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August 2007
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August 2009
 
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==Questions==
 
==Questions==
All questions are in this [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_07/MN-2%20QE%2007.pdf link]
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All questions are in this [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_09/MN-2%20QE%2009.pdf link]
  
 
=Solutions of all questions=
 
=Solutions of all questions=
  
1)  
+
1) a) Zinc blend crystal. 8 atoms/unit cell.
A) Forward
+
b) <math>1.12 eV</math>
  
B)   <math>10^{16}cm^{-3}</math>
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c) <math>\sim10^4 V/cm (??)
 
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C) <math>10^{14}cm^{-3}</math>
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+
D) <math>10^{9}\times10^{14} = n_i^2</math>
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<math>\implies n_i = 10^{23/2}</math>
+
 
+
E) Yes.
+
 
+
F) <math>
+
\begin{align*}
+
\triangle P_n&=10^{12}-10^9\approx 10^{12}\\
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\triangle P_n&=\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)=10^{12}\\
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&\implies e^{qV_A/kT} = 10^3\\
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&\implies V_A = 0.026\times\ln(10^3) = 0.026\times6.9\\
+
&= 0.17 V
+
\end{align*}
+
 
</math>
 
</math>
 +
 +
d) Chemistry exam question. (some other language) ??
  
 
   ------------------------------------------------------------------------------------
 
   ------------------------------------------------------------------------------------
 
2)
 
2)
 +
a)
 +
<math>
 +
n = N_Ce^{(E_F-E_C)/kT}
 +
p = N_Ve^{(E_V-E_F)/kT}
 +
</math>
 +
 +
b) If <math>E_F = E_i</math> then <math>n = n_i, p = n_i</math>
 
<math>
 
<math>
 
\begin{align*}
 
\begin{align*}
|I_E| & = qD_p\frac{10^{10}}{0.1\times10^{-4}} = 1.8\times10^{16}q\\
+
n_i &= N_Ce^{(E_i-E_C)/kT} =n\\
|I_B| & = qD_n\frac{10^{8}}{0.2\times10^{-4}} = 1.8\times10^{14}q\\
+
n_i &= N_Ve^{(E_V-E_i)/kT} =p\\
\beta + 1 &= \frac{|I_E|}{|I_B|}\approx 67\\
+
\therefore n_p&=n_i^2=N_CN_Ve^{-E_g/kT}
&\therefore \beta= 66 \text{ (chk)}
+
 
\end{align*}
 
\end{align*}
 
</math>
 
</math>
  
 +
c)
 +
[[Image:MN2_2009_1.png|Alt text|300x300px]]
 
  ------------------------------------------------------------------------------------
 
  ------------------------------------------------------------------------------------
3)
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3) a)
A)
+
 
  <math>
 
  <math>
\alpha_T = \frac{0.997J_0}{0.998J_0} = 0.99
+
\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*}
 
</math>
 
</math>
  
B)
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b)
 
  <math>
 
  <math>
\gamma = \frac{0.998J_0}{(0.998+0.002)J_0} = 0.998
+
\mu_p=\frac{1}{qp\cdot N_A}\sim 0.5\times10^3cm^2/V\cdot s \text{ (chk)}
</math>
+
</math>
   
+
  <math>\mu_p<\mu_n</math> as <math>p_p>p_n</math>
C)D)
+
because <math>m_p^*>m_n^*</math>
  <math>
+
 
\begin{align*}
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c) Ionized impurity
\alpha_{dc} &= \gamma\cdot\alpha_T\\
+
  &=0.98802
+
  \end{align*}
+
</math>  
+
 
  <math>
 
  <math>
\beta_{dc} = \frac{\alpha_{dc}}{1-\alpha_{dc}}\approx 82
+
\tau_n\sim\frac{T^{3/2}}{N_D}
</math>
+
</math>
 +
Photon scattering
 
  <math>
 
  <math>
\begin{align*}
+
\tau_n\sim T^{-3/2}
I_E &= (0.998+0.002)J_0 = J_0\\
+
I_C &= 0.997J_0\\
+
I_B &= I_E-I_C = 0.003J_0
+
\end{align*}
+
 
</math>
 
</math>
 +
<math>
 +
\mu_n = \frac{q\tau_n}{m^*}\sim\tau_n
 +
</math>
 +
[[Image:MN2_2009_2.png|Alt text|300x300px]]
  
 
  ------------------------------------------------------------------------------------
 
  ------------------------------------------------------------------------------------
  4)
+
  4) a) ??
  <math>
+
\text{Derivation of } \beta = \frac{D_n}{D_p}\cdot\frac{W_E}{W_B}\cdot\frac{N_E}{N_B}\cdot\frac{n_{iB}^2}{n_{iE}^2}
+
b) [[Image:MN2_2009_3.png|Alt text|300x300px]]
 +
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.
 +
 
 +
c) <math>
 +
I_D = \frac{1}{2}\mu_nC_{ox}\frac{W}{L}(V_{GS}-V_{th})^2
 +
</math>
 +
<math>
 +
g_m = \frac{\partial I_D}{\partial V_{GS}} = \mu_nC_{ox}\frac{W}{L}(V_{GS}-V_{th})
 
</math>
 
</math>
 +
If W=L then;  <math>\mu_n=\frac{g_m}{C_{ox}(V_{GS}-V_{th})}</math>
  
 
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Revision as of 21:19, 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) ??

 ------------------------------------------------------------------------------------

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

------------------------------------------------------------------------------------

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

------------------------------------------------------------------------------------
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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