# H C VERMA Solutions for Class 12-science Physics Chapter 23 - Semiconductors and Semiconducting Devices

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## Chapter 23 - Semiconductors and Semiconducting Devices Exercise 419

Question 1

Calculate the number of states per cubic meter of sodium in 3s band. The density of sodium is 1013kg/cm3, How many of them are empty?

Solution 1

We know ,

And no. of atoms is given as :

Total no. of states = 2n

Total no. of empty states =

Question 2

In a pure semiconductor, the number of conduction electrons is 6x1019Per cubic mertre. How many holes are there in a sample of size 1cm × 1cm × 1cm?

Solution 2

In case of pure semiconductor,

No. of electrons = No. of holes

Also,

And from 1 we know

Question 3

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

Solution 3

ABC

Question 4

The band gap for silicon is 1.1eV. (a) Find the ratio of the band gap to kT for silicon at room temperature 300k. (b) At what temperature does this ratio become one tenth of the value at 300k? (Silicon will not retain its structure at these high temperatures.)

Solution 4

a)

b)

T = 3000.47k

Question 5

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

Solution 5

Energy gap between two gaps is:

E=2kT

E = 50meV

Question 6

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

Solution 6

Band gap is given as :

Question 7

Suppose the energy liberated in the recombination of a ahole-electron pair is converted iinto electromagnetic radiation. If the maximum wavelength emitted is 820nm, what is the band gap?

Solution 7

We know,

Question 8

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

Solution 8

We know,

Question 9

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

Solution 9

We know

And energy required to overcome the gap is band gap.

Thus ,

E=2eV

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 . Find the ratio of the concentration of conduction electrons in diamond to that in silicon at room temperature 300K.ΔE for silicon is 1.1eV and for diamond is 6.0eV. How many conduction electrons are likely to be in one cubic metre of diamond?

Solution 10

As given :

And

For diamond ΔE is more the conduction electrons is apporxinately equal zero.

Question 11

The conductivity of a pure semiconductor is roughly proportional to  where is the band gap. The band gap for germanium is 0.74 eV at 4k and 0.67eV at 300k. By what factor does the conductivity of pure germanium increase as the temperature is raised from 4K to 300K?

Solution 11

As given :

And

Taking ratio or equation 1 and 2

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 holes per cubic metre. Density of silicon is  atoms per cubic metre.

Solution 12

Total no. Of charge carriers =

Also we know,

Solving we get

As value is more after doping

And no. Of atoms of boronadded

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  per cubic metre. When some phosphorus impurity is droped into a germanium sample, the concentration of conduction electrons increases to  per cubic metre. Find the concentration of the holes in the doped germanium.

Solution 13

Initially ,

After doping concentration becomes,

Product of concentration remains constant. Thus,

Question 14

The conductivity of an intrinsic semiconductor depends on temperature as  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=300K. Assume that the gap for germanium is 0.650eV and remains constant as the temperature is increased.

Solution 14

As given ,

And,

Taking log on both sides and solving we get;

Question 15

A semiconductor material has a band gap of 1eV. Acceptor impurities are doped into it which create acceptor levels 1 meV above the valence band. Assume 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 oK. 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 15

Net band gap is:

Also,

Now

Range is 23.2k to 231.8k

Question 16

In a p-n junction, the depletion region is 400nm wide and an electric field of V/m 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 16

a) Potential barrier is :

V =Ed

=0.2V

b) kinetic energy is :

=0.2eV

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 17

a)

=0.2eV (No bias condition)

b) When biasing is forward.

K.E. + Ve = 0.2e

K.E.=0.2e-Ve

=(0.2-0.1)e

=0.1e

c) When bias is reversed

K.E. - Ve = 0.2e

K.E.=0.2e+Ve

=(0.2+0.1)e

=0.3e

Question 18

In a p-n junction, a potential barrier of 250meV exists across the junction. A hole with a kinetic energy of 300meV 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 18

a) Kinetic energy of hole is decreased if junction is forward biased. Thus,

K.E. = 300-250

=50meV

b) Kinetic energy of hole is increased if junction is reverse biased. Thus,

K.E. = 300+250

=550meV

Question 19

When a p-n junction is reverse-biasedn the current becomes almost constant at 25 μA. When it is forward-biased at 200mV,a current of 75μA is obtained. Find the magnitude of diffusion current when the diode is (a) unbiased,(b) reverse-baised at 200mV and (c) forward-baised at 200mV.

Solution 19

a) unbiased condition:

diffusion current = drift current

Diffusion current

b) Reverse bias condition :

Diffusion current =OA

c) For forward bias condition :

## Chapter 23 - Semiconductors and Semiconducting Devices Exercise 420

Question 20

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

Solution 20

No. Of electrons is given as :

Question 21

The current-voltage characteristic of an ideal p-n junction diode is given by

Where the drift current i0 equals 10. A take the temperature T to be 300K. (a) Find the voltage V0 for which =100. One can neglect the term 1 for voltages greater than this value. (b) Find an expression for the dynamic resistance of the diode as a function of V for V > V0. (c) Find the voltage for which the dynamic resistance is 0.2Ω.

Solution 21

a) We know,

b) We know,

And

Differentiating w.r.t V

c) From above part

V = 0.25V

Question 22

Consider a p-n junction diode having the characteristic

Where i0=20μA. The diode is operated at T=300K. (a) Find the current through the diode when a voltage of 300mV is applied across it in forward bias. (b) At what voltage does the current double?

Solution 22

a) According to question :

i=2A

b)

Solving we get

V = 318mV

Question 23

Calculate the current through the circuit and the potential difference across the diode shown in figure (45-E1). The drift current for the diode is 20 μA.

Solution 23

a)

b)

Question 24

Each of the resistance shown in figure (45-E2) has a value of 20Ω. Find the equivalent resistance between A and B. Does it depends on whether the point A or B is at higher potential?

Solution 24

There is no current in diode by wheatstone bridge's principle

Now, net resistance is

R=20Ω

Question 25

Find the currents through the resistance in the circuits shown in figure (45-E3).

Solution 25

a) According to Ist figure both diodes are in forward bias

and current is given as :

b) In this case (according to IInd figure) one diode is in forward bias while other is in reverse bias.

Thus,

And current is given as :

I = OA

c) In this case (figure III) both diodes are forward biased

Thus,  and current is given as :

d) In this case (figure IV) one diode is forward bias and other is reverse bias but they are in parallel and thus current passes through diode which forward bias.

Question 26

What are the readings of the ammeters A1 and A2 shown in figure (45-E4). Neglect the resistances of the meters.

Solution 26

a) In A1 current is o because diode is reverse bias and thus resistance is infinite.

b) In A2 current is given as :

Question 27

Find the current through the battery in each of the circuits shown in figure (45-E5).

Solution 27

a) According to  figure , both diodes are forward bias

Net resistance offered by diodes is O.

Now

Now,

b) According to IInd figure one diode is forward bias and other diode is reverse bias thus current will pass through diode which is forward biased.

Thus,

Question 28

Find the current through the resistance R is figure (45-E6) if (a) R=12Ω (b) R =48Ω.

Solution 28

a)

I= 0.42A

b)

I= 0.16A

Question 29

Draw the current-voltage characteristics for the device shown in figure (45-E7) between the terminals A and B.

Solution 29

a)

b)

Question 30

Find the equivalent resistance of the network shown in figure (45-E8) between the points A and B.

Solution 30

There are two cases:

a) If , then diode is forward biased and resistance will be :

b) If , then diode is reverse biased and resistance will be:

Question 31

When the base current in a transistor is changed from 30μA to 80 μA, the collector is changed from 1.0mA to 3.5mA. find the current gain β.

Solution 31

We know,

Question 32

A load resistor of 2kΩ connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode. The current Β = 50.the input resistance of the transistor is 0.50k. If the input current is changed by 50 μA. (a) By what amount does the output voltage change, (b) by what amount does the input voltage change and (c) what is the power gain?

Solution 32

Now,

a) change

Change in V = 8000v

b) change in input voltage

=25mV

c) Power gain ,

Question 33

Let  . Evaluate X for

a) A=1,B=0,C=1,

b) A=B=C=1

c) A=B=C=0.

Solution 33

a) X=1 ( if A = 1, B = 0, C=1)

b) X=0 (if A=B=C=1)

c) X=0 (if A=B=C=0)

Question 34

Design a logical circuit using AND, or and NOT gates to evaluate

Solution 34

Circuit diagram for  is shown below :

Question 35

Show that AB+ is always 1.

Solution 35

To prove

a) If A=0

B=0

=0+1

=1

b) If A=0 B=1 or A=1 B=0

=0+1

=1

c) If A=1 B=1

=1+0

=1

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