Example 5.2 Consider a pnp bipolar transistor with emitter doping of 1018 cm-3
and base doping of 1017 cm-3. The quasi-neutral region width in
the emitter is 1 µm and 0.2 µm in the base. Use µn = 1000 cm2/V-s
and µp = 300 cm2/V-s . The minority carrier lifetime in the base is
10 ns.
Calculate the emitter efficiency, the base transport factor, and the
current gain of the transistor biased in the forward active mode.
Assume there is no recombination in the depletion region.
Solution The emitter efficiency is obtained from:
994.0
1
1
'
,
'
,
=
+
=
EEBn
BBEp
E
wND
wND
γ
The base transport factor equals:
9992.0
2
1,
2
'=−= nBn
B
TD
w
τ
α
The current gain then becomes:
5.147
1
=
−
=
α
α
β
where the transport factor, α, was calculated as the product of the
emitter efficiency and the base transport factor: 993.09992.0994.0=×== TEαγα
