SEMICONDUCTOR DEVICE PHYSICS AND DESIGN

7.2. MODULATION AND SWITCHING OF AP-NDIODE: AC RESPONSE 309

appear that the diffusion capacitance in figure 7.3b is given by equation 7.2.9. However, this
is not the case, since when an ac signal is applied,notalloftheinjectedminoritychargeis
reclaimedthroughthejunction. Some of the charge simply recombines in the neutral region. In
the forward-bias condition, the diffusion capacitance will dominate and we have the following
relation between the currentisand the applied voltage signalvs:

is=Gsvs+Cdif f

dvs

dt

(7.2.12)

If we assume an input voltage with frequencyω(vs∼vsoexp (jωt)), we get

is=Gsvs+jωCdif fvs (7.2.13)

and the admittance of the diode becomes

y=

is

vs

=Gs+jωCdif f (7.2.14)

To findGsandCdif fin equation 7.2.14. it is necessary to calculate the admittanceyby
solving the time-dependent continuity equation, and thenGsandCdif fcan be extracted. We
first solve the continuity equation to find the ac part of the injected charge distribution when a
biasV(t)=Vdc+vosexp (jωt)is applied to the diode. From this we can determine the ac
part of the current and thus calculate the admittance. The general form for the time-dependent
continuity equation is
∂p
∂t

=−

1

e

∇·J+G−R (7.2.15)

We assume here that we have a wide base diode(base>>(Dpτp)^1 /^2 ), and we are applying a
voltageV(t)=Vdc+vosexp (jωt). The continuity equation then takes the form

eA

∂(Δp)

∂t

=−eA

Δp

τp

−

∂Ip

∂x

(7.2.16)

Assuming our currentIpis purely diffusion(Ip=−eADpd(Δdxp)), the continuity equation be-
comes
∂(Δp)
∂t

=−

Δp

τp

+Dp

∂^2 (Δp)

∂x^2

(7.2.17)

Under dc bias, the left hand side of this equation is zero, and our solution was given by

(Δp)dc=Δp(0)e−x/Lp (7.2.18)

wherex=0is at our depletion region edge andLp=(Dpτp)^1 /^2. When an ac signal is added,
we assume a solution of the form

Δp=(Δp)dc+(Δp)ac=Δp(0)e−x/Lp+Δ ̃p(x)ejωt (7.2.19)