MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2010

Questions

All questions are in this link

Solutions of all questions

Part A a) $D_p\frac{d^2\triangle p}{dx^2} = 0$

b) $\triangle p = Ax +B$

c) $\triangle p(x_n) = \frac{n_i^2}{N_D}(e^{qV_A/kT}-1)$ $\triangle p(w_N) = 0$

d)$0=Aw_N+B$ \begin{align*} \frac{n_i^2}{N_D}(e^{qV_A/kT}-1)&=Ax_n+B\\ A(w_N+x_n)&=-\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)\\ \implies A&=-\frac{n_i^2}{N_D(w_N-x_n)}(e^{qV_A/kT}-1)\\ B=-Aw_N&=\frac{w_Nn_i^2}{N_D(w_N-x_n)}(e^{qV_A/kT}-1)\\ J_p = -qD_p\frac{d\triangle p}{dx}\bigg\vert_{x=x_n} &= -qD_p\bigg[-\frac{n_i^2}{N_D(w_N-x_n)}(e^{qV_A/kT}-1)\bigg]\\ &=\frac{qD_p}{w_N-x_n}\frac{n_i^2}{N_D}(e^{qV_A/kT}-1) \end{align*}

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Part B a) \begin{align*} I_{La}&=qA\int_0^{x_n}\frac{dn}{dt}\cdot dx\\ &=qAG_Lx_n \end{align*} $\frac{dn}{dt} = -R=G_L$

b)

\begin{align*} &D_p\frac{d^2\triangle p}{dx^2}+G_L =0\\ &\implies \frac{d^2\triangle p}{dx^2} =-\frac{G_L}{D_p}\\ &\implies \frac{d\triangle p}{dx} =-\frac{G_L}{D_p}x+A\\ &\implies \triangle p =-\frac{G_L}{D_p}\frac{x^2}{2}+Ax+B\\ &\triangle p(w_n) =0\\ &\triangle p(x_n) = \frac{n_i^2}{N_D}(e^{qV_A/kT}-1) (?)\\ &\text{or; }-D_p\frac{d\triangle p(x_n)}{dx}=G_Lx_n (?) \end{align*}


which one to use??

$I_{Lb} = -qAD_p\frac{d(\triangle p)}{dx}$

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Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.