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August 2007
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August 2012
 
</center>
 
</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_12/MN-2%20QE%2012.pdf link]
  
 
=Solutions of all questions=
 
=Solutions of all questions=
  
−
1)  
+
a) [[Image:MN2_2012_1.png|Alt text|300x300px]]
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A) Forward
+
  
−
B)   <math>10^{16}cm^{-3}</math>
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b) <math>x_n<1\mu m</math>
  
−
C) <math>10^{14}cm^{-3}</math>
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<math>\rho=q(p-n+N_D-N_A)</math>
  
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D) <math>10^{9}\times10^{14} = n_i^2</math>
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[[Image:MN2_2012_2.png|Alt text|300x300px]]
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<math>\implies n_i = 10^{23/2}</math>
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E) Yes.
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<math>x_n>1\mu m</math>
  
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F) <math>
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[[Image:MN2_2012_3.png|Alt text|300x300px]]
 +
 
 +
c) <math>
 
\begin{align*}
 
\begin{align*}
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\triangle P_n&=10^{12}-10^9\approx 10^{12}\\
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N_Ax_p&=N_{D1}\cdot1\mu m+N_{D2}(x_n-1)\\
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\triangle P_n&=\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)=10^{12}\\
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x_n&=1.25\mu m\text{ (chk)}
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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\\
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−
&= 0.17 V
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\end{align*}
 
\end{align*}
 
</math>
 
</math>
  
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  ------------------------------------------------------------------------------------
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d)<math>
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2)
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−
<math>
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\begin{align*}
 
\begin{align*}
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|I_E| & = qD_p\frac{10^{10}}{0.1\times10^{-4}} = 1.8\times10^{16}q\\
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E_{max}&=\frac{qN_Ax_p}{k_s\epsilon_0}\\
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|I_B| & = qD_n\frac{10^{8}}{0.2\times10^{-4}} = 1.8\times10^{14}q\\
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\text{where;}\\
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\beta + 1 &= \frac{|I_E|}{|I_B|}\approx 67\\
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x_p&=\sqrt{\frac{2k_s\epsilon_0}{q}\cdot \frac{N_{D1}}{N_A(N_{D1}+N_A)}\cdot V_{bi}}\\
−
&\therefore \beta= 66 \text{ (chk)}
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\text{and }V_{bi}&=\frac{kT}{q}\ln\frac{N_AN_{D1}}{n_i^2}
 
\end{align*}
 
\end{align*}
 
</math>
 
</math>
  
−
------------------------------------------------------------------------------------
+
e)<math>
−
3)
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\rho=\begin{cases}
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A)
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0 &x<-x_p\\
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<math>
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-qN_A &-x_p\le x\le \text{ region 1}\\
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\alpha_T = \frac{0.997J_0}{0.998J_0} = 0.99
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qN_{D1} & 0\le x\le x_{n1}\text{ region 2}\\
 +
qN_{D2} & x_{n1}\le x\le x_{n2}\text{ region 3}\\
 +
0 & x>x_{n2}
 +
\end{cases}
 
</math>
 
</math>
 +
need to solve; <math>\frac{dE}{dx}=\frac{\rho}{\epsilon}</math>
  
−
B)
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−
<math>
+
 
−
\gamma = \frac{0.998J_0}{(0.998+0.002)J_0} = 0.998
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* region 1
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</math>
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<math>
−
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\int_{E(-x_p)}^{E(x)}dE = \frac{-qN_A}{k_s\epsilon_0}\int_{-x_p}^xdx
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C)D)
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</math>
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  <math>
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<math>
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\begin{align*}
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\implies E(x)=\frac{-qN_A}{k_s\epsilon_0}(x+x_p)
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\alpha_{dc} &= \gamma\cdot\alpha_T\\
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</math>
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&=0.98802
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at <math>x=0</math>
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\end{align*}
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<math>
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</math>  
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E(0) = \frac{-qN_A}{k_s\epsilon_0}x_p
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<math>
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−
\beta_{dc} = \frac{\alpha_{dc}}{1-\alpha_{dc}}\approx 82
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−
</math>
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−
<math>
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−
\begin{align*}
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−
I_E &= (0.998+0.002)J_0 = J_0\\
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−
I_C &= 0.997J_0\\
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−
I_B &= I_E-I_C = 0.003J_0
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−
\end{align*}
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</math>
 
</math>
  
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------------------------------------------------------------------------------------
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* region 2
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4)
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<math>
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<math>
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\int_{E(0)}^{E(x)}dE = \frac{qN_{D1}}{k_s\epsilon_0}\int_{0}^xdx
−
\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}
+
</math>
 +
<math>
 +
\implies E(x)=\frac{qN_{D1}}{k_s\epsilon_0}x+\frac{qN_A}{k_s\epsilon_0}x_p
 
</math>
 
</math>
  
−
------------------------------------------------------------------------------------
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* region 3
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<math>
−
 
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\int_{E(x)}^{E(x_{n2})}dE = \frac{qN_{D2}}{k_s\epsilon_0}\int_{x}^{x_{n2}}dx
 +
</math>
 +
<math>
 +
\implies 0-E(x)=\frac{qN_{D2}}{k_s\epsilon_0}(x_{n2}-x)
 +
</math>
 +
<math>
 +
\implies E(x)=\frac{qN_{D2}}{k_s\epsilon_0}(x-x_{n2})
 +
</math>
  
 +
  ------------------------------------------------------------------------------------
 
----
 
----
 
[[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 09:52, 6 August 2017


ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2012

Questions

All questions are in this link

Solutions of all questions

a) Alt text

b) $ x_n<1\mu m $

$ \rho=q(p-n+N_D-N_A) $

Alt text

$ x_n>1\mu m $

Alt text

c) $ \begin{align*} N_Ax_p&=N_{D1}\cdot1\mu m+N_{D2}(x_n-1)\\ x_n&=1.25\mu m\text{ (chk)} \end{align*} $

d)$ \begin{align*} E_{max}&=\frac{qN_Ax_p}{k_s\epsilon_0}\\ \text{where;}\\ x_p&=\sqrt{\frac{2k_s\epsilon_0}{q}\cdot \frac{N_{D1}}{N_A(N_{D1}+N_A)}\cdot V_{bi}}\\ \text{and }V_{bi}&=\frac{kT}{q}\ln\frac{N_AN_{D1}}{n_i^2} \end{align*} $

e)$ \rho=\begin{cases} 0 &x<-x_p\\ -qN_A &-x_p\le x\le \text{ region 1}\\ qN_{D1} & 0\le x\le x_{n1}\text{ region 2}\\ qN_{D2} & x_{n1}\le x\le x_{n2}\text{ region 3}\\ 0 & x>x_{n2} \end{cases} $ need to solve; $ \frac{dE}{dx}=\frac{\rho}{\epsilon} $


  • region 1

$ \int_{E(-x_p)}^{E(x)}dE = \frac{-qN_A}{k_s\epsilon_0}\int_{-x_p}^xdx $ $ \implies E(x)=\frac{-qN_A}{k_s\epsilon_0}(x+x_p) $ at $ x=0 $ $ E(0) = \frac{-qN_A}{k_s\epsilon_0}x_p $

  • region 2

$ \int_{E(0)}^{E(x)}dE = \frac{qN_{D1}}{k_s\epsilon_0}\int_{0}^xdx $ $ \implies E(x)=\frac{qN_{D1}}{k_s\epsilon_0}x+\frac{qN_A}{k_s\epsilon_0}x_p $

  • region 3

$ \int_{E(x)}^{E(x_{n2})}dE = \frac{qN_{D2}}{k_s\epsilon_0}\int_{x}^{x_{n2}}dx $ $ \implies 0-E(x)=\frac{qN_{D2}}{k_s\epsilon_0}(x_{n2}-x) $ $ \implies E(x)=\frac{qN_{D2}}{k_s\epsilon_0}(x-x_{n2}) $ 

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

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