MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 1: Semiconductor Fundamentals

August 2011

## Questions

All questions are in this link

# Solutions of all questions

A) i) Metal ii) Semi-conductor iii) Semi-metal iv) Insulator

• If an atom has odd no. of electrons then the $E_F$ is likely to be in the conduction band to allow the odd electron to reside there.

Here only Al(13) has odd no. of electrons. Hence, it is definitely expected to be a metal.

• Even electron rule' $\rightarrow$ If a substance has odd no. of electrons then it is expected to be a metal as the Fermi level will be in the conduction band.
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B)

i) Here; doping density of p is way greater than Cu. So, we can ignore Cu for calculation of $n$.


\begin{align*} n&\cong N_d^+\approx N_d = 10^{17}cm^{-3}\\ n &= N_c e^{(E_F-E_C)/kT} = 10^{17}\\ \end{align*} \begin{align*} \implies e^{(E_F-E_C)/kT} &= \frac{10^{19}}{10^{17}} = 10^2\\ \implies E_C-E_F &= kT\cdot 2\ln10\\ &=4.6kT\\ &=4.6\times 0.025 eV\\ &=0.115eV \end{align*}

ii) As the Cu energy levels are much below the Fermi level; Cu is not fully ionized. P however is above $E_F$ and should be completely ionized.

iii) $\frac{1}{2}mV_{th}^2 = \frac{3}{2}kT$ \begin{align*} &\implies V_{th} = \sqrt{\frac{3kT}{m^*}} = \sqrt{\frac{3\times 0.025\times1.6\times10^{-19}}{0.5\times9.1\times10^{-31}}}\\ &\implies V_{th} = 1.6\times10^7 cm/s \end{align*}

iv)

\begin{align*} \tau&=\frac{1}{c_nN_T}\\ \tau&=\frac{1}{a_nV_{th}\cdot N_T}\\ \text{Here } \tau&=10^{-7}sec\\ N_T&=N_{Cu}=10^{15}cm^{-3} \end{align*}
(As Cu is located near the midgap, it is expected to be the recombination center)
\begin{align*} \therefore a_n=\frac{1}{\tau v_{th}N_T} &= \frac{1}{1.6\times10^{15}cm^{-2}}\\ &=6.25\times10^{-16}cm^2 \end{align*}

v) For Ge; lattice constant. $a=0.5mm$ distance between n nearest atom (zinc-blend structure)
$=a\sqrt{3}\times\frac{1}{4}$
$\implies d=2r=\frac{a\sqrt{3}}{4}\implies r=\frac{a\sqrt{3}}{8}$
\begin{align*} \therefore A=\pi r^2&=\pi\times\frac{3a^2}{64}\\ &=3.68\times10^{16}cm^2 \end{align*}
So; $a_n\approx A$; so the numbers look reasonable.

\#\underline{Just Absurd and Illogical problem to solve without a calculator.}
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