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C_G\approx C_{ox} = \frac{k_{ox}\epsilon_{ox}}{x_0} | C_G\approx C_{ox} = \frac{k_{ox}\epsilon_{ox}}{x_0} | ||
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So; here <math>EOT\approx x_0</math> | So; here <math>EOT\approx x_0</math> | ||
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to have the same EOT: | to have the same EOT: | ||
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Below <math>E_0 \rightarrow</math> Dominated by acceptor type (negative when full) | Below <math>E_0 \rightarrow</math> Dominated by acceptor type (negative when full) | ||
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Above <math>E_0 \rightarrow</math> Dominated by donor type (positive when empty) | Above <math>E_0 \rightarrow</math> Dominated by donor type (positive when empty) | ||
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\end{align*} | \end{align*} | ||
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* <math>S=80mV/dec</math> by using high k dielectric | * <math>S=80mV/dec</math> by using high k dielectric | ||
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[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Latest revision as of 23:36, 5 August 2017
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
