This fourteenth and second last chapter has eleven topics and makes you aware of the differentiation between extrinsic and intrinsic semiconductors, energy bands and their I-V characteristics, diodes, and types of diodes and their characteristics and uses of diodes, junction transistor and characteristics of a transistor and application. Here you also get knowledge about the concept of analog and digital signals, logic gates. Solving the question at the end of the chapter will tell you how well you have learned them with NCERT Solutions for Class 12 Physics.
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.
The statement is (d) true.
It is because 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.
The energy band gap is maximum for carbon, less for silicon and least for germanium out of the given three elements.
Hence answer is (c) is correct.
In an unbiased p-n junction, the diffusion of charge carriers across the junction takes place from higher concentration to lower concentration.
Thus answer (c) is correct.
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.
Hence, The correct statement is (c).
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.
The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.
Hence, the correct statement is (c).
Given, 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
Collector resistance, R C = 2 kΩ = 2000 Ω
Audio signal voltage across the collector resistance, V = 2 V
Current amplification factor of the transistor, β = 100
Base resistance, R B = 1 kΩ = 1000 Ω
Input signal voltage = V i
Base current = I B
We have the amplification relation as: V/Vi = β Rc/Rb
Voltage amplification, Vi = V Rb / β Rc
= 2x1000 / 100x2000 = 0.01 V
Therefore, the input signal voltage of the amplifier is 0.01 V.
Base resistance is given by the relation: RB = Vi / IB
= 0.01 / 1000 = 10 μA
Therefore, the base current of the amplifier is 10 μA.
Given that,
Voltage gain of the first amplifier, V 1 = 10
Voltage gain of the second amplifier, V 2 = 20
Voltage of input signal, V i = 0.01 V
Voltage of output AC signal = V o
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 = V 1 × V 2 = 10 × 20 = 200
It can be calculated by the relation: V = Vo/Vi
V 0 = V × V i = 200 × 0.01 = 2 V
Therefore, the output AC signal of the given amplifier is 2 V.
Given that,
Energy band gap of the given photodiode, E g = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10 −9 m
The energy of a signal is given by the relation: E = hc/λ
Where, h = Planck’s constant = 6.626 × 10 −34 Js
c = Speed of light = 3 × 10 8 m/s
E = 6.626 x 10-34 x 3 x 108 / 6000 x 10-9 = 3.313 x 10-20 J
But 1.6 × 10 −19 J = 1 eV
E = 3.313 × 10 −20 J
∴E = 3.313 × 10 −20 J = 3.313 x 10-20 / 1.6 x 10-19 = 0.207 eV
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.
Number of silicon atoms, N = 5 × 10 28 atoms/m 3
Number of arsenic atoms, n As = 5 × 10 22 atoms/m 3
Number of indium atoms, n In = 5 × 10 20 atoms/m 3
Number of thermally-generated electrons, n i = 1.5 × 10 16 electrons/m 3
Number of electrons, n e = 5 × 10 22 − 1.5 × 10 16 ≈ 4.99 × 10 22
Number of holes = n h
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as: n e n h = n i2
Therefore, nh = n i2 / ne = (1.5x1016)2 / 4.99 x 1022 ≈ 4.51 x 109
Therefore, the number of electrons is approximately 4.99 × 10 22 and the number of holes is about 4.51 × 10 9 . Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.