MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 3: Field Effect Devices

August 2011

## Questions

All questions are in this link

# Solutions of all questions

1)

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2) \begin{align*} I_D &=\mu_nC_{ox}\frac{Z}{L}(V_{GS}-V_t)V_{DS}\\ \implies G_d&=\frac{I_d}{V_{ds}} = 500\times3.45\times10^{-7}\times\frac{5}{0.25}\times0.5\\ &=0.0017 S (chk) \end{align*}

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3)

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4)
$C_G\approx C_{ox} = \frac{k_{ox}\epsilon_{ox}}{x_0}$

So; here $EOT\approx x_0$

\begin{align*} x_0 = \frac{k_{ox}\epsilon_{ox}}{C_{ox}}&=\frac{3.9\epsilon_0}{3.45\times10^{-7}}\\ &=10nm \end{align*}

to have the same EOT:

\begin{align*} \frac{1}{C_{ox}} &=\frac{1}{C_{SiO_2}}+\frac{1}{C_{HfO_2}}+\frac{1}{C_{SiO_2}}\\ \implies \frac{10nm}{\epsilon_{ox}}&=\frac{1nm}{\epsilon_{ox}}+\frac{x nm}{\epsilon_{HfO_2}}\\ \implies \frac{8nm}{3.9}&=\frac{x nm}{20}\\ \therefore x&\approx40nm \text{ chk} \end{align*}

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5) Below $E_0 \rightarrow$ Dominated by acceptor type (negative when full)

Above $E_0 \rightarrow$ Dominated by donor type (positive when empty)

• B)C) (chk)
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6) C-V stretch out

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7) * confusion

\begin{align*} S&=\ln10\frac{kT}{q}\bigg(1+\frac{C_{it}}{C_{ox}}\bigg)\\ \implies 100m&=\ln10\bigg(\frac{kT}{q}\bigg)\bigg(1+\frac{C_{it}}{C_{ox}}\bigg)\\ C_{it}&=0.67C_{ox}\\ ?D_{it}&=\frac{C_{it}}{q^2} = 9.03\times10^{30}cm^{-2} \text{ (Absurd??)} \end{align*}

• $S=80mV/dec$ by using high k dielectric

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