ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 1: Semiconductor Fundamentals

August 2008



Questions

All questions are in this link


Solutions of all questions

a)

$ \begin{align*} \frac{1}{m^*} &= \frac{1}{\hslash^2}\frac{d^2E}{dk^2} \\ &=\frac{1}{\hslash^2}\frac{d^2}{dk^2}(\alpha k^2) \\ &=2\alpha/\hslash^2 \\ \therefore m^* &= \frac{\hslash^2}{2\alpha} \end{align*} $

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b)

$ \begin{align*} g(E) &=\frac{\pi(k+\triangle k)^2 - \pi k^2}{\frac{2\pi}{W}\cdot\frac{2\pi}{L}}\cdot\frac{1}{\triangle E}\cdot\frac{1}{WL}\\ &\approx \frac{1}{4\pi^2}\cdot(\pi\cdot2k\triangle k)\cdot\frac{1}{\triangle E}\\ &=\frac{1}{2\pi}k\frac{\triangle k}{\triangle E}\approx = \frac{1}{2\pi}k\frac{dk}{dE} \end{align*} $

$ k = \bigg(E/\alpha\bigg)^{\frac{1}{2}} $

$ \therefore \frac{dk}{dE} = \frac{1}{2}\frac{E^{-\frac{1}{2}}}{\alpha^{-\frac{1}{2}}}\cdot\frac{1}{\alpha} = \frac{\alpha^{-\frac{1}{2}}}{2}E^{-\frac{1}{2}} $

$ \begin{align*} \therefore g(E) &= \frac{1}{2\pi}\cdot\frac{E^{\frac{1}{2}}}{\alpha^{\frac{1}{2}}}\cdot\frac{\alpha^{-\frac{1}{2}}}{2}E^{-\frac{1}{2}}\\ &=\frac{1}{4\pi \alpha} \end{align*} $

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c)

$ \begin{align*} F&=-qE=qE_x\:\:\:\:\:\:\:\:\:\:\text{(+x direction)}\\ F &=m^*a = qE_x\\ \therefore a_{max} &=\frac{qE_x}{m^*}=a \:\:\:\:\:\:\:\:\:\:\text{(constant)} \end{align*} $

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$ V(t) = \int_0^t a dt =at=\frac{qE_xt}{m^*} $

$ V_{max} = V(\tau_1) = a\tau_1=\frac{qE_x\tau_1}{m^*} $

at $\tau_1; V$ goes to 0. Then this process repeats in the next 2 cycles.

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$ x(t) = \int_0^tVdt=\int_0^t\frac{qE_x}{m^*}tdt = \frac{qE_x}{m^*}\cdot\frac{t^2}{2} $

$ \therefore x(\tau_1) =\frac{qE_x}{m^*}\cdot\frac{\tau_1^2}{2} $

At $ t = \tau_1 $; electron will be at $ x(\tau_1). $ For the next cycle; this position will be the initial one.

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d)

$   v_{avg} = \mu E \text{ (So, we use the $2^{nd}$ plot)}   $

$ \therefore \mu = \frac{v_{avg}}{E} $

So, we need to find avg. velocity and divide by the constant electric field value to find mobility.

$   v_{avg} = \frac{V(0)+V(\tau_1)}{2} = \frac{V(\tau_1)}{2} = \frac{qE_x\tau_1}{2m^*}    $
 

$ \therefore \mu = \frac{qE_x\tau_1}{2m^*E_x} = \frac{q\tau_1}{2m^*}. $

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e)

For elastic scattering, velocity cannot change. As the velocity goes to zero here, it cannot be elastic.\\
Ans: ii



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