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August 2007
+
August 2011
 
</center>
 
</center>
 
----
 
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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]
+
All questions are in this [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_11/MN-2%20QE%2011.pdf link]
  
 
=Solutions of all questions=
 
=Solutions of all questions=
  
−
1)  
+
1) a) Direct bandgap semiconductors used in Lasers and LEDs.
−
A) Forward
+
Photodetectors.
  
−
B)   <math>10^{16}cm^{-3}</math>
+
b) Si has a lattice matched <math>SiO_2</math> to reduce surface defects.
 +
Si is almost 99\% pure.
  
−
C) <math>10^{14}cm^{-3}</math>
+
c) <math>\sim 9eV</math>
  
−
D) <math>10^{9}\times10^{14} = n_i^2</math>
+
d) (2\:\:\:\:1\:\:\:\:0)
−
<math>\implies n_i = 10^{23/2}</math>
+
<math>\rightarrow \:\frac{1}{2}\:\:\:\:1\:\:\:\:\infty </math>
  
−
E) Yes.
+
<math>\rightarrow \:1\:\:\:\:2\:\:\:\:\infty</math>
 +
[[Image:MN2_2011_1.png|Alt text|300x300px]]
  
−
F) <math>
+
e)
−
\begin{align*}
+
<math>\sim 900-1200</math> deg C.
−
\triangle P_n&=10^{12}-10^9\approx 10^{12}\\
+
−
\triangle P_n&=\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)=10^{12}\\
+
−
&\implies e^{qV_A/kT} = 10^3\\
+
−
&\implies V_A = 0.026\times\ln(10^3) = 0.026\times6.9\\
+
−
&= 0.17 V
+
−
\end{align*}
+
−
</math>
+
  
 
   ------------------------------------------------------------------------------------
 
   ------------------------------------------------------------------------------------
 
2)
 
2)
 +
Neutral p
 
<math>
 
<math>
 
\begin{align*}
 
\begin{align*}
−
|I_E| & = qD_p\frac{10^{10}}{0.1\times10^{-4}} = 1.8\times10^{16}q\\
+
p_0&=10^{15}\\
−
|I_B| & = qD_n\frac{10^{8}}{0.2\times10^{-4}} = 1.8\times10^{14}q\\
+
n_0&=\frac{10^{20}}{10^{15}}=10^5
−
\beta + 1 &= \frac{|I_E|}{|I_B|}\approx 67\\
+
\end{align*}
−
&\therefore \beta= 66 \text{ (chk)}
+
</math>
 +
Neutral n
 +
<math>
 +
\begin{align*}
 +
n_0&=10^{18}\\
 +
p_0&=\frac{10^{20}}{10^{18}}=10^2
 
\end{align*}
 
\end{align*}
 
</math>
 
</math>
Line 60: Line 61:
 
  ------------------------------------------------------------------------------------
 
  ------------------------------------------------------------------------------------
 
3)
 
3)
−
A)
 
 
  <math>
 
  <math>
−
\alpha_T = \frac{0.997J_0}{0.998J_0} = 0.99
+
\begin{align*}
 +
x_{n0}\cdot 10^{18}&=x_{p0}\cdot10^{15}\\
 +
\implies x_{p0}&=10^3\times0.001\mu m = 1\mu m
 +
\end{align*}
 
</math>
 
</math>
  
−
B)
+
------------------------------------------------------------------------------------
−
  <math>
+
  4)  <math>
−
\gamma = \frac{0.998J_0}{(0.998+0.002)J_0} = 0.998
+
x_n = \sqrt{\frac{2\epsilon}{q}\frac{N_A}{N_D(N_A+N_D)}\cdot(V_{bi}+V)}
−
  </math>
+
−
+
−
C)D)
+
−
  <math>
+
−
\begin{align*}
+
−
\alpha_{dc} &= \gamma\cdot\alpha_T\\
+
−
&=0.98802
+
−
\end{align*}
+
−
</math>
+
−
<math>
+
−
\beta_{dc} = \frac{\alpha_{dc}}{1-\alpha_{dc}}\approx 82
+
−
</math>
+
−
<math>
+
−
\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>
  
 
  ------------------------------------------------------------------------------------
 
  ------------------------------------------------------------------------------------
−
4)
+
5)
 +
<math>
 +
\begin{align*}
 +
\rho &=\frac{1}{q\mu_nN_D} = \frac{1}{2\times10^{-19}\times500\times10^{18}}\\
 +
&=\frac{1}{1000\times10^{-1}}\Omega\cdot cm\\
 +
&=10^{-2}\Omega\cdot cm
 +
\end{align*}
 +
</math>
 +
<math>
 +
R_s=\frac{\rho}{L} =\frac{10^{-2}}{5\times10^{-4}} = 20\Omega/\Box
 +
</math>
 +
 
 +
------------------------------------------------------------------------------------
 +
6)
 
  <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}
+
\begin{align*}
 +
J_n&=q_n\mu_nE\\
 +
&=2\times10^{-19}\times10^{18}\times500\times1000\\
 +
&=10^5 A/cm^2
 +
\end{align*}
 
</math>
 
</math>
 +
 +
------------------------------------------------------------------------------------
 +
7)
 +
  <math>
 +
\begin{align*}
 +
J_{diff} &=qD_n\frac{d\triangle n}{dx}\\
 +
&=2\times10^{-19}\times500\times 25m\times\frac{\triangle n_p}{5\mu m}\\
 +
\end{align*}
 +
</math>
 +
 
 +
------------------------------------------------------------------------------------
 +
8)[[Image:MN2_2011_2.png|Alt text|300x300px]]
 +
------------------------------------------------------------------------------------
 +
9)[[Image:MN2_2011_3.png|Alt text|300x300px]]
 +
------------------------------------------------------------------------------------
 +
10) Forward Active Mode
 +
[[Image:MN2_2011_4.png|Alt text|300x300px]]
 +
------------------------------------------------------------------------------------
 +
11)Fig 11.15(SDF) Page 425
 +
[[Image:MN2_2011_5.png|Alt text|300x300px]]
 +
  <math>\beta  </math> is reduced by both phenomena.
  
 
  ------------------------------------------------------------------------------------
 
  ------------------------------------------------------------------------------------
 +
12)[[Image:MN2_2011_6.png|Alt text|300x300px]]
 
   
 
   
 +
  
  
 
----
 
----
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Revision as of 22:20, 5 August 2017


ECE Ph.D. Qualifying Exam

MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2011

Questions

All questions are in this link

Solutions of all questions

1) a) Direct bandgap semiconductors used in Lasers and LEDs. Photodetectors.

b) Si has a lattice matched $ SiO_2 $ to reduce surface defects. Si is almost 99\% pure.

c) $ \sim 9eV $

d) (2\:\:\:\:1\:\:\:\:0) $ \rightarrow \:\frac{1}{2}\:\:\:\:1\:\:\:\:\infty $

$ \rightarrow \:1\:\:\:\:2\:\:\:\:\infty $ Alt text

e) $ \sim 900-1200 $ deg C. 

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

2) Neutral p $ \begin{align*} p_0&=10^{15}\\ n_0&=\frac{10^{20}}{10^{15}}=10^5 \end{align*} $ Neutral n $ \begin{align*} n_0&=10^{18}\\ p_0&=\frac{10^{20}}{10^{18}}=10^2 \end{align*} $ 

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

3) 

$   \begin{align*} x_{n0}\cdot 10^{18}&=x_{p0}\cdot10^{15}\\ \implies x_{p0}&=10^3\times0.001\mu m = 1\mu m \end{align*}  $

------------------------------------------------------------------------------------
4)  $  x_n = \sqrt{\frac{2\epsilon}{q}\frac{N_A}{N_D(N_A+N_D)}\cdot(V_{bi}+V)}  $

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

5) $ \begin{align*} \rho &=\frac{1}{q\mu_nN_D} = \frac{1}{2\times10^{-19}\times500\times10^{18}}\\ &=\frac{1}{1000\times10^{-1}}\Omega\cdot cm\\ &=10^{-2}\Omega\cdot cm \end{align*} $ $ R_s=\frac{\rho}{L} =\frac{10^{-2}}{5\times10^{-4}} = 20\Omega/\Box $ 

------------------------------------------------------------------------------------
6)
$  \begin{align*} J_n&=q_n\mu_nE\\ &=2\times10^{-19}\times10^{18}\times500\times1000\\ &=10^5 A/cm^2 \end{align*}  $

------------------------------------------------------------------------------------
7)
 $  \begin{align*} J_{diff} &=qD_n\frac{d\triangle n}{dx}\\ &=2\times10^{-19}\times500\times 25m\times\frac{\triangle n_p}{5\mu m}\\ \end{align*}  $
 
------------------------------------------------------------------------------------
8)Alt text
------------------------------------------------------------------------------------
9)Alt text
------------------------------------------------------------------------------------
10) Forward Active Mode

Alt text 

------------------------------------------------------------------------------------
11)Fig 11.15(SDF) Page 425

Alt text 

 $ \beta   $ is reduced by both phenomena.

------------------------------------------------------------------------------------
12)Alt text


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