Chemistry 140 a Lecture 11 Surface, Bulk, and Depletion Region Recombination.

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Transcript of Chemistry 140 a Lecture 11 Surface, Bulk, and Depletion Region Recombination.
Chemistry 140 aLecture 11
Surface, Bulk, andDepletion RegionRecombination
Quasi Fermi Levels
For calculations, it would be convenient to assume flat QFLs within a certain region, Δx, under study. Then, the driving force for recombination would be equal everywhere.
When are QFLs flat?
When Δn or Δp is constant within Δx.
Flat QFLs
QFLs are flat in the bulk when:• Recombination at any position x is
slow compared to thermal diffusion
EF,n
EF,0
EF,p
Flat QFLs
QFLs are flat in the bulk when:• Light excitation is uniform or light excitation is
not uniform but diffusion of carriers flattens QFLs
e e e e
h+ h+ h+ h+
hν e e e e
h+ h+ h+ h+
t = 0nonuniform QFLs
t = t1
Flat QFLse e e e
h+ h+ h+ h+
hν e e e e
h+ h+ h+ h+
t = 0 t = t1
If t1, the time it takes for a uniform distribution of carriers to occur via diffusion, is less than τbr, τsr, or τdr,
then the QFLs are flat.
Diffusion Time in Si e e e e
h+ h+ h+ h+
d = 300 μm
μμs cm.
cm).(τ
DD
DDD
thicknessL
D
Lτ
τ
D L

pn
pn
eff
25938
0150
2
12
2
2
For recombination greater than about 25 μs, we can assume flat QFLs.
Depletion Region Recombination
EF,0
EF,n
EF,p EF,0
W W W’
Vapp
Let’s examine the depletion region after applying a bias Vapp:
Vapp
The new effective depletion region is W’ < x < 0.W to W’ is a quasineutral region, and there is no field there.
It is like the bulk, except EF,p changes with x.Quasi Fermi levels are flat within this new depletion region.
QFLs Not FlatFor flat QFLs in the depletion region, recombination at the surface
must be slow relative to diffusion of carriers. If surface recombination is fast relative to diffusion of carriers, QFLs will not be flat:
EF,n
EF,p
n(x) and p(x) Vary with x
EF,n
EF,p
W
ET
The recombination rate in the depletion region is not like in the bulk or on the surface.
We now need to plug in n(x) and p(x) and integrate over 0 to W.
Depletion Region RecombinationMost general form:
)p(x) (p(x)k)n(x) (n(x)k
)(x)n(x) (n(x)p(x) kk(x) NU(x)
pn
ipnT
11
2
Before, we just plugged in nb = n(x) or ns = n(x) and pb = p(x) or ps = p(x).
Now n(x) and p(x) change from 0 to W because of band bending.
We have to integrate over all n(x) in 0 < x < W:
dxU(x) UW
total 0
Depletion Region Recombination
Assumptions:
NT(x) = NT
n1 and p1 are constants with respect to x and do not change with Vbi for a given trap:
n1 = NCexp((ECET)/kT)p1 = NVexp((ETEV)/kT)
kn and kp are constant (this can fail since σ(ET) may vary due to ionized or unionized trap states, e.g. Zn2+ Zn+):
n(x) = NCexp[(EC(x)EF,n)/kT]p(x) = NVexp[(EF,pEV(x))/kT]
n(x)p(x) Not Dependent on x
If EF,n and EF,p are constant with x, then n(x)p(x) is a constant with x.
n(x)p(x) = NCNV exp[(EC(x)EF,n+EF,pEV(x))/kT] = NCNV exp[(EC(x)EV(x))/kT] exp[(EF,pEF,n)/kT]
EC(x)EV(x) = Eg(x) = Eg everywhere
n(x)p(x) = NCNV exp[Eg/kT] exp[(EF,pEF,n)/kT]n(x)p(x) = ni
2 exp[(EF,pEF,n)/kT]
(EF,pEF,n) = qVapp
n(x)p(x) = ni2 exp[qVapp/kT]
No dependence on x. Increases in –Vapp result in n(x)p(x) > ni2.
Depletion Region Recombination
Returning to U(x)…
))(())((
)1]/(exp[)(
))(())((
)]/exp[()(
11
2
11
22
pxpknxnk
kTqVnkkNxU
pxpknxnk
nkTqVnkkNxU
pn
appipnT
pn
iappipnT
We may ignore the “1” when Vapp > 0.75 V (Vapp > 3kT/q).
A harder assumption is to ignore n1 in the denominator for significant band bending. We must have either highlevel injection or large Vapp.
Determination of Utotal
Our assumptions:p1 and n1 are negligiblekn = kp = σν
Wappi
T
W
total
appiT
dxxpxn
kTqVnNdxxUU
xpxn
kTqVnNxU
0
2
0
2
)()(
]1)/[exp()()(
)()(
]1)/[exp()()(
Determination of Umax
There will be some Umax in this region (the depletion region) wheren(x) = p(x). This is where the denominator is a minimum.
)/exp()()( and )()( ,At
p(x)n(x)
p(x)n(x)
n(x)p(x)n(x)p(x)
)/exp(
n(x)p(x)
1by bottom and opMultiply t
)()(
])/exp([)(
2max
22
max
22
p(x)n(x) at x wheremax
kTqVnxpxnxpxnU
nkTqVnN
U
xpxn
nkTqVnNxUU
appi
iappiT
iappiT
Determination of Umax
2
)2/exp(
: then term, theexcludeyou If
)2/sinh(
: then, from term theincludeyou If
2
1sinhRemember
)]2/exp()2/exp([2
2
)/exp(n(x)p(x)
max
2
max
22
max
2
2
max
kTqVnNU
n
kTqVnNU
)n(n(x)p(x)n
)e(e (x)
kTqVnkTqVnN
U
kTqVn
nN
U
appiT
i
appiT
ii
xx
appiappiT
appi
iT
]/)(exp[)(
)()(
)()(
)(
maxmaxmaxkTxxqnxn
xpxn
xpxn
n(x)p(x)σνNxU
UUU
T
Analysis for x < xmax and x > xmax
EF,n
EF,p
W’ xm
1
2
0
1 2x < xm
exp(…) isnegative
n(x) < n(xm)
extra bandbending
x > xm
exp(…) ispositive
n(x) > n(xm)
V = εmx = (V/cm)*cm
]/)(cosh[)(
)]...exp()[exp(coshhat Remember t
])/)(exp[]/)((exp[)(
Since
]/)(2exp[]/)(2exp[
)(
)...2/exp( Since
]/)(exp[]/)(exp[
]/)(exp[]/)(exp[
)2/exp()(
)()(
)()(
)(
max
21
21
max
2/12/121
max
21
max
kTxxq
UxU
xx (x)
kTxxqkTxxq
UxU
...pn
kTxxqnp
kTxxqpn
σνNU
σνN
xU
kTqVnNU
kTxxqnkTxxqp
kTxxqpkTxxqn
kTqVσνnNxU
xpxn
xpxn
n(x)p(x)σνNxU
mm
mmmm
mm
mmm
mmm
m
m
TT
appiT
mmm
mmm
mmm
mmm
appiT
T
U(x) in Terms of Umax
Utotal in Terms of Umax
W
mkTq
W
total xx
dxUdxxUU
0
max
0 )](cosh[)(
For normally doped semiconductors, the maximum is strongly peaked away from W, so extend the
integral to .
max
0
max
0
0
max
2
)'cosh(
'
' and )('
2)'cosh(
'
)](cosh[
Uq
kTU
x
dxUU
dxdxxxx
x
dx
xx
dxUU
mtotal
kTqtotal
mkTq
mmkTq
mkTqtotal
m
)2/exp()(22
)(4
)2/exp()(
)(2
/)(
max
kTqVnNVV
kTWqUJ
VVq
kTqVnNkTWU
UVVq
kTWU
WVV
appiTappbi
total
appbi
appiTtotal
appbitotal
appbim
Utotal Replace exp(…) with sinh(…) if
you want to include the –ni
2 term we
neglected.
Important term. Not like thermionic emission, where it went like exp(qVapp/kT).
The e and h+ recombination isas if we lost half of the voltage
to the other carrier.
U vs. x
U
x
appbi VVq
kTW
1 …a dimensionless quantity
that is the ratio of the thermalto applied voltage.
End
Surface, Bulk, andDepletion RegionRecombination