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\end{align*}
 
\end{align*}
 
</math>
 
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[[Image:MN3_2013_1.png|Alt text|300x300px]]
 
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\end{align*}
 
\end{align*}
 
</math>
 
</math>
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<math>
 
<math>
 
\frac{d(\ln I_d)}{dv_{gs}}=\frac{qC_1}{kT}
 
\frac{d(\ln I_d)}{dv_{gs}}=\frac{qC_1}{kT}
 
</math>
 
</math>
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  <math>
 
  <math>
 
\begin{align*}
 
\begin{align*}
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\end{align*}
 
\end{align*}
 
</math>
 
</math>
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<math>
 
<math>
 
\therefore\frac{S}{S_{ideal}}=1.112  
 
\therefore\frac{S}{S_{ideal}}=1.112  
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\end{align*}
 
\end{align*}
 
</math>
 
</math>
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affected by c, f, h
 
affected by c, f, h
  
 
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10) Need <math>\triangle L\downarrow$\\
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10) Need <math>\triangle L\downarrow </math>
So; <math>
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So;
 +
 
 
<math>r_J\downarrow c</math>
 
<math>r_J\downarrow c</math>
  
<math>w_T\downarrow f</math>
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<math>w_T\downarrow f</math>
 
   
 
   
 
  <math>\downarrow h</math>
 
  <math>\downarrow h</math>
</math>
 
 
   
 
   
 
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Latest revision as of 00:42, 6 August 2017


ECE Ph.D. Qualifying Exam

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*} $

Alt text

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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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