ECE 440 Lecture 11 : Overview – Where we are and what we know

ECE 440Lecture 11 : Overview –

Where we are and what we know

Class Outline:

• Review of Material Covered So Far• Solve Example Problems

• Where do energy bands come from?

• What is thermal equilibrium?• What is extrinsic material?• What is the density of states?• What is the Fermi level?• How do I find carrier

concentrations?M. J. Gilbert ECE 440 – Lecture 1 1 2/1 2/1 0

Things you should already know…

Key Questions

M. J. Gilbert ECE 440 – Lecture 1 1 2/1 2/1 0

Energy BandsBy putting the atoms together, we get an energy gap

Eg

Energy

Ec

Ev

Distance (x)


Energy BandsLet’s use the band gap to classify these different classes of materials…

Eg

WIDE

Insulator Semiconductor Metal

Eg

Eg

Ev

Ec

Ec

Ev

Ev

Ev

EcEc


Thermal EquilibriumPutting temperature and band gap together to obtain carriers…

T = 0 K

T = 300 K

T = 300 KEHP


Thermal EquilibriumThe beginnings of conduction…

• Why doesn’t a semiconductor conduct at T = 0 K?

k

E

This is like a bottle filled most of the way with water!


Thermal EquilibriumRemember intrinsic material in thermal equilibrium…

Generation Rate:

Recombination Rate:

•N – number of electrons•P – number of holes


Effective MassAt the end of lecture 4, we talked about effective mass…

Electric Field

Electric Field


Effective MassDetermining the “effective” mass…

Silicon

GaAs


Effective MassWhich effective mass do I need?

kx

kz

ky

kx

ky

kz

ml

mt

mt

----- -- ----

• Use the “density of states” effective mass for calculating:• Density of states• Equilibrium carrier concentrations

•Use the “conductivity” effective mass for calculating•Binding energies•Electronic structure

ml = 0.98 m0mt = 0.19 m0


Effective Mass“density of states” vs. “conductivity” effective masses…Examine the different effective masses for silicon:

kx

kz

ky

kx

ky

kz

ml

mt

mt

ml = 0.98 m0mt = 0.19 m0Density of States:

Conductivity:


Extrinsic MaterialsHow does a donor work?

Silicon (Si) 4 valence electrons

Phosphorous (P) 5 valence electrons


Extrinsic MaterialsHow does an acceptor work?

Silicon (Si) 4 valence electrons

Boron (B) 3 valence electrons

Si

B


Extrinsic MaterialsHow tightly bound is the extra electron or hole?

r

Donor Acceptor

e-

h+

Donor in Si P As SbBinding energy (eV) 0.045 0.054 0.039

Acceptor in Si B Al Ga InBinding energy (eV) 0.045 0.067 0.072 0.16


Extrinsic MaterialsCalculate the donor and acceptor levels in Si and GaAs…

Silicon

GaAs

Shallow energy levels are in general given by:

The acceptor problem is much more complicated due to the complex valence band structure. Use the heavy-hole effective mass:


Extrinsic MaterialsVisualizing donors on the band diagram…

Ed

EaEv

Ec

Ev

Ec

Δx

Let’s take a look at Silicon with Phosphorus impurity atoms:

Ed

Ev

Ec

Eg = 1.12 eV

0.045 eV


Extrinsic MaterialsRevisiting the effect of temperature…

T = 0 K T = 50 K T = 300 K


Extrinsic MaterialsHow do we add electrons or holes to a compound semiconductor?

• In silicon or germanium, we just added impurity atoms where we wanted them.

• Nevertheless, when the electrons or holes are released, they leave behind charged impurities– Leads to increased

scattering.– Can spoil device

performance• Instead, use modulation

doping.

GaAs

GaAs SUBSTRATE

GaAs/AlGaAs SUPERLATTICE


Extrinsic MaterialsModulation doping of compound semiconductors…

n-AlGaAs UnderdopedGaAs

+ + + + +- - - - -

+ + +- - -

++--


Density of StatesWe are counting states…

kykx

kz

K

K + dK

Δk

The total number of states in a small region of volume Δk can be written as (states / volume):

Consider free electrons, the volume of the sphere between the two spheres:

The separation dK corresponds to an energy difference (for free electrons):

Above:•Constant energy surfaces in 3D k-space.•Δk is the volume of the sphere contained between the two spheres of radius k and k +dk.


Density of StatesBut it is in k-space…The number of states in the shell is given in terms of the density of states in energy by :

We now equate the two expressions for the total number of states:

Plug in what we know about how the free electron model changes in k as we change E then integrate to get the final result:

Ener

gy

Ec

Ev

Available Valence Band States

Available Conduction Band States

Eg


Density of StatesThree things to know about the density of states (DOS)…

1. DOS is like seating at a stadium with the number of seats a given distance from the field being equivalent to the number of states available in the range of energy E to E + dE.

2. In general, the number of states available in the conduction and valence band will not be equal.

• Differences in effective masses.

3. Considering closely spaced energies E and E + dE in the respective bands, we can state:

• gc(E)dE is the number of conduction band states lying in the energy range E to E + dE (if E > Ec ).

• gv(E)dE is the number of valence band states lying in the energy range E to E + dE (if E < Ev ).

Ener

gy

Ec

Ev

Eg


Density of StatesWhat about 2D?

kx

ky

k

k + dkΔk

The total number of states in a small region of volume Δk can be written as (states / volume):


Fermi Level Fine, but how many states are ACTUALLY filled?• The Fermi function tells us the probability of how many of the existing states

at any given energy will be filled with an electron.• It was derived based on three assumptions

– Each allowed state has a maximum of one electron (Pauli Exclusion Principle).– The probability of each allowed quantum state is the same.– All electrons are indistinguishable.

• When is it valid?– Under equilibrium conditions.– In ANY material (semiconductor, metal, insulator, etc…)

f(E)

1

0Ef E

Mathematically speaking the end result is a probability distribution function:

½

T = 0 K

Fermi Energy


Fermi LevelWhat about the temperature dependence?

Ef + 3kbT

Ef - 3kbT

T = 300 KEf = 0.5 eV


Fermi LevelVisualizing the Fermi level in intrinsic material…

• In intrinsic material:– Concentration of holes in valence band is equal

to the number of electrons in the conduction band.

– Electron probability tail, f(E), is symmetric with hole probability tail, 1 - f(E).

– No states in Eg despite non-zero occupation probability.

– Fermi level (Ef) lies in the middle of the energy gap.

Ec

Ev

Ef

Ener

gy

Ec

Ev

EgEf

Ener

gy

Ec

Ev

Eg

Carrier Distributions:


Fermi LevelVisualizing the Fermi level in n-type material…

• In n-type material:– The Fermi level now lies closer to

the conduction band.– There are many more electrons that

there are holes.– The difference between Ef and Ev

provides a measure of the strength of the doping.

Ev

Ef

Ener

gy

Ec

Ev

Eg

Ef

Ec

Ener

gy

Ec

Ev

Eg



Fermi LevelVisualizing the Fermi level in p-type material…

• In p-type material:– The Fermi level now lies

closer to the valence band.– There are many more holes

that there are electrons.

Ev

Ef

Ener

gy

Ec

Ev

Eg Ef

Ec

Ener

gy

Ec

Ev

Eg



Equilibrium Carrier DistributionsLet’s put together our knowledge of the density of states and the Fermi function…

We want to know how many carriers we have in our semiconductor for ANY energy or temperature.

For electrons

For holes

Focus on the electron relation.

Plug in the relationships for the density of states and the Fermi function:

Make some substitutions and bring the constants out:


Equilibrium Carrier DistributionsThat integral is awful! Isn’t there an easier way?

Fermi-Dirac integral of order ½.

Fermi-Dirac integral of order ½.Nc Nv

Effective conduction band density of states.

Effective valence band density of states.

23

0

*,

319

,11051.2

×=

mm

cmN pn

vc

3kbT

3kbT

Non-degenerate Semiconductor

Degenerate Semiconductor

Degenerate Semiconductor

Non-degenerate Semiconductor


Equilibrium Carrier ConcentrationsIs this the best and easiest??

In two regions of temperature and doping concentration the Fermi-Dirac integral is easily performed:

Low temperature, highly degenerate: ( ) ( ) 3

2

23

2/1 32

32

−==

TkEE

Fb

cfcc ηη

High temperature, non-degenerate: ( )

( )TkEE

cb

cf

eF−

=η2/1

But we can use curve fitting (no physics) to cover all of the ranges:

+−=

++=

vvbv

ccbcf N

pNpTkE

Nn

NnTkEE

81ln

81ln

Joyce-Dixon Approximation:


Equilibrium Carrier ConcentrationsLet’s compare Boltzmann with Joyce-Dixon:

A sample of GaAs has a free electron density of 1017 cm-3, where is the Fermi level at 300 K?

Boltzmann Approximation:



Equilibrium Carrier ConcentrationsLet’s compare Boltzmann with Joyce-Dixon:

The Fermi level in Si coincides with the conduction band edge at 300 K, calculate the carrier concentration:

Boltzmann Approximation:


Exact Answer:


Equilibrium Carrier ConcentrationsAssume that the closer than 3 kbT to the bands to be degenerate, what free electron concentration does this correspond to in Si and GaAs?

Use Joyce-Dixon:

We know that Ef – Ec = 3 kbT:

Using the exact Fermi-Dirac integral:


Equilibrium Carrier ConcentrationsThese are simple closed form expressions, but are they the simplest?

• There are other forms which you may find more useful in homeworks or tests.

• We can find the carrier concentrations relative to Ei , the Fermi level for an intrinsic semiconductor.

For intrinsic semiconductors, we know the following: n = p = ni and Ei = Ef

Then the relations for n and p become:

Solve for Nc and Nv

By combining equations, we arrive at two very important and useful results:

Solve for Ni


Equilibrium Carrier ConcentrationsRemember the intrinsic concentrations…

Siliconni ~ 1010 cm-3

Germaniumni ~ 2 x 1013 cm-3

GaAsni ~ 2 x 106 cm-3


Extrinsic Carrier ConcentrationsThe preceding analysis leads to the charge neutrality relationship…

The total charge inside any region of semiconductor must be zero:•Otherwise electric fields are present which break equilibrium.•Charge motion and currents would be present.

If there is sufficient thermal energy in the system, then we can simplify things further:

AA

DD

NNNN

=

=−

+


Extrinsic Carrier ConcentrationsLet’s use the charge neutrality relation…

Here, we make two critical assumptions in our analysis of extrinsic carrier concentrations…

•We are dealing with non-degenerate semiconductors.•All of the dopants are ionized.

Start with the np product expression:

Charge neutrality relationship

Solve quadratic equation


Extrinsic Carrier ConcentrationsWhere is the intrinsic level, Ei, really located?

In Lecture 8, we saw that the intrinsic Fermi level was located at mid-gap. Does our current knowledge change anything?

In intrinsic material, we know that n=p. Now use the Boltzmann equations:

TkEE

vTkEE

cb

if

b

fi

eNeN−−

=But,

23

*

*

=

n

p

c

v

mm

NN

So,Only when the masses are equal or T = 0 K does Ei lie at midgap!


Extrinsic Carrier ConcentrationsSo where does the Fermi level lie for a doped semiconductor??

Again, we make the assumption that the semiconductor is non-degenerate, maintained in equilibrium at temperatures where all of the dopants are ionized…

Solve for Ef – Ei:

In a typical doped semiconductor, where n ~ ND or p ~ NA…

ND >> NA , ND >> ni

NA >> ND , NA >> ni


Thermal Effects RevisitedWhat is the role of temperature??Temperature defines three distinct regions of conductivity…

ECE 440 Lecture 11 : Overview – Where we are and what we know

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Transcript of ECE 440 Lecture 11 : Overview – Where we are and what we know