Modeling defect level occupation
for recombination statistics
Adam Topaz and Tim Gfroerer
Davidson College
Mark Wanlass
National Renewable Energy Lab
Supported by the American Chemical
Society – Petroleum Research Fund
A semiconductor:
Energy
Conduction Band
Defect States
Valence Band
Equilibrium Occupation in a Low
Temperature Semiconductor.
Holes
Electron Trap
Hole Trap
Electrons
Photoexcitation
Photoexcitation
Photoexcitation
Photoexcitation
Radiative Recombination.
Radiative Recombination.
Radiative Recombination.
Electron Trapping.
Electron Trapping.
Defect Related Recombination.
Defect Related Recombination.
Defect Related Recombination.
What do we measure?
 Recombination rate includes radiative
and defect-related recombination.
 Measurements were taken of radiative
efficiency vs. recombination rate.
 (radRate)/(radRate+defRate)
vs. (radRate + defRate)
 Objective: Information about the
defect-related density of states.
The Defect-Related Density of
States (DOS) Function
Conduction Band
1.2
1
Energy
0.8
Defect States
0.6
0.4
0.2
0
-0.5
Ev
Valence Band
-0.3
-0.1
Energy
0.1
0.3
0.5
Ec
Band Density Of States

bandDOS  energy
Valence
Band
Energy
Conduction
Band
Energy
Looking at the Data…
Efficiency vs. Rate
1
Efficiency
0.8
0.6
77k
120k
165k
207k
250k
290k
0.4
0.2
0
1E+19
1E+20
1E+21
Recombination rate (#/s/cm^3)
1E+22
1E+23
radB  dPdN
efficiency  rate
efficiency 
 dPdN 
rate
radB
Efficiency vs. Rate
1.2
1
Rate vs. dPdN
Efficiency
0.8
0.6
77k
120k
165k
0.4
207k
250k
290k
0.2
0
1E+19
1E+23
1E+20
1E+21
1E+22
1E+23
Rate (#/s/cm^3)
Recombination rate (#/s/cm^3)
1E+22
77k
120k
165k
207k
250k
290k
1E+21
1E+20
Calculate x-Axis
Use Rate value
for y-Axis
1E+19
1E+25
•dP = hole concentration in valence band
•dN = electron concentration in conduction band
1E+27
1E+29
dPdN ((#/cm^3)^2)
1E+31
1E+33
The simple theory…


Assumptions:
 dP = dN = n

rate  defA  n  radB  n 2
 Defect states located near the middle of the gap
 No thermal excitation into bands.
Fitting the simple theory:
 radB is given.
 Find defA to minimize logarithmic error
 error 


2
|
log(
rate
)

log(
rate
)
|

measured
theoretical
defA is the defect related recombination constant
radB is the radiative recombination constant.
Simple Theory Fit…
Rate vs. dPdN
1.00E+23
Rate (#/s/cm^3)
1.00E+22
1.00E+21
77K
120K
165K
207K
250K
290K
1.00E+20
1.00E+19
1.00E+25
1.00E+27
1.00E+29
1.00E+31
dPdN ((#/cm^3)^2)
1.00E+33
A Better Model…
 Assumptions:
 rate  defA  (dP  dDn  dN  dDp )  radB  dPdN
 defA independent of temperature (and is related to
the carrier lifetime)
 Calculations:
 Calculate Ef for a given temperature, bandgap and
defect distribution
 Calculate QEfp / QEfn for a given exN (the value of
exN is chosen to match experimental dPdN)
 Calculate occupations (dP, dN, dDp, and dDn)





dDp = trapped hole concentration
dDn = trapped electron concentration
Ef is the Fermi energy
QEFp/n is the quasi-Fermi energy for holes and electrons respectively
exN is the number of excited carriers
Calculating Ef…

The Fermi energy Ef is the energy where:


(# empty states below Ef) = (# filled states above Ef)
Red area = Blue area
Ef
Carrier states
Filled with Holes
Filled with electrons
Valence
Band
Defect
States
Energy
Conduction
Band
Calculating QEFp and QEFn…
 Find QEFp and QEFn such that:
 exN = increased occupation (red area)
Increased hole occupation
Increased electron occupation
Ef
QEFp
exN
Ef
Filled
Hole
States
Non-eqfilling
already filled states
Energy
Filled
Electron
States
QEFn
exN
Non-Eq-filling
Already Filled states
Energy
Calculating band occupations…
 dP and dN depend on QEFp and QEFn,
respectively.
QEFp
QEFn
Conduction
Band
Valence
Band
dP
Band States
dN
Band States
dN
dP
Energy
Energy
Calculating defect occupation…
 dDp and dDn depend on Ef, and QEF’s
Trapped hole occupation
QEFp
Trapped electron occupation
Ef
dDp
Ef
Electron
Traps
Energy
dDn
Hole
Traps
defect states
QEFn
defect states
non-eq-dDp
non-eq-dDn
electron traps
hole traps
Energy
Note: graph represents an arbitrary midgap defect distribution
Symmetric vs. Asymmetric defect
distribution…
 Symmetric Defect DOS:
Defect DOS
1.2E+16
Number of States
1E+16
8E+15
6E+15
4E+15
2E+15
0
-0.5
Ev
-0.4
-0.3
-0.2
-0.1
0
0.1
Energy (% of Eg -- 0 is midGap)
0.2
0.3
0.4
0.5
Ec
Symmetric Defect Fit…
Rate vs. dPdN
1.00E+23
Rate (#/s/cm^3)
1.00E+22
1.00E+21
77K
120K
165K
207K
250K
290K
1.00E+20
1.00E+19
1.00E+25
1.00E+27
1.00E+29
1.00E+31
dPdN ((#/cm^3)^2)
1.00E+33
Asymmetric defect DOS…
 Using 2 Gaussians…(fit for 2 Gaussians)
Defect DOS
1E+16
9E+15
8E+15
Number of States
7E+15
6E+15
5E+15
4E+15
3E+15
2E+15
1E+15
0
-1E+15-0.5
Ev
-0.4
-0.3
-0.2
-0.1
0
0.1
Energy (% of Eg -- 0 is midGap)
0.2
0.3
0.4
0.5
Ec
2-Gaussian Asymmetric Fit…
Rate vs. dPdN
1.00E+23
Rate (#/s/cm^3)
1.00E+22
1.00E+21
77K
120K
165K
207K
250K
290K
1.00E+20
1.00E+19
1.00E+25
1.00E+27
1.00E+29
1.00E+31
dPdN ((#/cm^3)^2)
1.00E+33
3-Gaussian Asymmetric Fit.
Defect DOS
1.2E+16
Number of States
1E+16
8E+15
6E+15
4E+15
2E+15
0
-0.5
Ev
-0.4
-0.3
-0.2
-0.1
0
0.1
Energy (% of Eg -- 0 is midGap)
0.2
0.3
0.4
0.5
Ec
3-Gaussian Asymmetric Fit…
Rate vs. dPdN
1.00E+23
Rate (#/s/cm^3)
1.00E+22
1.00E+21
77K
120K
165K
207K
250K
290K
1.00E+20
1.00E+19
1.00E+25
1.00E+27
1.00E+29
1.00E+31
dPdN ((#/cm^3)^2)
1.00E+33
Conclusion…
 Simple Theory  Defect slope is too
steep and theory does not allow for
temperature dependence!
 Temperature dependence and shallow
defect slope can be modeled using:
 An occupation model that allows for
thermal defect-to-band excitation.
 An asymmetric defect level distribution
In-depth look at the model…
 Calculating DOS(e)
 DOS(e) = ValenceBand(e) +
ConductionBand(e) + defDos(e)
 ValenceBand(e) = 0 if e > Ev, if e >= Ev
 ConductionBand(e) = 0 if e < Ec, if e <=
Ec
 defDos(e) is an arbitrary function denoting
the defect density of states. defDos(e) = 0
when e <= Ev or e >= Ec
3/ 2
3
2

*
(
2
Me
)
*
h
 wcv 
(1.6 *10 19 ) 3 / 2 *10 6
2 * (2Mh) 3 / 2 * h 3
wcc 
(1.6 *10 19 ) 3 / 2 *10 6
Fermi Function, and calculating Ef…
 Fermi Function:
FF (e, f ) 
1
1  exp(( e  f ) / kT )
 To calculate Ef, find Ef where:
Ef


Ef
 DOS ( E ) * (1  FF ( E, Ef )) * dE   DOS ( E ) * FF ( E, Ef ) * dE
Calculating QEFp/n
 QEFp denotes the point where:

exN 
 DOS ( E ) * ( FF ( E, Ef )  FF ( E , QEFp)) * dE

 QEFn denotes the point where:

exN 
 DOS ( E ) * ( FF ( E, QEFn)  FF ( E , Ef )) * dE

Calculating Occupations…
Ev

dP 
 DOS ( E ) * (1  FF ( E, QEFp) * dE



dN   DOS ( E ) * FF ( E, QEFn) * dE
Ec
Ec

dDp   DOS ( E ) * ( FF ( E, Ef )  FF ( E , QEFp)) * dE
Ev
Ec
 dDn   DOS ( E ) * ( FF ( E, QEFn)  FF ( E, Ef )) * dE
Ev
Note: see slide 7 for rate value.
Numerical Infinite Integrals…
 Need: a bijection g :   (a, b)
g ( x)  a lim g ( x)  b
 And lim
x   
x  

f ( g 1 ( x))
 Then:  f ( x)dx  
dx
1


a g ( g ( x))
b

 /2

 / 2
 Using ArcTan,  f ( x)dx   f (tan( x)) * (1  tan( x))dx

 /2
k
arctan( k )
 f ( x)dx   f (tan( x)) * (1  tan( x))dx
k
arctan( k )

 /2
 f ( x)dx   f (tan( x)) * (1  tan( x))dx