# 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

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

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
ICEO=9.9505-199*0.0495=0.1mA==100µA.

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

Explanation: Given, ICBO=10µA, α=0.98 and IB =0.22µA. IC=α/ (1-α) IB+ 1/(1-α) ICBO
0.01078+0.5=0.51078mA.

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

Explanation: IC=V across RL/RL=5V/5KΩ=1mA.
IB=IC/β=1/50=0.02mA.

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

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

Explanation: Here, IC=0.8/800=1mA.
We know, VCE=VCC-ICRL
=10-0.8=9.2V.

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)

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

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

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

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.

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