# Microelectronics Questions and Answers – MOS Current Mirror

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

1. The drain current of M2 is 5mA. Neglecting channel length modulation, what should be the ratio of the aspect ratio of M2 upon M1?

a) 5
b) 4
c) 36
d) 2/9

Explanation: Neglecting C.L.M., we readily note that the ratio of drain current is simply equal to the ratio of aspect ratio of each transistor. Hence, the aspect ratio of M2 will have to be no less than 5 times that of M1 so that it can carry a drain current of 5mA.

2. If the overdrive voltage of M2 is 10 times of M1, what should be the ratio of the aspect ratio if the drain current of both the transistors is constant?

a) .1
b) 23
c) .01
d) 3

Explanation: It is said that the drain current of each MOSFET is same. If the ratio of overdrive voltage of the MOSFETs are in the ratio of 10:1, the aspect ratio of M2 will have to be .01 times of M1. Hence, the ratio is 1/100.

3. If both the transistors suffer from channel length modulation, λ1=1 & λ1=1/2 (in reality, the value is pretty low) and the drain voltage of both the transistors is 2V (a big assumption) – what should be the ratio of the aspect ratio if M3 has a drain current of 5 mA?

a) 5
b) 2
c) 6
d) 10

Explanation: Since both the transistors suffer from channel length modulation, the ratio of drain current is equal to the ratio of the product of aspect ratio and (1+λ*(VDS)) of each transistor. We can put the given values and determine that the ratio is 10. Note that the ratio has increased in presence of a higher channel length modulation parameter in M2.

4. If both the transistors suffer from channel length modulation, λ1=1 & λ1=1/2 (in reality, the value is pretty low) and the drain voltage of both the transistors is 2V (a big assumption) – what should be the ratio of the aspect ratio if the overdrive voltage of M3 is 5 times of M1?

a) 32
b) .015
c) .065
d) .013

Explanation: We note that if the same drain current flows through both the transistors, then the product of aspect ratio, the overdrive voltage and (1+λ*(VDS)) is equal for both the transistors. Hence, we can equate them for each transistor and we will find that the ratio of aspect ratio becomes .015.

5. If the channel length modulation parameter of only M2 is .1V-1, what is the impedance looking into the drain of M2 if the aspect ratio of each transistor is constant for a constant overdrive voltage but the drain current of M2 is 4 times that of M1?

a) 5KΩ
b) 5.1KΩ
c) 4.6KΩ
d) 4.8KΩ

Explanation: The impedance looking into the drain of M2 is r0=1/λID & λ=.1V-1. This leads us to the impedance, connected from the drain to the source of M2 as 5KΩ since the drain current are same for both the MOSFETs.

6. If the channel length modulation parameter of only M2 is .1V-1, what is the factor by which the aspect ratio of M2 is to be multiplied to make it equal to the aspect ratio of M1, when the drain current of M2 is 4 times that of M1?

a) .30702
b) .32895
c) .31465
d) .30769

Explanation: If the drain current of M2 is 4 times of M1, then for a constant overdrive voltage, the aspect ratio of M2 is 4/(1+λVDS2) since only M2 suffers from channel length modulation. Thus, the aspect ratio is greater than a factor of (.30769) since VDS2 is equal to 3V with the drain current being 2mA. Note that the magnitude of the drain current is given and we need not find the drain current separately. Moreover, the ratio of currents allows us to find the drain current separately and then we can check for convergence of the magnitude of drain current and voltages.

7. If the channel length modulation parameter of only M2 is .1V-1, what is the ratio of the transconductances of M2/M1, when the drain current of M2 is 4 times that of M1?

a) 1.7145
b) 1.7541
c) 1.7451
d) 1.7514

Explanation: The transconductance of a MOSFET in presence of channel length modulation becomes divided by a factor of √(1+λ VDS). Therefore, the transconductance, for a constant drain current, becomes proportional to √(W/L)/(1+λ VDS). We readily note that W/L will be in a ratio of 4:1 and the ratio of transconductance becomes 1.7541.

8. If the channel length modulation parameter of only M2 is .1V-1, what is the total impedance connected to the drain of M1, when the drain current of M2 is 4 times that of M1?

a) .836
b) .823
c) .833
d) .839

Explanation: The drain current of M2 is 2mA and the resistance due to channel length modulation in the MOSFET is 5KΩ. We note that the total resistance connected to the drain of the device is 1KΩ||5KΩ=.833KΩ.

9. By what factor should the aspect ratio of M2 be multiplied to get the aspect ratio of M1 such that the ratio of drain current in M2 and M1 is equal to 8/3 and proportional to ID0? Neglect Channel Length Modulation.

a) 48/71
b) 8/11
c) 18/21
d) 78/41

Explanation: If the drain current in M1 & M2 is proportional to that of M0, we can say that the drain current of M2 is 8/11 times that of M0. Then the aspect ratio of M2 is less than M1 by a factor of 8/11. Note that the gate current is almost 0 since the gate is made of the Oxide layer. The gate voltage of M0 is shorted to M1 and M2 and the drain currents become proportional to M0.

10. What is the ratio of transconductance of M1 and M0, for a constant overdrive voltage, if the ratio of drain current in M2 and M1 is equal to 8/3 and proportional to ID0? Neglect Channel Length Modulation.

a) .2772
b) .2745
c) .9295
d) .2727

Explanation: The transconductance is proportional to the product of the aspect ratio and the overdrive voltage for a MOSFET. Hence, the ratio of transconductance of M1/M0 is 3/11, i.e. 0.2727, since the drain current of M1 is 3/11 times that of M0.

11. If the transconductance of M2 & M0 is 4:1 for a constant overdrive voltage, what is the ratio of the resistances generated between the drain and the source if the ratio of their intrinsic gains is 16:1?

a) 4:99
b) 5:6
c) 3.99:1.1914
d) 4:1

Explanation: The intrinsic gain of the MOSFET is a product of its transconductance and the resistance between the source and the drain which arises due to channel length modulation. This leads us to the fact that the ratio of these resistances in M2 & M0 is 4:1

12. Why do we intend to use a Cascode Current Mirrors as follows?

a) To overcome the effect of channel length modulation
b) To increase the speed
c) To increase the aspect ratio
d) To increase the effective oxide thickness

Explanation: When we use a Cascode current mirror, we find that the drain voltages of M0 and M2 can be made equal and fixed to eliminate the channel length modulation. Henceforth, M3 does not reflect much change in its drain voltage when its source voltage is changing.

Sanfoundry Global Education & Learning Series – Microelectronics.

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