ECE Ph.D. Qualifying Exam

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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