This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Bipolar Current Mirror – Set 2”.

1. In the following circuit, we observe that the base current is not negligible and the current gain is β=8. If the collector current is 4 times the reference current, what should be the base width of Q_{3} if the base width of Q_{1} is W_{B}?

a) 83*W_{B}/9

b) W_{B}/98

c) 24*W_{B}/92

d) W_{B}/9

View Answer

Explanation: We can say that the collector current of Q

_{2}is p*(1/(1+(1/β)*(p+1))) where p is the total no. of transistors connected to Q

_{1}. From the given values, we can determine p to be 9. This implies that Q

_{3}is equivalent to 9 identical transistors in parallel and it’s base width is not more than (1/9) times the base width of Q

_{1}.

2. In the following circuit, we observe that the base current is not negligible and the current gain is β=64. If the collector current is 8 times the reference current, what should be the area of the emitter of Q_{3} if the emitter area of Q_{1} is A_{E}?

a) 35/2*A_{E}

b) 46/9*A_{E}

c) 65/7*A_{E}

d) 74/90*A_{E}

View Answer

Explanation: We can say that the collector current of Q

_{2}is p*(1/(1+(1/β)*(p+1))) where p is the total no. of transistors connected to Q

_{1}.From the given values, we can determine p to be 65/7. This implies that Q

_{3}is equivalent to a transistor whose emitter area is not less than 65/7 times the area of the emitter of Q

_{1}.

3. The current mirror overcomes all the problems of biasing which came up due to supply rail problems.

a) True

b) False

View Answer

Explanation: The current source, in a current mirror stage, itself proves to be a challenging design. However, the current mirror stage does seem to reduce a lot of problems which came up due to the effect of temperature in a V.D.B. stage. i.e. it does help us to improve the bias design significantly.

4. In the following circuit, Q_{3} is biased by a current mirror configuration while Q_{4} is biased at 1mA. If Q_{1} & Q_{6} are identical and only Q_{3} suffers from early effect, what is r_{o3} if V_{A}=5V?

a) 1KΩ

b) 40KΩ

c) 10KΩ

d) 30KΩ

View Answer

Explanation: Q

_{3}is biased by a current mirror and we find that the collector current of Q

_{3}is I

_{3}1/2=0.5mA since the total current due to Q

_{1}and Q

_{6}is equal to the reference current, the base-emitter voltage carries information to generate one-half of the reference current. Hence, r

_{o3}=(V

_{A}/I

_{3})=10KΩ.

5. In the following circuit, Q_{3} is biased by a current mirror configuration while Q_{4} is biased at 1mA. If Q_{1} & Q_{6} are identical and only Q_{3} suffers from early effect, what is the voltage gain if V_{A}=5V?

a) .9974

b) .9746

c) .3658

d) 1

View Answer

Explanation: We note that Q

_{4}is behaving as a follower stage. The transconductance of Q

_{4}is 1/26Ω. The output impedance of Q

_{3}is 10KΩ and thus, the voltage gain becomes .9974.

6. In the following circuit, Q_{3} is biased by a current mirror configuration while Q_{4} is biased at 1mA. If Q_{1} & Q_{6} are identical and only Q_{3} suffers from early effect, what is the voltage gain due to the degenerated transconductance of Q_{4} if V_{A}=5V?

a) .7481

b) .5698

c) .2379

d) .3654

View Answer

Explanation: The transconductance of a degenerated Q

_{4}will be g

_{m0}/(1+g

_{m0}(r

_{o3}||r

_{o4}). We have r

_{o4}=5KΩ and r

_{o3}=10KΩ. Initially, the transconductance of Q

_{4}i.e. g

_{m0}is 1/26Ω. Putting the values, we get the transconductance as .7481.

7. In the following circuit, Q_{3} is biased by a current mirror configuration while Q_{4} is biased at 1mA. If Q_{1} & Q_{6} are identical and only Q_{3} suffers from early effect, what is the voltage gain due to the degenerated transconductance of Q_{4} if V_{A}=5V?

a) -3.56

b) -2.56

c) -0.36

d) -3.74

View Answer

Explanation: We note that Q

_{3}generates an output impedance of 10KΩ while the degenerated transconductance of Q

_{4}is 2.9*10

^{-4}S. The voltage gain is the voltage gain of a degenerated C.E. stage and it is approximated to \(\frac {–R_C}{((\frac {1}{g_m})+r_{o3})}\) by substituting values we get -3.74.

8. What is the impedance of the following circuit, looking into the emitter of the transistor, in presence of early effect?

a) \(\frac {1}{((r_π+r_o) + g_m)}\)

b) \(\frac {1}{(\frac {3}{(r_π-r_o)} + g_m)}\)

c) \(\frac {1}{(\frac {1}{(r_π||r_o)} + g_m)}\)

d) \(\frac {1}{(\frac {2}{(r_o)} + g_m)}\)

View Answer

Explanation: We draw the small signal model and find the Thevenin’s resistance looking into the emitter of Q

_{1}. We note that the base and the collector are shorted to AC ground and we write a KCL at the emitter node. Henceforth, the resistance comes out to be \(\frac {1}{(\frac {1}{(r_π||r_o)} + g_m)}\). This is expected because the transistor behaves like a diode-connected transistor in this state ie a two-terminal device.

9. What is the impedance looking into the collector of the transistor in presence of early effect? Note that the base is tied to a voltage source.

a) r_{o}

b) 0

c) 3*r_{o}

d) Infinite

View Answer

Explanation: Looking into the collector, the resistance is simply 0. This can be verified from the small signal model but since we are finding the Thevenin’s resistance at the collector node, we note that the base-emitter voltage of the transistor is 0 since both the base & the emitter terminals will be connected to ground. We note that V

_{X}gets grounded.

10. What is the input impedance of the following circuit in presence of early effect? Note that the collector is connected to a voltage source.

a) \(\frac {1}{(\frac {1}{(r_π+r_o)} + g_m)}\)

b) \(\frac {1}{(\frac {1}{(r_π-r_o)} + g_m)}\)

c) \(\frac {1}{(\frac {1}{(r_π||r_o)} + g_m)}\)

d) 0

View Answer

Explanation: We draw the small signal model and readily note that the collector is shorted to ground. We conclude that any voltage at the input will get shorted to this ground and the input impedance calculated will be 0.

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