This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Bipolar Amplifiers – Common Base Stage – Set 2”.

1. Assume that the transistor has a very high current gain with negligible early effect. What is the output resistance of Q_{1}?

a) R_{1} || 1/g_{m}

b) (R_{1} || 1/g_{m}) + R_{2}

c) R_{1} || (1/g_{m} + R_{2})

d) R_{1}

View Answer

Explanation: The output resistance is only R

_{1}. Observe that Q

_{2}is unbiased so it is in the cut-off state. The emitter doesn’t contribute to any impedance at the collector of Q

_{1}.

2. Assume that the transistor has a very high current gain with negligible early effect. What is the output resistance of Q_{1}?

a) R_{1} || 1/g_{m}

b) (R_{1} + 1/g_{m}) + R_{2}

c) R_{1} + (1/g_{m} || R_{2})

d) 2*R_{1}

View Answer

Explanation: Q

_{2}is behaving as a common base stage. It offers an input impedance of 1/g

_{m}at the emitter of Q

_{1}and this make the output impedance of Q

_{1}, R

_{1}|| 1/g

_{m}.

3. Assume that the transistor has a very high current gain with only Q_{2} has early effect. What is the output resistance of Q_{2}?

a) {r_{O2}+(1+g_{m2}r_{O2})(R_{1}||r_{π2})}

b) (1+g_{m2}r_{O2})(R_{1}||r_{π2})

c) {r_{O2}+(1+g_{m2}r_{O2})(R_{1}||r_{π2})}||R_{2}

d) R_{2}

View Answer

Explanation: Q

_{2}is degenerated by R

_{1}and Q

_{1}doesn’t contribute due to negligible early effect. This leads us to the output resistance as {r

_{O2}+(1+g

_{m}r

_{O})(R

_{1}||r

_{π2})}||R

_{2}. Observe that even though Q

_{2}is behaving as a common base stage, the output impedance is quite similar to a degenerated common emitter stage.

4. Assume that the transistor has a very high current gain while both has early effect. What is the output resistance of Q_{2}?

a) R_{2}

b) {r_{O2}+(1+g_{m2}r_{O2})(R_{1}||r_{O1}||r_{π2})}||R_{2}

c) r_{O2}

d) R_{1}||r_{O1}||r_{π2}

View Answer

Explanation: Q

_{2}is degenerated by R

_{1}and Q

_{1}which contributes r

_{O1}due to early effect. These two resistances come in parallel and are connected to the emitter of Q

_{2}. This leads us to the output resistance as b) {r

_{O2}+(1+g

_{m2}r

_{O2})(R

_{1}||r

_{O1}||r

_{π2})}||R

_{2}.

5. Assume that the transistor has a very high current gain while both has early effect. What is the input resistance of Q_{2}?

a) 1/g_{m2} || r_{o2}

b) 1/g_{m2} || r_{o1}

c) 1/g_{m1} || r_{o2}

d) 1/g_{m2}

View Answer

Explanation: The input resistance of Q

_{2}, behaving as a common base stage, is 1/g

_{m2}|| r

_{o2}. This input impedance is independent of the previous stage. Note that in absence of early effect, the answer would’ve been 1/g

_{m2}.

6. Assume that the transistor has a very high current gain while both has early effect. What is the voltage gain till Q_{1}?

a) -g_{m1} * (R_{1} || r_{O1} || 1/g_{m2} || r_{O2})

b) g_{m1} * (R_{1} || r_{O1})

c) -g_{m1} * (R_{1} || r_{O1} + 1/g_{m2} || r_{O2})

d) -g_{m1} * (R_{1} || r_{O1})

View Answer

Explanation: The voltage gain till Q

_{1}is similar to that of a C.E. stage and that is g

_{m1}* (The total resistance connected to the collector of Q

_{1}). Note that three resistances are connected to the collector of Q

_{1}. First is R

_{1}, second is Q

_{1}itself which offers r

_{O1}due to early effect and third is the emitter of Q

_{3}which offers a resistance of 1/g

_{m2}|| r

_{O2}. All these resistances are connected from the collector to ground and the total resistance connected to the collector is (R

_{1}|| r

_{O1}|| 1/g

_{m2}|| r

_{O2}).

7. Assume that the transistor has a very high current gain while both has early effect. What is the voltage gain till Q_{2}?

a) -(R_{1} || r_{O1} || 1/g_{m2} || r_{O2}) * {r_{O2} + (1 + g_{m2}r_{O2})(R_{1} || r_{O1} || r_{π2})} || R_{2}

b) -g_{m1} * (R_{1} || r_{O1} || 1/g_{m2} || r_{O2}) * {r_{O2} + (1 + g_{m2}r_{O2})(R_{1} || r_{O1} || r_{π2})} || R_{2}

c) g_{m1} * (R_{1} || r_{O1} || 1/g_{m2} || r_{O2})

d) -g_{m1} * (R_{1} || r_{O1} || 1/g_{m2} || r_{O2}) * g_{m2} * {r_{O2} + (1 + g_{m2}r_{O2})(R_{1} || r_{O1} || r_{π2})} || R_{2}

View Answer

Explanation: The voltage gain till Q

_{2}is the product of the voltage gain from the input to Q

_{1}and from Q

_{1}to Q

_{2}. The first voltage gain is g

_{m1}* (R

_{1}|| r

_{O1}|| 1/g

_{m2}|| r

_{O2}). The second voltage gain is g

_{m2}* {r

_{O2}+ (1 + g

_{m2}r

_{O2})(R

_{1}|| r

_{O1}|| r

_{π2})} || R

_{2}since the output impedance of common base stage is degenerated by Q

_{1}and R

_{1}. The total voltage gain thus becomes

g

_{m1}* (R

_{1}|| r

_{O1}|| 1/g

_{m2}|| r

_{O2}) * g

_{m2}* {r

_{O2}+ (1 + g

_{m2}r

_{O2})(R

_{1}|| r

_{O1}|| r

_{π2})} || R

_{2}.

8. Assume that the transistor has a very high current gain while only Q_{2} has early effect. What is the voltage gain till Q_{1}?

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

b) g_{m1} * (R_{1} || r_{O2})

c) g_{m1} * (R1/g_{m2} || r_{O2})

d) g_{m1} * (R_{1} || 1/g_{m2} || r_{O2})

View Answer

Explanation: The voltage gain till Q

_{1}is g

_{m1}* the total gain connected to the collector of Q

_{1}. The total resistance connected to it is R

_{1}|| 1/g

_{m2}|| r

_{O2}. The overall voltage gain is g

_{m1}* (R

_{1}|| 1/g

_{m2}|| r

_{O2}).

9. Assume that the transistor has a very high current gain while only Q_{2} has early effect. What is the voltage gain till Q_{1}?

a) g_{m1}*(R_{1} || -1/g_{m2})

b) -g_{m1}*(R_{1} || 1/g_{m2})

c) g_{m1}*(R_{1} || 1/g_{m2})

d) -g_{m1}*(R_{1} || 1/g_{m2})

View Answer

Explanation: The voltage gain till Q

_{1}is g

_{m1}* the total resistance connected at the collector of Q

_{1}. The total resistance is R

_{1}|| 1/g

_{m2}|| r

_{O2}and the overall voltage gain is g

_{m1}*(R

_{1}|| 1/g

_{m2}|| r

_{O2}).

10. Assume that the transistor has a very high current gain while only Q_{2} has early effect. What is the voltage gain till Q_{2}?

a) g_{m2}*{r_{O2}+(1+g_{m2}r_{O2})(R_{1}||r_{O1}||r_{π2})}||R_{2}*g_{m1}*(R_{1}||1/g_{m2}||r_{O2})

b) g_{m1}*(R_{1}||1/g_{m2}||r_{O2})

c) g_{m2}*{r_{O2}+(1+g_{m2}r_{O2})(R_{1}||r_{O1}||r_{π2})}||R_{2}

d) -g_{m2}*{r_{O2}+(1+g_{m2}r_{O2})(R_{1}||r_{O1}||r_{π2})}||R_{2}

View Answer

Explanation: The voltage gain till Q

_{2}will be the product of voltage gains from input to Q

_{1}and from Q

_{1}to ground. The first gain is g

_{m1}*(R

_{1}||1/g

_{m2}||r

_{O2}) while the second gain is g

_{m2}*{r

_{O2}+(1+g

_{m2}r

_{O2})(R

_{1}||r

_{O1}||r

_{π2})}||R

_{2}. Multiplying both the gains yield the overall gain as

g

_{m2}*{r

_{O2}+(1+g

_{m2}r

_{O2})(R

_{1}||r

_{O1}||r

_{π2})}||R

_{2}*g

_{m1}*(R

_{1}||1/g

_{m2}||r

_{O2}).

11. Assume that the transistor has a very high current gain with negligible early effect. What is the input resistance of the following circuit?

a) \(\frac {(R_4 || R_5)}{(\beta+1)}\)

b) 1/g_{m}+R_{5}/(β+1)

c) 1/g_{m}+(R_{4})/(β+1)

d) 1/g_{m}+(R_{4}||R_{5})/(β+1)

View Answer

Explanation: The input impedance looking into the base of the emitter is 1/g

_{m}+ (The base resistance)/(β+1). The base resistance is approximated to R

_{4}|| R

_{5}since the current gain is very high and the base current is nearly 0. The overall input impedance is 1/g

_{m}+(R

_{4}||R

_{5})/(β+1).

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