MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 1: Semiconductor Fundamentals

August 2013

## Questions

All questions are in this link

# Solutions of all questions

1) \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*}

 Let;


\begin{align*} \eta &=\frac{E-E_c}{kT}\:\:\:\:\:\:\therefore dE = kTd\eta\\ \eta_c &=\frac{E_F-E_c}{kT} \end{align*}

 \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*}

 ------------------------------------------------------------------------------------


2)

At $T = 0\:\:\:\:\:f(E) = 1$ for $E\le E_F$

\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*}

------------------------------------------------------------------------------------\\


3)

For Maxwell Boltzmann Statistics


$F_1(\eta_c)\to e^{\eta_c}$

if

$\eta_c\le-3$

$E_F-E_c\le-3kT$

$E_c-E_F\ge3kT$

$\therefore n = \frac{2(kT)^2}{\pi\hslash^2V_F^2}\cdot e^{(E_F-E_c)/kT}$

------------------------------------------------------------------------------------\\
4)


$\bar{u} = \frac{\int_{E_c}^\infty D(E)f(E)(E-E_c)dE}{\int_{E_c}^\infty D(E)f(E)dE}$

from (1);

\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*}


$\therefore\bar{u} = 2kT\frac{F_2(\eta_c)}{F_1(\eta_c)}$

------------------------------------------------------------------------------------\\


5) At $T=0$

$\bar{u} = \frac{\int_{E_c}^E D(E)(E-E_c)dE}{\int_{E_c}^{E_F} D(E)dE}$

\begin{align*} n &= D(E)f(E)\\ &=\frac{\int(E-E_c)D(E)f(E)}{\int D(E)f(E)dE} \end{align*}

\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*}


$\therefore\bar{u} = \frac{2}{3}(E_F - E_c)$

------------------------------------------------------------------------------------\\


6) For Maxwell-Boltzmann statistics At $T=0$

$F_1(\eta_c) = F_2(\eta_c)\to e^{\eta_c}$

$\therefore \bar{u} = 2kT.$

------------------------------------------------------------------------------------\\



## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva