Transistor Switching Network MCQ’s

Analog Electronic Circuits Electronics & Communication Engineering

This set of Analog Electronic Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Transistor Switching Network”.

1. For a transistor in saturation, which is true?
a) IC = βIB
b) IC > βIB
c) IC < βIB
d) IC = (β+1)IB

2. Given that the BJT is completely saturated, what is the overdrive?
a) Overdrive = 1
b) Overdrive < 1
c) Overdrive > 1
d) Overdrive > 0

3. At saturation, which of these is not true for a BJT?
a) The collector current IC cannot increase further
b) The base current IB, cannot increase further
c) The collector-to-emitter voltage, VCE is due to the non-zero internal resistance of BJT
d) VCE(saturation) is the minimum voltage drop between C and E

4. Consider the graph of IC vs VI shown below for a transistor. Find the correct relation for region 3 in the diagram.

analog-circuits-questions-answers-transistor-switching-network-q4

a) IC = IC(sat) and VCE = VCE(sat)
b) IC = IC(sat) and VCE = VCC
c) IC = βIB and VCE = VCE(sat)
d) IC = βIB and VCE = VCC

5. For the graph which depicts collector current, find the ON time.

analog-circuits-questions-answers-transistor-switching-network-q6

t1 = 1ms
t2 = 2ms
t3 = 4ms
t4 = 6ms
t5 = 16ms
t6 = 18ms
a) 3ms
b) 1ms
c) 2ms
d) 5ms

6. Which of these relations is true always for the BJT as a switch?
a) Off time >> On time
b) Off time = Storage time – Rise time
c) Off time << On time
d) Off time = Storage time + Delay time

7. What is the ON resistance of a transistor?
a) RON = VCEsat/βIB
b) RON = VCEsat + VA/ICsat
c) RON = VCEsat/(β+1)IB
d) RON = VCEsat/ICsat

8. Find the storage time for the current variation shown below.

analog-circuits-questions-answers-transistor-switching-network-q6

t1 = 2ms
t2 = 3ms
t3 = 4ms
t4 = 6ms
t5 = 19ms
t6 = 20ms
a) 1ms
b) 13ms
c) 3ms
d) 2ms

9. How is BJT used as a faster switch?
a) By operating it in the saturation and cut-off region
b) By operating it in the active and cut-off region
c) By using it in strong saturation
d) By decreasing its ON resistance

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