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 = 0non-uniform 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 quasi-neutral 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(-(EC-ET)/kT)p1 = NVexp(-(ET-EV)/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,p-EV(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,p-EV(x))/kT] = NCNV exp[-(EC(x)-EV(x))/kT] exp[-(EF,p-EF,n)/kT]

EC(x)-EV(x) = Eg(x) = Eg everywhere

n(x)p(x) = NCNV exp[-Eg/kT] exp[-(EF,p-EF,n)/kT]n(x)p(x) = ni

2 exp[-(EF,p-EF,n)/kT]

-(EF,p-EF,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 high-level 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