MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 3: Field Effect Devices

August 2013

## Questions

All questions are in this link

# Solutions of all questions

1)\begin{align*} Q&=C_{ox}(V_{gs}-V_t)\\ &=\frac{\epsilon_0\epsilon_r}{t_{ox}}\times0.5\\ &=8.9\times10^{-7}C/cm^2 \text{ (chk)} \end{align*}

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2) $I_d = \mu_nC_{ox}\frac{W}{L}(V_{GS-V_t}V_{ds})$

\begin{align*} \implies R_d=\frac{V_{ds}}{I_d}&=\frac{1}{500\times\frac{\epsilon_0\epsilon_r}{t_{ox}}\times\frac{0.5\mu}{200n}\times0.5}\\ &=\frac{8}{8.9}\times10^3\Omega\\ &=0.89k\Omega \text {(chk)} \end{align*}

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3) \begin{align*} I_{d1}&=\frac{V_{ds}}{R_d}=\frac{10mV}{0.89k\Omega}=11.23\mu A\\ I_{d2}&=\frac{V_{ds}}{R_d+R_s+R_d'}=\frac{10mV}{2.89k\Omega}=3.4\mu A\\ \triangle I_{D2}&=7.7\mu A \end{align*}

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4) For ideal MOSFET; $g_d=0$ in saturation \begin{align*} \end{align*}

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5) \begin{align*} V_{SF}&=\frac{C_{ox}}{C_{ox}+(C_{it}+C_{dep})}\cdot V_{gs}\\ &=C1\cdot v_{gs} \end{align*}

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6)\begin{align*} \ln I_d&=\ln A+\frac{qV_{SF}}{kT}\\ &=\ln A+ \frac{qC_1V_{gs}}{kT} \end{align*}


$\frac{d(\ln I_d)}{dv_{gs}}=\frac{qC_1}{kT}$

\begin{align*} S&=\ln 10\cdot\frac{kT}{qC_1}\\ &=\ln 10\cdot\frac{kT(C_{ox}+C_{it}+C_{dep})}{qC_{ox}}.....1\\ &=\ln 10\cdot\frac{kT}{q}\cdot\frac{v_{gs}}{V_{SF}} \end{align*}

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7) Eqn 1.

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8)\begin{align*} \frac{S}{S_{ideal}}&=\frac{C_{ox}+C_{it}}{C_{ox}}; C_{dep}=0\\ C_{ox}&=\frac{\epsilon_0\cdot\epsilon_r}{t_{ox}}\times WL\\ &=1.78\times10^{-15}F \end{align*}

$\therefore\frac{S}{S_{ideal}}=1.112$

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9)\begin{align*} L'&=L=2r;\bigg(\sqrt{1+\frac{2w_T}{r;}}-1\bigg)\\ &=L-\triangle L \end{align*}

affected by c, f, h

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10) Need $\triangle L\downarrow$ So;

$r_J\downarrow c$

$w_T\downarrow f$

$\downarrow h$

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