Semiconductor
246 Questions
Start JEE Mains Test
2021
Q151
JEE Mains
Numerical
10 Mar 2026
A zener diode having zener voltage 8 V and power dissipation rating of 0.5 W is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance Rp is .................... $\Omega$.
Correct Answer: 192
Explanation:
P = Vi
0.5 = 8i
i = ${1 \over {16}}$A
E = 20 = 8 + i Rp
Rp = 12 $\times$ 16 = 192$\Omega$
0.5 = 8i
i = ${1 \over {16}}$A
E = 20 = 8 + i Rp
Rp = 12 $\times$ 16 = 192$\Omega$
2021
Q152
JEE Mains
Numerical
10 Mar 2026
For the forward biased diode characteristics shown in the figure, the dynamic resistance at ID = 3 mA will be __________ $\Omega$.
Correct Answer: 25
Explanation:
${R_d} = {{dV} \over {di}} = {1 \over {{{di} \over {dv}}}} = {1 \over {{{5 - 1 \times {{10}^{ - 3}}} \over {0.75 - 0.65}}}}$
$ = {{100} \over 4} = 25\Omega $
$ = {{100} \over 4} = 25\Omega $
2021
Q153
JEE Mains
Numerical
10 Mar 2026
The typical output characteristics curve for a transistor working in the common-emitter configuration is shown in the figure.
The estimated current gain from the figure is __________.
The estimated current gain from the figure is __________.
Correct Answer: 200
Explanation:
For common emitter configuration
$\beta $ = ${{\Delta {I_C}} \over {\Delta {I_B}}} = {{\left( {4 - 2} \right)\,mA} \over {\left( {20 - 10} \right)\,\mu A}}$
= ${2 \over {10}} \times {{{{10}^{ - 3}}} \over {{{10}^{ - 6}}}}$ = 200
$\beta $ = ${{\Delta {I_C}} \over {\Delta {I_B}}} = {{\left( {4 - 2} \right)\,mA} \over {\left( {20 - 10} \right)\,\mu A}}$
= ${2 \over {10}} \times {{{{10}^{ - 3}}} \over {{{10}^{ - 6}}}}$ = 200
2021
Q154
JEE Mains
Numerical
10 Mar 2026
An npn transistor operates as a common emitter amplifier with a power gain of 106. The input circuit resistance is 100$\Omega$ and the output load resistance is 10 K$\Omega$. The common emitter current gain '$\beta$' will be ________. (Round off to the Nearest Integer).
Correct Answer: 100
Explanation:
Power gain = 106
Input resistance = 100$\Omega$
Output load resistance = 10K$\Omega$
Power gain = ${\beta^2} \times {{{r_{out}}} \over {{R_{in}}}}$
$ \Rightarrow $ ${10^6} = {\beta ^2} \times {{10 \times {{10}^3}} \over {100}}$
$ \Rightarrow $ $\beta$2 = 104
$ \Rightarrow $ $\beta$ = 100
Input resistance = 100$\Omega$
Output load resistance = 10K$\Omega$
Power gain = ${\beta^2} \times {{{r_{out}}} \over {{R_{in}}}}$
$ \Rightarrow $ ${10^6} = {\beta ^2} \times {{10 \times {{10}^3}} \over {100}}$
$ \Rightarrow $ $\beta$2 = 104
$ \Rightarrow $ $\beta$ = 100
2021
Q155
JEE Mains
Numerical
10 Mar 2026
The value of power dissipated across the zener diode (Vz = 15V) connected in the circuit as shown in the figure is x $\times$ 10$-$1 watt.
The value of x, to the nearest integer, is __________.
The value of x, to the nearest integer, is __________.
Correct Answer: 5
Explanation:
$i = {7 \over {35}} = {1 \over 5}A$
${i_1} = {{15} \over {90}} = {1 \over 6}A$
${i_2} = i - {i_1}$
${i_2} = {1 \over 5} - {1 \over 6}$
${i_2} = {1 \over {30}}A$
Power across diode; P = V2 i2
$P = 15 \times {1 \over {30}}$
P = 0.5 W
$ \therefore $ P = 5 $\times$ 10$-$1 W
2021
Q156
JEE Mains
Numerical
10 Mar 2026
In the logic circuit shown in the figure, if input A and B are 0 to 1 respectively, the output at Y would be 'x'.
The value of x is ____________.
The value of x is ____________.
Correct Answer: 0
Explanation:
2021
Q157
JEE Mains
Numerical
10 Mar 2026
The zener diode has a Vz = 30V. The current passing through the diode for the following circuit is ________ mA.
Correct Answer: 9
Explanation:
$I = {{90 - 30} \over 4} = 15mA$
${I_1} = {{30} \over {5K\Omega }} = 6mA$
${I_2} = 15mA - 6mA = 9mA$
2021
Q158
JEE Mains
Numerical
10 Mar 2026
The circuit contains two diodes each with a forward resistance of 50$\Omega$ and with infinite reverse resistance. If the battery voltage is 6V, the current through the 120$\Omega$ resistance is ____________ mA.
Correct Answer: 20
Explanation:
Given, forward resistance, R1 = 50 $\Omega$
Reverse resistance, R2 = infinity
Battery voltage = 6V
According to circuit diagram,
In this case, diode D1 is forward biased, whereas diode D2 is reverse biased.
So, D2 will act as open circuit.
$6 - 50I - 130I - 120I = 0$
$ \Rightarrow 6 = 300I$
$ \Rightarrow I = {6 \over {300}} = {1 \over {50}}$
$ = {2 \over {100}} = 0.02$ A = 20 mA
Hence, current through 120 $\Omega$ = 20 mA
2021
Q159
JEE Mains
Numerical
10 Mar 2026
In connection with the circuit drawn below, the value of current flowing through 2 k$\Omega$ resistor is _________ $\times$ 10$-$4 A.
Correct Answer: 25
Explanation:
Given, resistance, R = 2 k$\Omega$ = 2 $\times$ 103 $\Omega$
In Zener breakdown,
$i = {V \over R} = {5 \over {2 \times {{10}^3}}} = 2.5 \times {10^{ - 3}}$
$\because$ $x \times {10^{ - 4}} = 25 \times {10^{ - 4}}$
$\therefore$ x = 25
In Zener breakdown,
$i = {V \over R} = {5 \over {2 \times {{10}^3}}} = 2.5 \times {10^{ - 3}}$
$\because$ $x \times {10^{ - 4}} = 25 \times {10^{ - 4}}$
$\therefore$ x = 25
2020
Q160
JEE Mains
MCQ
10 Mar 2026
Identify the correct output signal Y in the given combination of gates (as shown) for the given
inputs A and B.
A.
B.
C.
D.
2020
Q161
JEE Mains
MCQ
10 Mar 2026
Two Zener diodes (A and B) having breakdown
voltages of 6 V and 4 V respectively, are
connected as shown in the circuit below. The
output voltage V0 variation with input voltage
linearly increasing with time, is given by :
(Vinput = 0 V at t = 0)
(figures are qualitative)
(Vinput = 0 V at t = 0)
(figures are qualitative)
A.
B.
C.
D.
2020
Q162
JEE Mains
MCQ
10 Mar 2026
With increasing biasing voltage of a photodiode,
the photocurrent magnitude :
A.
Increases initially and after attaining
certain value, it decreases
B.
Increases linearly
C.
Increases initially and saturates finally
D.
Remains constant
2020
Q163
JEE Mains
MCQ
10 Mar 2026
Identify the operation performed by the circuit given below :
A.
AND
B.
NAND
C.
OR
D.
NOT
2020
Q164
JEE Mains
MCQ
10 Mar 2026
Take the breakdown voltage of the
zener diode used in the given circuit as 6V.
For the input voltage shown in figure below,
the time variation of the output voltage is:
(Graphs drawn are schematic and not to scale)
zener diode used in the given circuit as 6V.
For the input voltage shown in figure below,
the time variation of the output voltage is:
(Graphs drawn are schematic and not to scale)
A.
B.
C.
D.
2020
Q165
JEE Mains
MCQ
10 Mar 2026
If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then
its band gap energy is :
Planck’s constant h = 6.63 $ \times $ 10–34 J.s. Speed of light c = 3 $ \times $ 108 m/s
Planck’s constant h = 6.63 $ \times $ 10–34 J.s. Speed of light c = 3 $ \times $ 108 m/s
A.
1.5 eV
B.
2.0 eV
C.
3.1 eV
D.
1.1 eV
2020
Q166
JEE Mains
MCQ
10 Mar 2026
Which of the following will NOT be observed when a multimeter (operating in resistance measuring
mode) probes connected across a component, are just reversed?
A.
Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the
chosen component is metal wire.
B.
Multimeter shows a deflection, accompanied by a splash of light out of connected component in one direction and NO deflection on reversing the probes if the chosen component is LED.
C.
Multimeter shows an equal deflection in both cases i.e. before and after reversing the probes if
the chosen component is resistor.
D.
Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the
chosen component is capacitor.
2020
Q167
JEE Mains
MCQ
10 Mar 2026
When a diode is forward biased, it has a
voltage drop of 0.5 V. The safe limit of current
through the diode is 10 mA. If a battery of emf
1.5 V is used in the circuit, the value of
minimum resistance to be connected in series
with the diode so that the current does not
exceed the safe limit is
A.
50 $\Omega $
B.
200 $\Omega $
C.
300 $\Omega $
D.
100 $\Omega $
2020
Q168
JEE Mains
MCQ
10 Mar 2026
In the following digital circuit, what will be the
output at ‘Z’, when the
input (A, B) are (1, 0), (0, 0), (1, 1,), (0, 1)
input (A, B) are (1, 0), (0, 0), (1, 1,), (0, 1)
A.
1, 1, 0, 1
B.
0, 1, 0, 0
C.
1, 0, 1, 1
D.
0, 0, 1, 0
2020
Q169
JEE Mains
MCQ
10 Mar 2026
The current i in the network is :
A.
0.6 A
B.
0.3 A
C.
0 A
D.
0.2 A
2020
Q170
JEE Mains
MCQ
10 Mar 2026
Two identical capacitors A and B, charged to
the same potential 5V are connected in two
different circuits as shown below at time t = 0.
If the charge on capacitors A and B at time
t = CR is QA and QB respectively, then (Here
e is the base of natural logarithm)
A.
QA = ${{CV} \over e}$, QB = ${{VC} \over 2}$
B.
QA = ${{CV} \over 2}$, QB = ${{VC} \over e}$
C.
QA = VC, QB = ${{VC} \over e}$
D.
QA = VC, QB = CV
2020
Q171
JEE Mains
MCQ
10 Mar 2026
In the given circuit, value of Y is :
A.
toggles between 0 and 1
B.
1
C.
will not execute
D.
0
2020
Q172
JEE Mains
MCQ
10 Mar 2026
Boolean relation at the output stage-Y for the
following circuit is :
A.
A.B
B.
$\overline A + \overline B $
C.
$\overline A .\overline B $
D.
A + B
2020
Q173
JEE Mains
MCQ
10 Mar 2026
In the figure, potential difference between A and B is :
A.
10 V
B.
5 V
C.
zero
D.
15 V
2020
Q174
JEE Mains
MCQ
10 Mar 2026
Which of the following gives a reversible operation?
A.
B.
C.
D.
2020
Q175
JEE Mains
Numerical
10 Mar 2026
The output characteristics of a transistor is shown in the figure.
When VCE is 10V and IC = 4.0 mA, then value of $\beta $ac is __________.
When VCE is 10V and IC = 4.0 mA, then value of $\beta $ac is __________.
Correct Answer: 150
Explanation:
We know, $\beta $ac = ${{\Delta {I_C}} \over {\Delta {I_B}}}$
$\Delta $IB = (30 - 20) = 10 $\mu $A
$\Delta $IC = (4.5 - 3) = 1.5 mA
$ \therefore $ $\beta $ac = ${{1.5 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$ = 150
$\Delta $IB = (30 - 20) = 10 $\mu $A
$\Delta $IC = (4.5 - 3) = 1.5 mA
$ \therefore $ $\beta $ac = ${{1.5 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$ = 150
2020
Q176
JEE Mains
Numerical
10 Mar 2026
The circuit shown below is working as a 8 V
dc regulated voltage source. When 12 V is used
as input, the power dissipated (in mW) in each
diode is; (considering both zener diodes are
identical) _________.
Correct Answer: 40
Explanation:
Current in circuit = ${{12 - 8} \over {200 + 200}}$ = 10-2 A
So power dissipited in each diode = VI
= 4 $ \times $ 10-2
= 40 mW
So power dissipited in each diode = VI
= 4 $ \times $ 10-2
= 40 mW
2020
Q177
JEE Mains
Numerical
10 Mar 2026
Both the diodes used in the circuit shown are
assumed to be ideal and have negligible
resistance when these are forward biased. Built
in potential in each diode is 0.7 V. For the input
voltages shown in the figure, the voltage
(in Volts) at point A is __________.
Correct Answer: 12
Explanation:
VA = 12.7 – 0.7 = 12 Volt
Diode left hand diode is forward biased and right hand diode is reverse biased.
Diode left hand diode is forward biased and right hand diode is reverse biased.
2019
Q178
JEE Mains
MCQ
10 Mar 2026
Figure shows a DC voltage regulator circuit, with a Zener diode of breakdown voltage = 6V. If the
unregulated input voltage varies between 10 V to 16 V, then what is maximum Zener current?
A.
3.5 mA
B.
1.5 mA
C.
2.5 mA
D.
7.5 mA
2019
Q179
JEE Mains
MCQ
10 Mar 2026
The transfer characteristic curve of a transistor, having input and output resistance 100 $\Omega $ and 100 k$\Omega $
respectively is shown in the figure. The voltage and power gain, are respectively:
A.
5 × 104, 5 × 105
B.
5 × 104, 5 × 106
C.
5 × 104, 2.5 × 106
D.
2.5 × 104, 2.5 × 106
2019
Q180
JEE Mains
MCQ
10 Mar 2026
The truth table for the circuit given in the fig. is:
A.
$\left| {\matrix{
A & B & Y \cr
0 & 0 & 1 \cr
0 & 1 & 1 \cr
1 & 0 & 1 \cr
1 & 1 & 1 \cr
} } \right|$
B.
$\left| {\matrix{
A & B & Y \cr
0 & 0 & 1 \cr
0 & 1 & 0 \cr
1 & 0 & 0 \cr
1 & 1 & 0 \cr
} } \right|$
C.
$\left| {\matrix{
A & B & Y \cr
0 & 0 & 1 \cr
0 & 1 & 1 \cr
1 & 0 & 0 \cr
1 & 1 & 0 \cr
} } \right|$
D.
$\left| {\matrix{
A & B & Y \cr
0 & 0 & 0 \cr
0 & 1 & 0 \cr
1 & 0 & 1 \cr
1 & 1 & 1 \cr
} } \right|$
2019
Q181
JEE Mains
MCQ
10 Mar 2026
The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener
diode is 6 V and the load resistance is, RL = 4k$\Omega $. The series resistance of the circuit is Ri
= 1 k$\Omega $. If the
battery voltage VB varies from 8 V to 16 V, what are the minimum and maximum values of the current
through Zener diode?
A.
0.5 mA; 8.5 mA
B.
1.5 mA; 8.5 mA
C.
1 mA; 8.5 mA
D.
0.5 mA; 6 mA
Vbreakdwon = 6V, RL = 4k$\Omega $, Ri = 1 k$\Omega $
2019
Q182
JEE Mains
MCQ
10 Mar 2026
An n-p-n transistor operates as a common emitter
amplifier, with a power gain of 60 dB. The input
circuit resistance is 100$\Omega $ and the output load
resistance is 10 k$\Omega $. The common emitter
current gain $\beta $ is :
A.
104
B.
102
C.
6 × 102
D.
60
2019
Q183
JEE Mains
MCQ
10 Mar 2026
The logic gate equivalent to the given logic
circuit is :-
A.
NAND
B.
AND
C.
NOR
D.
OR
2019
Q184
JEE Mains
MCQ
10 Mar 2026
An NPN transistor is used in common emitter
configuration as an amplifier with 1 k$\Omega $ load
resistance. Signal voltage of 10 mV is applied
across the base-emitter. This produces a 3 mA
change in the collector current and 15μA
change in the base current of the amplifier. The
input resistance and voltage gain are :
A.
0.67 kW, 200
B.
0.33 kW, 1.5
C.
0.67 kW, 300
D.
0.33 kW, 300
2019
Q185
JEE Mains
MCQ
10 Mar 2026
A common emitter amplifier circuit, built using
an npn transistor, is shown in the figure. Its dc
current gain is 250, RC = 1k$\Omega$ and VCC = 10 V.
What is the minimum base current for VCE to
reach saturation ?
A.
100 $\mu $A
B.
7 $\mu $A
C.
10 $\mu $A
D.
40 $\mu $A
2019
Q186
JEE Mains
MCQ
10 Mar 2026
The reverse breakdown voltage of a Zener
diode is 5.6 V in the given circuit.
The current IZ through the Zener is :
The current IZ through the Zener is :
A.
7 mA
B.
17 mA
C.
15mA
D.
10 mA
2019
Q187
JEE Mains
MCQ
10 Mar 2026
In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC = 5V, $\beta $dc = 200, RB = 100 k$\Omega $, RC = 1 k$\Omega $ and VBE = 1.0 V. The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively :
A.
20 $\mu $A and 2.8 V
B.
25 $\mu $A and 2.8 V
C.
20 $\mu $A and 3.5 V
D.
25 $\mu $A and 3.5 V
2019
Q188
JEE Mains
MCQ
10 Mar 2026
The output of the given logic circuit is :
A.
$\overline A B$
B.
$AB + \overline {AB} $
C.
$A\overline B + \overline A B$
D.
$A\overline B $
2019
Q189
JEE Mains
MCQ
10 Mar 2026
The circuit shown below contains two ideal diodes, each with a forward resistance of 50 $\Omega $. If the battery voltage is 6 V, the current through the 100 $\Omega $ resistance (in Amperes) is :
A.
0.027
B.
0.030
C.
0.036
D.
0.020
2019
Q190
JEE Mains
MCQ
10 Mar 2026
In the given circuit the current through Zener Diode is close to:
A.
0.0 mA
B.
6.7 mA
C.
4.0 mA
D.
6.0 mA
2019
Q191
JEE Mains
MCQ
10 Mar 2026
For the circuit shown below, the current through the Zener diode is -
A.
5 mA
B.
zero
C.
14 mA
D.
9 mA
2019
Q192
JEE Mains
MCQ
10 Mar 2026
To get output 1 at R, for the given logic gate circuit the input values must be
A.
x = 0, y = 0
B.
x = 1, y = 0
C.
x = 0, y = 1
D.
x = 1, y = 1
2019
Q193
JEE Mains
MCQ
10 Mar 2026
Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of V0 changes by : (assume that the Ge diode has large breakdown voltage)
A.
0.8 V
B.
0.6 V
C.
0.2 V
D.
0.4 V
2019
Q194
JEE Mains
MCQ
10 Mar 2026
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m$-$3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to :
A.
$2\,\Omega $m
B.
4$\,\Omega $m
C.
0.4 $\,\Omega $m
D.
0.2 $\,\Omega $m
2018
Q195
JEE Mains
MCQ
10 Mar 2026
In the given circuit, the current through zener diode is :
A.
5.5 mA
B.
6.7 mA
C.
2.5 mA
D.
3.3 mA
2018
Q196
JEE Mains
MCQ
10 Mar 2026
The reading of the ammeter for a silicon diode in the given circuit
is :
A.
13.5 mA
B.
0
C.
15 mA
D.
11.5 mA
2018
Q197
JEE Mains
MCQ
10 Mar 2026
Truth table for the following digital circuit will be :
A.
B.
C.
D.
2018
Q198
JEE Mains
MCQ
10 Mar 2026
In a common emitter configuration with suitable bias, it is given that ${R_L}$ is the load resistance and ${R_{BE}}$ is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by :
$\beta $ is curret gain, ${{\rm I}_B},{{\rm I}_C}$ and ${{\rm I}_E}$ are respectively base, collector and emitter currents.
$\beta $ is curret gain, ${{\rm I}_B},{{\rm I}_C}$ and ${{\rm I}_E}$ are respectively base, collector and emitter currents.
A.
$\beta {{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_B}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}$
B.
$\beta {{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_E}} \over {\Delta {{\rm I}_B}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}$
C.
${\beta ^2}{{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_E}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}$
D.
${\beta ^2}{{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_B}}},\beta {{{R_L}} \over {{R_{BE}}}}$
2017
Q199
JEE Mains
MCQ
10 Mar 2026
The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is :
A.
9.6 mA
B.
6.9 mA
C.
0.69 mA
D.
69 mA
2017
Q200
JEE Mains
MCQ
10 Mar 2026
The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias
resistance is :
A.
10
B.
10$-$6
C.
106
D.
100