Ans: 1
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.
Ans: 2
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.
Ans: 3
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)C> (Eg)Si> (Eg)Ge
Ans: 4
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.
Ans: 5
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.
Ans: 6
The correct statement is (b), (c).
For a transistor action, the junction must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reverse-biased.
Ans: 7
The correct statement is (c).
The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.
Ans: 8
Input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
∴Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
∴Output frequency = 2 × 50 = 100 Hz
Ans: 9
Collector resistance,RC = 2 kΩ = 2000 Ω
Audio signal voltage across the collector resistance, V = 2 V
Current amplification factor of the transistor, β = 100
Base resistance, RB= 1 kΩ = 1000 Ω
Input signal voltage =Vi
Base current = IB
We have the amplification relation as:
Voltage amplification
Therefore, the input signal voltage of the amplifier is 0.01 V.
Base resistance is given by the relation:
Therefore, the base current of the amplifier is 10 μA.
Ans: 10
Voltage gain of the first amplifier, V1 = 10
Voltage gain of the second amplifier, V2 = 20
Input signal voltage,Vi = 0.01 V
Output AC signal voltage = Vo
The total voltage gain of a two-stage cascaded amplifier is given by the product of voltage gains of both the stages, i.e.,
V = V1× V2
= 10 × 20 = 200
We have the relation:
V0 =V × Vi
= 200 × 0.01 = 2 V
Therefore, the output AC signal of the given amplifier is 2 V.
Ans: 11
Energy band gap of the given photodiode, Eg= 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9m
The energy of a signal is given by the relation:
E =
Where,
h = Planck’s constant
= 6.626 × 10−34 Js
c = Speed of light
= 3 × 108 m/s
E
= 3.313 × 10−20 J
But 1.6 × 10−19 J = 1 eV
∴E = 3.313 × 10−20 J
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.
Ans: 12
Number of silicon atoms, N = 5 × 1028 atoms/m3
Number of arsenic atoms, nAs = 5 × 1022 atoms/m3
Number of indium atoms,nIn = 5 × 1020 atoms/m3
Number of thermally-generated electrons, ni = 1.5 × 1016 electrons/m3
Number of electrons, ne= 5 × 1022 − 1.5 × 1016 ≈ 4.99 × 1022
Number of holes = nh
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:
nenh= ni2
Therefore, the number of electrons is approximately 4.99 × 1022 and the number of holes is about 4.51 × 109. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.
Ans: 13
Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV
The temperature dependence of the intrinsic carrier-concentration is written as:
Where,
kB = Boltzmann constant = 8.62 × 10−5 eV/K
T = Temperature
n0 = Constant
Initial temperature, T1= 300 K
The intrinsic carrier-concentration at this temperature can be written as:
…(1)
Final temperature, T2= 600 K
The intrinsic carrier-concentration at this temperature can be written as:
…(2)
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.
Therefore, the ratio between the conductivities is 1.09 × 105.
Ans: 14
In a p-n junction diode, the expression for current is given as:
Where,
I0 = Reverse saturation current = 5 × 10−12 A
T = Absolute temperature = 300 K
kB = Boltzmann constant = 8.6 × 10−5 eV/K = 1.376 × 10−23 J K−1
V = Voltage across the diode
(a) Forward voltage, V = 0.6 V
∴Current,I
Therefore, the forward current is about 0.0256 A.
(b) For forward voltage, V’ = 0.7 V, we can write:
Hence, the increase in current, ΔI = I' −I
= 1.257 −0.0256 = 1.23 A
(c) Dynamic resistance
(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I0in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.
Ans: 15
(a) A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in the following figure.
Hence, the output of the NOR Gate =
This will be the input for the NOT Gate. Its output will be
= 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 figure 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:
Hence, this circuit functions as an AND Gate.
Ans: 16
A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.
Hence, the output can be written as:
The truth table for equation (i) can be drawn as:
Ans: 17
In both the given circuits, A and B are the inputs andY is the output.
(a) The output of the left NAND gate will be
, as shown in the following figure.
Hence, the output of the combination of the two NAND gates is given as:
Hence, this circuit functions as an AND gate.
(b)
is the output of the upper left of the NAND gate and 
is the output of the lower half of the NAND gate, as shown in the following figure.
Hence, the output of the combination of the NAND gates will be given as:
Hence, this circuit functions as an OR gate.
Ans: 18
A and Bare the inputs of the given circuit. The output of the first NOR gate is
. It can be observed from the following figure that the inputs of the second NOR gate become the out put of the first one.
Hence, the output of the combination is given as:
The truth table for this operation is given as:
Ans: 19
(a) A acts as the two inputs of the NOR gate and Yis the output, as shown in the following figure. Hence, the output of the circuit is
.
The truth table for the same is given as:
(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 first two NOR gates are
as shown in the following figure.
are the inputs for the last NOR gate. Hence, the output for the circuit can be written as:
The truth table for the same can be written as:
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.
Ans: 2
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.
Ans: 3
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)C> (Eg)Si> (Eg)Ge
Ans: 4
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.
Ans: 5
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.
Ans: 6
The correct statement is (b), (c).
For a transistor action, the junction must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reverse-biased.
Ans: 7
The correct statement is (c).
The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.
Ans: 8
Input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
∴Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
∴Output frequency = 2 × 50 = 100 Hz
Ans: 9
Collector resistance,RC = 2 kΩ = 2000 Ω
Audio signal voltage across the collector resistance, V = 2 V
Current amplification factor of the transistor, β = 100
Base resistance, RB= 1 kΩ = 1000 Ω
Input signal voltage =Vi
Base current = IB
We have the amplification relation as:
Voltage amplification
Therefore, the input signal voltage of the amplifier is 0.01 V.
Base resistance is given by the relation:
Therefore, the base current of the amplifier is 10 μA.
Ans: 10
Voltage gain of the first amplifier, V1 = 10
Voltage gain of the second amplifier, V2 = 20
Input signal voltage,Vi = 0.01 V
Output AC signal voltage = Vo
The total voltage gain of a two-stage cascaded amplifier is given by the product of voltage gains of both the stages, i.e.,
V = V1× V2
= 10 × 20 = 200
We have the relation:
V0 =V × Vi
= 200 × 0.01 = 2 V
Therefore, the output AC signal of the given amplifier is 2 V.
Ans: 11
Energy band gap of the given photodiode, Eg= 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9m
The energy of a signal is given by the relation:
E =
Where,
h = Planck’s constant
= 6.626 × 10−34 Js
c = Speed of light
= 3 × 108 m/s
E
= 3.313 × 10−20 J
But 1.6 × 10−19 J = 1 eV
∴E = 3.313 × 10−20 J
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.
Ans: 12
Number of silicon atoms, N = 5 × 1028 atoms/m3
Number of arsenic atoms, nAs = 5 × 1022 atoms/m3
Number of indium atoms,nIn = 5 × 1020 atoms/m3
Number of thermally-generated electrons, ni = 1.5 × 1016 electrons/m3
Number of electrons, ne= 5 × 1022 − 1.5 × 1016 ≈ 4.99 × 1022
Number of holes = nh
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:
nenh= ni2
Therefore, the number of electrons is approximately 4.99 × 1022 and the number of holes is about 4.51 × 109. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.
Ans: 13
Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV
The temperature dependence of the intrinsic carrier-concentration is written as:
Where,
kB = Boltzmann constant = 8.62 × 10−5 eV/K
T = Temperature
n0 = Constant
Initial temperature, T1= 300 K
The intrinsic carrier-concentration at this temperature can be written as:
…(1)Final temperature, T2= 600 K
The intrinsic carrier-concentration at this temperature can be written as:
…(2)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.
Therefore, the ratio between the conductivities is 1.09 × 105.
Ans: 14
In a p-n junction diode, the expression for current is given as:
Where,
I0 = Reverse saturation current = 5 × 10−12 A
T = Absolute temperature = 300 K
kB = Boltzmann constant = 8.6 × 10−5 eV/K = 1.376 × 10−23 J K−1
V = Voltage across the diode
(a) Forward voltage, V = 0.6 V
∴Current,I
Therefore, the forward current is about 0.0256 A.
(b) For forward voltage, V’ = 0.7 V, we can write:
Hence, the increase in current, ΔI = I' −I
= 1.257 −0.0256 = 1.23 A
(c) Dynamic resistance
(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I0in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.
Ans: 15
(a) A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in the following figure.
Hence, the output of the NOR Gate =
This will be the input for the NOT Gate. Its output will be
= 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 figure that the inputs of the right half NOR Gate are the outputs of the two NOT Gates.
Hence, this circuit functions as an AND Gate.
Ans: 16
A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.
Hence, the output can be written as:
The truth table for equation (i) can be drawn as:
- AY0110
Ans: 17
In both the given circuits, A and B are the inputs andY is the output.
(a) The output of the left NAND gate will be
, as shown in the following figure.Hence, the output of the combination of the two NAND gates is given as:
Hence, this circuit functions as an AND gate.
(b)
is the output of the upper left of the NAND gate and is the output of the lower half of the NAND gate, as shown in the following figure.Hence, the output of the combination of the NAND gates will be given as:
Hence, this circuit functions as an OR gate.
Ans: 18
A and Bare the inputs of the given circuit. The output of the first NOR gate is
. It can be observed from the following figure that the inputs of the second NOR gate become the out put of the first one.Hence, the output of the combination is given as:
The truth table for this operation is given as:
- ABY (=A + B)000011101111
Ans: 19
(a) A acts as the two inputs of the NOR gate and Yis the output, as shown in the following figure. Hence, the output of the circuit is
.The truth table for the same is given as:
- AY0110
(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 first two NOR gates are
as shown in the following figure.are the inputs for the last NOR gate. Hence, the output for the circuit can be written as:The truth table for the same can be written as:
- ABY (=A⋅B)000010100111
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