SEMICONDUCTOR DEVICE PHYSICS AND DESIGN

7.2. MODULATION AND SWITCHING OF AP-NDIODE: AC RESPONSE 311

Δ ̃p(0)can be calculated by noting that the total injected charge atx=0,Δptot(0),isthesum

of the dc and ac components

(

Δpdc(0) =pn 0 exp

[

eVdc

kBT

])

.

Δptot(0) = Δpdc(0) +Δ ̃p(0) =pn 0 exp

(

e(Vdc+vs)

kBT

)

= pn 0 exp

(

eVdc

kBT

)

exp

(

evs

kBT

)

=Δpdc(0) exp

(

evs

kBT

)

(7.2.26)

Since we are assumingvsto be small,exp

(

evs

kBT

)

1+kevBsT. Inserting this into equation

7.2.25 and solving forΔ ̃p(0)gives us

Δptot(0) =pn 0 e

eVdc

kBT

(

1+

evs

kBT

)

(7.2.27)

Δ ̃p(0) = Δpdc(0)· evs

kBT

(7.2.28)

Notice that the time-independent part of the ac injected charge in equation 7.2.25 has the same
form as the dc injected charge, only with a complex diffusion length,λ. It is interesting to note
thatλis frequency dependent; whenω=0,λ=Lp,andasωincreases,λdecreases, since the
injected charge can be reduced via reclamation through the junction during the negative swing
of the ac voltage.
Using the results of equation 7.2.25 and equation 7.2.28, we can find the currentisthat results
from our applied voltage signalvs. The total currentisis given by the diffusion current atx=0.

is = −eADp

dΔ ̃p(x)

dx

∣

∣∣

∣∣

x=0

=eADpΔ ̃p(0)

√

jω

Dp

+

1

Dpτp

(7.2.29)

=

eaDp

Lp

Δpdc(0)·

evs

kBT

√

1+jωτp (7.2.30)

=

eIvs

kBT

√

1+jωτp (7.2.31)

From this equation, we calculate the small signal admittance

y=

is

vs

=

eI

kBT

√

1+jωτp (7.2.32)

The admittance, as well as the small signal parametersCdif fandGs, take different forms at
low frequencies (ωτp< 1 ) and at high frequencies (ωτp> 1 ). At low frequency, the admittance
is

y

eI

kBT

[

1+

jωτp

2

]

(7.2.33)