Basic Semiconductor Physics

Question 1
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
A
18.02 meV
B
9.01 meV
C
13.45 meV
D
26.90 meV
GATE EC 2020   Electronic Devices
Question 1 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
Question 2
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 _________.
A
10
B
16
C
20
D
25
GATE EC 2017-SET-1   Electronic Devices
Question 2 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}
Question 3
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 _________.
A
1
B
1.2
C
1.6
D
2
GATE EC 2017-SET-1   Electronic Devices
Question 3 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 4
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?
A
Silicon atoms act as p-type dopants in Arsenic sites and n-type dopants in Gallium sites
B
Silicon atoms act as n-type dopants in Arsenic sites and p-type dopants in Gallium sites
C
Silicon atoms act as p-type dopants in Arsenic as well as Gallium sites
D
Silicon atoms act as n-type dopants in Arsenic as well as Gallium sites
GATE EC 2017-SET-1   Electronic Devices
Question 4 Explanation: 
Si acts as p-type dopant GA sites.
Si acts as n-type dopant GA sites.
Question 5
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
A
1.5 \times 10^{13}cm^{-3} \; and \; 7.47 \times 10^{11}cm^{-3}
B
1.5 \times 10^{13}cm^{-3} \; and \; 8.23 \times 10^{11}cm^{-3}
C
7.5 \times 10^{13}cm^{-3} \; and \; 3.73 \times 10^{11}cm^{-3}
D
7.5 \times 10^{13}cm^{-3} \; and \; 4.12 \times 10^{11}cm^{-3}
GATE EC 2016-SET-1   Electronic Devices
Question 5 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}
Question 6
The figure below shows the doping distribution in a p-type semiconductor in log scale. The magnitude of the electric field (in kV/cm) in the semiconductor due to non uniform doping is _________
A
1.20
B
2.25
C
0.25
D
0.5
GATE EC 2016-SET-1   Electronic Devices
Question 6 Explanation: 


Applying the current density equation
\begin{aligned} J&=J_{\text {Dritt }}+J_{\text {Diff }} \\ J&=0\\ 0 &=-q D_{p} \frac{d p}{d x}+q\mu_{p} p E \\ q D_{p} \frac{d p}{d x} &=a \mu_{p} p E \\ E &=\frac{V_{T}}{p(x)} \frac{\partial p}{d r} \\ \therefore \quad |E| &=\frac{V_{T}}{\rho(x)} \frac{\partial p}{d x} \\ E &=v_{7} \frac{d}{d x} \ln \left[N_{A}(x)\right] \\ E &=V_{T}\left[\frac{\ln \left(10^{16}\right)-\ln \left(10^{14}\right)}{(2-1) \times 10^{-4}}\right] \\ &=0.026\left[\frac{36-32.23}{1 \times 10^{-4}}\right] \\ &=1198.6 \mathrm{V} / \mathrm{cm} \\ &=1.198 \mathrm{kV} / \mathrm{cm} \end{aligned}
Question 7
A small percentage of impurity is added to an intrinsic semiconductor at 300 K. Which one of the following statements is true for the energy band diagram shown in the following figure?
A
Intrinsic semiconductor doped with pentavalent atoms to form n-type semiconductor
B
Intrinsic semiconductor doped with trivalent atoms to form n-type semiconductor
C
Intrinsic semiconductor doped with pentavalent atoms to form p-type semiconductor
D
Intrinsic semiconductor doped with trivalent atoms to form p-type semiconductor
GATE EC 2016-SET-1   Electronic Devices
Question 8
The energy band diagram and the electron density profile n(x) in a semiconductor are shown in the figures. Assume that n(x)=10^{15}e^{(\frac{qax}{kT})}cm^{-3}, with \alpha = 0.1 V/cm and x expressed in cm. Given \frac{kT}{q} = 0.026V, D_{n}=36 cm^{2}s^{-1}, and \frac{D}{\mu}=\frac{kT}{q}. The electron current density (in A/cm^{2} ) at x = 0 is

A
-4.4 \times 10^{-2}
B
-2.2 \times 10^{-2}
C
0
D
2.2 \times 10^{-2}
GATE EC 2015-SET-2   Electronic Devices
Question 8 Explanation: 
The concentration of doping is not uniform throughout the semiconductor, thus to maintain equilibrium internal electric field is generated due to which the band of the semiconductor is varying with slope-0.1 eV/cm.
The generated electric field opposes the diffusion of carriers due to concentration gradient and thus there will be no current flow inside the semiconductor.
Question 9
A dc voltage of 10 V is applied across an n-type silicon bar having a rectangular cross-section and a length of 1 cm as shown in figure. The donor doping concentration N_{D} and the mobility of electrons \mu_{n} are 10^{16} cm^{-3} and 1000 cm^{2}V^{-1}s^{-1}, respectively. The average time (in \mu s) taken by the electrons to move from one end of the bar to other end is ________.

A
90
B
100
C
150
D
200
GATE EC 2015-SET-2   Electronic Devices
Question 9 Explanation: 
Electric field inside semiconductor bar,
E=10 \mathrm{V} / \mathrm{cm}
Velocity of electron, V=\mu E=1 \times 10^{4} \mathrm{cm} / \mathrm{sec}
The average time (in \mu \mathrm{s} ) taken by the electrons to move from one end of the bar to other end is
\frac{L}{V}=100 \mu \mathrm{sec}
Question 10
A piece of silicon is doped uniformly with phosphorous with a doping concentration of 10^{16}/cm^{3}. The expected value of mobility versus doping concentration for silicon assuming full dopant ionization is shown below. The charge of an electron is 1.6 \times 10^{-19}C. The conductivity( in S cm^{-1}) of the silicon sample at 300 K is _______.

A
1.45
B
1.92
C
3.35
D
4.22
GATE EC 2015-SET-2   Electronic Devices
Question 10 Explanation: 
As per the graph, mobility of electrons at the concentration 10^{16}/cm^{3} \text{ is } 1200 cm^{2}/V-s.
\begin{aligned} \text { So, } \mu_{n} &=1200 \mathrm{cm}^{2} / \mathrm{V}-\mathrm{s} \\ \sigma_{N} &=N_{D} \mathrm{q} \mu_{n} \\ &=10^{16} \times 1.6 \times 10^{-19} \times 1200=1.92 \mathrm{Scm}^{-1} \end{aligned}
There are 10 questions to complete.
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