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Test: The Common Emitter Configuration - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: The Common Emitter Configuration

Test: The Common Emitter Configuration for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: The Common Emitter Configuration questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: The Common Emitter Configuration MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: The Common Emitter Configuration below.
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Test: The Common Emitter Configuration - Question 1

The base current amplification factor β is given by_________

Detailed Solution for Test: The Common Emitter Configuration - Question 1

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.

Test: The Common Emitter Configuration - Question 2

In an NPN silicon transistor, α=0.995, IE=10mA and leakage current ICBO=0.5µA. Determine ICEO.

Detailed Solution for Test: The Common Emitter Configuration - Question 2

 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.

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Test: The Common Emitter Configuration - Question 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.

Detailed Solution for Test: The Common Emitter Configuration - Question 3

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.

Test: The Common Emitter Configuration - Question 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.

Detailed Solution for Test: The Common Emitter Configuration - Question 4

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

Test: The Common Emitter Configuration - Question 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?

Detailed Solution for Test: The Common Emitter Configuration - Question 5

Here, IC=0.8/800=1mA
β= α/ (1-α)=0.96/1-0.96=24.
Now, IB=IC/ β=1/24=41.67µA.

Test: The Common Emitter Configuration - Question 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.

Detailed Solution for Test: The Common Emitter Configuration - Question 6

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

Test: The Common Emitter Configuration - Question 7

 The relation between α and β is_________

Detailed Solution for Test: The Common Emitter Configuration - Question 7

Test: The Common Emitter Configuration - Question 8

In ICEO, wt does the subscript ‘CEO’ mean?

Detailed Solution for Test: The Common Emitter Configuration - Question 8

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.

Test: The Common Emitter Configuration - Question 9

When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called_________

Detailed Solution for Test: The Common Emitter Configuration - Question 9

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.

Test: The Common Emitter Configuration - Question 10

 The range of β is _________

Detailed Solution for Test: The Common Emitter Configuration - Question 10

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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