NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits
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Page 1
NCERT
SOLUTIONS
CLASS - 12th
aglase .co
SOLUTIONS
CLASS - 12th
aglase .co
Page 2
Class : 12th
Subject : Physics
Chapter : 14
Chapter Name : Semiconductor Electronics: Materials, Devices And Simple Circuits
Q14.1 In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer. The correct statement is (c).
In an n-type silicon, the electrons are the majority carriers, while the holes are the minority
carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are
doped in silicon atoms.
Page : 497 , Block Name : Exercise
Q14.2 Which of the statements given in question 1 is true for p-type semiconductors.
Answer. The correct statement is (d).
In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority
carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped
in silicon atoms.
Page : 497 , Block Name : Exercise
Q14.3 Carbon, silicon and germanium have four valence electrons each. These are characterised by
valence and conduction bands separated by energy band gap respectively equal to
(E ) , (E )
g
c
and (E )
g . Which of the following statements is true?
si
g
Ge
(a) (E ) g
Si
< (Eg )
Ge
< (Eg )
C
(b) (E ) g
C
< (Eg )
Ge
> (Eg )
si
(c) (E ) g
c
> (Eg )
si
> (Eg )
ce
(d) (E ) g
C
= (Eg )
si
= (Eg )
Ge
Answer. The correct statement is (c).
Of the three given elements, the energy band gap of carbon is the maximum and that of
germanium is the least. The energy band gap of these elements are related as:
(Eg ) > (Eg ) > (Eg )
c si ce
Subject : Physics
Chapter : 14
Chapter Name : Semiconductor Electronics: Materials, Devices And Simple Circuits
Q14.1 In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer. The correct statement is (c).
In an n-type silicon, the electrons are the majority carriers, while the holes are the minority
carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are
doped in silicon atoms.
Page : 497 , Block Name : Exercise
Q14.2 Which of the statements given in question 1 is true for p-type semiconductors.
Answer. The correct statement is (d).
In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority
carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped
in silicon atoms.
Page : 497 , Block Name : Exercise
Q14.3 Carbon, silicon and germanium have four valence electrons each. These are characterised by
valence and conduction bands separated by energy band gap respectively equal to
(E ) , (E )
g
c
and (E )
g . Which of the following statements is true?
si
g
Ge
(a) (E ) g
Si
< (Eg )
Ge
< (Eg )
C
(b) (E ) g
C
< (Eg )
Ge
> (Eg )
si
(c) (E ) g
c
> (Eg )
si
> (Eg )
ce
(d) (E ) g
C
= (Eg )
si
= (Eg )
Ge
Answer. The correct statement is (c).
Of the three given elements, the energy band gap of carbon is the maximum and that of
germanium is the least. The energy band gap of these elements are related as:
(Eg ) > (Eg ) > (Eg )
c si ce
Page 3
Page : 497 , Block Name : Exercise
Q14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Answer. The correct statement is (c)
The diffusion of charge carriers across a junction takes place from the region of higher
concentration to the region of lower concentration. In this case, the p-region has greater
concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from
the p region to the n-region.
Page : 497 , Block Name : Exercise
Q14.5 When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above
Answer. The correct statement is (c).
When a forward bias is applied to a p-n junction, it lowers the value Of potential barrier. In the
case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential
barrier across the junction gets reduced.
Page : 497 , Block Name : Exercise
Q14.6 In half-wave recti cation, what is the output frequency if the input frequency is 50 Hz.
What is the output frequency of a full-wave recti er for the same input frequency.
Answer. Input frequency = 50 Hz
For a half-wave recti er, the output frequency is equal to the input frequency.
∴ Output frequency = 50 Hz
For a full-wave recti er, the output frequency is twice the input frequency.
∴ Output frequency = 2 × 50 = 100Hz
Page : 497 , Block Name : Exercise
Q14.7 A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect
a wavelength of 6000 nm?
Answer. Energy band gap of the given photodiode, E g = 2.8eV
Q14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Answer. The correct statement is (c)
The diffusion of charge carriers across a junction takes place from the region of higher
concentration to the region of lower concentration. In this case, the p-region has greater
concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from
the p region to the n-region.
Page : 497 , Block Name : Exercise
Q14.5 When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above
Answer. The correct statement is (c).
When a forward bias is applied to a p-n junction, it lowers the value Of potential barrier. In the
case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential
barrier across the junction gets reduced.
Page : 497 , Block Name : Exercise
Q14.6 In half-wave recti cation, what is the output frequency if the input frequency is 50 Hz.
What is the output frequency of a full-wave recti er for the same input frequency.
Answer. Input frequency = 50 Hz
For a half-wave recti er, the output frequency is equal to the input frequency.
∴ Output frequency = 50 Hz
For a full-wave recti er, the output frequency is twice the input frequency.
∴ Output frequency = 2 × 50 = 100Hz
Page : 497 , Block Name : Exercise
Q14.7 A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect
a wavelength of 6000 nm?
Answer. Energy band gap of the given photodiode, E g = 2.8eV
Page 4
Wavelength, λ = 6000nm = 6000 × 10 m −9
The energy of a signal is given by the relation:
hc
E =
λ
Where,
h = Planck’s constant
−34
= 6.626 × 10 Js
c = Speed of light
8
= 3 × 10 m/s
−34 8
6.626×10 ×3×10
= −9
6000×10
E = = 3.313 × 10 J −20
But 1.6 × 10 J = 1eV −19
−20
∴ E = 3.313 × 10 J
−20
3.313×10
= −19
= 0.207eV
1.6×10
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV - the energy
band gap of a photodiode. Hence, the photodiode cannot detect the signal.
Page : 497 , Block Name : Exercise
Q14.8 The number of silicon atoms per m is 5 × 10 . This is doped simultaneously with 5×
3 28
atoms per m of Arsenic and 5 × 10 per m atoms of Indium. Calculate the number of
22 3 20 3
10
electrons and holes. Given thatn = 1.5 × 10 m . Is the material n-type or p-type?
i
16 −3
Answer. Number of silicon atoms, N = 5 × 10 28 3
atoms/m
Number of arsenic atoms, n 22 3
As
= 5 × 10 atoms/m
Number of indium atoms, n
20 3
ln = 5 × 10 atoms/m
Number of thermally-generated electrons, n = 1.5 × 10 electrons/m
16 3
i
Number of electrons, n = 5 × 10 − 1.5 × 10 ≈ 4.99 × 10
e
22 16 22
Number of holes = n h
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related
as:
2
ne nh = n
i
2
n
i
∴ nh =
ne
16 2
(1.5×10 )
9
= 22
≈ 4.51 × 10
4.99×10
Therefore, the number of electrons is approximately 4.99 × 10 and the number of holes is about 22
4.51 × 10 . Since the number of electrons is more than the number of holes, the material is an n-
9
type semiconductor.
Page : 498 , Block Name : Additional Exercise
Q14.9 In an intrinsic semiconductor the energy gap E is 1.2eV. Its hole mobility is much smaller g
than electron mobility and independent of temperature. What is the ratio between conductivity at
The energy of a signal is given by the relation:
hc
E =
λ
Where,
h = Planck’s constant
−34
= 6.626 × 10 Js
c = Speed of light
8
= 3 × 10 m/s
−34 8
6.626×10 ×3×10
= −9
6000×10
E = = 3.313 × 10 J −20
But 1.6 × 10 J = 1eV −19
−20
∴ E = 3.313 × 10 J
−20
3.313×10
= −19
= 0.207eV
1.6×10
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV - the energy
band gap of a photodiode. Hence, the photodiode cannot detect the signal.
Page : 497 , Block Name : Exercise
Q14.8 The number of silicon atoms per m is 5 × 10 . This is doped simultaneously with 5×
3 28
atoms per m of Arsenic and 5 × 10 per m atoms of Indium. Calculate the number of
22 3 20 3
10
electrons and holes. Given thatn = 1.5 × 10 m . Is the material n-type or p-type?
i
16 −3
Answer. Number of silicon atoms, N = 5 × 10 28 3
atoms/m
Number of arsenic atoms, n 22 3
As
= 5 × 10 atoms/m
Number of indium atoms, n
20 3
ln = 5 × 10 atoms/m
Number of thermally-generated electrons, n = 1.5 × 10 electrons/m
16 3
i
Number of electrons, n = 5 × 10 − 1.5 × 10 ≈ 4.99 × 10
e
22 16 22
Number of holes = n h
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related
as:
2
ne nh = n
i
2
n
i
∴ nh =
ne
16 2
(1.5×10 )
9
= 22
≈ 4.51 × 10
4.99×10
Therefore, the number of electrons is approximately 4.99 × 10 and the number of holes is about 22
4.51 × 10 . Since the number of electrons is more than the number of holes, the material is an n-
9
type semiconductor.
Page : 498 , Block Name : Additional Exercise
Q14.9 In an intrinsic semiconductor the energy gap E is 1.2eV. Its hole mobility is much smaller g
than electron mobility and independent of temperature. What is the ratio between conductivity at
Page 5
600K and that at 300K? Assume that the temperature dependence of intrinsic carrier
concentration n is given by
i
Eg
ni = n0 exp[− ]
2kn T
Where n is a constant.
0
Answer. Energy gap of the given intrinsic semiconductor, E g = 1.2eV
The temperature dependence of the intrinsic carrier-concentration is written as:
Ex
ni = n0 exp[− ]
2kB T
Where,
= Boltzmann constant = 8.62 × 10
−5
kB eV/K
T = Temperature
n = Constant
0
Initial temperature, T = 300K 1
The intrinsic carrier-concentration at this temperature can be written as:
..(1)
Ex
ni1 = n0 exp[− ]
2kb ×300
Final temperature, T = 600k 2
The intrinsic carrier-concentration at this temperature can be written as:
.. (2)
Ex
ni2 = n0 exp[− ]
2kB ×600
The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between
the respective intrinsic carrier-concentrations at these temperatures.
Ex
n0 exp[− ]
ni2 2k
B
600
= Eg
nn
n0 exp[− ]
2k 300
B
Ex 1 1 1.2 2−1
= exp [ − ] = exp[ −5
× ]
2kB 300 600 2×8.62×10 600
5
= exp[11.6] = 1.09 × 10
Therefore, the ratio between the conductivities is 1.09 × 10 5
Page : 498 , Block Name : Additional Exercise
Q14.10 In a p-n junction diode, the current I can be expressed as I = I exp( 0
eV
− 1) where I 0
2kB T
is called the reverse saturation current, V is the voltage across the diode and is positive for forward
bias and negative for reverse bias, and I is the current through the diode, k is the Boltzmann B
constant (8.6 × 10 eV/K) and T is the absolute temperature. If for a given diode
−5
I0 = 5 × 10
−12
, then
A and T = 300k
(a) What will be the forward current at a forward voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Answer. In a p-n junction diode, the expression for current is given as:
concentration n is given by
i
Eg
ni = n0 exp[− ]
2kn T
Where n is a constant.
0
Answer. Energy gap of the given intrinsic semiconductor, E g = 1.2eV
The temperature dependence of the intrinsic carrier-concentration is written as:
Ex
ni = n0 exp[− ]
2kB T
Where,
= Boltzmann constant = 8.62 × 10
−5
kB eV/K
T = Temperature
n = Constant
0
Initial temperature, T = 300K 1
The intrinsic carrier-concentration at this temperature can be written as:
..(1)
Ex
ni1 = n0 exp[− ]
2kb ×300
Final temperature, T = 600k 2
The intrinsic carrier-concentration at this temperature can be written as:
.. (2)
Ex
ni2 = n0 exp[− ]
2kB ×600
The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between
the respective intrinsic carrier-concentrations at these temperatures.
Ex
n0 exp[− ]
ni2 2k
B
600
= Eg
nn
n0 exp[− ]
2k 300
B
Ex 1 1 1.2 2−1
= exp [ − ] = exp[ −5
× ]
2kB 300 600 2×8.62×10 600
5
= exp[11.6] = 1.09 × 10
Therefore, the ratio between the conductivities is 1.09 × 10 5
Page : 498 , Block Name : Additional Exercise
Q14.10 In a p-n junction diode, the current I can be expressed as I = I exp( 0
eV
− 1) where I 0
2kB T
is called the reverse saturation current, V is the voltage across the diode and is positive for forward
bias and negative for reverse bias, and I is the current through the diode, k is the Boltzmann B
constant (8.6 × 10 eV/K) and T is the absolute temperature. If for a given diode
−5
I0 = 5 × 10
−12
, then
A and T = 300k
(a) What will be the forward current at a forward voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Answer. In a p-n junction diode, the expression for current is given as:
Page 6
eV
I = I0 exp( − 1)
2kB T
Where,
−12
I0 = Reverse saturation current = 5 × 10 A
T = Absolute temperature = 300K
−5 −23 −1
kB = Boltzmann constant = 8.6 × 10 eV/K = 1.376 × 10 JK
V = Voltage across the diode
(a) Forward voltage, V = 0.6V
−19
−12 1.6×10 ×0.6
= 5 × 10 [exp( −23
) − 1]
1.376×10 ×300
∴ Current, I
−12
= 5 × 10 × exp[22.36] = 0.0256A
Therefore, the forward current is about 0.0256A.
(b) For forward voltage, V = 0.7V, we can write:
−19
′ −12 1.6×10 ×0.7
I = 5 × 10 [exp( − 1)]
−23
1.376×10 ×300
−12
= 5 × 10 × exp[26.25] = 1.257A
Hence, the increase in current, ΔI = I − I ′
= 1.257 − 0.0256 = 1.23A
Change in voltage
=
Change in current
(c) Dynamic resistance
0.7−0.6 0.1
= = = 0.081Ω
1.23 1.23
(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I ) will almost remain
equal to I in both cases. Therefore, the dynamic resistance in the reverse bias will be in nite.
0
Page : 498 , Block Name : Additional Exercise
Q14.11 You are given the two circuits as shown in Figure that circuit (a) acts as OR gate while the
circuit (b) acts as AND gate.
Answer. (a) and B are the inputs and Y is the output of the given circuit. The left half of the given
gure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in the
I = I0 exp( − 1)
2kB T
Where,
−12
I0 = Reverse saturation current = 5 × 10 A
T = Absolute temperature = 300K
−5 −23 −1
kB = Boltzmann constant = 8.6 × 10 eV/K = 1.376 × 10 JK
V = Voltage across the diode
(a) Forward voltage, V = 0.6V
−19
−12 1.6×10 ×0.6
= 5 × 10 [exp( −23
) − 1]
1.376×10 ×300
∴ Current, I
−12
= 5 × 10 × exp[22.36] = 0.0256A
Therefore, the forward current is about 0.0256A.
(b) For forward voltage, V = 0.7V, we can write:
−19
′ −12 1.6×10 ×0.7
I = 5 × 10 [exp( − 1)]
−23
1.376×10 ×300
−12
= 5 × 10 × exp[26.25] = 1.257A
Hence, the increase in current, ΔI = I − I ′
= 1.257 − 0.0256 = 1.23A
Change in voltage
=
Change in current
(c) Dynamic resistance
0.7−0.6 0.1
= = = 0.081Ω
1.23 1.23
(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I ) will almost remain
equal to I in both cases. Therefore, the dynamic resistance in the reverse bias will be in nite.
0
Page : 498 , Block Name : Additional Exercise
Q14.11 You are given the two circuits as shown in Figure that circuit (a) acts as OR gate while the
circuit (b) acts as AND gate.
Answer. (a) and B are the inputs and Y is the output of the given circuit. The left half of the given
gure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in the
Page 7
following gure.
Hence, the output of the NOR Gate = A + B
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
This will be the input for the NOT Gate. Its output will be A + B = A + B
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
∴ Y = A + B
Hence, this circuit functions as an OR Gate.
(b) A and B are the inputs and Y is the output of the given circuit. It can be observed from the
following gure that the inputs of the right half NOR Gate are the outputs of the two NOT Gates.
Hence, the output of the given circuit can be written as:
¯¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯ ¯
¯¯¯ ¯
¯¯¯
¯¯
¯¯ ¯¯¯¯ ¯
¯¯¯ ¯
¯¯¯
Y = A + B = A ⋅ B = A ⋅ B
Hence, this circuit functions as an AND Gate.
Page : 498 , Block Name : Additional Exercise
Q14.12 Write the truth table for a NAND gate connected as given in Figure.
Hence identify the exact logic operation carried out by this circuit.
Answer. A acts as the two inputs of the NAND gate and Y is the output, as shown in the following
gure.
Hence, the output can be written as:
… (i)
¯¯¯¯¯¯¯¯¯¯¯¯ ¯
¯¯¯ ¯
¯¯¯ ¯
¯¯¯
Y = A ⋅ A = A + A = A
The truth table for equation (i) can be drawn as:
Hence, the output of the NOR Gate = A + B
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
This will be the input for the NOT Gate. Its output will be A + B = A + B
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
∴ Y = A + B
Hence, this circuit functions as an OR Gate.
(b) A and B are the inputs and Y is the output of the given circuit. It can be observed from the
following gure that the inputs of the right half NOR Gate are the outputs of the two NOT Gates.
Hence, the output of the given circuit can be written as:
¯¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯ ¯
¯¯¯ ¯
¯¯¯
¯¯
¯¯ ¯¯¯¯ ¯
¯¯¯ ¯
¯¯¯
Y = A + B = A ⋅ B = A ⋅ B
Hence, this circuit functions as an AND Gate.
Page : 498 , Block Name : Additional Exercise
Q14.12 Write the truth table for a NAND gate connected as given in Figure.
Hence identify the exact logic operation carried out by this circuit.
Answer. A acts as the two inputs of the NAND gate and Y is the output, as shown in the following
gure.
Hence, the output can be written as:
… (i)
¯¯¯¯¯¯¯¯¯¯¯¯ ¯
¯¯¯ ¯
¯¯¯ ¯
¯¯¯
Y = A ⋅ A = A + A = A
The truth table for equation (i) can be drawn as:
Page 8
This current functions as a NOT gate. The symbol for this logic circuit is shown as:
Page : 498 , Block Name : Additional Exercise
Q14.13 Write the truth table for circuit given in Figure below consisting of NOR gates and identify
the logic operation (OR, AND, NOT) which this circuit is performing.
(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work
out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND,
NOT gates and nd the correct one.)
Answer. A and B are the inputs of the given circuit. The output of the rst NOR gate is A + B. It
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
can be observed from the following gure that the inputs of the second NOR gate become the
output of the rst one.
Hence, the output of the combination is given as:
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯ ¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
Y = A + B + A + B
¯¯
¯¯¯
¯¯¯
¯¯¯ ¯
¯¯¯
¯¯
¯¯ ¯
¯¯¯ ¯
¯¯¯
= A ⋅ B = A + B = A + B
The truth table for this operation is given as:
Page : 498 , Block Name : Additional Exercise
Q14.13 Write the truth table for circuit given in Figure below consisting of NOR gates and identify
the logic operation (OR, AND, NOT) which this circuit is performing.
(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work
out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND,
NOT gates and nd the correct one.)
Answer. A and B are the inputs of the given circuit. The output of the rst NOR gate is A + B. It
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
can be observed from the following gure that the inputs of the second NOR gate become the
output of the rst one.
Hence, the output of the combination is given as:
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯ ¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
Y = A + B + A + B
¯¯
¯¯¯
¯¯¯
¯¯¯ ¯
¯¯¯
¯¯
¯¯ ¯
¯¯¯ ¯
¯¯¯
= A ⋅ B = A + B = A + B
The truth table for this operation is given as:
Page 9
This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.
Page : 499 , Block Name : Additional Exercise
Q14.14 Write the truth table for circuit given in Fig below consisting of NOR gates and identify the
logic operation (OR, AND, NOT) which this circuit is performing.
Answer. A acts as the two inputs of the NOR gate and Y is the output, as shown in the following
gure. Hence, the output of the circuit is A + A.
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
Output, Y = A + A = A
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯ ¯
¯¯¯
The truth table for the same is given as:
This is the truth table of a NOT gate. Hence, this circuit functions as NOT gate.
Page : 499 , Block Name : Additional Exercise
Q14.14 Write the truth table for circuit given in Fig below consisting of NOR gates and identify the
logic operation (OR, AND, NOT) which this circuit is performing.
Answer. A acts as the two inputs of the NOR gate and Y is the output, as shown in the following
gure. Hence, the output of the circuit is A + A.
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯
Output, Y = A + A = A
¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯¯¯ ¯
¯¯¯
The truth table for the same is given as:
This is the truth table of a NOT gate. Hence, this circuit functions as NOT gate.
Page 10
(b) A and B are the inputs and Y is the output of the given circuit. By using the result
Obtained in solution (a), we can infer that the outputs Of the rst two NOR gates are A and B, as
¯
¯¯¯ ¯
¯¯¯
shown in the following gure.
are the inputs for the last NOR gate. Hence, the output for the circuit can be written as:
¯
¯¯¯ ¯
¯¯¯
A and B
¯¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯ ¯
¯¯¯ ¯
¯¯¯
¯¯
¯¯ ¯¯¯¯ ¯
¯¯¯ ¯
¯¯¯
Y = A + B = A ⋅ B = A ⋅ B
The truth table for the same can be written as:
This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.
Page : 499 , Block Name : Additional Exercise
Q14.15 You are given two circuits as shown in Figure, which consist of NAND gates. Identify the
logic operation carried out by the two circuits.
Answer. In both the given circuits, A and B are the inputs and Y is the output.
(a) The output of the left NAND gate will be A. B, as shown in the following gure.
¯
¯¯¯
¯¯¯
¯¯¯¯
Obtained in solution (a), we can infer that the outputs Of the rst two NOR gates are A and B, as
¯
¯¯¯ ¯
¯¯¯
shown in the following gure.
are the inputs for the last NOR gate. Hence, the output for the circuit can be written as:
¯
¯¯¯ ¯
¯¯¯
A and B
¯¯¯
¯¯¯
¯¯¯
¯¯¯
¯¯ ¯
¯¯¯ ¯
¯¯¯
¯¯
¯¯ ¯¯¯¯ ¯
¯¯¯ ¯
¯¯¯
Y = A + B = A ⋅ B = A ⋅ B
The truth table for the same can be written as:
This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.
Page : 499 , Block Name : Additional Exercise
Q14.15 You are given two circuits as shown in Figure, which consist of NAND gates. Identify the
logic operation carried out by the two circuits.
Answer. In both the given circuits, A and B are the inputs and Y is the output.
(a) The output of the left NAND gate will be A. B, as shown in the following gure.
¯
¯¯¯
¯¯¯
¯¯¯¯
Page 11
Hence, the output of the combination of the two NAND gates is given as:
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯
Y = (A ⋅ B) ⋅ (A ⋅ B) = AB + AB = AB
Hence, this circuit functions as an AND gate.
(b) A is the output of the upper left of the NAND gate and B is the output of the lower half of the
¯
¯¯¯ ¯
¯¯¯
NAND gate, as shown in the following gure.
Hence, the output of the combination of the NAND gates will be given as:
¯
¯¯¯
¯
¯¯¯ ¯
¯¯¯ ¯
¯¯¯ ¯
¯¯¯
Y = A ⋅ B = A + B = A + B
Hence, this circuit functions as an OR gate.
Page : 499 , Block Name : Additional Exercise
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯
Y = (A ⋅ B) ⋅ (A ⋅ B) = AB + AB = AB
Hence, this circuit functions as an AND gate.
(b) A is the output of the upper left of the NAND gate and B is the output of the lower half of the
¯
¯¯¯ ¯
¯¯¯
NAND gate, as shown in the following gure.
Hence, the output of the combination of the NAND gates will be given as:
¯
¯¯¯
¯
¯¯¯ ¯
¯¯¯ ¯
¯¯¯ ¯
¯¯¯
Y = A ⋅ B = A + B = A + B
Hence, this circuit functions as an OR gate.
Page : 499 , Block Name : Additional Exercise
Document Details
| Board / Org | NCERT |
| Exam | Class 12 |
| Type | Solution |
| Pages | 11 |
| Updated | 30 Apr 2026 |