Latest revision as of 21:22, 5 August 2017

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

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 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 $

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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