6.6. SECONDARY EFFECTS IN REAL DEVICES 289
To solve this problem we need to calculate the neutral base width in the device. Also note
that since the emitter thickness is small compared to the carrier diffusion length in the
emitter, we will use the narrow diode theory to calculate the emitter efficiency..
Using the parameters given, the built-in voltage in the BCJ is
Vbi=
kBT
e
ln
(
1017. 1016
2. 25 × 1020
)
=0.757 V
The depletion width on the base side of the BCJ is found to be
δW(5.0V)=8. 296 × 10 −^6 cm
and
δW(6.0V)=8. 981 × 10 −^6 cm
Thus the neutral base width is
Wbn(5.0V)=9. 17 × 10 −^5 cm
The emitter efficiency is (for a narrow emitter of widthWe)
γe=1−
pe 0 DeWbn
nb 0 DbWe
=0. 969
We find that the base transport factor is
B=1−
Wbn^2
2 L^2 b
=0. 996
This gives
α=γeB=0. 9656
and the current gain is
β=
α
1 −α
=28
The collector current is
IC=
eADbnb 0
Wbn
[
exp
(
eVBE
kBT
)
− 1
]
−
eADbnb 0 Wbn
2 L^2 b
[
exp
(
eVBE
kBT
)
− 1
]
with the second part being negligible.
We find that
IC(5.0V)=23.79 A
We now calculate the neutral base width when the BCJ is reverse biased at 6.0 V. This is
Wbn(6. 0 V)=9. 1 × 10 −^5 cm
This gives
IC(6. 0 V)=23.973 A
The output conductance is now
go=0.183 Ω−^1
