Revision as of 18:41, 30 July 2017

ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 1: Semiconductor Fundamentals

August 2007

Questions

All questions are in this link

Solutions of all questions

Back to ECE Qualifying Exams (QE) page

@@ Line 25: / Line 25: @@
 =Solutions of all questions=
-)
-<math>
-\begin{align*}
-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;
-<math>
-\begin{align*}
-\eta &=\frac{E-E_c}{kT}\:\:\:\:\:\:\therefore dE = kTd\eta\\
-\eta_c &=\frac{E_F-E_c}{kT}
-\end{align*}
-</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}}\\
-&=\frac{2(kT)^2}{\pi\hslash^2V_F^2}\cdot\cancelto{1!}{\Gamma 2}\cdot F_1(\eta_c)\\
-&=\frac{2(kT)^2}{\pi\hslash^2V_F^2} F_1(\eta_c)\\
-\end{align*}
-</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\\
- &=\int_{E_c}^{E_F}\frac{2(E-E_c)}{\pi\hslash^2V_F^2}dE\\
- &=\frac{2}{\pi\hslash^2V_F^2}\cdot\frac{(E-E_c)^2}{2}\bigg\vert_{E_c}^{E_F}\\
- &=\frac{(E_F-E_c)^2}{\pi\hslash^2V_F^2}
-\end{align*}
-</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*}
-\text{Denominator} &= \frac{2(kT)^2}{\pi\hslash^2V_F^2}F_1(\eta_c)\\
-\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}}\\
-&=\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*}
-n &= D(E)f(E)\\
-&=\frac{\int(E-E_c)D(E)f(E)}{\int D(E)f(E)dE}
- \end{align*}
-</math>
- <math>
- \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>
- ------------------------------------------------------------------------------------\\
-)
-For Maxwell-Boltzmann statistics
-At <math>T=0</math>
-<math>F_1(\eta_c) = F_2(\eta_c)\to e^{\eta_c}</math>
-<math>\therefore \bar{u} = 2kT.</math>
- ------------------------------------------------------------------------------------\\

Difference between revisions of "ECE PhD QE MN1 2007" - Rhea

Revision as of 18:41, 30 July 2017

Questions

Solutions of all questions

Alumni Liaison