JEE Main Numericals: Semiconductor Electronics - Materials, Devices & Simple Circuits

 Page 1


1 . A pure semiconductor has equal electron and hole
concentration of 10
16
 m
–3
. Doping by indium increases
number of hole concentration n
h
 to 5 × 10
22
 m
–3
. Then, the
value of number of electron concentration n
e
 in the doped
semiconductor is
(a) 10
6
/m
3
(b) 10
22
/m
3
(c) 2 × 10
6
/m
3
(d) 2 × 10
9
/m
3
2 . A change of 8.0 mA in the emitter current bring a change of
7.9 mA in the collector current. The values of parameters a
and b are respectively
(a) 0.99, 90 (b) 0.96,79 (c) 0.97,99 (d) 0.99,79
3 . The following circut diagram represents
Y
B
A
(a) OR gate (b) XOR gate
(c) AND gate (d) NAND gate
4 . Which of the following statements is incorrect?
(a) The resistance of intrinsic semiconductors decrease
with increase of temperature
(b) Doping pure Si with trivalent impurities give p-type
semiconductors
(c) The majority carriers in n-type semiconductors are holes
(d) A p-n junction can act as a semiconductor diode
5 . In a npn transistor 10
10
 electrons enter the emitter in
10
–6
 s. 4% of the electrons are lost in the base. The current
transfer ratio will be
(a) 0.98 (b) 0.97
(c) 0.96 (d) 0.94
r o t c u d n o c i m e S 	 s c i n o r t c e l E :	 s l a i r e t a M , s e c i v e D 	
d n a 	 e l p m i S 	 s t i u c r i C
J E E 	 n i a M 	 s l a c i r e m u N
Page 2


1 . A pure semiconductor has equal electron and hole
concentration of 10
16
 m
–3
. Doping by indium increases
number of hole concentration n
h
 to 5 × 10
22
 m
–3
. Then, the
value of number of electron concentration n
e
 in the doped
semiconductor is
(a) 10
6
/m
3
(b) 10
22
/m
3
(c) 2 × 10
6
/m
3
(d) 2 × 10
9
/m
3
2 . A change of 8.0 mA in the emitter current bring a change of
7.9 mA in the collector current. The values of parameters a
and b are respectively
(a) 0.99, 90 (b) 0.96,79 (c) 0.97,99 (d) 0.99,79
3 . The following circut diagram represents
Y
B
A
(a) OR gate (b) XOR gate
(c) AND gate (d) NAND gate
4 . Which of the following statements is incorrect?
(a) The resistance of intrinsic semiconductors decrease
with increase of temperature
(b) Doping pure Si with trivalent impurities give p-type
semiconductors
(c) The majority carriers in n-type semiconductors are holes
(d) A p-n junction can act as a semiconductor diode
5 . In a npn transistor 10
10
 electrons enter the emitter in
10
–6
 s. 4% of the electrons are lost in the base. The current
transfer ratio will be
(a) 0.98 (b) 0.97
(c) 0.96 (d) 0.94
r o t c u d n o c i m e S 	 s c i n o r t c e l E :	 s l a i r e t a M , s e c i v e D 	
d n a 	 e l p m i S 	 s t i u c r i C
J E E 	 n i a M 	 s l a c i r e m u N
6 . The correct truth table for system of four NAND gates as
shown in figure is :
Y
A
B
( a)
A B Y
0 0 0
0 1 1
1 0 1
1 1 0
(b)
A B Y
0 0 0
0 1 0
1 0 1
1 1 1
(c)
A B Y
0 0 1
0 1 1
1 0 0
1 1 0
(d)
A B Y
0 0 1
0 1 0
1 0 1
1 1 1
7 . A Zener diode is connected to a battery and a load as show
below:
A
B
I
I
L
60 V
R = 2k
L
W
4 kW
I
Z
10 V = V
Z
The currents, I, I
Z
 and I
L
 are respectively.
(a) 15 mA, 5 mA, 10 mA
(b) 15 mA, 7.5 mA, 7.5 mA
(c) 12.5 mA, 5 mA, 7.5 mA
(d) 12.5 mA, 7.5 mA, 5 mA
8 . In common emitter amplifier, the current gain is 62. The
collector resistance and input resistance are 5 kW  an 500W
respectively . If the input voltage is 0.01 V , the output voltage
is
(a) 0.62 V (b) 6.2 V (c) 62 V (d) 620 V
9 . The circuit diagram shows a logic combination with the
states of outputs X, Y and Z given for inputs P, Q, R and S
all at state 1. When inputs P and R change to state 0 with
inputs Q and S still at 1, the states of outputs X, Y and Z
change to
P(1)
Q(1)
R(1)
S(1)
Y(1)
X(1)
Z(0)
(a) 1, 0, 0 (b) 1, 1, 1 (c) 0, 1, 0 (d) 0, 0, 1
10. A sinusoidal voltage of amplitude 25 volt and frequency
50Hz is applied to a half wave rectifier using P-n junction
diode. No filter is used and the load resistance is 1000W.
The forward resistance R
f
 of ideal diode is 10W. The
percentage efficiency of rectifier is
(a) 40% (b) 20% (c) 30% (d) 15%
11. The intrinsic conductivity of germanium at 27° is 2.13 mho
m
–1
 and mobilities of electrons and holes are 0.38 and 0.18
m
2
V
–1
s
–1
 respectively. The density of charge carriers is
(a) 2.37 × 10
19
 m
–3
(b) 3.28 × 10
19
 m
–3
(c) 7.83 × 10
19
 m
–3
(d) 8.47 × 10
19
 m
–3
12. The circuit has two oppositively connected ideal diodes in
parallel. The current flowing in the circuit is
4W
D
1
D
2
2W 3W
12V
(a) 1.71 A (b) 2.00 A (c) 2.31 A (d) 1.33 A
Page 3


1 . A pure semiconductor has equal electron and hole
concentration of 10
16
 m
–3
. Doping by indium increases
number of hole concentration n
h
 to 5 × 10
22
 m
–3
. Then, the
value of number of electron concentration n
e
 in the doped
semiconductor is
(a) 10
6
/m
3
(b) 10
22
/m
3
(c) 2 × 10
6
/m
3
(d) 2 × 10
9
/m
3
2 . A change of 8.0 mA in the emitter current bring a change of
7.9 mA in the collector current. The values of parameters a
and b are respectively
(a) 0.99, 90 (b) 0.96,79 (c) 0.97,99 (d) 0.99,79
3 . The following circut diagram represents
Y
B
A
(a) OR gate (b) XOR gate
(c) AND gate (d) NAND gate
4 . Which of the following statements is incorrect?
(a) The resistance of intrinsic semiconductors decrease
with increase of temperature
(b) Doping pure Si with trivalent impurities give p-type
semiconductors
(c) The majority carriers in n-type semiconductors are holes
(d) A p-n junction can act as a semiconductor diode
5 . In a npn transistor 10
10
 electrons enter the emitter in
10
–6
 s. 4% of the electrons are lost in the base. The current
transfer ratio will be
(a) 0.98 (b) 0.97
(c) 0.96 (d) 0.94
r o t c u d n o c i m e S 	 s c i n o r t c e l E :	 s l a i r e t a M , s e c i v e D 	
d n a 	 e l p m i S 	 s t i u c r i C
J E E 	 n i a M 	 s l a c i r e m u N
6 . The correct truth table for system of four NAND gates as
shown in figure is :
Y
A
B
( a)
A B Y
0 0 0
0 1 1
1 0 1
1 1 0
(b)
A B Y
0 0 0
0 1 0
1 0 1
1 1 1
(c)
A B Y
0 0 1
0 1 1
1 0 0
1 1 0
(d)
A B Y
0 0 1
0 1 0
1 0 1
1 1 1
7 . A Zener diode is connected to a battery and a load as show
below:
A
B
I
I
L
60 V
R = 2k
L
W
4 kW
I
Z
10 V = V
Z
The currents, I, I
Z
 and I
L
 are respectively.
(a) 15 mA, 5 mA, 10 mA
(b) 15 mA, 7.5 mA, 7.5 mA
(c) 12.5 mA, 5 mA, 7.5 mA
(d) 12.5 mA, 7.5 mA, 5 mA
8 . In common emitter amplifier, the current gain is 62. The
collector resistance and input resistance are 5 kW  an 500W
respectively . If the input voltage is 0.01 V , the output voltage
is
(a) 0.62 V (b) 6.2 V (c) 62 V (d) 620 V
9 . The circuit diagram shows a logic combination with the
states of outputs X, Y and Z given for inputs P, Q, R and S
all at state 1. When inputs P and R change to state 0 with
inputs Q and S still at 1, the states of outputs X, Y and Z
change to
P(1)
Q(1)
R(1)
S(1)
Y(1)
X(1)
Z(0)
(a) 1, 0, 0 (b) 1, 1, 1 (c) 0, 1, 0 (d) 0, 0, 1
10. A sinusoidal voltage of amplitude 25 volt and frequency
50Hz is applied to a half wave rectifier using P-n junction
diode. No filter is used and the load resistance is 1000W.
The forward resistance R
f
 of ideal diode is 10W. The
percentage efficiency of rectifier is
(a) 40% (b) 20% (c) 30% (d) 15%
11. The intrinsic conductivity of germanium at 27° is 2.13 mho
m
–1
 and mobilities of electrons and holes are 0.38 and 0.18
m
2
V
–1
s
–1
 respectively. The density of charge carriers is
(a) 2.37 × 10
19
 m
–3
(b) 3.28 × 10
19
 m
–3
(c) 7.83 × 10
19
 m
–3
(d) 8.47 × 10
19
 m
–3
12. The circuit has two oppositively connected ideal diodes in
parallel. The current flowing in the circuit is
4W
D
1
D
2
2W 3W
12V
(a) 1.71 A (b) 2.00 A (c) 2.31 A (d) 1.33 A
13. A working transistor with its three legs marked P, Q and R is
tested using a multimeter. No conduction is found between
P and Q. By connecting the common (negative) terminal of
the multimeter to R and the other (positive) terminal to P o r
Q, some resistance is seen on the multimeter. Which of the
following is true for the transistor?
(a) It is an npn transistor with R as base
(b) It is a pnp transistor with R as base
(c) It is a pnp transistor with R as emitter
(d) It is an npn transistor with R as collector
14. The diagram of a logic circuit is given below. The output F
of the circuit is represented by
X
W
Y
F
W
(a) W . (X + Y) (b) W . (X . Y)
(c) W + (X . Y) (d) W + (X + Y)
15. The concentration of hole-electron pairs in pure silicon at
T = 300 K is 7 × 10
15
 per cubic meter. Antimony is doped
into silicon in a proportion of 1 atom in 10
7
 Si atoms.
Assuming half of the impurity atoms contribute electron in
the conduction band, calculate the factor by which the
number of charge carriers increases due to doping. The
number of silicon atoms per cubic meter is 5 × 10
28
(a) 2.8 × 10
5
(b) 3.1 × 10
2
(c) 4.2 × 10
5
(d) 1.8 × 10
5
16. For a transistor amplifier in common emitter configuration
for load impedance of 1kW (h
f e
 = 50 and h
0e
 = 25) the current
gain is
(a) – 24.8 (b) – 15.7 (c) – 5.2 (d) – 48.78
17. A PN-junction has a thickness of the order of
(a) 1 cm (b) 1 mm (c) 10
–6
 m (d) 10
–12
 cm
18. In a common emitter (CE) amplifier having a voltage gain G ,
the transistor used has transconductance 0.03 mho and
current gain 25. If the above transistor is replaced with
another one with transconductance 0.02 mho and current
gain 20, the voltage gain will be
(a) 1.5 G (b)
1
3
 G (c)
5
4
 G (d)
2
3
 G
19. Which of the junction diodes shown below are forward
biased?
( a)
–10 V
R
–5 V
(b)
+10 V
R
+5 V
(c)
–10 V
R
(d)
–5 V
R
20. For LED’s to emit light in visible region of electromagnetic
light, it should have energy band gap in the range of:
(a) 0.1 eV to 0.4 eV (b) 0.5 eV to 0.8 eV
(c) 0.9 eV to 1.6 eV (d) 1.7 eV to 3.0 eV
21. In a C E transistor amplifier, the audio signal voltage across
th e c o llec to r resista n c e o f 2 kW  is 2V .  If the base resistance
is 1kW  and the  current amplification of the transistor is 100,
the input signal voltage is :
(a) 0.1 V (b) 1.0 V (c) 1 mV (d) 10  mV
22. The current gain in the common emitter mode of a transistor
is 10. The input impedance is 20k W and load of resistance is
100kW. The power gain is
(a) 300 (b) 500 (c) 200 (d) 100
23. In the circuit given below, A and B represent two inputs and
C represents the output.
C
A
B
The circuit represents
(a) NOR gate (b) AND gate
(c) NAND gate (d) OR gate
Page 4


1 . A pure semiconductor has equal electron and hole
concentration of 10
16
 m
–3
. Doping by indium increases
number of hole concentration n
h
 to 5 × 10
22
 m
–3
. Then, the
value of number of electron concentration n
e
 in the doped
semiconductor is
(a) 10
6
/m
3
(b) 10
22
/m
3
(c) 2 × 10
6
/m
3
(d) 2 × 10
9
/m
3
2 . A change of 8.0 mA in the emitter current bring a change of
7.9 mA in the collector current. The values of parameters a
and b are respectively
(a) 0.99, 90 (b) 0.96,79 (c) 0.97,99 (d) 0.99,79
3 . The following circut diagram represents
Y
B
A
(a) OR gate (b) XOR gate
(c) AND gate (d) NAND gate
4 . Which of the following statements is incorrect?
(a) The resistance of intrinsic semiconductors decrease
with increase of temperature
(b) Doping pure Si with trivalent impurities give p-type
semiconductors
(c) The majority carriers in n-type semiconductors are holes
(d) A p-n junction can act as a semiconductor diode
5 . In a npn transistor 10
10
 electrons enter the emitter in
10
–6
 s. 4% of the electrons are lost in the base. The current
transfer ratio will be
(a) 0.98 (b) 0.97
(c) 0.96 (d) 0.94
r o t c u d n o c i m e S 	 s c i n o r t c e l E :	 s l a i r e t a M , s e c i v e D 	
d n a 	 e l p m i S 	 s t i u c r i C
J E E 	 n i a M 	 s l a c i r e m u N
6 . The correct truth table for system of four NAND gates as
shown in figure is :
Y
A
B
( a)
A B Y
0 0 0
0 1 1
1 0 1
1 1 0
(b)
A B Y
0 0 0
0 1 0
1 0 1
1 1 1
(c)
A B Y
0 0 1
0 1 1
1 0 0
1 1 0
(d)
A B Y
0 0 1
0 1 0
1 0 1
1 1 1
7 . A Zener diode is connected to a battery and a load as show
below:
A
B
I
I
L
60 V
R = 2k
L
W
4 kW
I
Z
10 V = V
Z
The currents, I, I
Z
 and I
L
 are respectively.
(a) 15 mA, 5 mA, 10 mA
(b) 15 mA, 7.5 mA, 7.5 mA
(c) 12.5 mA, 5 mA, 7.5 mA
(d) 12.5 mA, 7.5 mA, 5 mA
8 . In common emitter amplifier, the current gain is 62. The
collector resistance and input resistance are 5 kW  an 500W
respectively . If the input voltage is 0.01 V , the output voltage
is
(a) 0.62 V (b) 6.2 V (c) 62 V (d) 620 V
9 . The circuit diagram shows a logic combination with the
states of outputs X, Y and Z given for inputs P, Q, R and S
all at state 1. When inputs P and R change to state 0 with
inputs Q and S still at 1, the states of outputs X, Y and Z
change to
P(1)
Q(1)
R(1)
S(1)
Y(1)
X(1)
Z(0)
(a) 1, 0, 0 (b) 1, 1, 1 (c) 0, 1, 0 (d) 0, 0, 1
10. A sinusoidal voltage of amplitude 25 volt and frequency
50Hz is applied to a half wave rectifier using P-n junction
diode. No filter is used and the load resistance is 1000W.
The forward resistance R
f
 of ideal diode is 10W. The
percentage efficiency of rectifier is
(a) 40% (b) 20% (c) 30% (d) 15%
11. The intrinsic conductivity of germanium at 27° is 2.13 mho
m
–1
 and mobilities of electrons and holes are 0.38 and 0.18
m
2
V
–1
s
–1
 respectively. The density of charge carriers is
(a) 2.37 × 10
19
 m
–3
(b) 3.28 × 10
19
 m
–3
(c) 7.83 × 10
19
 m
–3
(d) 8.47 × 10
19
 m
–3
12. The circuit has two oppositively connected ideal diodes in
parallel. The current flowing in the circuit is
4W
D
1
D
2
2W 3W
12V
(a) 1.71 A (b) 2.00 A (c) 2.31 A (d) 1.33 A
13. A working transistor with its three legs marked P, Q and R is
tested using a multimeter. No conduction is found between
P and Q. By connecting the common (negative) terminal of
the multimeter to R and the other (positive) terminal to P o r
Q, some resistance is seen on the multimeter. Which of the
following is true for the transistor?
(a) It is an npn transistor with R as base
(b) It is a pnp transistor with R as base
(c) It is a pnp transistor with R as emitter
(d) It is an npn transistor with R as collector
14. The diagram of a logic circuit is given below. The output F
of the circuit is represented by
X
W
Y
F
W
(a) W . (X + Y) (b) W . (X . Y)
(c) W + (X . Y) (d) W + (X + Y)
15. The concentration of hole-electron pairs in pure silicon at
T = 300 K is 7 × 10
15
 per cubic meter. Antimony is doped
into silicon in a proportion of 1 atom in 10
7
 Si atoms.
Assuming half of the impurity atoms contribute electron in
the conduction band, calculate the factor by which the
number of charge carriers increases due to doping. The
number of silicon atoms per cubic meter is 5 × 10
28
(a) 2.8 × 10
5
(b) 3.1 × 10
2
(c) 4.2 × 10
5
(d) 1.8 × 10
5
16. For a transistor amplifier in common emitter configuration
for load impedance of 1kW (h
f e
 = 50 and h
0e
 = 25) the current
gain is
(a) – 24.8 (b) – 15.7 (c) – 5.2 (d) – 48.78
17. A PN-junction has a thickness of the order of
(a) 1 cm (b) 1 mm (c) 10
–6
 m (d) 10
–12
 cm
18. In a common emitter (CE) amplifier having a voltage gain G ,
the transistor used has transconductance 0.03 mho and
current gain 25. If the above transistor is replaced with
another one with transconductance 0.02 mho and current
gain 20, the voltage gain will be
(a) 1.5 G (b)
1
3
 G (c)
5
4
 G (d)
2
3
 G
19. Which of the junction diodes shown below are forward
biased?
( a)
–10 V
R
–5 V
(b)
+10 V
R
+5 V
(c)
–10 V
R
(d)
–5 V
R
20. For LED’s to emit light in visible region of electromagnetic
light, it should have energy band gap in the range of:
(a) 0.1 eV to 0.4 eV (b) 0.5 eV to 0.8 eV
(c) 0.9 eV to 1.6 eV (d) 1.7 eV to 3.0 eV
21. In a C E transistor amplifier, the audio signal voltage across
th e c o llec to r resista n c e o f 2 kW  is 2V .  If the base resistance
is 1kW  and the  current amplification of the transistor is 100,
the input signal voltage is :
(a) 0.1 V (b) 1.0 V (c) 1 mV (d) 10  mV
22. The current gain in the common emitter mode of a transistor
is 10. The input impedance is 20k W and load of resistance is
100kW. The power gain is
(a) 300 (b) 500 (c) 200 (d) 100
23. In the circuit given below, A and B represent two inputs and
C represents the output.
C
A
B
The circuit represents
(a) NOR gate (b) AND gate
(c) NAND gate (d) OR gate
   
24. If the lattice constant of this semiconductor is decreased,
then which of the following is correct?
E
g
E
c
E
v
conduction
band width
band gap
valence
band width
(a) All E
c
, E
g
, E
v
 increase
(b) E
c
 and E
v
 increase, but E
g
 decreases
(c) E
c
 and E
v
 decrease, but E
g
 increases
(d) All E
c
, E
g
, E
v
 decrease
25. The ratio of electron and hole currents in a semiconductor
is 7/4 and the ratio of drift velocities of electrons and holes
is 5/4, then the ratio of concentrations of electrons and holes
will be
(a) 5/7 (b) 7/5 (c) 25/49 (d) 49/25
26. The I-V characteristic of a P-N junction diode is shown
below. The approximate dynamic resistance of the p-n
junction when a forward bias voltage of 2 volt is applied is
400
800
2 2.1
I (mA)
V (volt)
(a) 1 W (b) 0.25 W (c) 0.5 W (d) 5 W
27. The following configuration of gate is equivalent to
OR
NAND
AND
Y
A
B
(a) NAND gate (b) XOR gate
(c) OR gate (d) NOR gate
28. A piece of copper and another of germanium are cooled
from room temperature to 77K. The resistance of
(a) copper increases and germanium decreases
(b) each of them decreases
(c) each of them increases
(d) copper decreases and germanium increases
29. Figure shows a circuit in which three identical diodes are
used. Each diode has forward resistance of 20 W and infinite
backward resistance. Resistors R
1
 = R
2
 = R
3
 = 50 W. Battery
voltage is 6 V . The current through R
3
 is :
D
1
D
2
D
3
R
1
R
2
R
3 6 V
+–
(a) 50 mA (b) 100 mA (c) 60 mA (d) 25 mA
30. The current gain for a transistor working as common-base
amplifier is 0.96. If the emitter current is 7.2 mA, then the
base current is
(a) 0.29 mA (b) 0.35 mA (c) 0.39 mA (d) 0.43 mA
Page 5


1 . A pure semiconductor has equal electron and hole
concentration of 10
16
 m
–3
. Doping by indium increases
number of hole concentration n
h
 to 5 × 10
22
 m
–3
. Then, the
value of number of electron concentration n
e
 in the doped
semiconductor is
(a) 10
6
/m
3
(b) 10
22
/m
3
(c) 2 × 10
6
/m
3
(d) 2 × 10
9
/m
3
2 . A change of 8.0 mA in the emitter current bring a change of
7.9 mA in the collector current. The values of parameters a
and b are respectively
(a) 0.99, 90 (b) 0.96,79 (c) 0.97,99 (d) 0.99,79
3 . The following circut diagram represents
Y
B
A
(a) OR gate (b) XOR gate
(c) AND gate (d) NAND gate
4 . Which of the following statements is incorrect?
(a) The resistance of intrinsic semiconductors decrease
with increase of temperature
(b) Doping pure Si with trivalent impurities give p-type
semiconductors
(c) The majority carriers in n-type semiconductors are holes
(d) A p-n junction can act as a semiconductor diode
5 . In a npn transistor 10
10
 electrons enter the emitter in
10
–6
 s. 4% of the electrons are lost in the base. The current
transfer ratio will be
(a) 0.98 (b) 0.97
(c) 0.96 (d) 0.94
r o t c u d n o c i m e S 	 s c i n o r t c e l E :	 s l a i r e t a M , s e c i v e D 	
d n a 	 e l p m i S 	 s t i u c r i C
J E E 	 n i a M 	 s l a c i r e m u N
6 . The correct truth table for system of four NAND gates as
shown in figure is :
Y
A
B
( a)
A B Y
0 0 0
0 1 1
1 0 1
1 1 0
(b)
A B Y
0 0 0
0 1 0
1 0 1
1 1 1
(c)
A B Y
0 0 1
0 1 1
1 0 0
1 1 0
(d)
A B Y
0 0 1
0 1 0
1 0 1
1 1 1
7 . A Zener diode is connected to a battery and a load as show
below:
A
B
I
I
L
60 V
R = 2k
L
W
4 kW
I
Z
10 V = V
Z
The currents, I, I
Z
 and I
L
 are respectively.
(a) 15 mA, 5 mA, 10 mA
(b) 15 mA, 7.5 mA, 7.5 mA
(c) 12.5 mA, 5 mA, 7.5 mA
(d) 12.5 mA, 7.5 mA, 5 mA
8 . In common emitter amplifier, the current gain is 62. The
collector resistance and input resistance are 5 kW  an 500W
respectively . If the input voltage is 0.01 V , the output voltage
is
(a) 0.62 V (b) 6.2 V (c) 62 V (d) 620 V
9 . The circuit diagram shows a logic combination with the
states of outputs X, Y and Z given for inputs P, Q, R and S
all at state 1. When inputs P and R change to state 0 with
inputs Q and S still at 1, the states of outputs X, Y and Z
change to
P(1)
Q(1)
R(1)
S(1)
Y(1)
X(1)
Z(0)
(a) 1, 0, 0 (b) 1, 1, 1 (c) 0, 1, 0 (d) 0, 0, 1
10. A sinusoidal voltage of amplitude 25 volt and frequency
50Hz is applied to a half wave rectifier using P-n junction
diode. No filter is used and the load resistance is 1000W.
The forward resistance R
f
 of ideal diode is 10W. The
percentage efficiency of rectifier is
(a) 40% (b) 20% (c) 30% (d) 15%
11. The intrinsic conductivity of germanium at 27° is 2.13 mho
m
–1
 and mobilities of electrons and holes are 0.38 and 0.18
m
2
V
–1
s
–1
 respectively. The density of charge carriers is
(a) 2.37 × 10
19
 m
–3
(b) 3.28 × 10
19
 m
–3
(c) 7.83 × 10
19
 m
–3
(d) 8.47 × 10
19
 m
–3
12. The circuit has two oppositively connected ideal diodes in
parallel. The current flowing in the circuit is
4W
D
1
D
2
2W 3W
12V
(a) 1.71 A (b) 2.00 A (c) 2.31 A (d) 1.33 A
13. A working transistor with its three legs marked P, Q and R is
tested using a multimeter. No conduction is found between
P and Q. By connecting the common (negative) terminal of
the multimeter to R and the other (positive) terminal to P o r
Q, some resistance is seen on the multimeter. Which of the
following is true for the transistor?
(a) It is an npn transistor with R as base
(b) It is a pnp transistor with R as base
(c) It is a pnp transistor with R as emitter
(d) It is an npn transistor with R as collector
14. The diagram of a logic circuit is given below. The output F
of the circuit is represented by
X
W
Y
F
W
(a) W . (X + Y) (b) W . (X . Y)
(c) W + (X . Y) (d) W + (X + Y)
15. The concentration of hole-electron pairs in pure silicon at
T = 300 K is 7 × 10
15
 per cubic meter. Antimony is doped
into silicon in a proportion of 1 atom in 10
7
 Si atoms.
Assuming half of the impurity atoms contribute electron in
the conduction band, calculate the factor by which the
number of charge carriers increases due to doping. The
number of silicon atoms per cubic meter is 5 × 10
28
(a) 2.8 × 10
5
(b) 3.1 × 10
2
(c) 4.2 × 10
5
(d) 1.8 × 10
5
16. For a transistor amplifier in common emitter configuration
for load impedance of 1kW (h
f e
 = 50 and h
0e
 = 25) the current
gain is
(a) – 24.8 (b) – 15.7 (c) – 5.2 (d) – 48.78
17. A PN-junction has a thickness of the order of
(a) 1 cm (b) 1 mm (c) 10
–6
 m (d) 10
–12
 cm
18. In a common emitter (CE) amplifier having a voltage gain G ,
the transistor used has transconductance 0.03 mho and
current gain 25. If the above transistor is replaced with
another one with transconductance 0.02 mho and current
gain 20, the voltage gain will be
(a) 1.5 G (b)
1
3
 G (c)
5
4
 G (d)
2
3
 G
19. Which of the junction diodes shown below are forward
biased?
( a)
–10 V
R
–5 V
(b)
+10 V
R
+5 V
(c)
–10 V
R
(d)
–5 V
R
20. For LED’s to emit light in visible region of electromagnetic
light, it should have energy band gap in the range of:
(a) 0.1 eV to 0.4 eV (b) 0.5 eV to 0.8 eV
(c) 0.9 eV to 1.6 eV (d) 1.7 eV to 3.0 eV
21. In a C E transistor amplifier, the audio signal voltage across
th e c o llec to r resista n c e o f 2 kW  is 2V .  If the base resistance
is 1kW  and the  current amplification of the transistor is 100,
the input signal voltage is :
(a) 0.1 V (b) 1.0 V (c) 1 mV (d) 10  mV
22. The current gain in the common emitter mode of a transistor
is 10. The input impedance is 20k W and load of resistance is
100kW. The power gain is
(a) 300 (b) 500 (c) 200 (d) 100
23. In the circuit given below, A and B represent two inputs and
C represents the output.
C
A
B
The circuit represents
(a) NOR gate (b) AND gate
(c) NAND gate (d) OR gate
   
24. If the lattice constant of this semiconductor is decreased,
then which of the following is correct?
E
g
E
c
E
v
conduction
band width
band gap
valence
band width
(a) All E
c
, E
g
, E
v
 increase
(b) E
c
 and E
v
 increase, but E
g
 decreases
(c) E
c
 and E
v
 decrease, but E
g
 increases
(d) All E
c
, E
g
, E
v
 decrease
25. The ratio of electron and hole currents in a semiconductor
is 7/4 and the ratio of drift velocities of electrons and holes
is 5/4, then the ratio of concentrations of electrons and holes
will be
(a) 5/7 (b) 7/5 (c) 25/49 (d) 49/25
26. The I-V characteristic of a P-N junction diode is shown
below. The approximate dynamic resistance of the p-n
junction when a forward bias voltage of 2 volt is applied is
400
800
2 2.1
I (mA)
V (volt)
(a) 1 W (b) 0.25 W (c) 0.5 W (d) 5 W
27. The following configuration of gate is equivalent to
OR
NAND
AND
Y
A
B
(a) NAND gate (b) XOR gate
(c) OR gate (d) NOR gate
28. A piece of copper and another of germanium are cooled
from room temperature to 77K. The resistance of
(a) copper increases and germanium decreases
(b) each of them decreases
(c) each of them increases
(d) copper decreases and germanium increases
29. Figure shows a circuit in which three identical diodes are
used. Each diode has forward resistance of 20 W and infinite
backward resistance. Resistors R
1
 = R
2
 = R
3
 = 50 W. Battery
voltage is 6 V . The current through R
3
 is :
D
1
D
2
D
3
R
1
R
2
R
3 6 V
+–
(a) 50 mA (b) 100 mA (c) 60 mA (d) 25 mA
30. The current gain for a transistor working as common-base
amplifier is 0.96. If the emitter current is 7.2 mA, then the
base current is
(a) 0.29 mA (b) 0.35 mA (c) 0.39 mA (d) 0.43 mA
1. (d) Here, n
i
 = 10
16
 m
–3
, n
h
 = 5 × 10
22
 m
–3
As n
e
n
h
 = n
i
2
2 16 3 2
93 i
e
223
h
n(10m)
n 210m
n
510m
-
-
-
\== =´
´
2. (d) DI
E
 = 8.0 mA
DI
C
 = 7.9 mA
C
E
I 7.9
0.9875 0.99
I 8.0
D
a===
D
;
Also,  
0.9875
79
1 (1 0.9875)
a
b===
-a-
3. (b) Output of upper AND gate = B A
Output of lower AND gate = AB
\ Output of OR gate, A B B A Y + =
This is boolean expression for XOR gate.
4. (c) In n-type semiconductors, electrons are the majority
charge carriers.
5. (c) No. of electrons reaching the collector,
10 10
C
10 96 . 0 10
100
96
n ´ = ´ =
Emitter current, 
E
E
ne
I
t
´
=
Collector current, 
C
C
ne
I
t
´
=
\ Current transfer ratio,
CC
EE
In
In
a== = 
96 . 0
10
10 96 . 0
10
10
=
´
6. (a)
A
B
Y = A.AB
2
Y = AB
1
Y = B.AB
3
Y =  A.AB B.AB
By expanding this Boolen expression
.. YABBA=+
Thus the truth table for this expression should be (a).
7. (d) Here, R = 4 kW = 4 × 10
3 
W
V
i
 = 60 V
Zener voltage V
z
 = 10 V
R
L
 = 2 kW = 2 × 10
3
 W
Load current, I
L
 = 
3
10
2 10
Z
L
V
R
=
´
 = 5 mA
Current through R
= 
33
60 10 50
4 10 4 10
-
=
´´
 = 12.5 mA
From circuit diagram,
I = I
Z
 + I
L
Þ 12.5 = I
Z
 + 5
Þ I
Z 
= 12.5 – 5 = 7.5 mA.
8. (b)
3
oo
in in
V R 5 10 62
10 62 620
V R 500
´´
=´b==´=
V
o
 = 620 × V
in
= 620 × 0.01 = 6.2 V
\ V
o
 = 6.2 volt.
9. (c)P() 0
Q(1)
R(0)
S(1)
Y(1)
X(0)
Z(0)
0
0 1
10. (a) I
m
 = 
m
fL
V
RR +
= 
25
(10 1000) +
 = 24.75 mA
I
dc
 = 
m
I
p
= 
24.75
3.14
 = 7.87 mA
I
rms
 = 
m
I
2
 = 
24.75
2
= 12.37 mA
P
dc
  = I
dc
2
 × R
L
  = (7.87 × 10
–3
)
2
 × 10
3
 = 61.9 mW
P
ac
 = I
rms
 
2
( R
f
 + R
L
) = (12.37 × 10
–3
)
2
 × (10 + 1000)
      = 154.54 mW
Rectifier efficiency
h = 
dc
ac
P
P
 × 100 = 
61.9
154.54
 × 100 = 40.05%
11. (a) Conductivity, 
eehh
1
s==e(nµ+nµ)
?
2.13 = 1.6 × 10
–19
(0.38 + 0.18) n
i
(Since in intrinsic semi-conductor, n
e
 = n
h
= n
i
)
\   density of charge carriers, n
i
193
19
2.13
2.37 10 m
1.6 10 0.56
-
-
==´
´´
12. (b) D
2
 is forward biased whereas D
1
 is reversed biased.
So effective resistance of the circuit
R = 4 + 2 = 6W
12
2A
6
i\==.
13. (b) It is a p-n-p transistor with R as base.
14. (c) (W + X) × (W + Y) = W + (X × Y)
15. (d) In pure semiconductor electron-hole pair = 7 × 10
15
/m
3
n
initial 
= n
h
 + n
e
 = 14 × 10
15
 after doping donor Impurity
21 
and  n
e
 = 
D
N
2
 = 2.5 × 10
21
HINTS & SOLUTIONS