Electronic Devices


Question 1
In a semiconductor device, the Fermi-energy level is 0.35 \mathrm{eV} above the valence band energy. The effective density of states in the valence band at T=300 \mathrm{~K} is 1 \times 10^{19} \mathrm{~cm}^{-3}. The thermal equilibrium hole concentration in silicon at 400 \mathrm{~K} is ___ \_\times 10^{13} \mathrm{~cm}^{-3}. (rounded off to two decimal places).
Given \mathrm{kT} at 300 \mathrm{~K} is 0.026 \mathrm{eV}.
A
63.36
B
25.36
C
45.25
D
98.36
GATE EC 2023      Basic Semiconductor Physics
Question 1 Explanation: 
Given,
E_{F}-E_{V}=0.35 \mathrm{eV} \quad \text { [Considering it is given at } 400 \mathrm{~K} \text { ] }
Also, V_{T_{1}}=K T_{1}=0.026 \mathrm{eV} at T_{1}=300 \mathrm{~K}
\therefore \frac{V_{T_{1}}}{V_{T_{2}}}=\frac{T_{1}}{T_{2}} \Rightarrow V_{T_{2}}=\frac{T_{2}}{T_{1}} \times V_{T_{1}}
\therefore V_{T_{2}}=\frac{400}{300} \times 0.026
V_{T_{2}}=0.03466 \mathrm{eV} at T_{2}=400 \mathrm{~K}

Now, N_{v}=1 \times 10^{19} / \mathrm{cm}^{3} at T_{1}=300 \mathrm{~K}
N_{v} \propto T^{3 / 2}
\frac{N_{V_{2}}}{N_{V_{1}}}=\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2}
N_{V_{2}}=\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2} N V_{1}
\left(\because T_{2}=400 \mathrm{~K}\right) =\left(\frac{400}{300}\right)^{3 / 2} N_{V_{1}}
N_{V_{2}}=1.5396 \times 10^{19} / \mathrm{cm}^{3}

Now, hole concentration at 400 \mathrm{~K} is given as
\begin{aligned} & p=N_{V} e^{-\left(E_{F}-E_{V}\right) / k T_{2}}=1.5396 \times 10^{19} \times e^{-0.35 \mathrm{eV} / 0.03466 \mathrm{eV}} \\ & p=63.36 \times 10^{13} \mathrm{~cm}^{-3} \end{aligned}
Question 2
In an extrinsic semiconductor, the hole concentration is given to be 1.5 n_{i} where n_{i} is the intrinsic carrier concentration of 1 \times 10^{10} \mathrm{~cm}^{-3}. The ratio of electron to hole mobility for equal hole and electron drift current is given as ___
(rounded off to two decimal places).
A
1.15
B
2.25
C
2.85
D
3.36
GATE EC 2023      Basic Semiconductor Physics
Question 2 Explanation: 
Given, intrinsic carrier concentration n_{i}=1 \times 10^{10} \mathrm{~cm}^{-3}
Hole concentration,
\quad p=1.5 \times n_{i}
p=1.5 \times 10^{10} \mathrm{~cm}^{-3}

Given, electron and hole current are equal
\begin{aligned} I_{p \text { dritt }} & =I_{n \text { dritt }} \\ p q \mu_{p} E A & =n q \mu_{n} E A \\ 1.5 \times 10^{10} \mu_{p} & =n \mu_{n} \quad ...(i) \end{aligned}

But according to mass action law,
\begin{aligned} n p & =n_{i}^{2} \\ n & =\frac{n_{i}}{1.5}=\frac{10^{10}}{1.5} \mathrm{~cm}^{-3} \end{aligned}

Put in equation (i)
\therefore \quad 1.5 \times 10^{10} \mu_{p}=\frac{10^{10}}{1.5} \times \mu_{n}
\frac{\mu_{n}}{\mu_{p}}=2.25


Question 3
For an intrinsic semiconductor at temperature T=0 \mathrm{~K}, which of the following statement is true?
A
All energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons.
B
All energy states in the valence band are empty of electrons and all energy states in the conduction band are filled with electrons.
C
All energy states in the valence and conduction band are filled with holes.
D
All energy states in the valence and conduction band are filled with electrons.
GATE EC 2023      Basic Semiconductor Physics
Question 3 Explanation: 
Intrinsic semiconductor at \mathrm{T}=0 \mathrm{~K} behaves as an insulator.
Hence, valence band is completely filled with electron and conduction band is completely empty.
Question 4
In a semiconductor, if the Fermi energy level lies in the conduction band, then the semiconductor is known as
A
degenerate n-type.
B
degenerate p-type.
C
non-degenerate n-type.
D
non-degenerate p-type.
GATE EC 2023      Basic Semiconductor Physics
Question 4 Explanation: 
As the Fermi lies inside the conduction band hence it is degenerate n-type semiconductor.
Question 5
A p-type semiconductor with zero electric field is under illumination (low level injection) in steady state condition. Excess minority carrier density is zero at x=\pm 2l_n, where l_n=10^{-4} cm is the diffusion length of electrons. Assume electronic charge, q=-1.6 \times 10^{-19}C. The profiles of photo-generation rate of carriers and the recombination rate of excess minority carriers (R) are shown. Under these conditions, the magnitude of the current density due to the photo-generated electrons at x=+2l_n is _________ mA/cm^2 (rounded off to two decimal places).

A
0.44
B
0.59
C
0.77
D
0.83
GATE EC 2022      PN-Junction Diodes and Special Diodes
Question 5 Explanation: 
Given,
\begin{aligned} \delta _n(x)&=R\tau _n=10^{20}e^{-|x|/l_n}\cdot \tau _n\\ \delta (l_n)&=10^{20}e^{-1}\tau _n \;\;\;...(i)\\ for \;\; l_n\leq x\leq 2l_n \end{aligned}
Continuity equation is given by
D_n\frac{\partial^2 \delta _n}{\partial x^2}+G-R \;\;\;...(ii)
G and R both are zero - for [l_n\leq x\leq 2l_n]
Hence Equation (i) reduced to D_n \frac{\partial^n \delta _n }{\partial x^2}=0
\Rightarrow \delta _n(x)=Ax+B
For calculating A and B we use Boundary condition
\delta _n(2l_n)=0\; \Rightarrow A=\frac{-B}{2l_n}
\therefore \delta _n(x)=\frac{-B}{2l_n}x+B=B\left [ 1-\frac{x}{2l_n} \right ] \;\;\;...(iii)
At x=l_n
\begin{aligned} 10^{20}e^{-1}\tau _n&=B\left [ 1-\frac{l_n}{2l_n} \right ] \\ B&= 2 \times 10^{20}e^{-1}\tau _n\\ \therefore \; \delta _n(x) &=2 \times 10^{20} e^{-1}\tau _n\left [ 1-\frac{x}{2l_n} \right ] \end{aligned}
for l_n \leq x \leq 2l_n
Electron diffusion current density is given by |J_n|_{diff}=qD_n\frac{d\eta }{dx}=qD_n \times 2 \times 10^{20} \times e^{-1} \times \tau _n\left [ 0-\frac{1}{2l_n} \right ]
=\frac{1.6 \times 10^{-19} \times l_n ^2 \times 2 \times 10^{20} \times e^{-1}}{2l_n}
=1.6 \times 10^{-19} \times l_n \times 10^{20} \times e^{-1}
=1.6 \times 10 \times 1 \times 10^{-4} \times e^{-1}
=0.59mA/cm^2




There are 5 questions to complete.