# Microelectronics Questions and Answers – Bipolar Current Mirror

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
d) Cascode Stage

Explanation: The above circuit is called a current mirror stage. The current source provides a current of I Ampere and this is biasing Q1. The bias voltage from Q1 is being taken and carried to the base of Q2. Hence, the base-emitter voltage of Q2 is also VBE & IC2=f(VBE)=I. Therefore, the collector current from Q2 is now I.

2. Why do we have to short the base and collector terminals of Q1?

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

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 VBE
• VBE is, now, a function of I1.
Q1 is biased by the current source. Hence, it’s base-emitter voltage is given by VTln(I/IS) ie I=IC. 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 – Q1 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

Explanation: The above configuration is not a current source. Firstly, Q1 generates a collector current proportional to VB and Q2 also does the same. Moreover, the current coming out from Q2’s collector terminal is proportional to VB 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 Q1 to the collector of Q2
b) Short the base of Q2 to the collector of Q1
c) Short the base of Q2 to the collector of Q2
d) Short the base of Q1 to the collector of Q1

Explanation: If we short the base of Q1 to the collector of Q1, 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 Q1 from the current source I1.

5. If gm is the transconductance of Q1, what is the transconductance of the combination of Q2, Q3 & Q4? Assume that all of them are identical. Ignore the base current for now.

a) 3*gm
b) gm
c) 2*gm
d) 0

Explanation: We note that Q2, Q3 & Q4 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 IX/VT=3*I/VT. Hence, the transconductance is 3*gm.

6. If the emitter area of Q2 is 4 times Q1 while the base-width of Q2 is twice that of Q1, what is the collector current of Q2? Ignore the base current for now.

a) I
b) 3*I
c) 2*I
d) 8*I

Explanation: The collector current of an NPN transistor is given by AEqDNni2/NBWB. Now, we observe that Q2 mirrors the current flowing though Q1 ie I. But since the emitter area is 4*Q1 while the base width is twice of Q1, 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 Q2 and Q3 is 5:6 and it is proportional to Q1. What should be the area of Q4 so that the total current flowing though node X is 1mA? Ignore the base current for now.

a) 17/4*AE1
b) 89/91*AE1
c) 127/42*AE1
d) 235/246*AE1

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 Q2, Q3 & Q4 iS 4 times of Q1. So we write that 4*I=IC2+IC3+IC4. Now, we note that all the transistors have the same voltage and only there is emitter area is different. We can now write 4*AE1=AE4+5/6*AE1+6/7AE1 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 AE4=127/42*AE1 ie the emitter area of Q4 should be 127/42 times that of Q1.

8. Suppose I=25µA in the following circuit. The ratio of base width of Q2 and Q3 is 3:4 and it is proportional to Q1. What should be the base width of Q4 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

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 Q2, Q3 & Q4 is 1/4 times of Q1. So, we write that 4*I=IC2+IC3+IC4. Now, we note that all the transistors have the same voltage and only there is emitter area is different. We can now write 4/WB1=4/3*WB2+3/4*AE3+1/WB4 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 WB4=12/13*WB1 ie the base width of Q4 should be 127/42 times that of Q1.

9. Q1, Q2 & Q3 have an emitter area proportional to AE. If I=8Ma &the collector current of Q4 is 16mA, what is the emitter area with respect to AE. Ignore the base current for now.

a) 10*AE
b) 7*AE
c) 9*AE
d) 24*AE

Explanation: We observe that Q1, Q2 & Q3 are in parallel to each other. Hence, the total collector current coming out from them is 3*IC which is equal to I & hence, their base-emitter voltage is a function of I/3. Now, this voltage goes to Q4. 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 Q4. Thus the total area of Q4 is 24*AE.

10. Q1, Q2 & Q3 have a base width proportional to WB. If I=2mA & the collector current of Q4 is 18mA, what is the base width with respect to WB. Ignore the base current for now.

a) 10*WB
b) 9*WB
c) WB
d) 12*WB

Explanation: We observe that Q1, Q2 & Q3 are in parallel to each other. Hence, the total collector current coming out from them is 3*IC which is equal to I & hence, their base-emitter voltage is a function of I/3. Now, this voltage goes to Q4. 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 Q4. Thus, the base width of Q4 is 9*WB.

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

a) 1/(1/rπ+gm)
b) 1/(2/rπ+gm)
c) 2/(1/rπ+gm)
d) 1/(1/rπ+2/gm)

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/rπ+gm).

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