2/19/14 Things you should know when you leave…
ECE 340
Lecture 13 : Optical
Absorption and Luminescence
Class Outline:
• Band Bending
• Optical Absorption
Key Questions
•  How do I calculate kinetic and
potential energy from the
bands?
•  What is direct recombination?
•  How do I describe the process
of direct recombination?
M.J. Gilbert
ECE 340 – Lecture 13
Band Bending
Band Bending
The energy bands are not constant in fields!
We can use elementary physics to determine the potential energy…
Electron Kinetic Energy (E – Ec)
Total
Electron
Energy
T.E. = K.E. + P.E.
P.E.
Potential Energy = Ec - Eref
Hole Kinetic Energy (Ev – E)
Instead, they move as a function of position.
x
Eref
•  We need energy equal to the band gap to break
bonds and excite carriers to the conduction band.
•  If we only impart enough energy to promote an
electron then it simply sits in the conduction band.
•  Extra energy allows carriers to move.
M.J. Gilbert
ECE 340 – Lecture 13
Electron Kinetic Energy (E – Ec)
Total
Electron
Energy
T.E. = K.E. + P.E.
Potential Energy = Ec - Eref
Hole Kinetic Energy (Ev – E)
P.E.
The potential energy:
− qV
But previously we said:
Ec − Eref
x
Eref
V =−
1
(Ec − Eref )
q
V
Electrostatic potential
x
M.J. Gilbert
ECE 340 – Lecture 13
1 2/19/14 Band Bending
Band Bending
But we can still determine more…
Total
Electron
Energy
Electron Kinetic Energy (E – Ec)
T.E. = K.E. + P.E.
Potential Energy = Ec - Eref
Hole Kinetic Energy (Ev – E)
P.E.
By definition, we know:
V
E = −∇V
1 dEc 1 dEv 1 dEi
=
=
=
q dx q dx q dx
E
0
-L
We know that : Ef – Ei = Eg/4 = 0.28 eV
From the bands, we know that in this region it is ptype.
E −E
F
cm 2
µ p = 459
V − sec
k bT
0.28
10 0.0259
3
-L
0
L
0
L
V
1
(Ec − Eref )
q
x
-L
ECE 340 – Lecture 13
Finally, let’s determine the kinetic and potential energies at the different
3
2
points…
Ec
E
0
L
For electrons:
• Total energy increases as we move up
in a band.
• KE = E – Ec.
• PE = Ec – Eref = Ec – Ef.
For holes:
• Total energy increases as we move up
in a band.
• KE = Ev – E.
Carrier
• PE = Eref – Ev = Ef – Ev.
Electron 1
= 10 e
M.J. Gilbert
2
x
Band Bending
x
What is the resistivity in the x > L portion of the
semiconductor?
= 4.96 × 1014 cm −3
1
Let’s sketch the electrostatic
potential inside the semiconductor:
M.J. Gilbert
There is still more information that we can determine…
E = −∇V
We can find the
1 dEc 1 dEv 1 dEi
electric field…
=
=
=
q dx q dx q dx
N A = p = ni e
Ev
• Choose the Fermi level as the
reference energy.
x
ECE 340 – Lecture 13
Band Bending
i
1
Ei
Ef
• It is silicon maintained at 300 K.
V =−
x
M.J. Gilbert
Assume that:
3
2
Ec
Consider the following band diagram:
• Ef – Ei = Eg/4 at ± L and Ef – Ei =
Eg/4 at x = 0.
x
Eref
Electrostatic potential
Let’s try an example…
ρ=
1
= 27.5Ω − cm
qµ p N A
ECE 340 – Lecture 13
M.J. Gilbert
1
Ei
Ef
Ev
1
2
3
KE (eV)
PE (eV)
0
0.28
Electron 2
0.56
0.28
Electron 3
0
0.84
Hole 1
0
0.28
Hole 2
0.56
0.28
Hole 3
0
0.84
ECE 340 – Lecture 13
2 2/19/14 Optical Absorption
Optical Absorption
We can use optical absorption to determine the band gap…
Who cares what the
band gap is anyway?
Recall the two different bandstructures…
Direct Gap
Indirect Gap
• Many materials
properties are very
sensitive to the value of
the band gap (carrier
concentration).
• Knowledge of the band
gap allows us the
possibility of designing
devices which use
transitions across the
band gap to generate
photons of specific
wavelengths.
• Can we use the opposite
process to measure the
gap?
M.J. Gilbert
ECE 340 – Lecture 13
M.J. Gilbert
ECE 340 – Lecture 13
Optical Absorption
Optical Absorption
How can we use optical absorption to measure the band gap?
If a beam of photons hits a semiconductor, there should be some
predictable amount of absorption…
Life of an excess carrier…
This should depend on materials
parameters:
• Thickness
• Energy of photons
(a)  Absorb a photon to create an
electron-hole pair.
(b) The electron loses its excess
energy back to the lattice.
(c)  The electron recombines with
the hole in the valence band
giving off a photon.
Monochromator
•
ħν > EG can be absorbed in the semiconductor.
•
Why are some materials transparent to photons at certain energies?
–  Excites electron from the full valence band to the relatively empty conduction
band.
–  May initially have excess energy but will eventually thermalize to the thermal
equilibrium velocity.
–  Eventually will recombine.
–  If the band gap is too large, then no photons are absorbed and no electrons are
promoted from the valence band to the conduction band.
–  Photons not absorbed simply pass through the material.
M.J. Gilbert
ECE 340 – Lecture 13
I0
IT
Detector
• A beam of photons (photons/cm2*s) is directed at the sample of thickness, l,
where the beam only contains photons of wavelength, λ.
• The intensity of the transmitted beam can be calculated by considering the
absorption in any increment dx.
• Since the photon has no memory of how far it has travelled, the probability of
absorption is constant.
−
dI (x )
= αI ( x )
dx
M.J. Gilbert
Degradation of intensity is proportional
to the intensity remaining at x.
ECE 340 – Lecture 13
3 2/19/14 Optical Absorption
Optical Absorption
This equation is easy to solve…
I0
IT
Monochromator
Detector
I ( x ) = I 0 e −αx
Optical processes are not only important for characterization but also for
device technology. Let’s begin to think about some simple optical processes
in semiconductors…
We could use light to excite and EHP in a direct gap material (fluorescence):
• Light would need energy of at least Eg.
• Electrons lose energy and move to conduction band edge.
• Electrons recombine giving off a photon.
• Process is very quick.
• 1 photon absorbed = 1 photon emitted.
•  α is the absorption
coefficient and has
units of cm-1.
•  There is negligible
absorption at long
wavelengths (hν small)
and considerable
absorption of photons
with energies greater
than EG.
M.J. Gilbert
Process can be slower…
(phosphorescence)
When light is turned off the
luminescesce can last for
seconds or even minutes!
ECE 340 – Lecture 13
M.J. Gilbert
ECE 340 – Lecture 13
Direct Recombination
Direct Recombination
Depending on the bandstructure, electrons may recombine directly or
indirectly…
This equation is hard to solve, let’s simplify…
• Direct recombination occurs spontaneously – constant in time.
• Decay of excess carriers should be exponential.
• Decay of electrons is proportional to the number of electrons and holes remaining
at time t and a proportionality constant for recombination, αr.
But we want the net rate of change in the conduction band electron
concentration…
dn t
= α r ni2 − α r n t p t
Thermal generation
Recombination rate
dt
rate
()
g i = α r n0 p0 = α r ni2
Let’s make some assumptions:
1.  Excess e-h pairs created by
short flash of light.
2.  Initial electron and hole
concentrations are equal.
3.  They recombine at an equal
rate. (δn = δp)
M.J. Gilbert
() ()
Rewrite this in terms of equilibrium n and p and
excess n and p…
dδn(t )
= α r ni2 − α r [n0 + δn(t )][ p0 + δp(t )]
dt
= −α r (n0 + p0 )δn(t ) + δn 2 (t )
[
ECE 340 – Lecture 13
]
• Assume low level injection and
a small amount of excess
carriers.
• Assume the material is
extrinsic (p-type).
Now we can easily solve this equation…
dδn(t )
= α r ni2 − α r [n0 + δn(t )][ p0 + δp(t )]
dt
= −α r (n0 + p0 )δn(t ) + δn 2 (t )
[
δn(t ) = Δne
]
−α r p0t
= Δne
−
t
τn
Where we define the recombination lifetime, or minority carrier lifetime…
τn =
1
α n p0
1
τp =
α n n0
or
τn =
M.J. Gilbert
1
α n (n0 + p0 )
• Excess majority carriers decay at
same rate as minority carriers.
• There is a large change in the
minority carrier concentration, but a
small one in the majority carrier
concentration.
ECE 340 – Lecture 13
4 2/19/14 Direct Recombination
Let’s visualize this decay of carriers…
•  This sample is doped with
1015 cm-3 acceptor atoms.
•  The intrinsic carrier
concentration is about 106
cm-3 in GaAs.
•  Minority carrier
concentration is ~ 103 cm-3.
•  1014 cm-3 EHP are added to
the system.
•  Figure shows the decay
lifetime of the excess
populations for a carrier
recombination lifetime of 10
ns.
M.J. Gilbert
ECE 340 – Lecture 13
5