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\mu_n = \frac{q\tau_n}{m^*}\sim\tau_n
 
\mu_n = \frac{q\tau_n}{m^*}\sim\tau_n
 
</math>
 
</math>
[[Image:MN2_2009_2.png|Alt text|300x300px]]
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[[Image:MN2_2009_2.png|Alt text|558x336px]]
  
 
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  4) a) ??
 
  4) a) ??
 
   
 
   
b) [[Image:MN2_2009_3.png|Alt text|300x300px]]
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b) [[Image:MN2_2009_3.png|Alt text|678x364px]]
 
Depletion mode device is ON without applied  <math>V_{GS}</math>. So,  <math>V_{GS}</math> is lower for Dep. mode device. Hence low current.
 
Depletion mode device is ON without applied  <math>V_{GS}</math>. So,  <math>V_{GS}</math> is lower for Dep. mode device. Hence low current.
  

Latest revision as of 21:28, 5 August 2017


ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2009



Questions

All questions are in this link

Solutions of all questions

1) a) Zinc blend crystal. 8 atoms/unit cell. b) $ 1.12 eV $

c) $ \sim10^4 V/cm (??) $

d) Chemistry exam question. (some other language) ??

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2) a) $ n = N_Ce^{(E_F-E_C)/kT} p = N_Ve^{(E_V-E_F)/kT} $

b) If $ E_F = E_i $ then $ n = n_i, p = n_i $ $ \begin{align*} n_i &= N_Ce^{(E_i-E_C)/kT} =n\\ n_i &= N_Ve^{(E_V-E_i)/kT} =p\\ \therefore n_p&=n_i^2=N_CN_Ve^{-E_g/kT} \end{align*} $

c) Alt text

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3) a)

$   \begin{align*} p &=\frac{1}{q\mu_nN_D}\\ \implies\mu_n&=\frac{1}{qp\cdot N_D}\sim 1.5\times10^3cm^2/V\cdot s \text{ (chk)} \end{align*}  $

b)

$  \mu_p=\frac{1}{qp\cdot N_A}\sim 0.5\times10^3cm^2/V\cdot s \text{ (chk)}  $
$ \mu_p<\mu_n $ as  $ p_p>p_n $

because $ m_p^*>m_n^* $

c) Ionized impurity

$  \tau_n\sim\frac{T^{3/2}}{N_D}  $

Photon scattering

$  \tau_n\sim T^{-3/2}  $
$  \mu_n = \frac{q\tau_n}{m^*}\sim\tau_n  $

Alt text

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4) a) ??

b) Alt text Depletion mode device is ON without applied $ V_{GS} $. So, $ V_{GS} $ is lower for Dep. mode device. Hence low current.

c) $ I_D = \frac{1}{2}\mu_nC_{ox}\frac{W}{L}(V_{GS}-V_{th})^2 $ $ g_m = \frac{\partial I_D}{\partial V_{GS}} = \mu_nC_{ox}\frac{W}{L}(V_{GS}-V_{th}) $ If W=L then; $ \mu_n=\frac{g_m}{C_{ox}(V_{GS}-V_{th})} $

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