“Man perfected by society is the best of
all animals; he is the most terrible of all when he lives without law and
without justice.”
– Aristotle
Questions on semiconductor
electronics are generally simple and you can answer most of the questions
correctly without consuming much time. But you need to understand the simple
fundamental principles of operation of various semiconductor devices so that common
mistakes are avoided.
Here are a few
multiple choice questions involving semiconductor electronics:
(1) A square wave of
peak to peak voltage 100 volt is applied between the input terminals A and B of
the circuit shown in the adjoining figure. What is the voltage across the
capacitor C?
(a) 100 V
(b) 0 V
(c) 50 V
(d) 100√2 V
(e) 50√2 V
The diode will pass
only the negative half cycles of the square voltage wave so that the voltage
across the capacitor will be 50 V.
(2) If a sine wave of
peak to peak voltage 100 volt is applied between the input terminals A and B of
the circuit shown in the above question, what will be the voltage across the
capacitor C?
(a) 100√2 V
(b) 50√2 V
(c) 50 V
(d) 100 V
(e) 0 V
In this case also the
diode will pass only the negative half cycles of the applied voltage wave. Even
though the voltage varies during the half cycle, the capacitor can retain the
charge on it so that the voltage across the capacitor will be 50 V itself.
(3) If a 230 volt
house supply alternating voltage is applied between the terminals A and B of
the circuit shown in Question No. 1, what will be the voltage across the
capacitor C?
(a) 230√2 V
(b) 115√2 V
(c) 230 V
(d) 150 V
(e) 0 V
Alternating
voltages and alternating currents (A.C.) supplied to houses are specified in
terms of root mean square (RMS) values. Thus a specified value of 230 volt a.c.
means that the root mean square value of the voltage is 230 V. Therefore, the
peak value of the alternating voltage is 230√2 V.
The diode will conduct during negative half
cycles only and the capacitor will be charged to the peak value of the applied
voltage. Assuming that the capacitor is ideal, the charge will be retained
through out the entire cycle of the applied voltage so that the voltage across
the capacitor will be 230√2 V.
(4) The truth table
for the digital circuit shown in the figure above is
When both A and B
are zero, the NAND gate on the input side
produces an input at level 1 for the NOR gate on the output side. Therefore,
the final output Y = 0.
When A = 0 and B = 1,
the NAND gate on the input side again produces
an input at level 1 for the NOR gate on the output side. In this case also the
final output Y = 0.
When A = 1 and B =
0, the NAND gate on the input side once again
produces an input at level 1 for the NOR gate on the output side. Once again
the final output Y = 0.
When both A and B
are at level 1, the NAND gate on the input side
produces an input at level 0 for the NOR gate on the output side.
The inverter
connected to the other input of the NOR
gate also supplies an input at level 0 for the NOR gate on the output side.
Since both inputs of the NOR gate are at zero level, its output Y = 1.
The correct truth
table is (b).
(5) The transfer
characteristic curve of an npn transistor used as a switch is shown in the
figure. Vi is the base bias voltage and Vo is the collector voltage. Which portions of the characteristic curve are used
for operating the transistor as a switch?
(a) AB and EF
(b) BC, CD and DE
(c) CD only
(d) AB only
(e) EF only
When a transistor
is use as a switch, it operates in the saturation region EF for full conduction
(switch on) and in the cut off region AB for no conduction (switch off). The
correct option is (a).
[When the transistor
operates in the saturation region, the base current is large so that the
collector current also is large. Almost the entire collector supply voltage will then be dropped across the resistance
connected to the collector lead so that the emitter to collector voltage will
be very small. The transistor will operate in the cut off region when the
emitter to base voltage is insufficient to forward bias the emitter base
junction and to make the transistor conduct. Consequently there is no voltage
drop across the resistance connected to the collector lead so that the entire
collector supply voltage appears on the
collector].
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