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| − | + | MICROELECTRONICS and NANOTECHNOLOGY (MN) | |
| − | Question | + | Question 1: Semiconductor Fundamentals |
</font size> | </font size> | ||
| − | August | + | August 2011 |
</center> | </center> | ||
---- | ---- | ||
---- | ---- | ||
==Questions== | ==Questions== | ||
| − | All questions are in this [https://engineering.purdue.edu/ECE/Academics/Graduates/ | + | All questions are in this [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_11/MN-1%20QE%2011.pdf link] |
| + | |||
=Solutions of all questions= | =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 <math>E_F</math> 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' <math>\rightarrow</math> 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. | ||
| + | |||
| + | ------------------------------------------------------------------------------------ | ||
| + | B) | ||
| + | i) Here; doping density of p is way greater than Cu. So, we can ignore Cu for calculation of <math>n</math>. | ||
| + | <math> | ||
| + | \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*} | ||
| + | </math> | ||
| + | <math> | ||
| + | \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*} | ||
| + | </math> | ||
| + | |||
| + | ii) As the Cu energy levels are much below the Fermi level; Cu is not fully ionized. P however is above <math>E_F</math> and should be completely ionized. | ||
| + | |||
| + | iii) | ||
| + | <math> | ||
| + | \frac{1}{2}mV_{th}^2 = \frac{3}{2}kT | ||
| + | </math> | ||
| + | <math> | ||
| + | \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*} | ||
| + | </math> | ||
| + | |||
| + | iv) | ||
| + | <math> | ||
| + | \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*} | ||
| + | </math> | ||
| + | (As Cu is located near the midgap, it is expected to be the recombination center) | ||
| + | <math> | ||
| + | \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*} | ||
| + | </math> | ||
| + | |||
| + | v) For Ge; lattice constant. <math>a=0.5mm</math> distance between n nearest atom (zinc-blend structure) | ||
| + | <math> | ||
| + | =a\sqrt{3}\times\frac{1}{4} | ||
| + | </math> | ||
| + | <math> | ||
| + | \implies d=2r=\frac{a\sqrt{3}}{4}\implies r=\frac{a\sqrt{3}}{8} | ||
| + | </math> | ||
| + | <math> | ||
| + | \begin{align*} | ||
| + | \therefore A=\pi r^2&=\pi\times\frac{3a^2}{64}\\ | ||
| + | &=3.68\times10^{16}cm^2 | ||
| + | \end{align*} | ||
| + | </math> | ||
| + | So; <math>a_n\approx A</math>; so the numbers look reasonable. | ||
| + | |||
| + | \#\underline{Just Absurd and Illogical problem to solve without a calculator.} | ||
---- | ---- | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Latest revision as of 22:05, 5 August 2017
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.
------------------------------------------------------------------------------------
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.}
