MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2011

Questions

All questions are in this link

Solutions of all questions

1) a) Direct bandgap semiconductors used in Lasers and LEDs. Photodetectors.

b) Si has a lattice matched $SiO_2$ to reduce surface defects. Si is almost 99\% pure.

c) $\sim 9eV$

d) (2 1 0) $\rightarrow \:\frac{1}{2}\:\:\:\:1\:\:\:\:\infty$

$\rightarrow \:1\:\:\:\:2\:\:\:\:\infty$

e) $\sim 900-1200$ deg C.

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2) Neutral p \begin{align*} p_0&=10^{15}\\ n_0&=\frac{10^{20}}{10^{15}}=10^5 \end{align*} Neutral n \begin{align*} n_0&=10^{18}\\ p_0&=\frac{10^{20}}{10^{18}}=10^2 \end{align*}

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3)

\begin{align*} x_{n0}\cdot 10^{18}&=x_{p0}\cdot10^{15}\\ \implies x_{p0}&=10^3\times0.001\mu m = 1\mu m \end{align*}

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4)  $x_n = \sqrt{\frac{2\epsilon}{q}\frac{N_A}{N_D(N_A+N_D)}\cdot(V_{bi}+V)}$

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5) \begin{align*} \rho &=\frac{1}{q\mu_nN_D} = \frac{1}{2\times10^{-19}\times500\times10^{18}}\\ &=\frac{1}{1000\times10^{-1}}\Omega\cdot cm\\ &=10^{-2}\Omega\cdot cm \end{align*} $R_s=\frac{\rho}{L} =\frac{10^{-2}}{5\times10^{-4}} = 20\Omega/\Box$

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6)
\begin{align*} J_n&=q_n\mu_nE\\ &=2\times10^{-19}\times10^{18}\times500\times1000\\ &=10^5 A/cm^2 \end{align*}

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7)
\begin{align*} J_{diff} &=qD_n\frac{d\triangle n}{dx}\\ &=2\times10^{-19}\times500\times 25m\times\frac{\triangle n_p}{5\mu m}\\ \end{align*}

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8)
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9)
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10) Forward Active Mode


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11)Fig 11.15(SDF) Page 425


 $\beta$ is reduced by both phenomena.

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12)



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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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