HC Verma Solutions Class 12 Chapter 23 Semiconductor And Semiconductor Devices

HC Verma Solutions Class 12 Chapter 23 Semiconductor and Semiconductor Devices can be used as a source of background information on topics where student’s might lack proficiency such as finding the number of conduction electrons, holes, unoccupied states in semiconductors etc. The solutions are mainly designed for class 12 students and cover the complete JEE Physics topics as well as the exercises discussed in the chapter. Some of the common questions asked in this unit are;

  • There are questions related to finding the energy band between the acceptor band and valence band.
  • Questions that deal with finding the energy needed to overcome the band gap.
  • Students will solve questions based on unbiased, reverse and forward biasing of current and many more.

The whole idea behind preparing and offering the solutions is to help students get familiar with the question-solving pattern and integrate them into their study activities. With the solutions, students will further get an insight into important topics and understand them more clearly in the process. Students can use the solutions to also prepare well for competitive exams like JEE.

Download This Solution As PDF: HC Verma Solutions Chapter 23 PDF

Key Topics Related To Semiconductor In Physics

  • Energy Bands in Solids
  • p-type and n-type Semiconductors
  • The Density of Charge Carriers and Conductivity
  • p-n Junction
  • p-n Junction Diode
  • p-n Junction as a Rectifier
  • Junction Transistors
  • Logic Gates

Class 12 Important Questions In Chapter 23

1. If the conduction band of a solid is partially filled at 0 K then which category will it belong to. Will it be a semiconductor, a conductor or an insulator?

2. When we increase the temperature of the junction the drift current in a reverse-biased p-n junction also increases in magnitude. On the basis of creation of hole-electron pairs can you explain why and how this occurs?

3. In a semiconductor, you will normally find valence electrons and conduction electrons. Does a semiconductor also have ‘conduction holes’ and ‘valence holes’?

4. What condition will you find a p-type semiconductor always in?

(a) It is positively charged (b) has a negative charge (c) remains uncharged at 0 K but gets charged at high temperatures (d) no charge exists

5. When will the diffusion current in a p-n junction be greater than the drift current in magnitude?

(a) when the junction is unbiased (b) if the junction is forward-biased (c) if the junction is reverse-biased (d) such criteria do not exist

HC Verma Solutions Vol 2 Semiconductor and Semiconductor Devices Chapter 23

Question 1: Calculate the number of states per cubic metre of sodium is 3s band. The density of sodium is 1013 kg m-3. How many of them are empty?

Solution:

Given: Mass of the sodium in 1 m3 = 1013 kgm-3

Atomic mass of sodium = 23

Total number of atoms = N = [mass x avogadro’s number]/[atomic mass]

= [1013×103x6.022×1023]/[23]

= 265.22 x 1026 atoms

In N atoms of sodium there will be 2N states possible states and N states will be empty.

Total number of states = 2 x N = 2 x 265.22 x 1026 = 5.30 x 1028 states

As atomic number of sodium = 11

Its electronic configuration = 1s2, 2s2, 2p6, 3s1

=> The 3s band is half-filed in case of sodium, the total number of unoccupied states is 2.65 x 1028.

Question 2: In a pure semiconductor, the number of conduction electrons is 6 × 1019 per cubic metre. How many holes are there in a sample of size 1 cm × 1cm × 1 mm?

Solution:

Since in pure semiconductor the number of holes = number of conduction electron

=> Number of holes = 6 × 1019 m-3

Therefore, Number of holes in 1 cm × 1cm × 1 mm = 6 × 1019 x 10-7 = 6 x 1012 holes

[Using 1mm = 10-7 m3]

Question 3: Indium antimonide has a band gap of 0.23 eV between the valence and the conduction band. Find the temperature at which kT equals the band gap.

Solution:

We know, kT = 0.23 eV

Where, k = Boltzmann constant and t = temperature

=> T = 0.23 eV/k = 0.23 eV/[8.62×10-5eVK-1] = 2670 K (approx)

Question 4: The band gap for silicon is 1.1 eV.

(a) Find the ratio of the band gap to kT for silicon at room temperature 300 K.

(b) At what temperature does this ratio become one tenth of the value of 300 K? (Silicon will not retain its structure at these high temperatures)

Solution:

The band gap for silicon is 1.1 eV.

kT at room temp = 0.026 eV

(a) Ratio of band gap to kT = 1.1/0.026 = 43 (approx)

(b) Consider new ratio is 1/10th of the Part (a) result.

Now, Band gap/kT = 4.3

=> 1.1/[8.62×10-5xT] = 4.3

=> T = 3000 K(approx)

Question 5: When a semiconducting material is doped with an impurity, new acceptor levels are created. In a particular thermal collision, a valence electron revives an energy equal to 2kT and just reaches one of the acceptor levels. Assuming that the energy of the electron was at the top of edge of the valence band and that the temperature T is equal to 300 K, find the energy of the acceptor levels above the valence band.

Solution:

The electron was at the top of the edge of valance band and reaches acceptor level on receiving the energy given.

Energy gap between the acceptor level and valence band must be 2kT.

Where, k = Boltzmann constant and t = temperature

=> E = 2 x 8.62 x 10-5 x 300 = 51.72 meV

Question 6: The band gap between the valence and the conduction bands in zinc oxide (ZnO) is 3.2 eV. Suppose an electron in the conduction band combines with a hole in the valence band and the excess energy is released in the form of electromagnetic radiation. Find the maximum wavelength that can be emitted in this process.

Solution:

For the electron to move from the conduction band to valance band will have to lose energy same as the energy band gap of ZnO OR The max energy released in this process will be equal to the band gap of the material.

=> E = 3.2 eV

=> 3.2 = [1242eV-nm]/λ

=> λ = 388.1 nm

Question 7: Suppose the energy liberated in the recombination of a hole-electron pair is converted into electromagnetic radiation. If the maximum wavelength emitted is 820 nm, what is the band gap?

Solution:

The min energy release in the recombination of the conduction band electron with valence bond hole = band gap of the material

We know, E = hc/λ

=> E = [1242eV-nm]/820nm

[given λ = 820nm]

=> E = 1.5 eV

Question 8: Find the maximum wavelength of electromagnetic radiation which can create a hole-electron pair in germanium. The band gap in germanium is 0.65 eV.

Solution:

The band gap in germanium is 0.65 eV.

The energy required to produce the hole-electron pair = energy band gap

We know, E = hc/λ

=> λ = hc/E = [4.14 x10-15x3x108]/0.65

= 1.9 x 10-6 m

Which is required wavelength.

Question 9: In a photodiode, the conductivity increases when the material is exposed to light. It is found that the conductivity changes only if the wavelength is less than 620 nm. What is the band gap?

Solution:

Maximum wavelength at which the conduction starts = 620 nm

and this wavelength provides minimum energy to the electron to move to conduction band.

we know, Energy of this radiation = Energy band gap

=> E = [4.14×10-15x3x108]/[620×10-9] = 2.003 eV = 2 eV (approx)

Which is the required band gap.

Question 10: Let ΔE denote the energy gap between the valence band and the conduction band. The population of conduction electrons (and of the holes) is roughly proportional to e–ΔE/2kT. Find the ratio of the concentration of conduction electrons in diamond to that in silicon at room temperature 300 K. ΔE for silicon is 1.1 eV and for diamond is 6.0 eV. How many conduction electrons are likely to be in one cubic meter of diamond?

Solution:

Let ΔE1 and ΔE2 be the energy band gap for silicon and diamond.

=> ΔE1 = 1.1 eV and ΔE2 = 6.0 eV (Given)

Also, we are given that, population of conduction electrons (and of the holes) is roughly proportional to e–ΔE/2kT.

Therefore,

HC Verma Class 12 Ch 23 Solution 10

Because of more band gap, there will be almost zero conduction electrons will be there in the diamond.

Question 11: The conductivity of a pure semiconductor is roughly proportional to T3/2 e–ΔE/2kT where ΔE is the band gap. The band gap for germanium is 0.74 eV at 4 K and 0.67 eV at 300 K. By what factor does the conductivity of pure germanium increase as the temperature is raised from 4 K to 300 K.

Solution:

Let E1 = 0.74 eV and E2 = 0.67 eV

Let σ1 and σ2 be the conductivity at 4 K and 300K.

Therefore,

HC Verma Class 12 Ch 23 Solution 11

or σ2 = 0.148 x 10463 x σ1 = 10463 σ1 (approx)

Which shows that conductivity increases by the factor 10463.

Question 12: Estimate the proportion of boron impurity which will increase the conductivity of a pure silicon sample by a factor of 100. Assume that each boron atom creates a hole and the concentration of holes in pure silicon at the same temperature is 7 × 1015 holes per cubic meter. Density of silicon is 5 × 1028 atoms per cubic meter.

Solution:

Initially, the number of charge carriers = 7×1015 + 7×1015 = 14 x 1015

[As number conduction electron and holes are the same]

Total charge carriers after adding the impurity to the pure silicon = 14 x 1015 x 100 = 14 x 1017

let “y” be holes after adding the impurity and (14 x 1017 – y) will be the electrons.

Now,

Product of the total conduction electrons and holes remains nearly constant.

=> 7×1015 x 7×1015 = y x (14 x 1017 – y)

Solving above equation for y, we get

y = 13.982 x 1015 or y = 1.8 x 1015

As, initial number of holes must be smaller than the number of holes after doping, so neglecting 2nd value.

Thus, the number of boron atoms added = 13.982 x 1017 – 7×1015

= 1398.2 x 1015 – 7×1015

= 13.912 x 1017

13.912 x 1017 atoms are added per 5 × 1028 atoms of Silicon per cubic meter.

So, 1 atom of boron is added per [5 × 1028]/[13.912 x 1017] = 3.59 x 1010 atoms of Silicon per cubic meter.

Therefore, 3.59 x 1010 is the proportion of born impurity.

Question 13: The product of the hole concentration and the conduction electron concentration turns out to be independent of the amount of any impurity doped. The concentration of conduction electrons in germanium is 6 × 1019 per cubic meter. When some phosphorus impurity is doped into a germanium sample, the concentration of conduction electrons increases to 2 × 1023 per cubic meter. Find the concentration of the holes in the doped germanium.

Solution:

Concentration of holes = 6 x 1019

Product of concentration of conduction electron and that of holes remains constant.

Let nh be the Concentration of holes.

=>(6 x 1019) x (6 x 1019) = (2 x 1023) x nh

As per question, 2×1023 be the concentration of conduction electron after doping. And

6 x 1019 be the concentration of conduction electron before doping.

=> nh = [36×1038]/[2×1023]

=> nh = 18 x 1015

Thus, number of holes after doping is 18 x 1015 per cubic metre.

Question 14: The conductivity of an intrinsic semiconductor depends on temperature as σ = σ0 e–ΔE/2kT, where σ0 is a constant. Find the temperature at which the conductivity of an intrinsic germanium semiconductor will be double of its value at T = 300 K. Assume that the gap for germanium is 0.650 eV and remains constant as the temperature is increased.

Solution:

ΔE (Band gap) = 0.650 eV

Given: σ = σ0 e–ΔE/2kT …(1)

Let the conductivities at temperatures T1 and T2 be σ1 and σ2 respectively.

Given T1 = 300K and T2 = temperature at which the conductivity is double of its value at T1.

=> 2σ1 = σ2

(1)=>

HC Verma Class 12 Ch 23 Solution 14

Question 15: A semiconducting material has a band gap of 1 eV. Acceptor impurities are doped into it which create acceptor levels 1 meV above the valence band. Assumed that the transition from one energy level to the other is almost forbidden if kT is less than 1/50 of the energy gap. Also, if kT is more than twice the gap, the upper levels have maximum population. The temperature of the semiconductor is increased from 0K. The concentration of the holes increases with temperature and after a certain temperature it becomes approximately constant. As the temperature is further increased, the hole concentration again starts increasing at a certain temperature. Find the order of the temperature range in which the hole concentration remains approximately constant.

Solution:

A semiconducting material has a band gap of 1 eV. Acceptor impurities are doped into it which create acceptor levels 1 meV above the valence band.

Band gap after doping = 1eV -1 meV = (1 – 0.001) = 0.999 eV

Given Condition 1: Any transition from one energy level to the other is almost forbidden if kT is less than 1/50 of the energy gap.

=> kT1 = 0.999/50

Where, k = Boltzmann constant

T1 be the upper limit for temperature over which transition becomes a forbidden transition.

=> T1 = 0.999/[50×8.62×10-5] = 232.8 K (approx)

Given Condition 2: If kT is more than twice the gap, the upper levels have maximum population.

For max limit

kT2 = 2 x 10-3

=> T2 = [2 x 10-3]/[8.62×10-5] = 23.2 K

Required temperature range = (23.2K – 232.8K)

Question 16: In a p-n junction, the depletion region is 400 nm wide and an electric field of 5 × 105 V m–1 exists in it.

(a) Find the height of the potential barrier (b) What should be the minimum kinetic energy of a conduction electron which can diffuse from the n-side to the p-side.

Solution:

(a) We know the relation between electric field and potential,

E = V/d

or V = Ed = 5×105 x 4×10-7 = 0.2 volts

Given E = 5×105 Vm-1 and d = 4×10-7 m

(b) Minimum K.E. = Potential barrier x charge on an electron

= 0.2 x 1.6 x 10-19

= 0.2 eV

Question 17: The potential barrier existing across an unbiased p-n junction is 0.2 volt. What minimum kinetic energy a hole should have to diffuse from the p-side to the n-side if

(a) the junction is unbiased,

(b) the junction is forward biased at 0.1 volt and

(c) the junction is reverse biased at 0.1 volt?

Solution:

Minimum K.E. = [Potential barrier – biasing voltage] x charge on hole

(a) Biasing voltage = 0 V

Minimum K.E. = 0.2V x e = 0.2 eV

(b) Biasing Voltage for forward biased = 0.1V

Minimum K.E. = (0.2-0.1) × e = 0.1

(c) Biasing Voltage for reverse biased = -0.1

Minimum K.E. = (0.2 + 0.1) × e = 0.3 eV

Question 18: In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction

(a) from the p-side and (b) from the n-side

Solution:

(a) A junction will act like forward biasing when the hole approaches the junction from p-side

K.E. of the hole will decrease.

Final kinetic energy of the hole = (300-250) = 50 meV

(b) A junction will act like reverse biasing, when the hole approaches the junction from n-side

K.E. of the hole in this case will increase.

Final kinetic energy of the hole = (300+250)= 550 meV

Question 19: When a p-n junction is reverse biased, the current becomes almost constant at 25 μA. When it is forward biased at 200 mV, a current of 75 μA is obtained. Find the magnitude of diffusion current when the diode is

(a) unbiased (b) reverse biased at 200 mV (c) forward biased at 200 mV.

Solution:

When a p-n junction is reverse biased, the current becomes almost constant at 25 μA. When it is forward biased at 200 mV, a current of 75 μA is obtained.

(a) When the diode is unbiased,

Diffusion current = drift current = 25μA

(b) When the diode is reverse biased, then diffusion current = 0

(c) When the diode is forward biased at 200 mV, then

Diffusion current – Drift current = Forward biasing current

Diffusion current = (75 + 25) μA = 100 μA

Question 20: The drift current in a p-n junction is 20.0 μA. Estimate the number of electrons crossing a cross section per second in the depletion region.

Solution:

The drift current (id) in a p-n junction is 20.0 μA = 20.0×10-6A

we know, id = (Ne + Nh) x e

where, Nh = Number of holes crossing a cross section per second

Ne= Number of electrons crossing a cross section per second

e = magnitude of charge on a hole = 1.6×10-19 C

Without any biasing applied on the junction, Nh = Ne = N

=> id = 2N x e

or N = id/2e = [20×10-6]/[2×1.6×10-19] = 6.25 x 1013.