SEMICONDUCTOR DEVICE PHYSICS AND DESIGN

334 CHAPTER 7. TEMPORAL RESPONSE OF DIODES AND BIPOLAR TRANSISTORS

We assume solutions of the form

n(x, t)=ndc(x)+nω(x)ejωt (7.5.31)

wherendc(x)is the dc component of the current calculated above andnω(x, t)=nω(x)ejωt.If
we insert the ac part of equation 7.5.30 back into equation 7.5.29, the result can be written in the
form
d^2 nω(x)
dx^2

=

nω(x)

λ^2 e

(7.5.32)

whereλeis thefrequencydependent diffusion length and is given by

1

λe

=

√

jω

Dn

=(1+j)

√

ω

2 Dn

(7.5.33)

Assumingnω(wB)=0(Shockley boundary conditions), the solution fornω(x)in equation
7.5.31 is given as

nω(x)=nω(0)

sinh [(wB−x)/λe]

sinh [wB/λe]

(7.5.34)

wherenω(0)is the amplitude of the the ac portion of the electron concentration atx=0.
The value ofnω(0)obviously depends on the magnitude of the ac voltage, which we have
calledvω. Again assuming Shockley boundary conditions,n(0,t)can be written as

n(0,t)=np 0 exp

[

eVBE(t)

kBT

]

=np 0 exp

(

eVdc

kBT

)

exp

(

evωejωt

kBT

)

(7.5.35)

Assumingvωis small, we can linearize the second exponential in this equation, such that

exp

(

evωejωt

kBT

)

1+

evω

kBT

ejωt (7.5.36)

We may then write equation 7.5.35 as

n(0,t)=ndc(0) +nω(0)ejωt (7.5.37)

where

ndc(0) =np 0 exp

(

eVdc

kBT

)

(7.5.38)

and

nω(0) =np 0 exp

(

eVdc

kBT

)

·

evω

kBT

=ndc(0)

evω

kBT

(7.5.39)

Now that we know our ac charge distributionnω(x), we can calculate the amplitudes of the
ac currentsiω(0)andiω(wB).Ifweinsertnω(x)from equation 7.5.34 into equation 7.5.29 and
evaluate the derivative atx=0,weget

iω(0) =−

eAEDnnω(0)

λe

coth

[

wB

λe

]

(7.5.40)