A) \[{{l}_{3}}<{{l}_{2}}<{{l}_{1}}\]
B) \[{{l}_{1}}={{l}_{2}}={{l}_{3}}\]
C) \[{{l}_{3}}<{{l}_{1}}<{{l}_{2}}\]
D) \[{{l}_{3}}<{{l}_{1}}<{{l}_{2}}\]
Correct Answer: C
Solution :
Ebers-Moll equation for transistor parameter. \[{{I}_{E}}={{I}_{pn}}(o)+{{I}_{np}}(o)\] \[{{a}_{11}}=Ae\left( \frac{{{D}_{p}}{{P}_{no}}}{W}+\frac{{{D}_{n}}n{{E}_{o}}}{{{L}_{E}}} \right)\] \[{{a}_{12}}=-\frac{Ae{{D}_{P}}{{P}_{no}}}{W}={{a}_{21}}\] \[{{a}_{22}}=Ae\left( \frac{{{D}_{P}}{{P}_{no}}}{W}+\frac{{{D}_{n}}{{n}_{Co}}}{{{L}_{C}}} \right)\] Here \[{{D}_{p}}({{D}_{n}})=\] diffusion constant for hole (e) m /sec. W = base width \[{{L}_{E}},{{L}_{C}},{{L}_{B}}=\] Length of minority carries in E,C and B. Here\[{{L}_{B}}<{{L}_{E}}<{{L}_{C}}\] \[{{l}_{3}}>{{l}_{1}}>{{l}_{2}}\] Collector - is in 1-5 mm. Base - in \[\mu m\]. Emitter - 1 to 2 mm.
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