Question 1 |
Select the CORRECT statement(s) regarding semiconductor devices.
Electrons and holes are of equal density in an intrinsic semiconductor at equilibrium. | |
Collector region is generally more heavily doped than Base region in a BJT. | |
Total current is spatially constant in a two terminal electronic device in dark under
steady state condition. | |
Mobility of electrons always increases with temperature in Silicon beyond 300 K. |
Question 1 Explanation:
At equilibrium n = p = n_i
for intrinsic
semiconductor
Collector region is generally lightly doped then base region in BJT. Hence option B is wrong.
By increasing temperature above 300K, mobality of electrons decreases hence option (D) is also wrong
Collector region is generally lightly doped then base region in BJT. Hence option B is wrong.
By increasing temperature above 300K, mobality of electrons decreases hence option (D) is also wrong
Question 2 |
In a non-degenerate bulk semiconductor with electron density n=10^{16}cm^{-3}, the
value of E_C-E_{Fn}=200meV, where E_C and E_{Fn} denote the bottom of the
conduction band energy and electron Fermi level energy, respectively. Assume
thermal voltage as 26 meV and the intrinsic carrier concentration is 10^{10}cm^{-3}. For n=0.5 \times 10^{16}cm^{-3}, the closest approximation of the value of (E_C-E_{Fn}), among
the given options, is ______.
226 meV | |
174 meV | |
218 meV | |
182 meV |
Question 2 Explanation:
Here we have to find the value of
E_c-E_{fn}
As we know,
E_C-E_F=kT \ln\left ( \frac{N_c}{n} \right ) \;\;\;...(i)
E_C-E_{F1}=kT \ln\left ( \frac{N_c}{n_1} \right ) \;\;\;...(ii)
E_C-E_{F2}=kT \ln\left ( \frac{N_c}{n_2} \right ) \;\;\;...(iii)
Equation (ii) - Equation (iii)
(E_C-E_{F1})-(E_C-E_{F2})=kT \ln \left ( \frac{\frac{N_c}{n_1}}{\frac{N_c}{n_2}} \right )=kT \ln \frac{n_2}{n_1}
\Rightarrow 200meV-(E_C-E_{F2})=26meV \times \ln \left ( \frac{0.5 \times 10^{16}}{1 \times 10^{16}} \right )
200meV-(E_C-E_{F2})=+26meV \ln (0.5)=-18
(E_C-E_{F2})=200+8=218meV
As we know,
E_C-E_F=kT \ln\left ( \frac{N_c}{n} \right ) \;\;\;...(i)
E_C-E_{F1}=kT \ln\left ( \frac{N_c}{n_1} \right ) \;\;\;...(ii)
E_C-E_{F2}=kT \ln\left ( \frac{N_c}{n_2} \right ) \;\;\;...(iii)
Equation (ii) - Equation (iii)
(E_C-E_{F1})-(E_C-E_{F2})=kT \ln \left ( \frac{\frac{N_c}{n_1}}{\frac{N_c}{n_2}} \right )=kT \ln \frac{n_2}{n_1}
\Rightarrow 200meV-(E_C-E_{F2})=26meV \times \ln \left ( \frac{0.5 \times 10^{16}}{1 \times 10^{16}} \right )
200meV-(E_C-E_{F2})=+26meV \ln (0.5)=-18
(E_C-E_{F2})=200+8=218meV
Question 3 |
Consider a long rectangular bar of direct bandgap p-type semiconductor. The
equilibrium hole density is 10^{17}cm^{-3} and the intrinsic carrier concentration is 10^{10}cm^{-3}. Electron and hole diffusion lengths are 2\mu mand 1\mu m, respectively.
The left side of the bar (x=0) is uniformly illuminated with a laser having photon
energy greater than the bandgap of the semiconductor. Excess electron-hole pairs
are generated ONLY at x=0 because of the laser. The steady state electron density
at x=0 is 10^{14}cm^{-3}
due to laser illumination. Under these conditions and ignoring
electric field, the closest approximation (among the given options) of the steady state
electron density at x=2 \mu m, is _____
0.37 \times 10^{14} cm^{-3} | |
0.63 \times 10^{13} cm^{-3} | |
3.7 \times 10^{14} cm^{-3} | |
0^{3} cm^{-3} |
Question 3 Explanation:
From continuity equation of electrons
\frac{dn}{dt}=n\mu _n\frac{dE}{dx}+\mu _nE\frac{dn}{dx}+G_n-R_n+x_n\frac{d^2x}{dx^2} \;\;\;...(i)
[Because \vec{E} is not mentioned hence
\frac{dE}{dx}=0
For x \gt 0, G_n is also zero
n=\frac{n_i^2}{N_A}=\frac{10^{20}}{10^{17}}=10^3
n=n_0+\delta n=10^3+10^{14}=10^{14}
at steady state, \frac{db}{dt}=0
Hence equation (i) becomes:
O=D_n\frac{d^2\delta n}{dx^2}-\frac{\delta n}{\tau _n}
\frac{d^2\delta n}{dx^2}=\frac{\delta n}{L_n^2} \;\;\;...(ii)
From solving equation (ii)
\delta _n(x)=\delta _n(0)e^{-x/L_n}
at x=2\mu m
\delta _n(2\mu m)=10^{14}e^{-2/2}=10^{14}e^{-1}=0.37 \times 10^{14}
Question 4 |
The energy band diagram of a p-type semiconductor bar of length L under equilibrium condition (i.e.. the Fermi energy level E_{F}
is constant) is shown in the figure. The valance band E_{V}
is sloped since doping is non-uniform along the bar. The difference between the energy levels of the valence band at the two edges of the bar is \Delta.
If the charge of an electron is q, then the magnitude of the electric field developed inside this semiconductor bar is
If the charge of an electron is q, then the magnitude of the electric field developed inside this semiconductor bar is
\frac{\Delta }{qL} | |
\frac{2\Delta }{qL} | |
\frac{\Delta }{2qL} | |
\frac{3\Delta }{2qL} |
Question 4 Explanation:
The built-in electric field is due to non-uniform doping (the semiconductor is under equilibrium)
\begin{aligned} E &=\frac{1}{q}\frac{ d E_{v}}{d x} \\ &=\frac{1}{q} \frac{\Delta}{L} \\ &=\frac{\Delta}{q L} \end{aligned}
\begin{aligned} E &=\frac{1}{q}\frac{ d E_{v}}{d x} \\ &=\frac{1}{q} \frac{\Delta}{L} \\ &=\frac{\Delta}{q L} \end{aligned}
Question 5 |
A bar of silicon is doped with boron concentration of 10^{16} \text{cm}^{-3} and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of 10^{20} \text{cm}^{-3} s^{-1}. If the recombination lifetime is 100 \;\mu s, intrinsic carrier concentration of silicon is 10^{10} \text{cm}^{-3} and assuming 100\% ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is
10^{20} \text{cm}^{-6} | |
2 \times 10^{20} \text{cm}^{-6} | |
10^{32} \text{cm}^{-6} | |
2 \times 10^{32} \text{cm}^{-6} |
Question 5 Explanation:
Boron \rightarrow Acceptor type doping
\begin{aligned} N_{A} &=10^{16} \mathrm{~cm}^{-3} \\ g_{0 p} &=1020 \mathrm{~cm}^{-3} \mathrm{~s}^{-1} \\ \tau &=100 \mu \mathrm{s} \\ n_{i} &=10^{10} \mathrm{~cm}^{-3} \end{aligned}
Product of steady state electron-hole concentration =?
At thermal equilibrium (before shining light)
\begin{array}{ll} \text { Hole concentration, } & p_{o} \simeq N_{A}=10^{16} \mathrm{~cm}^{-3} \\ \text { Electron concentration, } & n_{0}=\frac{n_{i}^{2}}{p_{0}}=\frac{10^{20}}{10^{16}}=10^{4} \mathrm{~cm}^{-3} \end{array}
After, illumination of light,
Hole concentration, p=p_{o}+\delta p
Electron concentration, \quad n=n_{o}+\delta n
Due to shining light, excess carrier concentration,
\begin{aligned} \delta p &=\delta n=g_{o p} \cdot \tau=10^{20} \times 100 \times 10^{-6}=10^{16} \mathrm{~cm}^{-3} \\ \therefore \qquad p &=10^{16}+10^{16}=2 \times 10^{16} \mathrm{~cm}^{-3}\\ n&=10^{4}+10^{16} \simeq 10^{16} \mathrm{~cm}^{-3} \end{aligned}
So, product of steady state electron-hole concentration
\begin{aligned} &=n p=10^{16} \times 2 \times 10^{16} \\ &=2 \times 10^{32} \mathrm{~cm}^{-6} \end{aligned}
\begin{aligned} N_{A} &=10^{16} \mathrm{~cm}^{-3} \\ g_{0 p} &=1020 \mathrm{~cm}^{-3} \mathrm{~s}^{-1} \\ \tau &=100 \mu \mathrm{s} \\ n_{i} &=10^{10} \mathrm{~cm}^{-3} \end{aligned}
Product of steady state electron-hole concentration =?
At thermal equilibrium (before shining light)
\begin{array}{ll} \text { Hole concentration, } & p_{o} \simeq N_{A}=10^{16} \mathrm{~cm}^{-3} \\ \text { Electron concentration, } & n_{0}=\frac{n_{i}^{2}}{p_{0}}=\frac{10^{20}}{10^{16}}=10^{4} \mathrm{~cm}^{-3} \end{array}
After, illumination of light,
Hole concentration, p=p_{o}+\delta p
Electron concentration, \quad n=n_{o}+\delta n
Due to shining light, excess carrier concentration,
\begin{aligned} \delta p &=\delta n=g_{o p} \cdot \tau=10^{20} \times 100 \times 10^{-6}=10^{16} \mathrm{~cm}^{-3} \\ \therefore \qquad p &=10^{16}+10^{16}=2 \times 10^{16} \mathrm{~cm}^{-3}\\ n&=10^{4}+10^{16} \simeq 10^{16} \mathrm{~cm}^{-3} \end{aligned}
So, product of steady state electron-hole concentration
\begin{aligned} &=n p=10^{16} \times 2 \times 10^{16} \\ &=2 \times 10^{32} \mathrm{~cm}^{-6} \end{aligned}
Question 6 |
A single crystal intrinsic semiconductor is at a temperature of 300 K with effective density
of states for holes twice that of electrons. The thermal voltage is 26 mV. The intrinsic
Fermi level is shifted from mid-bandgap energy level by
18.02 meV | |
9.01 meV | |
13.45 meV | |
26.90 meV |
Question 6 Explanation:
\frac{E_{c}+E_{v}}{2}-E_{F_{i}}=\frac{KT}{2}\ln \left ( \frac{N_{C}}{N_{V}}\right )\, \, \, \, \, \, \left ( \because N_{C}=\frac{N_{V}}{2} \right )
=\frac{0.026}{2}\ln 0.5=-9.01\, meV
=\frac{0.026}{2}\ln 0.5=-9.01\, meV
Question 7 |
As shown a uniformly doped Silicon (Si) bar of length L = 0.1 \mum with a donor concentration N_{D}=10^{16}cm^{-3} is illuminated at x = 0 such that electron and hole pairs are generated at the rate of G_{L}=G_{L0}(1-\frac{x}{L}
), 0 \leq x \leq L , where G_{L0}=10^{17} cm^{-3}s^{-1}. Hole lifetime is 10^{-4}s,
electronic charge q=1.6\times 10^{-19}C, hole diffusion coefficient D_{P}=100 cm^{2}/s and low level injection condition prevails. Assuming a linearly decaying steady state excess hole concentration that goes to 0 at x = L, the magnitude of the diffusion current density at x = L/2, in A/cm^{2}, is _________.
10 | |
16 | |
20 | |
25 |
Question 7 Explanation:
Net hole density varying in the direction of x is,
\begin{aligned} p_{n}(x) &=p_{n 0}+\Delta p=p_{n 0}+G_{L} \tau_{p} \\ &=p_{n 0}+G_{L o} \tau_{p}\left(1-\frac{x}{L}\right) \\ J_{p, \text { diff }} &=-e D_{p} \frac{d p}{d x}=-e D_{p}\left[\frac{-G_{L o} \tau_{p}}{L}\right] \\ &=\frac{1.6 \times 10^{-19} \times 100 \times 10^{17} \times 10^{-4}}{0.1 \times 10^{-4}} \mathrm{A} / \mathrm{cm}^{2} \\ &=16 \mathrm{A} / \mathrm{cm}^{2} \end{aligned}
\begin{aligned} p_{n}(x) &=p_{n 0}+\Delta p=p_{n 0}+G_{L} \tau_{p} \\ &=p_{n 0}+G_{L o} \tau_{p}\left(1-\frac{x}{L}\right) \\ J_{p, \text { diff }} &=-e D_{p} \frac{d p}{d x}=-e D_{p}\left[\frac{-G_{L o} \tau_{p}}{L}\right] \\ &=\frac{1.6 \times 10^{-19} \times 100 \times 10^{17} \times 10^{-4}}{0.1 \times 10^{-4}} \mathrm{A} / \mathrm{cm}^{2} \\ &=16 \mathrm{A} / \mathrm{cm}^{2} \end{aligned}
Question 8 |
The dependence of drift velocity of electrons on electric field in a semiconductor is shown below. The semiconductor has a uniform electron concentration of n=1\times 10^{16}cm^{-3} and electronic charge q=1.6\times 10^{-19}C. If a bias of 5V is applied across a 1 \mum region of this semiconductor, the resulting current density in this region, in kA/cm^{2}, is _________.
1 | |
1.2 | |
1.6 | |
2 |
Question 8 Explanation:
\begin{aligned} E&=\frac{V}{d}=\frac{5}{10^{-4}}=5 \times 10^{4} \mathrm{V} / \mathrm{cm}\\ &\text{Slope of the curve,}\\ m &=\frac{10^{7}-0}{5 \times 10^{5}}=20 \quad y=m_{x} \\ v_{d} &=20 \times E=20 \times 5 \times 10^{4} \\ &=10^{6} \mathrm{V} / \mathrm{cm} \\ J &=n e v_{d} \\ &=1 \times 10^{16} \times 1.6 \times 10^{-19} \times 10^{6} \\ &=1.6 \times 10^{3} \mathrm{A} / \mathrm{cm}^{2}=1.6 \mathrm{kA} / \mathrm{cm}^{2} \end{aligned}
Question 9 |
A bar of Gallium Arsenide (GaAs) is doped with Silicon such that the Silicon atoms occupy Gallium and Arsenic sites in the GaAs crystal. Which one of the following statement is true?
Silicon atoms act as p-type dopants in Arsenic sites and n-type dopants in Gallium sites | |
Silicon atoms act as n-type dopants in Arsenic sites and p-type dopants in Gallium sites | |
Silicon atoms act as p-type dopants in Arsenic as well as Gallium sites | |
Silicon atoms act as n-type dopants in Arsenic as well as Gallium sites |
Question 9 Explanation:
Si acts as p-type dopant GA sites.
Si acts as n-type dopant GA sites.
Si acts as n-type dopant GA sites.
Question 10 |
Consider a silicon sample at T = 300 K, with a uniform donor density N_{d}=5 \times 10^{16}cm^{-3} illuminated uniformly such that the optical generation rate is G_{opt}=1.5 \times 10^{20}cm^{-3}s^{-1}
throughout the sample. The incident radiation is turned off at t = 0. Assume low-level injection to be valid and ignore surface effects. The carrier lifetimes are \tau _{p0}=0.1 \mu s \; and \; \tau _{n0}=0.5\mu s
The hole concentration at t = 0 and the hole concentration at t= 0.3 \mus, respectively, are
The hole concentration at t = 0 and the hole concentration at t= 0.3 \mus, respectively, are
1.5 \times 10^{13}cm^{-3} \; and \; 7.47 \times 10^{11}cm^{-3} | |
1.5 \times 10^{13}cm^{-3} \; and \; 8.23 \times 10^{11}cm^{-3} | |
7.5 \times 10^{13}cm^{-3} \; and \; 3.73 \times 10^{11}cm^{-3} | |
7.5 \times 10^{13}cm^{-3} \; and \; 4.12 \times 10^{11}cm^{-3} |
Question 10 Explanation:
\begin{aligned} \text{Given}quad G_{opt} &=1.5 \times 10^{20} / \mathrm{cm}^{3} / \mathrm{sec} \\ G_{opt} &=R=\frac{N_{A}}{\tau_{P}} \\ \Rightarrow \quad 1.5 \times 10^{2} &=\frac{N_{A}}{0.1 \times 10^{-6}} \\ \therefore\quad N_{A} &=1.5 \times 10^{13} / \mathrm{cm}^{3} \\ P(t) &=P_{n0} e^{-t / \tau_{p}} \\ &=1.5 \times 10^{13} e^{\frac{-0.3}{0.1}} \\ &=7.46 \times 10^{11} / \mathrm{cm}^{3}\\ \end{aligned}
There are 10 questions to complete.