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}.
Given \mathrm{kT} at 300 \mathrm{~K} is 0.026 \mathrm{eV}.
63.36 | |
25.36 | |
45.25 | |
98.36 |
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}
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).
(rounded off to two decimal places).
1.15 | |
2.25 | |
2.85 | |
3.36 |
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
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?
All energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons. | |
All energy states in the valence band are empty of electrons and all energy states in the conduction band are filled with electrons. | |
All energy states in the valence and conduction band are filled with holes. | |
All energy states in the valence and conduction band are filled with electrons. |
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.
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
degenerate n-type. | |
degenerate p-type. | |
non-degenerate n-type. | |
non-degenerate p-type. |
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).
0.44 | |
0.59 | |
0.77 | |
0.83 |
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
\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.