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

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}  $

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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*}  $

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$ \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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