The Fermi level in a semiconductor bar should
- The fermi level across the entire material will be sum and does not vary with distance.
- If there is any disturbance in the material, like junction contact, injection of impurities at any point, the charge carriers redistribute themselves such that the fermi-potential is same in entire material.
- The fermi level is uneven with gradient of charge distribution across distance, in material.
- It can be studied using quasi-fermi states, related to charge distribution.
Hole mobility in Ge at room temperature is 1900 cm2/V-sec. The diffusion coefficient is ________cm2/sec.
(Take kT = 25 mV)
Using the relation
D = 0.025 × 1900 cm2/sec
= 47.5 cm2/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.
Given that fermi level is 0.3 eV above the intrinsic level, hence the type of semiconductor is n-type.
In n-type the donor level is just below the conduction band and above the fermi level.
A hypothetical semiconductor has bandgap of 0.56 eV at 300 K. If the effective mass of electron in such semiconductor is 4 times that of hole. Then the probability of occupancy of top of valence band (upto 2 decimal) at 300 K is
(Take kT = 26 mV)
Position of Fermi level
EF = 0.28 – 0.027
EF = 0.252 eV
Probability of occupancy
This signifies that at room temperature the top of valence band is almost filled
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.
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
⇒ Diffusion length
Hole concentration at any distance x is sum of equillibrium and excess hole concentration
excess hole concentration varies with the distance x as Δp e-x/Lp
p = po + Δp e-x/Lp
Where po = Doping concentration
Δp = extra hole concentration
= (1+.379) x 1017
1.379 × 1017 cm-3
In an intrinsic semiconductor, the Fermi energy level EF doesn’t lie in the middle of the band gap cause
In an intrinsic semi-conductor the Fermi energy
Where NC and NV are density of state in valence and conduction bond respectively.
Since NC ∝ mn, effective mass of electron and Nv ∝ mp, effective mass of hole
Since mn and mp are not equal
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