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In a semiconductor which of the following carrier can contribute to the current?
In a semiconductor, two types of charges are there by which the flow of the current takes place. So, both the holes and electrons take part in the flow of the current.
Under equilibrium:
Under nonequilibrium:
Hole mobility in Ge at room temperature is 1900 cm^{2}/Vsec. The diffusion coefficient is ________cm^{2}/sec.
(Write answer to one decimal point.)
(Take kT = 25 mV)
Using the relation, D = 0.025 × 1900 cm^{2}/sec
= 47.5 cm^{2}/sec
In the band diagram of a semiconductor, the fermi level is 0.3 eV above the intrinsic level. The energy level E in the diagram represents:
The Difference between the donor energy level and fermi level in a ntype semiconductor in where 25% of the atoms are ionised at 300 k is:
As 25% of donor atoms are ionised, the occupation probability of donor level is 0.75.
E_{D}  E_{F} = 0.028 eV
A semiconductor with intrinsic carrier concentration 1 × 10^{10} cm^{3} at 300°K has both valence and conduction band effective densities of states N_{C} and N_{V} equal to 10^{19} cm^{3}. The band gap E_{g} is _____ eV.
The intrinsic concentration is related to bandgap by:
E_{g} = 0.0518 ln (10^{8}) eV
= 0.954 eV
In a very long ptype Si bar with doping concentration N_{a} = 10^{17} cm^{3}, excess holes are injected such that excessive concentration of holes at x = 0 is 5 × 10^{16} cm^{3}. The hole concentration at x = 1 μm is _____ × 10^{17} cm^{3}. Take μ_{p} = 500 cm^{2}/vs and recombination time constant τ_{p} = 10^{8} s, kT = 0.0259 eV
Diffusion constant:
Diffusion length:
In an intrinsic semiconductor, the Fermi energy level EF doesn’t lie in the middle of the band gap cause:
A silicon bar is doped with donor impurities ND = 2.25 × 10^{15} cm^{3}. If the electron mobility μn = 1000 cm2/vs then the approximate value of resistivity of silicon bar assuming partial ionization of 55% is __________ (Ω  cm)
Given ND = 2.25 × 1015 cm^{3}
Partial ionization of 55%, number of electrons is:
⇒ n = 0.55 × 2.25 × 1015 cm3
= 1.2375 × 1015 cm3
Electron mobility μn = 1000 cm2/vs
Resistivity of ntype silicon =
= 5.05 Ω – cm
Holes are injected into ntype Ge so that the at the surface of the semiconductor hole concentration is 10^{14}/cm^{3}. If diffusion constant of a hole in Ge is 49cm^{2}/sec and minority carrier lifetime is τ_{p} = 10^{3} sec. Then the hole concentration Δp at a distance of 4mm from the surface is ______10^{14}/cm^{3}.
= 1.6 × 10^{13}/cm^{3}
= 0.16 × 10^{14}/cm^{3}
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