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Test: Common Base Configuration - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test - Test: Common Base Configuration

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

In a Common Base Configuration, BJT has _______ input impedance and ________output impedance. 

Detailed Solution for Test: Common Base Configuration - Question 1

Common base configuration:

Input Terminal Emitter – Base (EB)
Output Terminal Collector – Base (CB)
Since for Amplification Application in BJT
EB Junction → Forward Biased (Low Impedance)
CB Junction → Reversed Biased (High Impedance)
Various transistor configurations and their characteristics are shown in the table

Test: Common Base Configuration - Question 2

Which of the following statement/s is /are correct regarding the CB Configuration amplifier?
A). Current gain of an ideal Common base amplifier is unity.
B). Voltage gain is unity.
C). The Phase between input and output is 180º.

Detailed Solution for Test: Common Base Configuration - Question 2

Current gain(α) of a Common base amplifier is unity (ideally).

  • The Input resistance is very low ( 30-150 Ω ).
  • The output resistance is very high ( ≈ 500K ).
  • The Voltage gain is high.
  • The Phase between input and output is the same.

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Test: Common Base Configuration - Question 3

For a BJT with common base connection, the emitter current is 1 mA. If the emitter circuit is open, the collector current is 50 µA. Find the total collector current, if α = 0.92.

Detailed Solution for Test: Common Base Configuration - Question 3

IE = IC + IB = αIE + ICBO
Where,
IE is emitter current
IC is collector current
IB is base current
Amplification factor (α) = IC/IE
ICBO is an open circuit current
Application:
We have,
IE = 1 mA
ICBO = 50 μA
α = 0.92
Hence,
IC = αIE + ICBO = 0.92 × 1 + 50 × 10-3 = 0.97 A

Test: Common Base Configuration - Question 4

Which of the following quantities is boosted by this circuit when a BJT is connected in a circuit in common base configuration? 

Detailed Solution for Test: Common Base Configuration - Question 4

The output resistance is boosted when a BJT is connected in a circuit in a common base configuration. 
Current gain(α) of a Common base amplifier is unity (ideally).
The Input resistance is very low ( 30-150 Ω ).
The output resistance is very high ( ≈ 500K ).
The Voltage gain is high.
The Power gain is high.
The Phase between input and output is same.

Test: Common Base Configuration - Question 5

For a transistor connected in common base connection, collector current is 0.95 mA and base current is 0.05 mA. Find the value of α :

Detailed Solution for Test: Common Base Configuration - Question 5

For a transistor, the base current, the emitter current, and the collector current are related as:
IE = IB + IC
where IC = αIE
α = Current gain of the transistor
Given,
IC = 0.95 mA
IB = 0.05 mA
∴ IE = IB + IC = 0.05 + 0.95 = 1 mA
From the above concept,
IC = αIE

Test: Common Base Configuration - Question 6

The current gain value of common base configuration is _______.

Detailed Solution for Test: Common Base Configuration - Question 6


Input Terminal Emitter–Base (EB)
Output Terminal Collector–Base (CB)
Since for Amplification Application in BJT
EB Junction → Forward Biased (Low Impedance)
CB Junction → Reversed Biased (High Impedance)
Various transistor configurations and their characteristics are shown in the table

Test: Common Base Configuration - Question 7

In a common base cenfiguration, the alpha of the transistor is 0.99, its collector current is 1 mA and the collector to base current with emitter open is 1 μA. The value of base current is  

Detailed Solution for Test: Common Base Configuration - Question 7

Given that,
α = .99
ICBO = 1 μA
IC = 1 mA

Test: Common Base Configuration - Question 8

In CB configuration of transistor, the output impedance is:

Detailed Solution for Test: Common Base Configuration - Question 8

Common base configuration:

Input Terminal Emitter – Base (EB)
Output Terminal Collector – Base (CB)
Since for Amplification Application in BJT
EB Junction → Forward Biased (Low Impedance)
CB Junction → Reversed Biased (High Impedance)

Test: Common Base Configuration - Question 9

In common base configuration, the current amplification factor is:

Detailed Solution for Test: Common Base Configuration - Question 9

Current amplification factor: It is defined as the ratio of the output current to the input current. In the common-base configuration, the output current is emitter current IC, whereas the input current is base current IE.
Thus, the ratio of change in collector current to the change in the emitter current is known as the current amplification factor. It is expressed by the α.

β is the current amplification factor in the common-emitter configuration.

​​​​​​​γ is the current amplification factor in the common collector configuration.

Test: Common Base Configuration - Question 10

In a common base transistor configuration, if IB = 0.25 mA and IE = 1.4 mA, then IC is:

Detailed Solution for Test: Common Base Configuration - Question 10

Current amplification factor: It is defined as the ratio of the output current to the input current. In the common-base configuration, the output current is emitter current IC, whereas the input current is base current IE.
Thus, the ratio of change in collector current to the change in the emitter current is known as the current amplification factor. It is expressed by the α.

Where, IE = IC + IB
Given,
IB = 0.25 mA and IE = 1.4 mA
We know that,
IE = IC + IB
⇒ IC = IE - IB = 1.4 - 0.25 = 1.15 mA

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