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


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11 Questions MCQ Test - Test: The Common Base Configuration

Test: The Common Base Configuration for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: The Common Base Configuration questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: The Common Base 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 Base Configuration below.
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Test: The Common Base Configuration - Question 1

 The AC current gain in a common base configuration is_________

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

The AC current gain is denoted by αac. The ratio of change in collector current to the change in emitter current at constant collector base voltage is defined as current amplification factor.

Test: The Common Base Configuration - Question 2

The value of αac for all practical purposes, for commercial transistors range from_________

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

 For all practical purposes, αac=αdc=α and practical values in commercial transistors range from 0.9-0.99. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.

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Test: The Common Base Configuration - Question 3

 A transistor has an IC of 100mA and IB of 0.5mA. What is the value of αdc?

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

 Emitter current IE=IC+IB =100+0.5=100.5mA
αdc=IC/IE=100/100.5=0.995.

Test: The Common Base Configuration - Question 4

In CB configuration, the value of α=0.98A. A voltage drop of 4.9V is obtained across the resistor of 5KΩ when connected in collector circuit. Find the base current.

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

Here, IC=4.9/5K=0.98mA
α = IC/IE .So,
IE=IC/α=0.98/0.98=1mA.
IB=IE-IC=1-0.98=0.02Ma.

Test: The Common Base Configuration - Question 5

The emitter current IE in a transistor is 3mA. If the leakage current ICBO is 5µA and α=0.98, calculate the collector and base current.

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

IC=αIE + ICBO
=0.98*3+0.005=2.945mA.
IE=IC+IB . So, IB=3-2.495=0.055mA=55µA.

Test: The Common Base Configuration - Question 6

Determine the value of emitter current and collector current of a transistor having α=0.98 and collector to base leakage current ICBO=4µA. The base current is 50µA.

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

Given, IB=50µA=0.05mA
ICBO=4µA=0.004Ma
IC=α/(1- α)IB+1/(1- α)ICBO=2.45+0.2=2.65Ma
IE=IC+IB=2.65+0.05=2.7mA.

Test: The Common Base Configuration - Question 7

The negative sign in the formula of amplification factor indicates_________

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

 When no signal is applied, the ratio of collector current to emitter current is called dc alpha, αdc of a transistor. αdc=-IC/IE. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.

Test: The Common Base Configuration - Question 8

 The relation between α and β is _________

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

The terms Alpha and Beta refer to BJTs, Bipolar Junction Transistors. In these devices current leaves the emitter and crosses to the base and most is diverted towards the collector. The remainder leaves the base terminal.

Using a convenient notation, Ie = Ic + Ib

So Ib = Ie - Ic

Alpha is the proportion of Ie that flows to the collector,

ɑ = Ic/Ie ————— Definition

So Ic = ɑ Ie

Also Ib = Ie - ɑ Ie = Ie(1 - ɑ)

Beta is the current gain, the ratio of Ic to Ib.

β = Ic/Ib ————— Definition

Substituting in this definition for Ib and Ic

β = (ɑ Ie)/ Ie(1 - ɑ)

Cancelling Ie leaves:

β = ɑ/(1 - ɑ)

Test: The Common Base Configuration - Question 9

A transistor has an IE of 0.9mA and amplification factor of 0.98. What will be the IC?

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

Given, IE = 0.9mA, α=0.98
We know, α= IC/IE
So, IC=0.98*0.9=0.882mA.

Test: The Common Base Configuration - Question 10

The collector current is 2.945A and α=0.98. The leakage current is 2µA. What is the emitter current and base current?

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

 (IC – ICBO)/α=IE
= (2.945-0.002)/0.98=3mA.
IE=IC+IB . So, IB=3-2.495=0.055mA=55µA.

Test: The Common Base Configuration - Question 11

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: The Common Base Configuration - Question 11

Formula Used:

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 =  50 × 10-3 mA

α = 0.92

Hence,

IC = αIE + ICBO = 0.92 × 1 + 50 × 10-3 = 0.97 A

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