MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 3: Field Effect Devices

August 2012

Questions

All questions are in this link

Solutions of all questions

a)\begin{align*} C_1&=C_{ox}=\frac{\epsilon_0\epsilon_{ox}}{t_{ox}}\\ \therefore t_{ox}&=\frac{20\epsilon_0}{C_1} \end{align*}

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b)i) $V_{GS} =0$

ii) $V_{GS} = 0.5V;$ dep/inv transition (from C-V curve)

iii) $V_{GS} = 1V;$ inversion

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c)$V_{th} = 0.5V$. Lowest value of C.

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d)

Drift current;

\begin{align*} J_1&=Q_1\mu\epsilon_1=Q_1\mu\frac{dV}{dy}\vert_1\\ J_2&=Q-2\mu\frac{dV}{dy}\vert_2\\ J_3&=Q_3\mu\frac{dV}{dy}\vert_3 \end{align*}

$\sum_{i=1}^N\frac{J_idy}{\mu} = \sum_{i=1}^NQ_idV$

\begin{align*} &\implies \frac{J_D}{\mu}\sum_{i=1}^Ndy = \int_0^{V_d}C_{ox}(V_G-V_{th}-mV)dV\\ &\implies \frac{J_D}{\mu}L=C_{ox}(V_G-V_{th})V_d-m\frac{V_d^2}{2}\\ &\therefore J_D=\frac{\mu C_{ox}}{L}\bigg[(V_G-V_{th})V_d-m\frac{V_d^2}{2}\bigg] \end{align*}

\begin{align*} \therefore I_D&=W\cdot J_D\\ &=\mu C_{ox}\frac{W}{L}\bigg[(V_G-V_{th})V_d-m\frac{V_d^2}{2}\bigg] \end{align*}

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e)*(confusion)

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f) \begin{align*} L_{min}&=\frac{V_d}{E_{crit}} = \frac{3V}{10^4 V/cm \text{ from curve}}\\ &=3\times10^{-4}cm \end{align*}

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g) $V_T \downarrow$ for short channel

$I_{DSTAT}\uparrow$ as $\sim(V_{GS}-V_{th})$

$V_{DSTAT}\uparrow$ as $\sim(V_{GS}-V_{th})$

(* quantitative how??)

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

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