Questions

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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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