250+ TOP MCQs on The AC Analysis of a Small-Signal Low-Frequency Common Emitter Transistor. and Answers

Electronic Devices and Circuits Mcqs on “The AC Analysis of a Small-Signal Low-Frequency Common Emitter Transistor”.

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1. The feature of an approximate model of a transistor is
A. it helps in quicker analysis
B. it provides individual analysis for different configurations
C. it helps in dc analysis
D. ac analysis is not possible
Answer: A
Clarification: The small signal model helps in quicker ac analysis of a transistor. The approximate model is applicable for all the configurations. The dc analysis is not obtained by using a small signal model of transistor.

2 A transistor has h_fe=100, h_ie=2kΩ, h_oe=0.005mmhos, h_re=0. Find the output impedance if the lad resistance is 5kΩ.
A. 5kΩ
B. 4kΩ
C. 20kΩ
D. 15kΩ
Answer: B
Clarification: R_O=I/h_oe=1/0.005m
=20kΩ.R_O^I= R_O || R_L^I=20||5
=4kΩ.

3. A CE amplifier when bypassed with a capacitor at the emitter resistance has
A. increased input resistance and increased voltage gain
B. increased input resistance and decreased voltage gain
C. decreased input resistance and increased voltage gain
D. decreased input resistance and decreased voltage gain
Answer: C
Clarification: When a transistor is bypassed with a capacitor, it short circuits in the small signal analysis of transistor and the resistor too shorts. The input resistance becomes RI=hie. The value of the input resistance is decreased and the gain now will be increasing.

4. A transistor has h_ie =2kΩ, h_oe=25µmhos and h_fe=60 with an unbypassed emitter resistor Re=1kΩ. What will be the input resistance and output resistance?
A. 90kΩ and 50kΩ respectively
B. 33kΩ and 45kΩ respectively
C. 6kΩ and 40kΩ respectively
D. 63kΩ and 40kΩ respectively
Answer: D
Clarification: As the emitter is unbypassed, the input resistance Ri=h_ie+(1+h_fe)Re
=2+61=63kΩ. The output resistance R_O=1/h_oe=1/25MΩ=40kΩ.

5. A transistor has h_ie =1KΩ and h_fe=60 with an bypassed emitter resistor R_e=1kΩ. What will be the input resistance and output resistance?
A. 90kΩ and 50kΩ respectively
B. 33kΩ and 45kΩ respectively
C. 6kΩ and 40kΩ respectively
D. 63kΩ and 40kΩ respectively
Answer: D
Clarification: As the emitter is bypassed, the input resistance R_i=h_ie
=1kΩ. The output resistance R_O=1/h_oe but the value is not given.
So, h_oe=0 and R_O=1/0=∞.

6. In the given circuit, find the equivalent resistance between A and B nodes.
2018/02/electronic-devices-circuits-questions-answers-mcqs-q6″>
A. 100kΩ
B. 50kΩ
C. 40kΩ
D. 60kΩ
Answer: B
Clarification: R_AB=R_O||100Ω
= (R_S^I+h_ie/1+h_fe)||100
=9+1/100||100=100||100=50Ω.

7. Which of the following acts as a buffer?
A. CC amplifier
B. CE amplifier
C. CB amplifier
D. cascaded amplifier
Answer: A
Clarification: The voltage gain of a common collector amplifier is unity. It is then used as a buffer. The CC amplifier is also called as an emitter follower. Though there is no amplification done, the output will be stabiliseC.

8. Which of the following is true?
A. CC amplifier has a large current gain
B. CE amplifier has a large current gain
C. CB amplifier has low voltage gain
D. CC amplifier has low current gain
Answer: b
Clarification: The CE amplifier has high current and voltage gains. The CC amplifier has unity voltage gain which cannot be regarded as high. The common base amplifier has a unity current gain and high voltage gain.

9. In an NPN silicon transistor, α=0.995, I_E=10mA and leakage current I_CBO=0.5µA. Determine I_CEO.
A. 10µA
B. 100µA
C. 90µA
D. 500µA
Answer: B
Clarification: I_C=α I_E +I_CBO =0.995*10mA+0.5µA=9.9505mA.
I_B=I_E-I_C=10-9.9505=0.0495mA. β=α/(1-α)=0.995/(1-0.995)=199
I_CEO=9.9505-199*0.0495=0.1mA==100µA.

10. In CB configuration, the value of α=0.98A. A voltage drop of 4.9V is obtained across the resistor of 5KΩ when connected in collector circuit. Find the base current.
A. 0.01mA
B. 0.07mA
C. 0.02mA
D. 0.05mA
Answer: C
Clarification: Here, I_C=4.9/5K=0.98mA
α = I_C/I_E .So,
I_E=I_C/α=0.98/0.98=1mA.
I_B=I_E-I_C=1-0.98=0.02mA.

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