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# Test: The Common Emitter Configuration

## 10 Questions MCQ Test Electronic Devices | Test: The Common Emitter Configuration

Description
This mock test of Test: The Common Emitter Configuration for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: The Common Emitter Configuration (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: The Common Emitter Configuration quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: The Common Emitter Configuration exercise for a better result in the exam. You can find other Test: The Common Emitter Configuration extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

### The base current amplification factor β is given by_________

Solution:

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.

QUESTION: 2

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

Solution:

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.

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.

Solution:

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.

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.

Solution:

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

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?

Solution:

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

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.

Solution:

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

QUESTION: 7

The relation between α and β is_________

Solution:

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.

QUESTION: 8

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

Solution:

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.

QUESTION: 9

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

Solution:

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.

QUESTION: 10

The range of β is _________

Solution:

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.