[1987: 2 M] Consider two energy levels: E1, E eV above the fermi level and E2, E eV below the Fermi level. P1 and P2 are respectively the probabilities of E1 being occupied by an electron and E2 being empty. Then
a) P1 > P2
b) P1 = P2
c) P1 < P2
d) P1 and P2 depend on number of free electrons
Ans: (b)
Explanation
By the definition of Fermi Level.
[1987: 2 M] In an intrinsic semiconductor the free electron concentration depends on
a) effective mass of electrons only
b) effective mass of holes only
c) temperature of the semiconductor
d) width of the forbidden energy band of the semiconductor
Ans: (c)
Explanation
In the intrinsic semiconductors, p= hole concentration and n= electron concentration are equal and known as intrinsic carrier concentration. And we know that,
n_i^2\propto T^3 n_i\propto T^{3/2}For intrinsic semiconductor,
n = p = ni
n\propto T^{3/2}[1987: 2M] According to the Einstein relation, for any semiconductor the ratio of diffusion constant to mobility of carriers
a) depends upon the temperature of the semiconductor
b) depends upon the type of the semi conductor
c) varies with life time of the semi conductor
d) is a universal constant
Ans: (a)
Explanation
By the Einstein’s relation,
\frac D\mu=\frac{D_n}{\mu_n}=\frac{D_p}{\mu_p}=V_T=\frac{T\left({}^oK\right)}{11600}where,
T= Temperature in {}^oK
VT= Thermal voltage
D= Diffusion constant
μ= Mobility
Dn=Electron diffusion constant
Dp= Hole diffusion constant
μn=Electron mobility
[1987: 2M] Direct band gap semi conductors
a) exhibit short carrier life time and they are used for fabricating BJT’s
b) exhibit long carrier life time and they are used for fabricating BJT’s
c) exhibit short carrier life time and they are used for fabricating lasers
d) exhibit long carrier life time and they are used for fabricating lasers
Ans: (c)
Explanation
DBG (Direct Band Gap) semiconductors exhibit short carrier life time they are used for fabricating lasers.
Also, the energy is released during the recombination in the form of light.
[1989: 2 M] Due to illumination by light, the electron and hole concentrations in a heavily doped N type semi conductor increases by \triangle_n and \triangle_P respectively if ni is the intrinsic concentration then,
a) \triangle_n<\triangle_P
b) \triangle_n>\triangle_P
c) \triangle_n=\triangle_P
d) \triangle_n\times\triangle_P=n_i^2
Ans: (c)
Explanation
Due to illumination by light EHP (electron-hole pair) generation occurs.
So, \triangle n\;=\;\triangle p
[1989: 2 M] The concentration of ionized acceptors and donors in a semi conductor are NA, ND respectively. If NA > ND and ni is the intrinsic concentration, the position of the fermi level with respect to the intrinsic level depends on
a) N_A-N_D
b) N_A+N_D
c) \frac{N_AN_D}{n_i^2}
d) n_i
Ans: (a)
[1990: 2 M] Under high electric fields, in a semiconductor with increasing electric field,
a) the mobility of charge carriers decreases
b) the mobility of the carries increases
c) the velocity of the charge carriers saturates
d) the velocity of the charge carriers increases
Ans: (a, c)
Explanation
Typical graph of mobility of charge carriers versus electric field is shown below.
Hence, the mobility of charge carriers decreases as electric field increases at the high E
\mu\propto\frac1EAlso, v_d=\mu E
vd = Drift velocity; μ = Mobility; E = Applied electric field
Hence, we can conclude, the velocity (drift velocity) of charge carriers saturates.
[1991: 2 M] A silicon sample is uniformly doped with 1016 phosphorus atoms/cm3 and 2 x 1016 boron atoms/cm3. If all the dopants are fully ionized, the material is
a) n-type with carrier concentration of 1016/cm3
b) p-type with carrier concentration of 1016/cm3
c) p-type with carrier of 2 x 1016/cm3
d) n-type with a carrier concentration of 2 x 1016/cm3
Ans: (b)
Explanation
We have,
ND = n = Phosphorus atoms = 1016/cm3
NA = p= Boron atoms = 2 x 1016/cm3
\because\;N_A>\;>N_DSo, the resultant material will be p-type semiconductor carrier concentration
= NA-ND
= 2 x 1016-1016
=1016/cm3
[1992: 2 M] A semi conductor is irradiated with light such that carriers are uniformly generated thr0ughtout its volume. The semiconductor is n-type with ND=1019/cm3. If the excess electron concentration in the steady state is \triangle n=10^{15}/cm^3 and if \tau_p=10\;\mu sec. (minority carriers life time) the generation rate due to irradiation
a) is 1020 e-h pairs/cm3/s
b) is 1024 e-h pairs/cm3/s
c) is 1010 e-h pairs/cm3/s
d) cannot be determined, the given data is insufficient
Ans: (a)
Explanation
\triangle n=10^{15}/cm^3 \tau_p=10\;\mu sec=10\;\times10^{-6}sec Generation\;rate=\frac{\triangle n}{\tau_p}=\frac{10^{15}}{10\times10^{-6}}= 1020 e-h pairs/cm3/s
[1994: 1 M] A p-type silicon sample has a higher conductivity compared to an n-type silicon sample having the same dopant concentration. (True/False)
Ans: False
Explanation
For a given semiconductor the electron mobility (μn) is always higher than the hole mobility (μp) i.e. μn > μp
And, the conductivity of a given n-type semiconductor σn=n qμn the conductivity of a given p-type semiconductor. σp=p qμp
given that n = p = same dopant concentration, we can easily conclude that, σn > σp