# 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.