ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 1: Semiconductor Fundamentals

August 2009



Questions

All questions are in this link

Solutions of all questions

1)

$ E = Ak^\alpha ; k=\sqrt{k_x^2+k_y^2} $

$ \begin{align*} D(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}{A}\\ &\approx \frac{\cancel{WL}}{4\pi^2}\cdot(\pi\cdot2k\triangle k)\cdot\frac{1}{\triangle E}\cdot\frac{1}{\cancel{A}}\\ &=\frac{1}{2\pi}k\frac{\triangle k}{\triangle E}\approx \frac{1}{2\pi}k\frac{dk}{dE} \end{align*} $

Now; $ k = \bigg(\frac{E}{A}\bigg)^\frac{1}{\alpha} $\\

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

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2)

At $ T=0; f(E)=1 $ upto $ E=E_F $ 

$ \begin{align*} \therefore n(E) &=\int_{E_C}^{E_F}D(E)dE\\ &=\int_{E_C}^{E_F}\frac{A^{\frac{-2}{\alpha}}}{2\pi \alpha}\cdot E^{\frac{2}{\alpha}-1}dE\\ &=\frac{A^{\frac{-2}{\alpha}}}{2\pi \alpha}\frac{E^{\frac{2}{\alpha}}}{2/\alpha}\bigg\vert_{E_C}^{E_F}\\ &= \frac{A^{\frac{-2}{\alpha}}}{4\pi}\cdot\bigg(E_F^{\frac{2}{\alpha}}-E_C^{\frac{2}{\alpha}}\bigg). \end{align*} $

If we take $ E_C $ as the reference energy i.e. $ E_C = 0 $ then;

$ n(E) = \frac{A^{\frac{-2}{\alpha}}}{4\pi}E_F^{\frac{2}{\alpha}}. $

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3)

Boltzmann statistics; $ f(E)\approx e^{-(E-E_F)/kT} $

$ \therefore n(E) = \int_{E_C}^{\infty}D(E)f(E)dE $

taking $ E_C $ as reference; i.e. $ E_C = 0 $

$ \begin{align*} n(E) &= \int_0^\infty\frac{A^{\frac{-2}{\alpha}}}{2\pi\alpha}e^{-(E-E_F)/kT}\cdot dE\cdot E^{\frac{2}{\alpha}-1}\\ &=\frac{A^{\frac{-2}{\alpha}}}{2\pi\alpha}\cdot e^{E_F/kT}\int_0^\infty e^{-E/kT}\cdot E^{\frac{2}{\alpha}-1}dE \end{align*} $

Let;

$ \frac{E}{kT} = t \:\:\:\:\:\:\:\:\:\:\therefore dE = kTdt $

$ \begin{align*} \therefore n(E)&= \frac{A^{\frac{-2}{\alpha}}\cdot e^{E_F/kT}}{2\pi\alpha}kT\int_0^\infty e^{-t}\cdot (tkT)^{\frac{2}{\alpha}-1}dt\\ &= \frac{A^{\frac{-2}{\alpha}}\cdot e^{E_F/kT}}{2\pi\alpha}\cdot(kT)\cdot(kt)^{\frac{2}{\alpha}-1}\int_0^\infty e^{-t}\cdot t^{\frac{2}{\alpha}-1}dt\\ \end{align*} $


Now; $ \Gamma (z) = \int_0^\infty dt e^{-t}t^{z-1} $

$   \therefore n(E)= \frac{A^{\frac{-2}{\alpha}}e^{E_F/kT}}{2\pi\alpha}(kT)^\frac{2}{\alpha}\cdot\Gamma(2/\alpha)   $

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d)

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