Ncert Solution Class 12th Physics Chapter – 14 Semiconductor Electronics: Materials, Devices and Simple Circuits Question & Answer

Ncert Solution Class 12th Physics Chapter – 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

Ncert Solution Class 12th Physics Chapter – 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

?Chapter – 14?

✍Semiconductor Electronics: Materials, Devices and Simple Circuits✍

?Question & Answer?

Q 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: Here, (c) is the correct option.

For n-type silicon, the majority carriers are electrons while the minority carriers are holes. An n-type semiconductor is obtained by dropping pentavalent atoms like phosphorus in silicon atoms.

 Q 2. Which of the statements given in Exercise 14.1 is true for p-type semiconductors?

Answer: Here, (d) is the correct explanation.

For p-type semiconductor, holes are the majority carriers while electrons are the minority carriers. p-type semiconductor is obtained by using trivalent atoms like aluminium in silicon atoms.

Q 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 _g ) _C , (E g ) Si and (E _g ) _Ge . Which of the following statements is true?

(a) (E _g ) _Si < (E _g ) _Ge < (E _g ) _C

(b) (E _g ) _C < (E _g ) _Ge > (E _g ) _Si

(c) (E _g ) _C > (E _g ) _Si > (E _g ) _Ge

(d) (E _g ) _C = (E _g ) _Si = (E _g ) _Ge

​Answer: 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.

Q 4. In an unbiased p – n junction, holes diffuse to n – region from p – 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 of the above.

Answer: (c) is the correct option.

The usual tendency of the charge carriers is to disperse towards the lower concentration region from the higher concentration region. So it can be said that in an unbiased p-n junction, holes disperse from p-region to the n-region as the p-region has a greater concentration of holes than in n-region.

Q 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) potential barrier is reduced.
(d) None of the above.

Answer: (c) is the correct option

The potential barrier reduces for a p-n junction when a forward bias is applied.

In the above case, the potential barrier across the junction reduces as the applied voltage is opposed by the potential barrier.

Q 6. In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency?

Answer: For a half-wave rectifier, the output frequency is equal to the input frequency, in this case, the input frequency of the half-wave rectifier is 50 Hz.

On the other hand, the output frequency for a full-wave rectifier is twice the input frequency.
Therefore, the output frequency is 2 × 50 = 100 Hz.

Q 7. A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm?

Answer: No, the photodiode cannot detect the wavelength of 6000 nm because of the following reason:

The energy bandgap of the given photodiode, E_g = 2.8 eV

The wavelength is given by λ = 6000 nm = 6000 × 10⁻⁹ m

We can find the energy of the signal from the following relation:

E = hc/λ

In the equation, h is the Planck’s constant = 6.626 × 10⁻³⁴ J and c is the speed of light = 3 × 10⁸ m/s.

Substituting the values in the equation, we get

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

Therefore, 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.

Q 8. The number of silicon atoms per m³ is 5 × 10²⁸. This is doped simultaneously with 5 × 10²² atoms per m³ of Arsenic and 5 × 10²⁰ per m³ atoms of Indium. Calculate the number of electrons and holes. Given that n_I = 1.5 × 10¹⁶m^–3. Is the material n-type or p-type?

Answer: Following values are given in the question:

Number of silicon atoms, N = 5 × 10 ²⁸ atoms/m³

Number of arsenic atoms, n_AS =5×10²²atoms/m³

Number of indium atoms, n_In=5×10²²atoms/m³

n_i=1.5×10¹⁶electrons/m³

n_e=5×10²²−1.5×10¹⁶=4.99×10²²

Let us consider the number of holes to be n_h

In the thermal equilibrium, n_en_h = n_i²

Calculating, we get

n_h=4.51×10⁹

Here, n_e>n_h, therefore the material is a n-type semiconductor.

Q 9. In an intrinsic semiconductor the energy gap E_g is 1.2eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration n_i is given by

where n₀ is constant.

Answer: Energy gap of the given intrinsic semiconductor, E_g = 1.2 eV

The temperature dependence of the intrinsic carrier-concentration is written as:

Where,

k_B = 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.

Q 10. In a p-n junction diode, the current I can be expressed as

where I₀ 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_B is the Boltzmann constant (8.6×10^–5 eV/K) and T is the absolute temperature. If for a given diode I₀ = 5 × 10^–12 A and T = 300 K, then
(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: Given: For a given diode, reverse saturation current is 5× 10 −12  A and the absolute temperature is 300 K.

(a)The expression for the current is given as,

I= I 0 exp( eV k B T −1 )

Where, I 0 is the reverse saturation current, V is the voltage across the diode, k B os the Boltzmann constant and T is the absolute temperature.

By substituting the given values in the above equation, we get

I=5× 10 −12 exp( 1.6× 10 −19 ×0.6 8.6× 10 −5 ×1.6× 10 −19 ×300 −1 ) =5× 10 −12 ×1.259× 10 10 =6.295× 10 −2  A =62.95 mA

Thus, the forward current is 62.95 mA.

(b) The forward current when the voltage across the diode is 0.7 V is,

I ′ =5× 10 −12 exp( 1.6× 10 −19 ×0.7 8.6× 10 −5 ×1.6× 10 −19 ×300 −1 ) =5× 10 −12 ×6.07× 10 11 =3.035 A

The increase in the current is,

I ′ −I=3.035−0.0629 =2.972 A

Thus, the increase in the current is 2.972 A.

(c) The dynamic resistance is given as,

R= ΔV ΔI

By substituting the values in the above equation, we get

R= 0.7−0.6 2.972 =33.65 mΩ

Thus, the dynamic resistance is 33.65 mΩ.

(d) The current in the diode be almost equal when voltage changes, which shows almost infinite dynamic resistance in reverse bias.

Hence, the current in the diode for both voltages will be,

I≈ I 0 =( −5× 10 −12 )Α

Q 11. You are given the two circuits as shown in Fig. 14.36. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

Answer: (a) The left half of the figure acts as NOR Gate, while right half as NOT Gate.

Q 12. Write the truth table for a NAND gate connected as given in
Fig. 14.37

Answer: Step 1: Find output for the given circuit.
Given, input for the circuit, =?
Output for the circuit, = Y
In the given circuit the shown gate is NAND gate.
Output for the circuit,
$Y=\overline{A.A}$
$Y=\overline{A}+\overline{A}$
$Y=\overline{A}$

The truth table is:
$\begin{array}{cc}\text{INPUTS}& OUTPUT\\ A& Y\\ 0& 1\\ 1& 0\end{array}$

Q 13. You are given two circuits as shown in Figure, which consists of NAND gates. Identify the logic operation carried out by the two circuits.

Answer: Given: A and B acts as the Two inputs of the circuit Y is the output.

(a) Let left side of the output is given as, Y 1 .

The output of the left of the NAND gate is given as,

Y 1 = A⋅B ¯

This output is fed through the another NAND Gate.

The circuit output is given as,

Y= Y 1 ⋅ Y 1 ¯

Substitute the value of Y 1 in above expression,

Y= ( A⋅B ¯ )⋅( A⋅B ¯ ) ¯ = ( AB ¯ ) ¯ + ( AB ¯ ) ¯ =AB

Thus, the circuit acts as AND Gate.

(b) A ¯ is output of the upper left of the NAND gate and B ¯ is the output of the lower half of the NAND gate.

The output of the circuit is given as,

Y= A ¯ ⋅ B ¯ = A ¯ ¯ + B ¯ ¯ =A+B

Thus, the circuit acts as OR gate.

Q 14. Write the truth table for circuit given in Fig. 14.39 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing

Answer: Let Y 1 be the output of first NOR gate.

Q 15. Write the truth table for the circuits given in Figure consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

Answer: (a) In the circuit, A is the inputs and Y is the output.

The output of the circuit is given as,

Y= A+A ¯ = A ¯

The truth table of the circuit is given as,

The truth table represents NOT gate.

Thus, the circuit behaves like a NOT gate.

(b) The output of two NOR gates are A ¯ and B ¯ respectively.

The output of the circuit is given as,

Y= A ¯ + B ¯ ¯ = A ¯ ¯ ⋅ B ¯ ¯ =A⋅B

The truth table of the circuit is given by,

The truth table represents AND gate.

Thus, the circuit behaves like an AND gate.