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Read the following statements regarding semiconductor and mark the incorrect answer.
There are two types of semiconductors:
The Fermi level in a semiconductor bar should:
Semiconductors have ______ conduction band and ______ valence band.
Property of Semiconductors:
Hole mobility in Ge at room temperature is 1900 cm2/V-sec. The diffusion coefficient is ________cm2/sec.
(Write answer to one decimal point.)
(Take kT = 25 mV)
Using the relation, D = 0.025 × 1900 cm2/sec
= 47.5 cm2/sec
The Difference between the donor energy level and fermi level in a n-type 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.
ED - EF = -0.028 eV
A semiconductor with intrinsic carrier concentration 1 × 1010 cm-3 at 300°K has both valence and conduction band effective densities of states NC and NV equal to 1019 cm-3. The band gap Eg is _____ eV. (Write answer to one decimal point.)
The intrinsic concentration is related to bandgap by:
Eg = 0.0518 ln (108) eV
= 0.954 eV
In a very long p-type Si bar with doping concentration Na = 1017 cm-3, excess holes are injected such that excessive concentration of holes at x = 0 is 5 × 1016 cm-3. The hole concentration at x = 1 μm is _____ × 1017 cm-3. Take μp = 500 cm2/v-s and recombination time constant τp = 10-8 s, kT = 0.0259 eV (Write answer to two decimal point.)
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 × 1015 cm-3. If the electron mobility μn = 1000 cm2/v-s 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 cm-3
= 1.2375 × 1015 cm-3
Electron mobility μn = 1000 cm2/v-s
Resistivity of n-type silicon =
= 5.05 Ω – cm
Holes are injected into n-type Ge so that the at the surface of the semiconductor hole concentration is 1014/cm3. If diffusion constant of a hole in Ge is 49cm2/sec and minority carrier lifetime is τp = 10-3 sec. Then the hole concentration Δp at a distance of 4mm from the surface is ______1014/cm3.
= 1.6 × 1013/cm3
= 0.16 × 1014/cm3
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