Download NOC: Fundamentals of electronic materials and devices

NOC: Fundamentals of electronic materials and devices
Assignment 2
The assignment consists of 10 questions, each question carries two marks.
1. Conductivity of intrinsic semiconductors is independent of
(a) Band gap
(b) Fermi level
(c) carrier mobility
perature
Ans: (b)
(d) tem-
2. An intrinsic direct band gap semiconductor has energy gap of 1.4
eV . Which of the following EM radiation can it absorb?
(a) 1500 nm
(b) 1300 nm
(c) 1100 nm
(d) 800 nm
Ans: (d). Use the expression E = hc/λ and convert 1.4 eV into
wavelength. The answer is around 890 nm. So, any wavelengths below
that will be absorbed.
3. Which of the following statements is always true for intrinsic semiconductors?
(a) Fermi level lies close to the midpoint of valence and conduction
band edges
(b) Fermi level does not depend on the carrier effective mass
(c) The electron-atom interactions are negligible compared to the electronelectron interactions
(d) Higher than band gap, higher is the electrical conductivity
Ans: (a). Fermi level need not be at the center of the band gap, but
is close to the center.
4. The hole concentration in intrinsic Ge is independent of
(a) temperature
(b) effective mass
(c) band gap
(d) mobility
Ans: (d). Mobility only affects conductivity, not the carrier concentration.
5. Conductivity in an intrinsic semiconductor is due to
(a) the movement of holes in conduction band
(b) the movement of electrons in valence band
(c) both (a) and (b)
(d) neither (a) and (b)
Ans: (d). Holes move the in the valence band and electrons in the
conduction band.
1
6. The band gap of intrinsic Si is 1.1 eV and its electron affinity is 4.1
eV . Assuming that the effective density of states at the valence and
conduction bands are equal (Nc = Nv ), the work function of the
material is approximately
(a) 1.1 eV
(b) 4.1 eV
(c) 4.7 eV
(d) 5.2 eV
Ans: (c). Fermi level lies in the middle of the band gap. Hence, work
function is 4.1 + 1.1/2 = 4.7 eV .
7. For intrinsic Si, take Nc = Nv = 2 × 1025 m−3 . The intrinsic carrier
concentration at 300 K, in cm−3 , is approximately
(a) 104
(b) 1010
(c) 1014
(d) 1016
Ans: (b). This can be got by direct
√ substitution of the formula, since
all parameters are known. ni = Nc Nv exp(−Eg /kB T )
8. If the intrinsic carrier concentration in GaAs is 2 × 106 cm−3 at 300 K,
at what temperature is the carrier concentration, 10 times higher,
i.e. 2 × 107 cm−3 . Take band gap of GaAs to be 1.4 eV and Nc and Nv
to be independent of temperature.
(a) 330 K
(b) 390 K
(c) 500 K
(d) 600 K
Ans: (a). Use the formula given in the previous question and take the
ratio at two different temperatures.
9. The conductivity of intrinsic GaAs at 300 K is 2.9 × 10−9 Ω−1 cm−1 . If
the ratio of electron to hole mobility is 19:1, then the electron mobility,
in cm2 V −1 s−1 , is approximately
(a) 1500
(b) 2500
(c) 5500
(d) 8500
(Use any required values from previous problems)
Ans: (d). Use the expression, σ = ni e(µe + muh ). The ratio of
the mobilities are given, so you can use this to solve for mue . Use the
value for ni from the previous problem.
10. CdSe is direct band gap II-VI semiconductor with Eg = 1.74 eV at
300 K. The electron effective mass is 0.13me and hole effective mass is
0.45me , where me is the rest mass of the electron. The intrinsic carrier
concentration, at 300 K, is (units cm−3 )
(a) 7.4 × 103
(b) 1.4 × 106
(c) 7.4 × 109
(d) 1.4 × 1012
Ans: (a). This problem requires you to calculate Nc and Nv first, from
the effective masses. Once, you get that ni can be calculated directly.
Just have to convert the units to cm−3 . Looking at the band gap and
the other choices, you can also use elimination to arrive at the right
answer.
2