ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 1: Semiconductor Fundamentals

August 2010



Questions

All questions are in this link

Solutions of all questions

1)

$ D_p\frac{d^2\triangle p}{dn^2} - \frac{\triangle p}{\tau_p} =0 $ $ \implies \triangle p(x) = Ae^{-\frac{x}{L_p}}+Be^{\frac{x}{L_p}} $ where;$ L_p^2 = D_p\tau_p $ At $ x=0; $ $ -D_p\frac{\triangle p}{\triangle x}\big\vert_{x=0} = G_0 $ $ \implies-D_p\bigg(-\frac{A}{L_p} + \frac{B}{L_p}\bigg) =G_0 $

  • Right boundary $ -D_p\frac{\delta\triangle p}{\delta x}\vert_w = 0 $

$ \begin{align*} B & = \frac{L_pG_0}{D_p}\cdot \frac{1}{1-\rho^{-\frac{2w}{L_p}}} \\ A &= \frac{L_pG_0}{D_p}\cdot \frac{e^{-\frac{2w}{L_p}}}{1-\rho^{-\frac{2w}{L_p}}} \end{align*} $

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2)

$ \begin{align*} r_N = -\frac{\delta n}{\delta t} = c_nnp_T - e_nn_T\\ r_p = -\frac{\delta p}{\delta t} = c_ppn_T - e_pp_T \end{align*} $ $ \begin{align*} \text{Let; } n_1 &=n_i\exp\bigg(\frac{E_T-E_i}{kT}\bigg)\\ p_1&=n_i\exp\bigg(\frac{E_i-E_T}{kT}\bigg) \end{align*} $ $ \begin{align*} r_N = c_nnp_T - c_nn_1n_T\\ r_p = c_ppn_T - c_pp_1p_T \end{align*} $

for detailed balance

$ r_N = r_p $

$ \begin{align*} &\implies c_nnp_T - c_nn_1n_T = c_ppn_T - c_pp_1p_T\\ &\implies c_nn(N_T-n_T) - c_Nn_1n_T = c_ppn_T-c_pp_1(N_T-n_T)\\ &\implies n_T(c_pp+c_pp_1+c_nn+c_nn_1) =c_nnN_T + c_pp_1N_T\\ &\implies n_T=\frac{N_T(c_nn+c_pp_1)}{c_n(n+n_1)+c_p(p+p_1)} \end{align*} $

Occupancy probability;
$   \begin{align*} f_T &= \frac{n_T}{N_T} = \frac{c_nn_i\exp\frac{F_n-E_i}{kT} + c_pn_i\exp\frac{E_i-E_t}{kT}} {c_n(n_i\exp\frac{F_n-E_i}{kT} +n_i\exp\frac{E_T-E_i}{kT} )}\\ &+c_p(n_i\exp\frac{E_i-F_p}{kT} +n_i\exp\frac{E_i-E_T}{kT} )  \end{align*}   $

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