SEMICONDUCTOR DEVICE PHYSICS AND DESIGN

178 CHAPTER 4. JUNCTIONS IN SEMICONDUCTORS:P-NDIODES

4.7 Avalanche Breakdown in ap-njunction ......................

Consider ap−i−njunction where the applied voltage is such that the electric field on the
intrinsic region which is a constant is assumed to be large enough to saturate the electron and
hole velocities. We assume in our analysis thatvse=vsh=vs. As is shown schematically
in figure 4.19, a few lucky electrons (minority carriers) injected from thep-side into the high
field region can gain enough kinetic energy (>Eg) to collide with the lattice creating electron-
hole pairs. This process is called impact ionization. These electrons and holes accelerate again
leading to more collisions and further generation. The same applies to holes injected from the
n−side. To analyze the resultant current due to impact ionization one solves the continuity
equation for electrons and holes:

∂n

∂t

=

1

e

∂Jn

∂x

+Gn(e)+Gh(e) (4.7.1)

whereGn(e)=the rate of generation of secondary electrons by accelerated electrons=αnn(x)vs,
whereα(cm−^1 ), the ionization coefficient of electrons, is the number of electron-hole pairs
generated per electron per cm, andn(x)(cm−^3 )is the local concentration of electrons, andvs
(cm/s) is the saturated electron velocity. The ionization coefficient is much less than 1 because
only lucky electrons create electron-hole pairs and the above equation assumes that all elec-
trons are participating in the process.Gh(e)is the rate of generation of secondary electrons by
accelerated holes:
Gh(e)=αp·p(x)·vs (4.7.2)

whereαpis the ionization coefficient for holes.p(x)

(

cm−^3

)

= local concentration of holes.vs
is the saturated hole velocity. Assumingαn=αp=αand thatα=f(E), the latter being a
good approximation in ap−i−nstructure,

∂n

∂t

=

1

e

∂n

∂x

+α(n(x)vs+p(x)vs)=0 (4.7.3)

in steady state which we now consider. The impact ionization process causes the electron current
to increase from its reverse saturation value,

Jn 0 =−e

Dn

Ln

np 0 (4.7.4)

to a larger value at thep-side. The same is true for holes as shown in figure 4.20. To solve the
steady state continuity equation we also note that

Jn(x)=−en(x)vs (4.7.5)

and
Jp(x)=ep(x)(−vs)=−ep(x)vs (4.7.6)

which gives
∂Jn
∂x

−α(envs+epvs)=0 (4.7.7)