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

1. What is the following stage called?

a) Mirror stage

b) Differential Stage

c) Cascade Stage

d) Cascode Stage

View Answer

Explanation: The above circuit is called a current mirror stage. The current source provides a current of I Ampere and this is biasing Q

_{1}. The bias voltage from Q

_{1}is being taken and carried to the base of Q

_{2}. Hence, the base-emitter voltage of Q

_{2}is also V

_{BE}& I

_{C2}=f(V

_{BE})=I. Therefore, the collector current from Q

_{2}is now I.

2. Why do we have to short the base and collector terminals of Q_{1}?

a) For active mode operation

b) To maintain the base-emitter voltage as a function of I

c) For saturation mode operation

d) A good follower

View Answer

Explanation: Note that the collector and the base terminals are connected for two reasons:

• The transistor always remains in active mode and is independent of changes in V

_{BE}

• V

_{BE}is, now, a function of I

_{1}.

Q

_{1}is biased by the current source. Hence, it’s base-emitter voltage is given by V

_{T}ln(I/I

_{S}) ie I=I

_{C}. We observe this the collector to base feedback biasing method and this keeps the transistor always in the active mode ie it has a very good amplification. Finally, if this wasn’t so – Q

_{1}wouldn’t have been able to generate a proper base-emitter voltage as a function of the current from the current source.

3. Is the following configuration a current source?

a) Yes

b) No

c) Cannot be determined

d) Only if both the emitter areas are same

View Answer

Explanation: The above configuration is not a current source. Firstly, Q

_{1}generates a collector current proportional to V

_{B}and Q

_{2}also does the same. Moreover, the current coming out from Q

_{2}’s collector terminal is proportional to V

_{B}and is not a copy of the current from the current source ie I.

4. How can we modify the following circuit so that it behaves like a current source?

a) Short the base of Q_{1} to the collector of Q_{2}

b) Short the base of Q_{2} to the collector of Q_{1}

c) Short the base of Q_{2} to the collector of Q_{2}

d) Short the base of Q_{1} to the collector of Q_{1}

View Answer

Explanation: If we short the base of Q

_{1}to the collector of Q

_{1}, we have modified the above circuit to behave as a current mirror stage. Now, we really don’t need the base-emitter voltage since the base-emitter voltage is a function of the bias current entering Q

_{1}from the current source I

_{1}.

5. If g_{m} is the transconductance of Q_{1}, what is the transconductance of the combination of Q_{2}, Q_{3} & Q_{4}? Assume that all of them are identical. Ignore the base current for now.

a) 3*g_{m}

b) g_{m}

c) 2*g_{m}

d) 0

View Answer

Explanation: We note that Q

_{2}, Q

_{3}& Q

_{4}are connected in parallel. This is because their bases are attached to the same terminal, their emitters are connected to the ground and their collector is also connected to the same terminal. Hence, the total current flowing through node X is 3*I where I is the collector current from each transistor. The overall transconductance is I

_{X}/V

_{T}=3*I/V

_{T}. Hence, the transconductance is 3*g

_{m}.

6. If the emitter area of Q_{2} is 4 times Q_{1} while the base-width of Q_{2} is twice that of Q_{1}, what is the collector current of Q_{2}? Ignore the base current for now.

a) I

b) 3*I

c) 2*I

d) 8*I

View Answer

Explanation: The collector current of an NPN transistor is given by A

_{E}qD

_{N}n

_{i}

^{2}/N

_{B}W

_{B}. Now, we observe that Q

_{2}mirrors the current flowing though Q

_{1}ie I. But since the emitter area is 4*Q

_{1}while the base width is twice of Q

_{1}, we readily observe that now – the collector current will be doubled i.e. the collector current is 2*I.

7. Suppose I=25µA in the following circuit. The ratio of emitter areas of Q_{2} and Q_{3} is 5:6 and it is proportional to Q_{1}. What should be the area of Q_{4} so that the total current flowing though node X is 1mA? Ignore the base current for now.

a) 17/4*A_{E1}

b) 89/91*A_{E1}

c) 127/42*A_{E1}

d) 235/246*A_{E1}

View Answer

Explanation: We note that the total current flowing through node X is 4*I. We go on to analyze the circuit by claiming that the total emitter area of Q

_{2}, Q

_{3}& Q

_{4}iS 4 times of Q

_{1}. So we write that 4*I=I

_{C2}+I

_{C3}+I

_{C4}. Now, we note that all the transistors have the same voltage and only there is emitter area is different. We can now write 4*A

_{E1}=A

_{E4}+5/6*A

_{E1}+6/7A

_{E1}by replacing I with the exponential relationship of the npn transistor and noting that all other parameters, except emitter are, is constant. After solving, we get A

_{E4}=127/42*A

_{E1}ie the emitter area of Q

_{4}should be 127/42 times that of Q

_{1}.

8. Suppose I=25µA in the following circuit. The ratio of base width of Q_{2} and Q_{3} is 3:4 and it is proportional to Q_{1}. What should be the base width of Q_{4} so that the total current flowing though node X is 1mA? Ignore the base current for now.

a) 12/17

b) 1/7

c) 2/7

d) 3/7

View Answer

Explanation: We note that the total current flowing through node X is 4*I. We go on to analyze the circuit by claiming that the total base width of Q

_{2}, Q

_{3}& Q

_{4}is 1/4 times of Q

_{1}. So, we write that 4*I=I

_{C2}+I

_{C3}+I

_{C4}. Now, we note that all the transistors have the same voltage and only there is emitter area is different. We can now write 4/W

_{B1}=4/3*W

_{B2}+3/4*A

_{E3}+1/W

_{B4}by replacing I with the exponential relationship of the npn transistor and noting that all other parameters, except emitter are, is constant. After solving, we get W

_{B4}=12/13*W

_{B1}ie the base width of Q

_{4}should be 127/42 times that of Q

_{1}.

9. Q_{1}, Q_{2} & Q_{3} have an emitter area proportional to A_{E}. If I=8Ma &the collector current of Q_{4} is 16mA, what is the emitter area with respect to A_{E}. Ignore the base current for now.

a) 10*A_{E}

b) 7*A_{E}

c) 9*A_{E}

d) 24*A_{E}

View Answer

Explanation: We observe that Q

_{1}, Q

_{2}& Q

_{3}are in parallel to each other. Hence, the total collector current coming out from them is 3*I

_{C}which is equal to I & hence, their base-emitter voltage is a function of I/3. Now, this voltage goes to Q

_{4}. But we seek a current of 8*I. Hence, the collector current is 8*3*I since the base-emitter voltage will only generate I/3 as the collector current from Q

_{4}. Thus the total area of Q

_{4}is 24*A

_{E}.

10. Q_{1}, Q_{2} & Q_{3} have a base width proportional to W_{B}. If I=2mA & the collector current of Q_{4} is 18mA, what is the base width with respect to W_{B}. Ignore the base current for now.

a) 10*W_{B}

b) 9*W_{B}

c) W_{B}

d) 12*W_{B}

View Answer

Explanation: We observe that Q

_{1}, Q

_{2}& Q

_{3}are in parallel to each other. Hence, the total collector current coming out from them is 3*I

_{C}which is equal to I & hence, their base-emitter voltage is a function of I/3. Now, this voltage goes to Q

_{4}. But we seek a current of 9*I. Hence, the collector current is 9*3*I since the base-emitter voltage will only generate I/3 as the collector current from Q

_{4}. Thus, the base width of Q

_{4}is 9*W

_{B}.

11.What is the input impedance of the following circuit? Ignore early effect.

a) 1/(1/r_{π}+g_{m})

b) 1/(2/r_{π}+g_{m})

c) 2/(1/r_{π}+g_{m})

d) 1/(1/r_{π}+2/g_{m})

View Answer

Explanation: We draw the small signal model of the transistor and calculate the Thevenin resistance looking into its base. A KCL does the trick and we find that the input resistance is 1/(1/r

_{π}+g

_{m}).

**Sanfoundry Global Education & Learning Series – Microelectronics**.

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