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MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2013

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1) Two effects: i) Kirk effect or base pushout ii) Current crowding

 ------------------------------------------------------------------------------------
 2)  $$ Eg(InP) > Eg(InGaAs) $$

 So; InP is in emitter and collector.

$\beta$ increases as:

$\begin{align*} \beta &=\frac{D_n}{D_p}\cdot\frac{W_E}{W_B}\cdot\frac{n_{iB}^2}{n_{iE}^2}\cdot\frac{B_E}{N_B}\\ &\approx\frac{n_{iB}^2}{n_{iE}^2}\cdot\frac{B_E}{N_B}\cdot\frac{V_{th}}{v_s} \end{align*}$ $\frac{n_{iB}^2}{n_{iE}^2} = e^{\triangle Eg/kT}$

So huge improvement

$V_{cb,br}$ increases as $V_{cb,br}\approx\frac{3}{2}$ Eg collector

$f_{max}$ increases

$f_{max}^{-1}\approx f_T^{-1} = \bigg(\frac{w_B^2}{2D_n}+\frac{w_{BC}}{2v_{sat}}\bigg)+\frac{kT}{qSe}[C_{jBC}+C_{jBE}]$ here; $w_B\downarrow I_C\uparrow C_J\downarrow$ all leads to $f_T\uparrow$

high current effect reduces

$$ V_cesat $$ (??)

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 3)a) $\begin{align*} V_a&=0\\ \therefore E&=0 \end{align*}$

$\therefore n_p(x,t) = \frac{N}{\sqrt{4\pi D_nt}}exp\bigg(\frac{-x^2}{4D_nt}-\frac{t}{\tau_n}\bigg)+n_{p0}$

 At  $$ x=0 $$

\begin{figure}[H] \noindent 2 unknowns $\to D_n,\tau_n$ selecting 2 specific times and using the excess carrier can find $$ D_n $$ and $\tau_n$ .

At $$ x=X $$ ; initially no excess carrier.

b) For applied field; decay will be much quicker.

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@@ Line 16: / Line 16: @@
 </font size>
-August 2007
+August 2013
 </center>
 ----
 ----
 ==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_13/MN-2%20QE%2013.pdf link]
 =Solutions of all questions=
-)
+) Two effects:
-A) Forward
+i) Kirk effect or base pushout
+ii) Current crowding
-B)   <math>10^{16}cm^{-3}</math>
+   ------------------------------------------------------------------------------------
+) <math>Eg(InP) > Eg(InGaAs)</math>
-C) <math>10^{14}cm^{-3}</math>
+  So; InP is in emitter and collector.
-D) <math>10^{9}\times10^{14} = n_i^2</math>
+[[Image:MN2_2013_1.png|Alt text|300x300px]]
-<math>\implies n_i = 10^{23/2}</math>
+[[Image:MN2_2013_2.png|Alt text|300x300px]]
-E) Yes.
+* <math>\beta</math> increases as:
+<math>
-F) <math>
 \begin{align*}
-\triangle P_n&=10^{12}-10^9\approx 10^{12}\\
+\beta &=\frac{D_n}{D_p}\cdot\frac{W_E}{W_B}\cdot\frac{n_{iB}^2}{n_{iE}^2}\cdot\frac{B_E}{N_B}\\
-\triangle P_n&=\frac{n_i^2}{N_D}(e^{qV_A/kT}-1)=10^{12}\\
+&\approx\frac{n_{iB}^2}{n_{iE}^2}\cdot\frac{B_E}{N_B}\cdot\frac{V_{th}}{v_s}
-&\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>
+<math>
+\frac{n_{iB}^2}{n_{iE}^2} = e^{\triangle Eg/kT}
+  </math>
+; So huge improvement
-  ------------------------------------------------------------------------------------
-)
+* <math>V_{cb,br}</math> increases as <math>V_{cb,br}\approx\frac{3}{2}</math> Eg collector
+* <math>f_{max}</math> increases
 <math>
+f_{max}^{-1}\approx f_T^{-1} = \bigg(\frac{w_B^2}{2D_n}+\frac{w_{BC}}{2v_{sat}}\bigg)+\frac{kT}{qSe}[C_{jBC}+C_{jBE}]
+</math>
+here; <math>w_B\downarrow I_C\uparrow C_J\downarrow</math> all leads to <math>f_T\uparrow</math>
+* high current effect reduces
+* <math>V_cesat</math> (??)
+  ------------------------------------------------------------------------------------
+)a)<math>
 \begin{align*}
-|I_E| & = qD_p\frac{10^{10}}{0.1\times10^{-4}} = 1.8\times10^{16}q\\
+V_a&=0\\
-|I_B| & = qD_n\frac{10^{8}}{0.2\times10^{-4}} = 1.8\times10^{14}q\\
+\therefore E&=0
-\beta + 1 &= \frac{|I_E|}{|I_B|}\approx 67\\
-&\therefore \beta= 66 \text{ (chk)}
 \end{align*}
 </math>
+<math>
- ------------------------------------------------------------------------------------
+\therefore n_p(x,t) = \frac{N}{\sqrt{4\pi D_nt}}exp\bigg(\frac{-x^2}{4D_nt}-\frac{t}{\tau_n}\bigg)+n_{p0}
-)
-A)
- <math>
-\alpha_T = \frac{0.997J_0}{0.998J_0} = 0.99
 </math>
+  At <math>x=0</math>
-B)
+\begin{figure}[H]
- <math>
+[[Image:MN2_2013_3.png|Alt text|300x300px]]
-\gamma = \frac{0.998J_0}{(0.998+0.002)J_0} = 0.998
+\noindent 2 unknowns <math>\to D_n,\tau_n</math> selecting 2 specific times and using the excess carrier can find <math>D_n</math> and <math>\tau_n</math>.
- </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>
- ------------------------------------------------------------------------------------
+At <math>x=X</math>; initially no excess carrier.
-)
+[[Image:MN2_2013_4.png|Alt text|300x300px]]
- <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}
-</math>
- ------------------------------------------------------------------------------------
+b) For applied field; decay will be much quicker.
+  ------------------------------------------------------------------------------------
 ----
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