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Read the following statements regarding semiconductor and mark the incorrect answer.
There are two types of semiconductors:
Ptype semiconductor:
Ntype semiconductor:
Under equilibrium:
Under nonequilibrium:
Semiconductors have ______ conduction band and ______ valence band.
Property of Semiconductors:
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
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. (Write answer to one decimal point.)
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 (Write answer to two decimal point.)
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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