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August 2007
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August 2010
 
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</center>
 
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==Questions==
 
==Questions==
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All questions are in this [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_07/MN-2%20QE%2007.pdf link]
+
All questions are in this [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_10/MN-2%20QE%2010.pdf link]
  
 
=Solutions of all questions=
 
=Solutions of all questions=
  
−
1)
+
Part A
−
A) Forward
+
a) <math>D_p\frac{d^2\triangle p}{dx^2} = 0</math>
  
−
B)   <math>10^{16}cm^{-3}</math>
+
b) <math>\triangle p = Ax +B</math>
  
−
C) <math>10^{14}cm^{-3}</math>
+
c) <math>\triangle p(x_n) = \frac{n_i^2}{N_D}(e^{qV_A/kT}-1)</math>
 +
<math>\triangle p(w_N) = 0
 +
</math>
  
−
D) <math>10^{9}\times10^{14} = n_i^2</math>
+
d)<math>0=Aw_N+B</math>
−
<math>\implies n_i = 10^{23/2}</math>
+
<math>
−
 
+
−
E) Yes.
+
−
 
+
−
F) <math>
+
 
\begin{align*}
 
\begin{align*}
−
\triangle P_n&=10^{12}-10^9\approx 10^{12}\\
+
\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)&=Ax_n+B\\
−
\triangle P_n&=\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)=10^{12}\\
+
A(w_N+x_n)&=-\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)\\
−
&\implies e^{qV_A/kT} = 10^3\\
+
\implies A&=-\frac{n_i^2}{N_D(w_N-x_n)}(e^{qV_A/kT}-1)\\
−
&\implies V_A = 0.026\times\ln(10^3) = 0.026\times6.9\\
+
B=-Aw_N&=\frac{w_Nn_i^2}{N_D(w_N-x_n)}(e^{qV_A/kT}-1)\\
−
&= 0.17 V
+
J_p = -qD_p\frac{d\triangle p}{dx}\bigg\vert_{x=x_n} &= -qD_p\bigg[-\frac{n_i^2}{N_D(w_N-x_n)}(e^{qV_A/kT}-1)\bigg]\\
 +
&=\frac{qD_p}{w_N-x_n}\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)
 
\end{align*}
 
\end{align*}
 
</math>
 
</math>
  
 
   ------------------------------------------------------------------------------------
 
   ------------------------------------------------------------------------------------
−
2)
+
Part B
 +
a)  
 
<math>
 
<math>
 
\begin{align*}
 
\begin{align*}
−
|I_E| & = qD_p\frac{10^{10}}{0.1\times10^{-4}} = 1.8\times10^{16}q\\
+
I_{La}&=qA\int_0^{x_n}\frac{dn}{dt}\cdot dx\\
−
|I_B| & = qD_n\frac{10^{8}}{0.2\times10^{-4}} = 1.8\times10^{14}q\\
+
&=qAG_Lx_n
−
\beta + 1 &= \frac{|I_E|}{|I_B|}\approx 67\\
+
−
&\therefore \beta= 66 \text{ (chk)}
+
 
\end{align*}
 
\end{align*}
 
</math>
 
</math>
−
 
+
<math>
−
------------------------------------------------------------------------------------
+
\frac{dn}{dt} = -R=G_L
−
3)
+
−
A)
+
−
<math>
+
−
\alpha_T = \frac{0.997J_0}{0.998J_0} = 0.99
+
 
</math>
 
</math>
  
−
B)
+
b)
 
  <math>
 
  <math>
−
\gamma = \frac{0.998J_0}{(0.998+0.002)J_0} = 0.998
 
−
</math>
 
−
 
−
C)D)
 
−
  <math>
 
 
  \begin{align*}
 
  \begin{align*}
−
\alpha_{dc} &= \gamma\cdot\alpha_T\\
+
&D_p\frac{d^2\triangle p}{dx^2}+G_L =0\\
−
&=0.98802
+
&\implies \frac{d^2\triangle p}{dx^2} =-\frac{G_L}{D_p}\\
−
\end{align*}
+
&\implies \frac{d\triangle p}{dx} =-\frac{G_L}{D_p}x+A\\
−
</math>
+
&\implies \triangle p =-\frac{G_L}{D_p}\frac{x^2}{2}+Ax+B\\
−
<math>
+
&\triangle p(w_n) =0\\
−
\beta_{dc} = \frac{\alpha_{dc}}{1-\alpha_{dc}}\approx 82
+
&\triangle p(x_n) = \frac{n_i^2}{N_D}(e^{qV_A/kT}-1) (?)\\
−
</math>
+
&\text{or; }-D_p\frac{d\triangle p(x_n)}{dx}=G_Lx_n (?)
−
<math>
+
\end{align*}
−
\begin{align*}
+
−
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>
−
 
+
which one to use??
−
------------------------------------------------------------------------------------
+
−
4)
+
 
  <math>
 
  <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}
+
I_{Lb} = -qAD_p\frac{d(\triangle p)}{dx}
 
</math>
 
</math>
  
 
  ------------------------------------------------------------------------------------
 
  ------------------------------------------------------------------------------------
−
 
  
  
 
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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 21:19, 5 August 2017


ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2010

Questions

All questions are in this link

Solutions of all questions

Part A a) $ D_p\frac{d^2\triangle p}{dx^2} = 0 $

b) $ \triangle p = Ax +B $

c) $ \triangle p(x_n) = \frac{n_i^2}{N_D}(e^{qV_A/kT}-1) $ $ \triangle p(w_N) = 0 $

d)$ 0=Aw_N+B $ $ \begin{align*} \frac{n_i^2}{N_D}(e^{qV_A/kT}-1)&=Ax_n+B\\ A(w_N+x_n)&=-\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)\\ \implies A&=-\frac{n_i^2}{N_D(w_N-x_n)}(e^{qV_A/kT}-1)\\ B=-Aw_N&=\frac{w_Nn_i^2}{N_D(w_N-x_n)}(e^{qV_A/kT}-1)\\ J_p = -qD_p\frac{d\triangle p}{dx}\bigg\vert_{x=x_n} &= -qD_p\bigg[-\frac{n_i^2}{N_D(w_N-x_n)}(e^{qV_A/kT}-1)\bigg]\\ &=\frac{qD_p}{w_N-x_n}\frac{n_i^2}{N_D}(e^{qV_A/kT}-1) \end{align*} $ 

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

Part B a) $ \begin{align*} I_{La}&=qA\int_0^{x_n}\frac{dn}{dt}\cdot dx\\ &=qAG_Lx_n \end{align*} $ $ \frac{dn}{dt} = -R=G_L $

b) 

$   \begin{align*} &D_p\frac{d^2\triangle p}{dx^2}+G_L =0\\ &\implies \frac{d^2\triangle p}{dx^2} =-\frac{G_L}{D_p}\\ &\implies \frac{d\triangle p}{dx} =-\frac{G_L}{D_p}x+A\\ &\implies \triangle p =-\frac{G_L}{D_p}\frac{x^2}{2}+Ax+B\\ &\triangle p(w_n) =0\\ &\triangle p(x_n) = \frac{n_i^2}{N_D}(e^{qV_A/kT}-1) (?)\\ &\text{or; }-D_p\frac{d\triangle p(x_n)}{dx}=G_Lx_n (?) \end{align*}  $

which one to use?? 

$  I_{Lb} = -qAD_p\frac{d(\triangle p)}{dx}  $

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