# Microelectronics Questions and Answers – Bipolar Amplifiers – Common Emitter Stage – Set 2

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

1. What happens to the collector current, in a C.E. stage, if the early voltage is very low? Assume that all other parameters are constant and we provide a simple bias voltage to the transistor.
a) It increases
b) It decreases
c) It remains unchanged
d) It is directly proportional to the early voltage

Explanation: In presence of early effect, the collector current gets multiplied by a factor of (1+VCE/VA). We observe that if the early voltage is very low, the current would increase.

2. We know that the voltage gain of a degenerated Common Emitter stage is (-RC/1/gm+RE). What will happen in presence of early effect?
a) The gain will increase
b) The gain will decrease
c) The gain will remain constant
d) The gain doubles

Explanation: The gain decreases significantly. The overall voltage gain becomes becomes [-RC || {r0 + (RE || rπ) * (1 + gm * r0)}]/(1/gm + RE) and we observe that the output resistance reduces significantly and the gain reduces.

3. A transistor is being used as a simple C.E. stage with negligible early effect. If the emitter area is doubled, what happens to the gain?
a) It doubles
b) It becomes half
c) It becomes 4 times the original gain
d) It becomes $$\frac {1}{4}$$ times the original gain

Explanation: The transconductance is directly proportional to the collector current which is directly proportional to the emitter area. If the emitters are doubles, the current doubles leading to the doubling of the transconductance and simultaneously, the voltage gain doubles.

4. A transistor is being used as a simple C.E. stage with negligible early effect. If the base width is doubled, what happens to the gain?
a) It doubles
b) It becomes half
c) It triples
d) It doesn’t affect the gain

Explanation: The collector current is inversely proportional to the base width. If the base width doubles, the collector current becomes half and the voltage gain effectively becomes half. The voltage gain is not to be confused with output voltage. The ratio of the output to the input voltage is the voltage gain.

5. A transistor is being used as a simple C.E. stage with negligible early effect. If the base width is doubled, what happens to the r0 for a constant early voltage?
a) It becomes 1/8th times the original
b) It becomes 1/4th times the original
c) It becomes 1/2 of the original
d) It doesn’t get affected

Explanation: The collector current is inversely proportional to the base width. This implies that if the base width is doubled, the collector current becomes half and ro becomes half. Note that the impact on ro is independent of the C.E. stage but purely dependent on the collector current.

6. Assuming that the transistor has negligible early effect and high current gain, what is the input resistance of the following circuit?

a) R + rπ + (β+1) * R
b) R
c) rπ + (β+1) * R
d) Infinite

Explanation: The overall resistance is the series combination of R3 & the input impedance of a degenerated C.E. stage which is simply rπ+(β+1)*R. Resistors, in series, get added and the overall resistance is (R + rπ + (β + 1) * R).

7. Assuming that the transistor has negligible early effect and high current gain, what is the voltage gain of the following circuit?

a) -RC*gm
b) -RC/(1/gm + R)
c) -RC/(1/gm + R + R/(β+1))
d) Infinite

Explanation: The voltage gain is to be determined by drawing the small signal model of the transistor. We will find that that the overall voltage gain is the product of the voltage gain from VIN to VBE and the voltage gain of a degenerated C.E. stage. The first voltage gain is $$\frac { \{ r_{\pi} + (\beta+1)*R \} }{ \{ R + r_{\pi} + (\beta+1)*R \} }$$ and the second voltage gain is $$\frac {- g_m*R}{(1+( r_{\pi} + \frac {1}{g_m})R)}$$. We multiply both the gains and multiply the denominator further by rπ and note that the denominator gets cancelled while gm*rπ = β. Thereby, the voltage gain is -RC/(1/gm + R + R/(β+1)) where β+1 is approximated to β.

8. Assuming that the transistor has negligible early effect and high current gain, what is the output resistance of the following circuit?

a) R
b) 2*R
c) 4*R
d) R + (β+1) * R

Explanation: The output resistance is R (R1). To verify this, we can draw the small signal model and use K.C.L. at the emitter node. We will find that vπ becomes 0 and the output resistance is independent of the input resistance in absence of early effect.

9. The supply voltage is 5V while the input voltage is 700mV. For a reverse saturation current of 10-16A and R1=1K, calculate the voltage gain with early effect.

a) 2.2
b) 0
c) 2.3*10-4
d) 200

Explanation: The total gain is simply gm*R1||ro. The transconductance turns out to be 188mS while the output resistance is 1.223mΩ. We observe that in presence of early effect, the gain has decreased considerably to 2.3*10-4.

10. The supply voltage is 5V while the input voltage is 700mV. For a reverse saturation current of 10-16A and R1=1K, calculate the internal resistance with early effect.

a) 1 kΩ
b) 3 kΩ
c) 2 kΩ
d) .5 kΩ

Explanation: The input resistance in presence of early effect is given by β/gm. The transconductance comes out to be 188mS while the internal resistance is approximated to .5 kΩ.

Sanfoundry Global Education & Learning Series – Microelectronics.

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