336 CHAPTER 7. TEMPORAL RESPONSE OF DIODES AND BIPOLAR TRANSISTORS
ΔJc
ΔJc=eΔn(x)Δx/τ
τ= Wc/vs
ΔQc
ΔQc
vs
Δn(x)
Wc
Δ x
x
BASE
Figure 7.15: The induced charges and collector current versus time for a sheet of charge traveling
at a constant velocityvs.
Now, to calculate the flux leaving the baseiω(wB),weinsertnω(x)from equation 7.5.34
into equation 7.5.29 and evaluate the derivative atx=wB. If we Taylor expand the result and
neglect higher order terms, we can expressiω(wB)as
iω(wB)=−
1
re
(
1 −jω
τB
3
)
vω (7.5.50)
Note that the reactive part ofiω(wB)corresponds to anegative capacitance. This behavior
results from the fact that any rise in current at the emitter end of the base appears at the collector
end with a delay of one transit timeτB, as can be seen by examination of equation 7.5.49 and
equation 7.5.50.
Now all that remains is to calculate the collector currentiC. We showed in our time delay
analysis (see figure 7.13) that electrons which are injected into the base-collector depletion re-
gion have a finite velocity and thus require a finite amount of time to transit this region. As the
electrons travel through the depletion region, they induce image charges at the depletion edges,
as illustrated in figure 7.13a. The collector current is equal to the time rate of change of the
induced charge at the collector end of the depletion region.
7.5.2 Attenuation and Phase Shift of a Traveling Electron Wave .......
To analyze the delay introduced due to velocity saturation in the collector, we first derive the
current,ΔJcinduced by a sheet of charge of areal densityΔn(x)·Δxtraveling with a velocity
vsand a distancexfrom the edge of the base as shown in figure 7.15.