key questions ece 340 lecture 13 : optical how do i calculate kinetic...

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ECE 340 Lecture 13 : Optical

Absorption and Luminescence

Class Outline: • Band Bending • Optical Absorption

• 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

Things you should know when you leave…

Key Questions

M.J. Gilbert ECE 340 – Lecture 13

Band Bending

The energy bands are not constant in fields!

Instead, they move as a function of position.

Total Electron Energy

x

Electron Kinetic Energy (E – Ec)

Hole Kinetic Energy (Ev – E)

•  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.

T.E. = K.E. + P.E. Potential Energy = Ec - Eref

P.E. Eref


Band Bending

We can use elementary physics to determine the potential energy…


x




P.E. Eref

The potential energy: qV−

But previously we said: refc EE −( )refc EEq

V −−=1

V

Electrostatic potential

x

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Band Bending

But we can still determine more…


x




P.E. Eref

V

Electrostatic potential

x

By definition, we know:

dxdE

qdxdE

qdxdE

q

VE

ivc 111===

−∇=

E

x

0


Band Bending

Let’s try an example…

Consider the following band diagram:

Ec

Ei

Ef

Ev

Let’s sketch the electrostatic potential inside the semiconductor:

Assume that: • It is silicon maintained at 300 K.

• Ef – Ei = Eg/4 at ± L and Ef – Ei = Eg/4 at x = 0.

• Choose the Fermi level as the reference energy.

x

-L L 0

( )refc EEq

V −−=1

x

-L L 0

V

2

1

3

1

2 3


Band Bending

There is still more information that we can determine…

dxdE

qdxdE

qdxdE

q

VE

ivc 111===

−∇=We can find the electric field…

x -L L 0

E

What is the resistivity in the x > L portion of the semiconductor?

We know that : Ef – Ei = Eg/4 = 0.28 eV

From the bands, we know that in this region it is p-type.

314

0259.028.0

10

1096.410

−

−

×=

=

==

cme

enpN TkEE

iAb

Fi

cmNq Ap

−Ω== 5.271µ

ρ

sec459

2

−=

Vcm

pµ


Band Bending

Finally, let’s determine the kinetic and potential energies at the different points… Ec

Ei

Ef

Ev

2

1

3

1

2 3

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. • PE = Eref – Ev = Ef – Ev. Carrier KE (eV) PE (eV)

Electron 1 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

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Optical Absorption

We can use optical absorption to determine the band gap…

Direct Gap

Indirect Gap

Recall the two different bandstructures…


Optical Absorption

Who cares what the band gap is anyway? • 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?


Optical Absorption

How can we use optical absorption to measure the band gap?

•  ħν > EG can be absorbed in the semiconductor. –  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.

•  Why are some materials transparent to photons at certain energies? –  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.

Life of an excess carrier… (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.


Optical Absorption

If a beam of photons hits a semiconductor, there should be some predictable amount of absorption… This should depend on materials parameters: • Thickness • Energy of photons

Monochromator Detector I0 IT

• 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.

( ) ( )xIdxxdI

α=− Degradation of intensity is proportional to the intensity remaining at x.

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Optical Absorption

This equation is easy to solve… Monochromator Detector

I0 IT

( ) xeIxI α−= 0

•  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.


Optical Absorption

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.

Process can be slower…(phosphorescence) When light is turned off the luminescesce can last for seconds or even minutes!


Direct Recombination

Depending on the bandstructure, electrons may recombine directly or indirectly… • 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.

200 irri npng αα ==

But we want the net rate of change in the conduction band electron concentration… ( ) ( ) ( )tptnn

dttdn

rir αα −= 2Thermal generation rate

Recombination rate

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)

Rewrite this in terms of equilibrium n and p and excess n and p…

( ) ( )[ ] ( )[ ]

( ) ( ) ( )[ ]tntnpn

tpptnnndttnd

r

rir

200

002

δδα

δδααδ

++−=

++−=



This equation is hard to solve, let’s simplify… ( ) ( )[ ] ( )[ ]

( ) ( ) ( )[ ]tntnpn

tpptnnndttnd

r

rir

200

002

δδα

δδααδ

++−=

++−=• 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… ( ) nr

ttp nenetn ταδ

−− Δ=Δ= 0

Where we define the recombination lifetime, or minority carrier lifetime…

0

0

1

1

n

p

np

nn

ατ

ατ

=

=

( )00

1pnn

n +=α

τor

• 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.

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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.

key questions ece 340 lecture 13 : optical how do i calculate kinetic...

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