(g

_{m2}+g

_{mb2})*r

_{o2})||r

_{o1}.

8. What is the output resistance at node A if both the MOSFETs suffer from channel length modulation?

a) R_{3}||[1+(g_{m2}+g_{mb2})*r_{o2}]*r_{o1}+r_{o2}

b) R_{3}||[1+g_{m1}*r_{o2}]*r_{o2}+r_{o1}

c) [1+g_{m2}*r_{o1}]*r_{o1}+r_{o2}

d) R_{3}||[1+(g_{mb2}*r_{o2})]*r_{o1}+r_{o1}

View Answer

Explanation: The output impedance at node A will be R

_{3}in parallel to the output impedance looking into the drain of M

_{2}. M

_{2}is degenerated by M

_{1}and the impedance looking into its drain terminal becomes [1+g

_{m2}*r

_{o2}]*r

_{o1}+r

_{o2}. The overall resistance is R

_{3}||[1+g

_{m2}*r

_{o2}]*r

_{o1}+r

_{o2}.

9. What is the output resistance at node A if both the MOSFETs suffer from channel length modulation?

a) R_{3}||[1+g_{m2}*r_{o2}]*r_{o1}+r_{o2}

b) R_{3}||[1+g_{m1}*r_{o2}]*r_{o2}+r_{o1}

c) [1+g_{m2}*r_{o1}]*r_{o1}+r_{o2}

d) R_{3}||[1+(g_{m1}+g_{mb2})*r_{o1}]*r_{o2}+r_{o1}

View Answer

Explanation: The output impedance at node A will be R

_{3}in parallel to the output impedance looking into the drain of M

_{2}. M

_{2}is degenerated by M

_{1}and the impedance looking into its drain terminal becomes [1+(g

_{m1}+g

_{mb2})*r

_{o2}]*r

_{o1}+r

_{o2}. The overall resistance is R

_{3}||[1+(g

_{m1}+g

_{mb2})*r

_{o2}]*r

_{o1}+r

_{o2}.

10. What is the ratio of Voltage at node A to the input voltage in the Cascode stage?

a) -g_{m2}*(R_{3}||r_{o2}||(R_{3}+r_{o2})/(1+(g_{mb2}*r_{o2}))

b) -g_{m1}*(R_{4}||r_{o1}||(R_{3}+r_{o2})/(1+(g_{m}*r_{o2}))

c) -g_{m1}*(R_{3}||r_{o2}||(R_{1}+r_{o2})/(1+(g_{mb2}*r_{o2}))

d) -g_{m2}*(R_{4}||r_{o1}||(R_{1}+r_{o2})/(1+(g_{mb2}*r_{o2}))

View Answer

Explanation: The ratio of the voltage at node B to the input voltage is the voltage gain of M

_{1}being used as a CS stage. Hence, we can perform a small signal analysis and observe that the voltage gain is -g

_{m1}*the total impedance connected to its drain. This will be the output impedance at the node B. Node B has 3 branches. One branch connects to the source of M

_{2}which offers an impedance of (R

_{1}+r

_{o2})/(1+(g

_{mb2}*r

_{o2}) while another branch connects to the drain of M

_{1}which offers a resistance of r

_{o1}. Finally, the last branch connects to R

_{4}& all these resistances come in parallel. The overall output impedance is (R

_{4}||r

_{o1}||(R

_{1}+r

_{o2})/(1+(g

_{mb2}*r

_{o2})) and the voltage gain becomes -g

_{m1}*(R

_{4}||r

_{o1}||(R

_{1}+r

_{o2})/(1+(g

_{mb2}*r

_{o2})). Note that we can also calculate the voltage gain if M

_{2}suffers from Body effect.

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