Semiconductor Electronics: Materials, Devices and
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New answer posted
8 months agoContributor-Level 10
14.9 Energy gap of the given intrinsic semiconductor, = 1.2 eV
The temperature dependence of the intrinsic carrier-concentration is written as:
Where,
= Boltzmann constant = 8.62
eV/K
T = Temperature
= constant
Initial temperature,
= 300 K
The intrinsic carrier-concentration at this temperature can be written as
……(1)
Final temperature, = 600 K
The intrinsic carrier-concentration at this temperature can be written as
……(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,
=&nbs
New answer posted
8 months agoContributor-Level 10
14.8 Number of silicone atoms, = 5 /
Number of arsenic atoms, = 5 /
Number of indium atoms, = 5 /
Number of thermally generated atoms, = 1.5 /
Hence number of electrons, = 5 1.5 = 4.99
Number of holes =
In thermal equilibrium, the concentration of electrons and holes in a semiconductor are related as:
=
Therefore, = = = 4.51
Since the number of electrons are more (4.99 than the number of holes (4.51 , the material is a
New answer posted
8 months agoContributor-Level 10
14.7 Energy band gap of the given photodiode, = 2.8 eV
Wavelength, = 6000 nm = 6000 m
The energy of a signal is given by the relation,
=
Where
h = Planck's constant = 6.626 Js
c = sped of light = 3 m/s
= = 3.313 J = 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.
New answer posted
8 months agoContributor-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
New answer posted
8 months agoContributor-Level 10
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.
New answer posted
8 months agoContributor-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.
New answer posted
8 months agoContributor-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.
New answer posted
8 months agoContributor-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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