This set of Class 12 Physics Chapter 14 Multiple Choice Questions & Answers (MCQs) focuses on “Semiconductor Electronics – Junction Transistor”.
1. How many types of transistors are there?
a) 2
b) 5
c) 1
d) 7
View Answer
Explanation: A transistor has three doped regions forming two p-n junctions between them. There are two types of transistors, namely n-p-n transistors and p-n-p transistors. In n-p-n transistors, the two segments of the n-type semiconductor are separated by a segment of p-type semiconductor, and in a p-n-p transistor, it’s just the opposite scenario.
2. Which of the following supplies charge carriers in a transistor?
a) Collector
b) Base
c) Emitter
d) Charger
View Answer
Explanation: Emitter is the section or region on one side of the transistor that supplies charge carriers. It is heavily doped and is always kept forward biased with respect to the base so that it can supply a large number of charge carriers to the base.
3. Which among the following is larger compared to the other regions of a transistor?
a) Emitter
b) Collector
c) Base
d) Charger
View Answer
Explanation: The collector is the section on the other side of the transistor that collects the charge carriers supplied by the emitter. It is moderately doped but large in size and is always kept in reverse bias with respect to the base.
4. The base forms two p-n junctions with the emitter and the collector.
a) True
b) False
View Answer
Explanation: Yes, this is a true statement. The base is the middle section of the transistor that forms two p-n junctions with the emitter and the collector. It is very thin and lightly doped so as to pass most of the emitter injected charge carriers to the collector.
5. On which of the following does base current not depend on?
a) The thickness of the base
b) The shape of the transistor
c) Doping levels
d) Number of charge carriers
View Answer
Explanation: In an n-p-n transistor, as the base is very thin and lightly doped, a very few electrons from the emitter combine with the holes of the base, giving rise to base current and the electrons finally collected by the positive terminal of the battery gives rise to collector current. This base current is a small fraction of collector current depending on the shape of a transistor, thickness of the base, doping levels, and bias voltage.
6. Identify the relationship between base current amplification (α) and emitter current amplification (β).
a) β=\(\frac {\alpha }{1- \alpha}\)
b) β=\(\frac {1 – \alpha }{\alpha}\)
c) β=\(\frac {\alpha }{1 + \alpha}\)
d) β=\(\frac {1 + \alpha }{1 – \alpha}\)
View Answer
Explanation: IE=IB+IC
\(\frac {I_E}{I_C} =\frac {I_B}{I_C}\) + 1
\(\frac {1}{\alpha }=\frac {1}{\beta }\) + 1
α = \(\frac {\beta }{1 + \beta}\)
Therefore, β=\(\frac {\alpha }{1- \alpha}\)
7. The current gain of a transistor in a common emitter configuration is 50. If the emitter current is 5.5 mA, find the base current.
a) 0.203 A
b) 0.107 mA
c) 0.107 A
d) 0.203 mA
View Answer
Explanation: The expression for current gain is given as:
β=\(\frac {I_C}{I_B}\)
IC=50IB
Since, IE=IB+IC ➔ IE=IB+50IB
IE=51IB
IB=\(\frac {I_E}{51}=\frac {5.5}{51}\)=0.107 mA
8. DC current gain is the ratio of change in collector current to the change in base current.
a) True
b) False
View Answer
Explanation: No, this is a false statement. DC current is defined as the ratio of the collector current (IC) to the base current (IB). The ratio of change in the collector current to the change in the base current is known as AC current gain.
9. A transistor having α = 0.90 is used in a common base amplifier. If the load resistance is 5.0 kΩ and the dynamic resistance of the emitter junction is 50 Ω, then calculate the voltage gain.
a) 50
b) 100
c) 70
d) 90
View Answer
Explanation: The expression of voltage gain is as follows:
AV=\(\frac {αR_L}{R_e}\)
Given: α = 0.90, RL = 5.0 kΩ = 5000 Ω, Re = 50 Ω
AV=\(\frac {0.90×5000}{50}\)
AV = 90
10. What will the power gain of a transistor if it’s α value is 0.80 and the voltage gain is 95?
a) 70
b) 80
c) 76
d) 75
View Answer
Explanation: Power gain is defined as the ratio of output power to input power. It can also be determined as the product of current gain and voltage gain.
Given: α = 0.80, voltage gain (AV) = 95
The required equation ➔ AP = α × AV
AP=0.80×95
AP=76
Sanfoundry Global Education & Learning Series – Physics – Class 12.
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