MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2007

## Questions

All questions are in this link

# Solutions of all questions

1) A) Forward

B) $10^{16}cm^{-3}$

C) $10^{14}cm^{-3}$

D) $10^{9}\times10^{14} = n_i^2$ $\implies n_i = 10^{23/2}$

E) Yes.

F) \begin{align*} \triangle P_n&=10^{12}-10^9\approx 10^{12}\\ \triangle P_n&=\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)=10^{12}\\ &\implies e^{qV_A/kT} = 10^3\\ &\implies V_A = 0.026\times\ln(10^3) = 0.026\times6.9\\ &= 0.17 V \end{align*}

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2) \begin{align*} |I_E| & = qD_p\frac{10^{10}}{0.1\times10^{-4}} = 1.8\times10^{16}q\\ |I_B| & = qD_n\frac{10^{8}}{0.2\times10^{-4}} = 1.8\times10^{14}q\\ \beta + 1 &= \frac{|I_E|}{|I_B|}\approx 67\\ &\therefore \beta= 66 \text{ (chk)} \end{align*}

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3) A)

$\alpha_T = \frac{0.997J_0}{0.998J_0} = 0.99$


B)

$\gamma = \frac{0.998J_0}{(0.998+0.002)J_0} = 0.998$

C)D)
\begin{align*} \alpha_{dc} &= \gamma\cdot\alpha_T\\ &=0.98802 \end{align*}
$\beta_{dc} = \frac{\alpha_{dc}}{1-\alpha_{dc}}\approx 82$
\begin{align*} I_E &= (0.998+0.002)J_0 = J_0\\ I_C &= 0.997J_0\\ I_B &= I_E-I_C = 0.003J_0 \end{align*}

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4)
$\text{Derivation of } \beta = \frac{D_n}{D_p}\cdot\frac{W_E}{W_B}\cdot\frac{N_E}{N_B}\cdot\frac{n_{iB}^2}{n_{iE}^2}$

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