This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “MOS Amplifiers – Common Gate Stage – Set 2”.

1. In presence of channel length modulation in M_{1}, what is the voltage gain of the following circuit?

a) \(\{ \frac { (\frac {1}{g_{m3}}||r_{o3}+r_{o1})}{((\frac {1}{g_{m3}}||r_{o3}+r_{o1})+R) } \}\)*(g_{m3}*R_{3})

b) \(\{ \frac { (\frac {1}{g_{m3}}||r_{o3}||r_{o1})}{((\frac {1}{g_{m3}}||r_{o3}||r_{o1})+R) } \}\)*(g_{m3}*R_{3})

c) \(\{ \frac { (\frac {1}{g_{m3}}||r_{o1})}{((\frac {1}{g_{m3}}||r_{o1})+R) } \}\)*(g_{m3}*R_{3})

d) \(\frac { \{ (\frac {1}{g_{m3}}||r_{o3}||r_{o1})}{((\frac {1}{g_{m3}}+r_{o3}+r_{o1})+R) \} }\)*(g_{m3}*R_{3})

View Answer

Explanation: M

_{1}offers an output impedance of r

_{o1}at the drain terminal while M

_{3}offers an output impedance of 1/g

_{m3}at the source terminal. Hence, the input to the source of M

_{3}is a result of voltage division between R and (1/g

_{m3}||r

_{o1}) and R – the input voltage becomes V

_{IN}*(1/g

_{m3}||r

_{o1})/((1/g

_{m3}||r

_{o1})+R). This voltage will get multiplied by a factor of g

_{m3}*R

_{3}due to the CG stage and we get the overall voltage gain as \(\{ \frac { (\frac {1}{g_{m3}}||r_{o1})}{((\frac {1}{g_{m3}}||r_{o1})+R) } \}\)*(g

_{m3}*R

_{3}).

2. In presence of Body effect and Channel Length Modulation in M_{3} and only channel length modulation in M_{1}, what is the voltage gain of the following circuit?

a) \(\frac { \{ (g_{mb}+ g_m)*r_o + 1 \}R_D}{(r_o + (g_{mb}+ g_m)*r_oR_S + R_S + R_D)}\)

b) \(\frac {(g_{mb}+ g_m)*r_o + 1 \}R_D}{R_D}\)

c) \(\frac {(g_{mb}+ g_m)*r_o + 1 \}R_D}{(r_o + 2R_D)}\)

d) \(\frac { \{ (g_{mb3}+ g_{m3})*r_{o3} + 1 \} R_3}{(r_o + (g_{mb3}+ g_{m3})*r_{o3}(R||r_{o1}) + (R||r_{o1}) + R_3)}\)

View Answer

Explanation: M

_{1}offers an output impedance of r

_{o1}at the drain terminal while M

_{3}offers an output impedance of 1/g

_{m3}at the source terminal. Hence, the input to the source of M

_{3}is a result of voltage division between R and (1/g

_{m3}||r

_{o1}) and R – the input voltage becomes V

_{IN}*(1/g

_{m3}||r

_{o1})/((1/g

_{m3}||r

_{o1})+R). The total resistance connected to the source terminal is R||r

_{o1}. We can calculate the voltage gain of a CG stage with a source resistance, say R

_{S}, which comes out to be \(\frac { \{ (g_{mb}+ g_m)*r_o + 1 \}R_D}{(r_o + (g_{mb}+ g_m)*r_oR_S + R_S + R_D)}\). We replace R

_{S}by R||r

_{o1}and voltage gain of the following circuit becomes \(\frac { \{ (g_{mb}+ g_{m})*r_{o} + 1 \} R_D}{(r_o + (g_{mb}+ g_{m})*r_{o}(R||r_{o1}) + (R||r_{o1}) + R_D)}\).

3. Will body effect manifest itself in M_{1}?

a) YES

b) NO

c) It may, depending upon V_{DD}

d) It may, depending upon V_{IN}

View Answer

Explanation: For body effect to manifest itself, the source voltage must change with respect to the substrate voltage. For M

_{1}, the source voltage doesn’t change since it is connected to the ground and body effect won’t occur in M

_{3}.

4. In presence of Body effect and Channel Length Modulation in M_{3}, what is the voltage gain of the following circuit?

a) Error

b) (1/g_{m2}||1/g_{m1}||r_{o1})/((1/g_{m2}||1/g_{m1}||r_{o1})+R)*{(g_{mb}+g_{m})*r_{o}+1}R_{d}/(r_{o1}+(g_{mb}+g_{m})*r_{o}*(R||1/g_{m2})+(R||1/g_{m2})+R_{d})

c) (1/g_{m2}+1/g_{m1}+r_{o1})/((1/g_{m2}+1/g_{m1}+r_{o1})+R)*{(g_{mb}+g_{m})*r_{o}+1}R_{D}/(r_{o1}+(g_{mb}+g_{m})*r_{o}*(R||1/g_{m2})+(R||1/g_{m2})+R_{D})

d) (1/g_{m2}+1/g_{m1}+r_{o1})/((1/g_{m2}+1/g_{m1}+r_{o1})+R)*{(g_{mb}+g_{m})*r_{o}+1}R_{D}/(r_{o1}+(g_{mb}+g_{m})*r_{o}*(R+1/g_{m2})+(R+1/g_{m2})+R_{D})

View Answer

Explanation: M

_{2}offers an output impedance of 1/g

_{m2}at the drain terminal while M

_{1}offers an output impedance of 1/g

_{m1}||r

_{o1}at the source terminal. Hence, the input to the source of M

_{3}is a result of voltage division between R and (1/g

_{m2}||1/g

_{m1}||r

_{o1}) and R – the input voltage becomes V

_{IN}*(1/g

_{m2}||1/g

_{m1}||r

_{o1})/((1/g

_{m2}||1/g

_{m1}||r

_{o1})+R). The total resistance connected to the source terminal is R||1/g

_{m2}. We can calculate the voltage gain of a CG stage with a source resistance, say R

_{S}, which comes out to be \(\frac { \{ (g_{mb}+ g_m)*r_o + 1 \}R_d}{(r_o + (g _{mb}+ g_m)*r_oR_S + R_S + R_d)}\). Hence, the output voltage for only the CG stage becomes V

_{S}*{(g

_{mb}+g

_{m})*r

_{o}+1}R

_{D}/(r

_{o}+(g

_{mb}+g

_{m})*r

_{o}*(R||1/g

_{m2}||r

_{o2})+(R||1/g

_{m2}||r

_{o2})+R

_{D}). The overall voltage gain is (1/g

_{m2}||1/g

_{m1}||r

_{o1})/((1/g

_{m2}||1/g

_{m1}||r

_{o1})+R)*{(g

_{mb}+g

_{m})*r

_{o}+1}R

_{D}/(r

_{o1}+(g

_{mb}+g

_{m})*r

_{o}*(R||1/g

_{m2})+(R||1/g

_{m2})+R

_{D}). Note that performing a small signal analysis will reveal to the same expression of voltage gain, but it becomes more complicated.

5. In presence of Body effect and Channel Length Modulation in M_{3} and only Channel Length Modulation in M_{1}, what is the voltage gain of the following circuit?

a) (1/g_{m2}+1/g_{m1}||r_{o1})/((1/g_{m2}+1/g_{m1}||r_{o1})+R)*{(g_{mb}+g_{m})*r_{o}+1}R_{d}/(r_{o}+(g_{mb}+g_{m})*r_{o}*(R||1/g_{m2})+(R||1/g_{m2})+R_{d})

b) 1/g_{m2}+g_{m2}||r_{o1})/(1/g_{m2}||1/g_{m1}||r_{o1})*{(g_{mb}+g_{m})*r_{o}+1}R_{d}/(r_{o}+(g_{mb}+g_{m})*r_{o}*(R||1/g_{m2})+(R||1/g_{m2})+R_{d})

c) 1/g_{m1}+g_{m1}||r_{o1})/((1/g_{m1}||1/g_{m2}||r_{o2})+R)*{(g_{mb}+g_{m})*r_{o}+1}R_{d}/(r_{o}+(g_{mb}+g_{m})*r_{o}*(R||1/g_{m2})+(R||1/g_{m2})+R_{d})

d) (1/g_{m2}||r_{o2}||1/g_{m1}||r_{o1})/((1/g_{m2}||r_{o2}||1/g_{m1}||r_{o1})+R)*{(g_{mb}+g_{m})*r_{o}+1}R_{d}/(r_{o}+(g_{mb}+g_{m})*r_{o}*(R||1/g_{m2}||r_{o2})+(R||1/g_{m2}||r_{o2})+R_{d})

View Answer

Explanation: M

_{2}offers an output impedance of 1/g

_{m2}||r

_{o2}at the drain terminal while M

_{1}offers an output impedance of 1/g

_{m1}||r

_{o1}at the source terminal. Hence, the input to the source of M

_{3}is a result of voltage division between R and (1/g

_{m2}||r

_{o2}||1/g

_{m1}||r

_{o1}) and R – the input voltage becomes V

_{IN}*(1/g

_{m2}||r

_{o2}||1/g

_{m1}||r

_{o1})/((1/g

_{m2}||r

_{o2}||1/g

_{m1}||r

_{o1})+R). The total resistance connected to the source terminal is R||1/g

_{m2}r

_{o2}||. We can calculate the voltage gain of a CG stage with a source resistance, say R

_{S}, which comes out to be \(\frac { \{ (g_{mb}+ g_m)*r_o + 1 \}R_d}{(r_o + (g _{mb}+ g_m)*r_oR_S + R_S + R_d)}\). Hence, the output voltage for only the CG stage becomes V

_{S}*{(g

_{mb}+ g

_{m})*r

_{o}+1}R

_{d}/(r

_{o1}+(g

_{mb}+ g

_{m})*r

_{o}*(R||1/g

_{m2}||r

_{o2})+(R||1/g

_{m2}||r

_{o2})+R

_{d}). The overall voltage gain is (1/g

_{m2}||r

_{o2}||1/g

_{m1}||r

_{o1})/((1/g

_{m2}||r

_{o2}||1/g

_{m1}||r

_{o1})+R)*{(g

_{mb}+g

_{m})*r

_{o}+1}R

_{d}/(r

_{o}+(g

_{mb}+g

_{m})*r

_{o}*(R||1/g

_{m2}||r

_{o2})+(R||1/g

_{m2}||r

_{o2})+R

_{d}). If body effect is also present in M

_{2}, the calculation is same and only the output resistance of M

_{2}changes.

6. What is the output impedance of the following circuit if only M_{1} suffers from channel length modulation but M_{2} doesn’t suffer from any second order effect?

a) R_{d}||{(1-g_{m1}*r_{o1})*(R||g_{m2})+r_{o1}}

b) R_{d}||{(1+g_{m1}*r_{o1})*(R-1/g_{m2})+r_{o1}}

c) R_{d}||{(1+g_{m1}*r_{o1})*(R||1/g_{m2})+r_{o1}}

d) R_{d}||{(1+g_{m1}*r_{o1})*(R+1/g_{m2})+r_{o1}}

View Answer

Explanation: The output impedance of the following circuit is R

_{d}in parallel to the output resistance of the degenerated M

_{1}. M

_{1}is degenerated by the resistance connected to the source of M

_{1}. The total resistance connected to the source of M

_{1}is R in parallel to the output resistance of M

_{2}, which is 1/g

_{m2}, and the total resistance connected to the source of M

_{1}is R||1/g

_{m2}. The degenerated resistance of M

_{1}is (1+g

_{m1}*r

_{o1})*(R||1/g

_{m2})+r

_{o1}and the overall output resistance becomes R

_{d}||{(1+g

_{m1}*r

_{o1})*(R||1/g

_{m2})+r

_{o1}}.

7. What is the output impedance of the following circuit if both the MOSFET suffer from channel length modulation and body effect?

a) R_{d}||{(1+(g_{m1}+g_{mb1})*r_{o1})*(R||1/(g_{mb2}+g_{m2})||r_{o2})+r_{o1}}

b) R_{d}||{(1+g_{mb1}*r_{o1})*(R||1/(g_{m2})||r_{o2})+r_{o1}}

c) R_{d}||{(1+(g_{m1}-g_{mb1})*r_{o1})*(R||1/(g_{mb2}+g_{m2})||r_{o2})+r_{o1}}

d) R_{d}||{(1+(g_{m1})*r_{o1})*(R||1/(g_{mb2})||r_{o2})+r_{o1}}

View Answer

Explanation: The output impedance of the following circuit is R

_{d}in parallel to the output resistance of the degenerated M

_{1}. M

_{1}is degenerated by the resistance connected to the source of M

_{1}. The total resistance connected to the source of M

_{1}is R in parallel to the output resistance of M

_{2}, which is 1/(g

_{mb2}+g

_{m2})||r

_{o2}, and the total resistance connected to the source of M

_{1}is R||1/(g

_{mb2}+g

_{m2})||r

_{o2}. The degenerated resistance of M

_{1}is (1+(g

_{m1}+g

_{mb1})*r

_{o1})*(R||1/(g

_{mb2}+g

_{m2})||r

_{o2})+r

_{o1}and the overall output resistance becomes R

_{d}||{(1+(g

_{m1}+g

_{mb1})*r

_{o1})*(R||1/(g

_{mb2}+g

_{m2})||r

_{o2})+r

_{o1}}.

8. How should we select V_{b} for M_{2} to work properly?

a) Always more than 0

b) Always less than 0

c) Equal to 0

d) Cannot be less than 0

View Answer

Explanation: M

_{2}is a PMOS. Depending on the output resistance we need from M

_{2}, we choose the magnitude of V

_{B}but it will always be less than 0.

9. In presence of channel length modulation and body effect in M_{1} and early effect in Q_{1}, what is the voltage gain?

a) (1/g_{m1}||r_{o1}||r_{o1(Q)})/((1/g_{m1}||r_{o1}||r_{o1(Q)})+R)*{(g_{mb})*r_{o}+1}R_{d}/(r_{o}+(g_{mb}+g_{m})*r_{o}*(r_{o1(Q)})+(R+r_{o1(Q)})+R_{d})

b) (1/g_{m1}+r_{o1}||r_{o1(Q)})/((1/g_{m1}-r_{o1}||r_{o1(Q)})+R)*{(g_{m})*r_{o}+1}R_{d}/(r_{o}+(g_{mb}+g_{m})*r_{o}*(R||r_{o1(Q)})+(R||r_{o1(Q)})+R_{d})

c) (1/g_{m1}||r_{o1}||r_{o1(Q)})/((1/g_{m1}||r_{o1}||r_{o1(Q)})+R)*{(g_{mb}+g_{m})*r_{o}+1}R_{d}/(r_{o}+(g_{mb}+g_{m})*r_{o}*(R||r_{o1(Q)})+(R||r_{o1(Q)})+R_{d})

d) Error

View Answer

Explanation: M

_{1}offers an output impedance of 1/g

_{m1}||r

_{o1}at the drain terminal while Q

_{1}offers an output impedance of r

_{o1(Q)}at the collector terminal. Hence, the input to the source of M

_{3}is a result of voltage division between R and (1/g

_{m1}||r

_{o1}||r

_{o1(Q)}) and R – the input voltage becomes V

_{IN}*(1/g

_{m1}||r

_{o1}||r

_{o1(Q)})/((1/g

_{m1}||r

_{o1}||r

_{o1(Q)})+R). The total resistance connected to the source terminal is R||r

_{o1(Q)}. We can calculate the voltage gain of a CG stage with a source resistance, say R

_{S}, which comes out to be \(\frac { \{ (g_{mb}+ g_m)*r_o + 1 \}R_d}{(r_o + (g _{mb}+ g_m)*r_oR_S + R_S + R_d)}\). Hence, the output voltage for only the CG stage becomes V

_{S}*{(g

_{mb}+g

_{m})*r

_{o}+1}R

_{d}/(r

_{o}+(g

_{mb}+g

_{m})*r

_{o}*(R||r

_{o1(Q)})+(R||r

_{o1(Q)})+R

_{d}). The overall voltage gain is (1/g

_{m1}||r

_{o1}||r

_{o1(Q)})/((1/g

_{m1}||r

_{o1}||r

_{o1(Q)})+R)*{(g

_{mb}+g

_{m})*r

_{o}+1}R

_{d}/(r

_{o}+(g

_{mb}+ g

_{m})*r

_{o}*(R||r

_{o1(Q)})+(R||r

_{o1(Q)})+R

_{d}).

10. In presence of channel length modulation in M_{1} & M_{3}, what is the voltage gain of the following circuit?

a) {(1/g_{m3}||r_{o3}+r_{o1})/((1/g_{m3}||r_{o3}+r_{o1})+R)}*(g_{m3}*R_{3})

b) {(1/g_{m3}||r_{o3}||r_{o1})/((1/g_{m3}||r_{o3}||r_{o1})+R)}*(g_{m3}*R_{3})

c) {(1/g_{m3}||r_{o3}|r_{o1})/((1/g_{m3}+r_{o3}+r_{o1})+R)}*(g_{m3}*R_{3})

d) {(1/g_{m3}+r_{o3}+r_{o1})/((1/g_{m3}||r_{o3}||r_{o1})+R)}*(g_{m3}*R_{3})

View Answer

Explanation: M

_{1}offers an output impedance of r

_{o1}at the drain terminal while M

_{3}offers an output impedance of 1/g

_{m3}||r

_{o3}at the source terminal. Hence, the input to the source of M

_{3}is a result of voltage division between R and (1/g

_{m3}||r

_{o3}||r

_{o1}) and R – the input voltage becomes V

_{IN}*(1/g

_{m3}||r

_{o3}||r

_{o1})/((1/g

_{m3}||r

_{o3}||r

_{o1})+R). This voltage will get multiplied by a factor of g

_{m3}*R

_{3}due to the CG stage and we get the overall voltage gain as {(1/g

_{m3}||r

_{o3}|r

_{o1})/((1/g

_{m3}+r

_{o3}+r

_{o1})+R)}*(g

_{m3}*R

_{3}).

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