Microelectronics Questions and Answers - Bipolar Amplifiers - Common Base Stage - Set 2

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 Q₁?

a) R₁ || 1/g_m
b) (R₁ || 1/g_m) + R₂
c) R₁ || (1/g_m + R₂)
d) R₁
View Answer

Answer: d
Explanation: The output resistance is only R₁. Observe that Q₂ is unbiased so it is in the cut-off state. The emitter doesn’t contribute to any impedance at the collector of Q₁.

2. Assume that the transistor has a very high current gain with negligible early effect. What is the output resistance of Q₁?

a) R₁ || 1/g_m
b) (R₁ + 1/g_m) + R₂
c) R₁ + (1/g_m || R₂)
d) 2*R₁
View Answer

Answer: a
Explanation: Q₂ is behaving as a common base stage. It offers an input impedance of 1/g_m at the emitter of Q₁ and this make the output impedance of Q₁, R₁ || 1/g_m.

3. Assume that the transistor has a very high current gain with only Q₂ has early effect. What is the output resistance of Q₂?

a) {r_O2+(1+g_m2r_O2)(R₁||r_π2)}
b) (1+g_m2r_O2)(R₁||r_π2)
c) {r_O2+(1+g_m2r_O2)(R₁||r_π2)}||R₂
d) R₂
View Answer

Answer: c
Explanation: Q₂ is degenerated by R₁ and Q₁ doesn’t contribute due to negligible early effect. This leads us to the output resistance as {r_O2+(1+g_mr_O)(R₁||r_π2)}||R₂. Observe that even though Q₂ 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 Q₂?

a) R₂
b) {r_O2+(1+g_m2r_O2)(R₁||r_O1||r_π2)}||R₂
c) r_O2
d) R₁||r_O1||r_π2
View Answer

Answer: b
Explanation: Q₂ is degenerated by R₁ and Q₁ which contributes r_O1 due to early effect. These two resistances come in parallel and are connected to the emitter of Q₂. This leads us to the output resistance as b) {r_O2+(1+g_m2r_O2)(R₁||r_O1||r_π2)}||R₂.

5. Assume that the transistor has a very high current gain while both has early effect. What is the input resistance of Q₂?

a) 1/g_m2 || r_o2
b) 1/g_m2 || r_o1
c) 1/g_m1 || r_o2
d) 1/g_m2
View Answer

Answer: a
Explanation: The input resistance of Q₂, 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 Q₁?

a) -g_m1 * (R₁ || r_O1 || 1/g_m2 || r_O2)
b) g_m1 * (R₁ || r_O1)
c) -g_m1 * (R₁ || r_O1 + 1/g_m2 || r_O2)
d) -g_m1 * (R₁ || r_O1)
View Answer

Answer: a
Explanation: The voltage gain till Q₁ is similar to that of a C.E. stage and that is g_m1 * (The total resistance connected to the collector of Q₁). Note that three resistances are connected to the collector of Q₁. First is R₁, second is Q₁ itself which offers r_O1 due to early effect and third is the emitter of Q₃ 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 (R₁ || 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 Q₂?

a) -(R₁ || r_O1 || 1/g_m2 || r_O2) * {r_O2 + (1 + g_m2r_O2)(R₁ || r_O1 || r_π2)} || R₂
b) -g_m1 * (R₁ || r_O1 || 1/g_m2 || r_O2) * {r_O2 + (1 + g_m2r_O2)(R₁ || r_O1 || r_π2)} || R₂
c) g_m1 * (R₁ || r_O1 || 1/g_m2 || r_O2)
d) -g_m1 * (R₁ || r_O1 || 1/g_m2 || r_O2) * g_m2 * {r_O2 + (1 + g_m2r_O2)(R₁ || r_O1 || r_π2)} || R₂
View Answer

Answer: d
Explanation: The voltage gain till Q₂ is the product of the voltage gain from the input to Q₁ and from Q₁ to Q₂. The first voltage gain is g_m1 * (R₁ || r_O1 || 1/g_m2 || r_O2). The second voltage gain is g_m2 * {r_O2 + (1 + g_m2r_O2)(R₁ || r_O1 || r_π2)} || R₂ since the output impedance of common base stage is degenerated by Q₁ and R₁ . The total voltage gain thus becomes
g_m1 * (R₁ || r_O1 || 1/g_m2 || r_O2) * g_m2 * {r_O2 + (1 + g_m2r_O2)(R₁ || r_O1 || r_π2)} || R₂.

8. Assume that the transistor has a very high current gain while only Q₂ has early effect. What is the voltage gain till Q₁?

a) g_m1 * (R₁ || 1/g_m2)
b) g_m1 * (R₁ || r_O2)
c) g_m1 * (R1/g_m2 || r_O2)
d) g_m1 * (R₁ || 1/g_m2 || r_O2)
View Answer

Answer: d
Explanation: The voltage gain till Q₁ is g_m1 * the total gain connected to the collector of Q₁. The total resistance connected to it is R₁ || 1/g_m2 || r_O2. The overall voltage gain is g_m1 * (R₁ || 1/g_m2 || r_O2).

9. Assume that the transistor has a very high current gain while only Q₂ has early effect. What is the voltage gain till Q₁?

a) g_m1*(R₁ || -1/g_m2)
b) -g_m1*(R₁ || 1/g_m2)
c) g_m1*(R₁ || 1/g_m2)
d) -g_m1*(R₁ || 1/g_m2)
View Answer

Answer: c
Explanation: The voltage gain till Q₁ is g_m1 * the total resistance connected at the collector of Q₁. The total resistance is R₁ || 1/g_m2 || r_O2 and the overall voltage gain is g_m1*(R₁ || 1/g_m2 || r_O2).

10. Assume that the transistor has a very high current gain while only Q₂ has early effect. What is the voltage gain till Q₂?

a) g_m2*{r_O2+(1+g_m2r_O2)(R₁||r_O1||r_π2)}||R₂*g_m1*(R₁||1/g_m2||r_O2)
b) g_m1*(R₁||1/g_m2||r_O2)
c) g_m2*{r_O2+(1+g_m2r_O2)(R₁||r_O1||r_π2)}||R₂
d) -g_m2*{r_O2+(1+g_m2r_O2)(R₁||r_O1||r_π2)}||R₂
View Answer

Answer: a
Explanation: The voltage gain till Q₂ will be the product of voltage gains from input to Q₁ and from Q₁ to ground. The first gain is g_m1*(R₁||1/g_m2||r_O2) while the second gain is g_m2*{r_O2+(1+g_m2r_O2)(R₁||r_O1||r_π2)}||R₂. Multiplying both the gains yield the overall gain as
g_m2*{r_O2+(1+g_m2r_O2)(R₁||r_O1||r_π2)}||R₂*g_m1*(R₁||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+R₅/(β+1)
c) 1/g_m+(R₄)/(β+1)
d) 1/g_m+(R₄||R₅)/(β+1)
View Answer

Answer: d
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 R₄ || R₅ since the current gain is very high and the base current is nearly 0. The overall input impedance is 1/g_m+(R₄||R₅)/(β+1).

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