This set of Analog Circuits Assessment Questions and Answers focuses on “MOSFET Amplifier with CG Configuration – 2”.

1. If the inverse of transconductance is .5KS^{-1}, what should be the value of R_{c} so that the overall voltage gain is 8?

a) About 3K

b) .4K

c) About 4K

d) Such a gain is not possible

View Answer

Explanation: The voltage gain is given by g

_{m}R

_{d}. So, g

_{m}become 2mS and a resistance of 4k Ω will be necessary.

2. What is the input impedance of the following circuit, if channel length modulation is present?

a) g_{m} || 1/r_{o}

b) g_{m} * r_{o}

c) g_{m} || 2/r_{o}

d) g_{m}

View Answer

Explanation: The input impedance can be found by performing a small signal analysis at the input node i.e. source of the MOSFET. Due to channel length modulation, the effect of r

_{o}would make the input impedance g

_{m}|| 1/r

_{o}.

3. The C.G. stage can be regarded as a current buffer.

a) True

b) False

View Answer

Explanation: The input current flows through the source of the MOSFET while the gate current is nearly equal to 0. Hence, the entire current coming out of the drain is due to the source current.

4. If M_{1} suffers from channel length modulation but M_{2} doesn’t, what is the voltage gain?

a) g_{m1} * R_{d} || 1/g_{m2} * g_{m2} * R_{d2}

b) g_{m1} * R_{d} || 1/g_{m2} * g_{m2} * 2R_{d2}

c) g_{m1} * R_{d} || 12/g_{m2} * g_{m2} * R_{d2 }

d) g_{m1} * R_{d} * g_{m2} * R_{d2}

View Answer

Explanation: The voltage gain due to M

_{1}is g

_{m1}* R

_{d}|| 1/g

_{m2}since we find that the drain of M

_{1}is connected to the source of M

_{2}and the impedance looking into the source of a MOSFET is 1/g

_{m2}. Now, the voltage gain due to the 2nd CG stage is simply g

_{m2}*R

_{d2}. Hence, the total voltage gain is a product of both the factors ie g

_{m1}* R

_{d}|| 1/g

_{m2}* g

_{m2}* R

_{d2}.

5. If M_{2} suffers from channel length modulation but M_{1} doesn’t, what is the voltage gain?

a) 2g_{m1} * R_{d} || 1/g_{m2} * g_{m2} * R_{d2}

b) 2g_{m1} * R_{d} || 1/g_{m2} * g_{m2} * 2R_{d2}

c) g_{m1} * R_{d} || 12/g_{m2} * g_{m2} * R_{d2}

d) g_{m1} * R_{d} || 1/g_{m2} || r_{o2} * g_{m2} * R_{d2}

View Answer

Explanation: The voltage gain due to M

_{1}is g

_{m1}* R

_{d}|| 1/g

_{m2}|| r

_{o2}since we find that the drain of M

_{1}is connected to the source of M

_{2}and the impedance looking into the source of a MOSFET is 1/g

_{m2}|| r

_{o2}due to the channel length modulation of M

_{2}. Now, the voltage gain due to the 2

^{nd}CG stage is simply g

_{m2}*R

_{d2}. Hence, the total voltage gain is a product of both the factors ie g

_{m1}* R

_{d}|| 1/g

_{m2}* g

_{m2}* R

_{d2}.

6. If channel length modulation is present, what is the total impedance at node X?

a) R_{d} || r_{o1} || 1/g_{m2}

b) R_{d} || r_{o1} || 1/g_{m2} || r_{o2}

c) R_{d} || 2r_{o1} || 1/g_{m2} || r_{o2 }

d) R_{d} || r_{o1} || 2/g_{m2} || r_{o2}

View Answer

Explanation: We develop a Thevenin’s equivalent w.r.t. node X and ground. But this is lengthy process. But then again, we can use the process as follows. For Thevenising, firstly we set V

_{dd}and V

_{in}to ground. Thereafter, we see that R

_{d}and r

_{o1}are in parallel to each other since they are connected from node X to ground. Next, we see that the impedance provided by M

_{2}is 1/g

_{m2}|| r

_{o2}but it again occurs from Node x to ground. Hence, all the impedances are parallel to each other and the total impedance at node X is R

_{d}|| r

_{o1}|| 1/g

_{m2}|| r

_{o2}.

7. For an ideal current source, what is the voltage gain of the following circuit?

a) g_{m1} * r_{d1} * g_{m2} * 2R_{d2}

b) – g_{m1} * r_{d1} * g_{m2} * R_{d2}

c) 2g_{m1} * r_{d1} * g_{m2} * R_{d2}

d) – g_{m1} * r_{d1} * g_{m2} * R_{d2}

View Answer

Explanation: The voltage gain for the C.G. stage is simply g

_{m}* R

_{d1}. The voltage gain of the next stage is – g

_{m}* R

_{d2}. Note that, the ideal current source exhibits infinite output impedance and doesn’t affect but only increases the linearity of operation.

8. How can we reduce the number of MOSFET’s in the following circuit?

a) Scale the aspect ratio

b) Increase the transconductance

c) Increase the Power supply

d) Not possible

View Answer

Explanation: increasing the transconductance or power supply is possible but we need to increase our power budget for both situations which is not possible. However, we can increase the aspect ratio of one MOSFET. This is because the total current entering into R

_{1}is a sum of the currents originating from the drain of each MOSFET and they are equal to each other. Hence, scaling the aspect ratio is a plausible situation.

9. What is the relation between V_{b} and V_{in} in the following circuit?

a) V_{b} > V_{in} by V_{th}

b) V_{b} = V_{in}

c) V_{b} < V_{in} by V_{th}

d) No relation

View Answer

Explanation: To turn an NMOS on, we need to keep the voltage difference between the gate and the source greater than the threshold voltage. Hence, the gate potential should greater than the source potential by V

_{th}.

10. What is the total output impedance at node x?

a) R_{1} || r_{o1} || 2 * r_{o2} || r_{o2} || 2*r_{o4} … || r_{on}

b) R_{1} || 2 * ( r_{o1} || r_{o2} … || r_{on})

c) R_{1} || r_{o1} || r_{o2 }… || r_{on}

d) R_{1} + (r_{o1} || r_{o2} … || r_{on})

View Answer

Explanation: We note that all the MOSFET’s are in parallel. The output impedance of the i

^{th}MOSFET is given by r

_{oi}. Hence, the total impedance at node x is simply R

_{1}|| r

_{o1}|| r

_{o2}…|| r

_{on}. Note that R

_{1}is parallel to each MOSFET. This is because, while dining the output impedance, we set V

_{in}and V

_{cc}to zero. Note that R

_{1}is connected from node X to ground, all the MOSFET’s are connected form node x to ground.

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