This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “The Common Base Configuration”.
1. The AC current gain in a common base configuration is_________
a) -∆IC/∆IE
b) ∆IC/∆IE
c) ∆IE/∆IC
d) -∆IE/∆IC
View Answer
Explanation: 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.
2. The value of αac for all practical purposes, for commercial transistors range from_________
a) 0.5-0.6
b) 0.7-0.77
c) 0.8-0.88
d) 0.9-0.99
View Answer
Explanation: 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.
3. A transistor has an IC of 100mA and IB of 0.5mA. What is the value of αdc?
a) 0.787
b) 0.995
c) 0.543
d) 0.659
View Answer
Explanation: Emitter current IE=IC+IB =100+0.5=100.5mA
αdc=IC/IE=100/100.5=0.995.
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.
a) 0.01mA
b) 0.07mA
c) 0.02mA
d) 0.05mA
View Answer
Explanation: 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.
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.
a) 3.64mA and 35µA
b) 2.945mA and 55µA
c) 3.64mA and 33µA
d) 5.89mA and 65µA
View Answer
Explanation: IC=αIE + ICBO
=0.98*3+0.005=2.945mA.
IE=IC+IB . So, IB=3-2.495=0.055mA=55µA.
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.
a) 1.5mA
b) 3.7mA
c) 2.7mA
d) 4.5mA
View Answer
Explanation: 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.
7. The negative sign in the formula of amplification factor indicates_________
a) that IE flows into transistor while IC flows out it
b) that IC flows into transistor while IE flows out it
c) that IB flows into transistor while IC flows out it
d) that IC flows into transistor while IB flows out it
View Answer
Explanation: 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.
8. The relation between α and β is _________
a) β=α/(1-α)
b) α= β/(1+β)
c) β=α/(1+α)
d) α= β/(1- β)
View Answer
Explanation: 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.
9. A transistor has an IE of 0.9mA and amplification factor of 0.98. What will be the IC?
a) 0.745mA
b) 0.564mA
c) 0.236mA
d) 0.882mA
View Answer
Explanation: Given, IE = 0.9mA, α=0.98
We know, α= IC/IE
So, IC=0.98*0.9=0.882mA.
10. The collector current is 2.945A and α=0.98. The leakage current is 2µA. What is the emitter current and base current?
a) 3mA and 55µA
b) 2.945mA and 55µA
c) 3.64mA and 33µA
d) 5.89mA and 65µA
View Answer
Explanation: (IC – ICBO)/α=IE
= (2.945-0.002)/0.98=3mA.
IE=IC+IB . So, IB=3-2.495=0.055mA=55µA.
Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.
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