258 CHAPTER 6. BIPOLAR JUNCTION TRANSISTORS
Using the results shown in the first part of equation 6.3.9 atxb=Wbn,we have
InBC= −eADbnbo
Lbsinh(Wbn/Lb)[
exp(
eVBE
kBT)
− 1
]
+
eADbnbo
Lbcoth(
Wbn
Lb)[
exp(
−
eVCB
kBT)
− 1
]
(6.3.13)
The hole current on the collector side is the same as for a reverse-biasedp-njunction:
IpBC=−eADcpco
Lc[
exp(
−
eVCB
kBT)
− 1
]
(6.3.14)
From the way we have defined the currents, the two current components flow along +xdirection.
If we defineICas the total current flowing from the collector into the base, we have
−IC =[
eADcpco
Lc+
eADbnbo
Lbcoth(
Wbn
Lb)] [
exp(
−
eVCB
kBT)
− 1
]
−
eADbnbo
Lbsinh(
Wbn
Lb)
[
exp(
eVBE
kBT)
− 1
]
(6.3.15)
The base current is the difference between the emitter and collector currents:(IB=IE−|IC|).
It is interesting to point out that if the base regionWbnis much smaller than the diffusion length,
the electron gradient in the base region can be simplified by using the approximations
sinh(α)=eα−e−α
2=α+α^3
3!+
α^5
5!+···
cosh(α)=eα+e−α
2=1+
α^2
2!+
α^4
4!···
For theforwardactivemodeifweignorethecurrentflowinthereverse-biasedBCJwegetIE =
−eADbnb 0
Lbcoth(
Wbn
Lb)[
exp(
eVBE
kBT)
− 1
]
−
eADepe 0
Le[
exp(
eVBE
kBT)
− 1
]
(6.3.16)
Here the first part is due to electron injection from the emitter into the base (III and IV) and the
second part is due to the hole injection from the base into the emitter (II). The collector current
is
IC=eADbnb 0
Lbsinh(
Wbn
Lb)
[
exp(
eVBE
kBT)
− 1
]
(6.3.17)
Assuming thatWbnLb, we can expand the hyperbolic functions as noted above. The base
current is the difference between the emitter and collector current. We find that
IB=eADepe 0
Le[
exp(
eVBE
kBT)
− 1
]
+
eADbnb 0 Wbn
2 L^2 b[
exp(
eVBE
kBT