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So consider p-channel MOSFET.  
 
So consider p-channel MOSFET.  
  
[[Image:MN3_09_1.png|Alt text|400x400px]]
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[[Image:MN3_09_1.png]]
  
 
<math>=\mu_pC_{ox}\frac{W}{L} = \mu_p\frac{\epsilon_{ox}}{t_{ox}}\frac{W}{L}</math>
 
<math>=\mu_pC_{ox}\frac{W}{L} = \mu_p\frac{\epsilon_{ox}}{t_{ox}}\frac{W}{L}</math>

Revision as of 09:46, 6 August 2017


ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 3: Field Effect Devices

August 2009



Questions

All questions are in this link

Solutions of all questions

a) Previous year question.\\ b) $ V_T = -1V $ So consider p-channel MOSFET.

MN3 09 1.png

$ =\mu_pC_{ox}\frac{W}{L} = \mu_p\frac{\epsilon_{ox}}{t_{ox}}\frac{W}{L} $


$ \begin{align*} & t_{ox} \uparrow k\downarrow\\ & \mu_p \uparrow k\uparrow\\ & L \uparrow k\downarrow\\ & W \uparrow k\uparrow\\ & N_A \uparrow \text{ No change} \end{align*} $

$ V_{TH} = \psi_s+\frac{k_s}{k_{ox}}x_0\sqrt{\frac{2qN_A}{k_s\epsilon_0}\psi_s} $

chk only 2

f) square law $ \[Q_i=C_{ox}(V_{GS} - V_{th}-V)\] $

n-channel

Bulk charge

$ Q_i = -C_{ox}(V_{GS} - V_{th}-V))+\underbrace{qN_A[w_T(V)-w_T(V=0)]}_{\text{variation of $w_T$ with $V$ is considered}} $

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3)i)

Alt text

Turned on without any applied gate voltage.

ii) Alt text

Better control on channel due to lowered body effect.

iii) Alt text

Very high speed device due to $e^-$ gas formation.

iv) Alt text

Can handle short channel effect much better

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