Latest revision as of 10:02, 6 August 2017

ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 1: Semiconductor Fundamentals

August 2007

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Solutions of all questions

1)

$E = \pm\alpha\sqrt{k_x^x+k_y^2}$

Let ; $k=\pm\sqrt{k_x^x+k_y^2}$

$\therefore E = \alpha k$

$\begin{align*} D(E) &= \frac{1}{2\pi}k\frac{dk}{dE}\\ &= \frac{1}{2\pi}\cdot\frac{E}{\alpha}\cdot\frac{1}{\alpha} = \frac{E}{2\pi\alpha^2} \end{align*}$

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2)

 $\begin{align*} n&=\int_{E_c}^{E_f}D(E)dE\\ &=\int_{E_c}^{E_f}\frac{E}{2\alpha^2}\cdot dE\\ &=\frac{1}{4\alpha^2}(E_F^2 - E_c^2) \end{align*}$

Taking $$ E_c = 0 $$

 $n = \frac{E_F^2}{4\alpha^2\pi}$

320x220px

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3)

 $\begin{align*} F &= -qE =qE_x \hspace{0.5cm} [\because E = -\hat{x}E_x]\\ F&= \frac{d(\hslash k)}{dt} \end{align*}$

400x400px

$\implies \int_0^{k_x}dk = \frac{1}{\hslash}\int_0^t Fdt$ $\implies k_x = \frac{qE_xt}{\hslash}$ $E = \alpha k$ $V = \frac{1}{\hslash}\frac{dE}{dk} = \frac{\alpha}{\hslash}$ $x = \int_0^tVdt = \frac{\alpha}{\hslash}t$

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@@ Line 26: / Line 26: @@
 )
 <math>
-\begin{align*}
+E = \pm\alpha\sqrt{k_x^x+k_y^2}
-n& = \int_{E_c}^\infty D(E)f(E)dE\\
-&=\int_{E_c}^\infty\frac{2(E - E_c)}{\pi\hslash^2V_F^2}\cdot\frac{1}{1+e^{(E-E_F)/kT}}dE
-\end{align*}
 </math>
-  Let;
+Let ; <math>k=\pm\sqrt{k_x^x+k_y^2}</math>
 <math>
-\begin{align*}
+\therefore E = \alpha k
-\eta &=\frac{E-E_c}{kT}\:\:\:\:\:\:\therefore dE = kTd\eta\\
-\eta_c &=\frac{E_F-E_c}{kT}
-\end{align*}
 </math>
-  <math>
+<math>
 \begin{align*}
-n& = \frac{2}{\pi\hslash^2V_F^2}\cdot(kT)^2\int_0^\infty\frac{\eta d\eta}{1+e^{\eta-\eta_c}}\\
+D(E) &= \frac{1}{2\pi}k\frac{dk}{dE}\\
-&=\frac{2(kT)^2}{\pi\hslash^2V_F^2}\cdot\cancelto{1!}{\Gamma 2}\cdot F_1(\eta_c)\\
+&= \frac{1}{2\pi}\cdot\frac{E}{\alpha}\cdot\frac{1}{\alpha} = \frac{E}{2\pi\alpha^2}
-&=\frac{2(kT)^2}{\pi\hslash^2V_F^2} F_1(\eta_c)\\
 \end{align*}
 </math>
-  ------------------------------------------------------------------------------------
+ ------------------------------------------------------------------------------------
+)
-)
+ <math>
-At <math>T = 0\:\:\:\:\:f(E) = 1</math> for <math>E\le E_F</math>
-<math>
 \begin{align*}
-\therefore n &=\int_{E_c}^{E_F}D(E)dE\\
+n&=\int_{E_c}^{E_f}D(E)dE\\
- &=\int_{E_c}^{E_F}\frac{2(E-E_c)}{\pi\hslash^2V_F^2}dE\\
+&=\int_{E_c}^{E_f}\frac{E}{2\alpha^2}\cdot dE\\
- &=\frac{2}{\pi\hslash^2V_F^2}\cdot\frac{(E-E_c)^2}{2}\bigg\vert_{E_c}^{E_F}\\
+&=\frac{1}{4\alpha^2}(E_F^2 - E_c^2)
- &=\frac{(E_F-E_c)^2}{\pi\hslash^2V_F^2}
 \end{align*}
 </math>
+Taking <math>E_c = 0</math>
- ------------------------------------------------------------------------------------\\
-)
- For Maxwell Boltzmann Statistics
-<math>F_1(\eta_c)\to e^{\eta_c}</math>
-if
-<math>\eta_c\le-3</math>
-<math>E_F-E_c\le-3kT</math>
-<math>E_c-E_F\ge3kT</math>
-<math>\therefore n = \frac{2(kT)^2}{\pi\hslash^2V_F^2}\cdot e^{(E_F-E_c)/kT}</math>
- ------------------------------------------------------------------------------------\\
-)
-<math>\bar{u} = \frac{\int_{E_c}^\infty D(E)f(E)(E-E_c)dE}{\int_{E_c}^\infty D(E)f(E)dE}</math>
-from (1);
   <math>
- \begin{align*}
+n = \frac{E_F^2}{4\alpha^2\pi}
-\text{Denominator} &= \frac{2(kT)^2}{\pi\hslash^2V_F^2}F_1(\eta_c)\\
+</math>
-\text{Numerator} &= \int_{E_c}^\infty\frac{2(E-E_c)^2}{\pi\hslash^2V_F^2}\cdot\frac{1}{1+e^{(E-E_F)/kT}}dE\\
-&=\frac{2}{\pi\hslash^2V_F^2}(kT)^3\int_0^\infty\frac{\eta^2d\eta}{1+e^{\eta-\eta_c}}\\
+[[Image:MN1_2007_1.png|Alt text|320x220px]]
-&=\frac{2(kT)^3}{\pi\hslash^2V_F^2}\cdot\cancelto{2!}{\Gamma 3}\cdot F_2(\eta_c)\\
-&=\frac{4(kT)^3}{\pi\hslash^2V_F^2} F_2(\eta_c)
- \end{align*}
- </math>
-<math>\therefore\bar{u} = 2kT\frac{F_2(\eta_c)}{F_1(\eta_c)}</math>
+  ------------------------------------------------------------------------------------
+)
-  ------------------------------------------------------------------------------------\\
-)
-At <math>T=0</math>
-<math>\bar{u} = \frac{\int_{E_c}^E D(E)(E-E_c)dE}{\int_{E_c}^{E_F} D(E)dE}</math>
   <math>
- \begin{align*}
+\begin{align*}
-n &= D(E)f(E)\\
+F &= -qE =qE_x \hspace{0.5cm} [\because E = -\hat{x}E_x]\\
-&=\frac{\int(E-E_c)D(E)f(E)}{\int D(E)f(E)dE}
+F&= \frac{d(\hslash k)}{dt}
- \end{align*}
+\end{align*}
 </math>
- <math>
+[[Image:MN1_2007_2.png|Alt text|400x400px]]
- \begin{align*}
-\text{Denominator} &= \frac{(E_F-E_c)^2}{\pi\hslash^2V_F^2}\\
-\text{Numerator} &= \int_{E_c}^{E_F}\frac{2(E-E_c)^2}{\pi\hslash^2V_F^2}dE\\
-&=\frac{2}{\pi\hslash^2V_F}\cdot\frac{(E-E_c)^3}{3}\bigg\vert_{E_c}^{E_F}\\
-&=\frac{2(E_F-E_c)^3}{3\pi\hslash^2V_F}
- \end{align*}
- </math>
-<math>\therefore\bar{u} = \frac{2}{3}(E_F - E_c)</math>
+<math>
+\implies \int_0^{k_x}dk = \frac{1}{\hslash}\int_0^t Fdt
+</math>
- ------------------------------------------------------------------------------------\\
+<math>
-)
+\implies k_x = \frac{qE_xt}{\hslash}
-For Maxwell-Boltzmann statistics
+</math>
-At <math>T=0</math>
+<math>
+E = \alpha k
-<math>F_1(\eta_c) = F_2(\eta_c)\to e^{\eta_c}</math>
+</math>
+<math>
-<math>\therefore \bar{u} = 2kT.</math>
+V = \frac{1}{\hslash}\frac{dE}{dk} = \frac{\alpha}{\hslash}
+</math>
+<math>
+x = \int_0^tVdt = \frac{\alpha}{\hslash}t
+</math>
- ------------------------------------------------------------------------------------\\

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