Semiconductor Electronics: Materials, Devices and

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Payal Gupta

Contributor-Level 10

14.9 Energy gap of the given intrinsic semiconductor,  E g  = 1.2 eV

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

n i = n 0 e x p - E g 2 k B T

Where,  k B = Boltzmann constant = 8.62 * 10 - 5
 eV/K

T = Temperature

n 0 = constant

Initial temperature, T 1
 = 300 K

The intrinsic carrier-concentration at this temperature can be written as

n i 1 = n 0 e x p - E g 2 k B * 300 ……(1)

Final temperature,  T 2 = 600 K

The intrinsic carrier-concentration at this temperature can be written as

n i 2 = n 0 e x p - E g 2 k B * 600 ……(2)

The ratio between conductivity at 600K and at 300 K is equal to the ratio between the respective intrinsic carrier concentration at these temperatures.

Therefore,

n i 2 n i 1 =&nbs

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New answer posted

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Payal Gupta

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14.8 Number of silicone atoms, nS = 5 *1028 / m3

Number of arsenic atoms, nA = 5 *1022 / m3

Number of indium atoms, nI = 5 *1020 / m3

Number of thermally generated atoms, ni = 1.5 *1016 / m3

Hence number of electrons, ne = 5 *1022- 1.5 *1016 = 4.99 *1022

Number of holes = nH

In thermal equilibrium, the concentration of electrons and holes in a semiconductor are related as:

nenH = ni2

Therefore, nH = ni2ne = (1.5*1016)24.99*1022 = 4.51 *109

Since the number of electrons are more (4.99 *1022) than the number of holes (4.51 *109) , the material is a

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Payal Gupta

Contributor-Level 10

14.7 Energy band gap of the given photodiode, Eg = 2.8 eV

Wavelength, λ = 6000 nm = 6000 *10-9 m

The energy of a signal is given by the relation,

E = hcλ

Where

h = Planck's constant = 6.626 *10-34 Js

c = sped of light = 3 *108 m/s

E = 6.626*10-34*3*1086000*10-9 = 3.313 *10-20 J = 3.313*10-201.6*10-19 eV = 0.207 eV

Therefore 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 the photodiode. Hence, the photodiode cannot detect the signal.

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Payal Gupta

Contributor-Level 10

14.6 Input frequency = 50 Hz

For a half-wave rectifier, the output frequency is equal to the input frequency.

Hence, the output frequency = 50 Hz

For a full-wave rectifier, the output frequency is twice the input frequency.

Hence the output frequency = 2 *50Hz=100Hz

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Payal Gupta

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14.5 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.

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Payal Gupta

Contributor-Level 10

14.4 The correct statement is (c).

The diffusion of charge carriers across a junction takes place from the regions 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.

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Payal Gupta

Contributor-Level 10

14.2 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.

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Payal Gupta

Contributor-Level 10

14.1 The correct statement is (c ).

In an n-type silicon, the electrons are the majority carrier, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorous, are doped in silicon atoms.

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