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==Questions== | ==Questions== | ||
| − | All questions are in this [https://engineering.purdue.edu/ECE/Academics/Graduates/ | + | All questions are in this [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_08/MN-2%20QE%2008.pdf link] |
=Solutions of all questions= | =Solutions of all questions= | ||
| − | 1 | + | 1) <math> |
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\begin{align*} | \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 |
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\end{align*} | \end{align*} | ||
</math> | </math> | ||
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<math> | <math> | ||
\begin{align*} | \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 | |
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\end{align*} | \end{align*} | ||
</math> | </math> | ||
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\begin{align*} | \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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</math> | </math> | ||
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| − | 4) | + | 4) (chk, confusion) |
<math> | <math> | ||
| − | \ | + | \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*} | ||
</math> | </math> | ||
Latest revision as of 22:18, 5 August 2017
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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