Test: The Common Base Configuration


10 Questions MCQ Test Electronic Devices | Test: The Common Base Configuration


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QUESTION: 1

 The AC current gain in a common base configuration is_________

Solution:

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.

QUESTION: 2

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

Solution:

 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.

QUESTION: 3

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

Solution:

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

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.

Solution:

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.

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.

Solution:

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

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.

Solution:

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.

QUESTION: 7

The negative sign in the formula of amplification factor indicates_________

Solution:

 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.

QUESTION: 8

 The relation between α and β is _________

Solution:

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)

QUESTION: 9

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

Solution:

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

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?

Solution:

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