Electronic Devices and Circuits Questions and Answers – The Common Emitter Configuration

This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “The Common Emitter Configuration”.

1. The base current amplification factor β is given by_________
a) IC/IB
b) IB/IC
c) IE/IB
d) IB/IE
View Answer

Answer: a
Explanation: The current amplification factor (β) is given by IC//IB. When no signal is applied, then the ratio of collector current to the base current is called current amplification factor of a transistor.

2. In an NPN silicon transistor, α=0.995, IE=10mA and leakage current ICBO=0.5µA. Determine ICEO.
a) 10µA
b) 100µA
c) 90µA
d) 500µA
View Answer

Answer: b
Explanation: IC=α IE +ICBO =0.995*10mA+0.5µA=9.9505mA.
IB=IE-IC=10-9.9505=0.0495mA. β=α/(1-α)=0.995/(1-0.995)=199

3. A germanium transistor with α=0.98 gives a reverse saturation current ICBO=10µA in a CB configuration. When it is used in CE configuration with a base current of 0.22µA, calculate the collector current.
a) 0.9867mA
b) 0.7654mA
c) 0.51078mA
d) 0.23456mA
View Answer

Answer: c
Explanation: Given, ICBO=10µA, α=0.98 and IB =0.22µA. IC=α/ (1-α) IB+ 1/(1-α) ICBO

4. In CE configuration, if the voltage drop across 5kΩ resistor connected in the collector circuit is 5V. Find the value of IB when β=50.
a) 0.01mA
b) 0.25mA
c) 0.03mA
d) 0.02mA
View Answer

Answer: d
Explanation: IC=V across RL/RL=5V/5KΩ=1mA.

5. A transistor is connected in CE configuration. Collector supply voltage Vcc=10V, RL=800Ω, voltage drop across RL=0.8V, α=0.96. What is base current?
a) 41.97µA
b) 56.78µA
c) 67.67µA
d) 78.54µA
View Answer

Answer: a
Explanation: Here, IC=0.8/800=1mA
β= α/ (1-α)=0.96/1-0.96=24.
Now, IB=IC/ β=1/24=41.67µA.
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6. The collector supply voltage for a CE configured transistor is 10V. The resistance RL=800Ω. The voltage drop across RL is 0.8V. Find the value of collector emitter voltage.
a) 3.7V
b) 9.2V
c) 6.5V
d) 9.8V
View Answer

Answer: b
Explanation: Here, IC=0.8/800=1mA.

7. The relation between α and β is_________
a) β = α/ (1-α)
b) α = β/(1+β)
c) β = α/ (1+α)
d) α = β/(1- β)
View Answer

Answer: b
Explanation: β is an ac base amplification factor. α is called as current amplification factor. The relation of IC and IB change as IC= βIB+ (1+ β) ICBO.

8. In ICEO, wt does the subscript ‘CEO’ mean?
a) collector to base emitter open
b) emitter to base collector open
c) collector to emitter base open
d) emitter to collector base open
View Answer

Answer: c
Explanation: The subscript ‘CEO’ means that it is collector to emitter base open. It is called as the leakage current. It occurs in a reverse bias in PNP transistor. The total current can be calculated by IC=βIB+IC.

9. When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called_________
a) dc current gain
b) base current amplification factor
c) emitter current amplification factor
d) ac current gain
View Answer

Answer: d
Explanation: The ac current gain is given by β=∆IC/∆IB. When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called ac current gain.

10. The range of β is _________
a) 20 to 500
b) 50 to 300
c) 30 to 400
d) 10 to 20
View Answer

Answer: a
Explanation: Almost in all the transistors, the base current is less than 5% of the emitter current. Due to this fact, it is generally greater than 20. Usually it ranges from 20 to 500. Hence this configuration is frequently used when appreciable current gain as well as voltage gain is required.

Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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