ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 1: Semiconductor Fundamentals

August 2012



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2) Alt text

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1) 
Huge theory. Covered mainly in lec 5 and somewhat in lec 6. Can also follow chapter 3 of ASF(Pierret)
 
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 3)

$ \begin{align*} P&=\int_{E_{bottom}}^{E_v}g_v(E)(1-f(E))dE\\ &=\int_{-\infty}^{E_v}\frac{m_P^*\sqrt{2m_p^*(E_v-E)}}{\pi^2\hslash^3}\cdot\bigg(1-\frac{1}{1+e^{(E-E_F)/kT}}\bigg)dE\\ &=\int_{-\infty}^{E_v}\frac{m_P^*\sqrt{2m_p^*}}{\pi^2\hslash^3}\cdot\sqrt{E_v-E}\cdot\frac{e^{(E-E_F)/kT}}{1+e^{(E-E_F)/kT}}dE\\ Let\:\:\:\:\:\: \eta=(E_v-E)/kT\\ \eta_v=(E_v-E_F)/kT\\ dE=-kTd\eta\\ &=\frac{m_P^*\sqrt{2m_p^*}}{\pi^2\hslash^3}\cdot\int_{-\infty}^{E_v}\frac{\sqrt{E_v-E}dE}{1+e^{(E_F-E)/kT}}\\ &=\frac{m_P^*\sqrt{2m_p^*}}{\pi^2\hslash^3}\cdot(kT)^{\frac{3}{2}}\int_0^\infty\frac{\eta^{\frac{1}{2}}d\eta}{1+e^{\eta-\eta_we}}\\ &=\frac{2}{\sqrt{\pi}}\times\frac{\sqrt{2}(m_p^*kT)^{\frac{3}{2}}}{\pi^2\times\frac{\hslash^3}{8\pi^3}}\times\frac{\sqrt{\pi}}{2}\int_0^\infty\frac{\eta^{\frac{1}{2}}d\eta}{1+e^{\eta-\eta_v}}\\ &=\frac{\sqrt{\pi}}{2}\times\frac{\sqrt{2}\cdot 8\pi(m_p^*kT)^{\frac{3}{2}}}{\hslash^3}\times\frac{2}{\sqrt{\pi}}\int_0^\infty\frac{\eta^{\frac{1}{2}}d\eta}{1+e^{\eta-\eta_v}}\\ &=2\times\frac{2\sqrt{2}\cdot \pi^{\frac{3}{2}}(m_p^*kT)^{\frac{3}{2}}}{\hslash^3}\cdot \mathcal{F}_{1/2}(\eta_v)\\ &=2\times\bigg(\frac{2\pi m_p^*kT}{\hslash^2}\bigg)^{\frac{3}{2}}\cdot \mathcal{F}_{1/2}(\eta_v)\\ &=N_v\mathcal{F}_{1/2}(\eta_v) \end{align*} $ For non-degenerate condition $ (E_F-E_v\ge3kT) $

$ \[\mathcal{F}_{1/2}(\eta_v)\to e^{\eta_v}=e^{(E_v-E_F)/kT}\] $ $ \[\therefore P = N_ve^{(E_v-E_F)/kT}\] $ $ \[P-n+N_D^+-N_A^- =0\] $ $ \[P-\frac{n_i^2}{P}+N_D-N_A =0\] $

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4)

$  \[R = (c_pp+c_nn)(p_n-n_i^2)\]  $

Let; $ \triangle n\to $ extra no. of generated carrier then;

$   \[R = (c_p(p_0+\triangle n)+c_n(n_0+\triangle n))((p_0+\triangle r)(n_0+\triangle n)-n_i^2)\]   $
High injection condition means $ \triangle n >> n_0,p_0 $
$   \begin{align*}  \implies R&=\triangle n (c_p+c_n)\cdot \triangle n^2\\  &=(c_p+c_n) \triangle n^3\\ \end{align*}  $

Now;

$   \begin{align*} n&=n_ie^{(F_n-E_i)/kT}\\ p&=n_ie^{(E_i-F_p)/kT}\\ \end{align*}  $
$   \begin{align*} &\therefore n_p =n_i^2e^{(F_n-E_i)/kT} = n_i^2e^{qV/kT}\\ &\implies (n_0+\triangle n)(p_0+\triangle n) = n_i^2e^{qV/kT}\\ &\implies \triangle n^2 =  n_i^2e^{qV/kT}\\ &\implies \triangle n =  n_ie^{qV/2kT}\\ \end{align*}  $
$  \[\therefore R= (c_p+c_n)n_i^3\cdot e^{3qV/2kT}\]  $
 $  \[\implies qV = \frac{2}{3}kT\ln\frac{R}{(c_p+c_n)n_i^3}\]  $
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