MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2008

## Questions

All questions are in this link

# Solutions of all questions

1) \begin{align*} D_p &= \frac{kT}{q}\cdot\mu_p\\ &= 0.025\times 400 = 10cm^2/sec\\ D_n&= \frac{kT}{q}\cdot\mu_n = 0.025\times1000=25cm^2/s \end{align*}

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2) \begin{align*} \beta &\approx \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}\\ &=\frac{25}{10}\times\frac{1}{0.5}\times\frac{3\times10^{18}}{3\times10^{17}}\\ &= 50 \end{align*}

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3)

\begin{align*} \tau_b &= \frac{W_B^2}{2D_n} = \frac{(0.5\times10^{-4})^2}{2\times25}\\ &=5\times10^{-11}sec \end{align*}

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4) (chk, confusion)
\begin{align*} \beta &= \frac{D_n}{D_p}\cdot\frac{W_E}{L_B}\cdot\frac{N_E}{N_B}\cdot\frac{n_{iB}^2}{n_{iE}^2}\\ &=\beta_{initial}\times\frac{W_B}{L_B}\\ L_B&=\sqrt{D_n\tau_b} = \sqrt{25\times250\times10^{-12}}\\ &=25\times10^{-11/2}\\ \therefore\beta&=50\times0.6\\ &=30 \end{align*}

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## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva