Lecture 6

Semiconductor physics IV The Semiconductor in Equilibrium

Equilibrium, or thermal equilibrium

No external forces such as voltages, electric fields.

Magnetic fields, or temperature gradients are acting

on the semiconductor.

All properties of the semiconductor will be

independent of time in this case.

Goal

The concentration of electrons and holes in

the conduction and valence bands with the

Fermi-Dirac probability function and the

density of quantum states.

The properties of an intrinsic semiconductor.

The properties of an semiconductor with

impurities (dopants).

3

CHARGE CARRIERS IN SEMICONDUCTORS

Two types of charge carrier, the electron and the hole.

The current in a semiconductor is determined largely by the

number of electrons in the conduction band and the number of

holes in the valence hand.

4

The distribution (with respect to energy) of electrons in the

conduction band is

the density of quantum states

in the conduction band

the probability that a state is

occupied by an electron

The total electron concentration per unit volume in the conduction

band is

( ) ( ) ( )c Fn E g E f E

( )n n E dE

The distribution (with respect to energy) of holes in the valence

band is

the density of allowed quantum

states in the valence band

the probability that a state is not

occupied by an electron

The total hole concentration per unit volume is found by

integrating this function over the entire valance-band energy.

5

( ) ( )[1 ( )]Fvp E g E f E

( )p p E dE

Fig (b) The splitting of the 3s

and 3p states of silicon into the

allowed and forbidden energy

bands.

At T=0K, valence band is full and the

conduction band is empty in an intrinsic

semiconductor (no impurities and no

lattice damage in crystal).

The location of Fermi energy EF

v F cE E E

At T>0K, the valence electrons gain

energy and a few move to conduction band

and leave empty states.

Electrons and holes are created in pairs by

the thermal energy.

The number of electrons in the conduction

band is equal to the number of holes in the

valence band.

Fig Density of states functions, Fermi-Dirac

probability function, and areas representing

electron and hole concentrations for the case

when E, is near the midgap energy7

If we assume that the electron and

hole effective masses are equal, then

gc(E) and gv ( E ) are symmetrical

functions about the midgap energy.

The function fF(E) for E > EF is

symmetrical to 1 - fF(E) for E < EF

about the energy E = EF.

The areas representing electron and

hole concentrations are equal

* 3/2

3

4 (2 )( ) n

c c

mg E E E

h

* 3/2

3

4 (2 )( )

p

v v

mg E E E

h

1( )

1 exp( )F

F

f EE E

kT

in the middle of bandgap energyFE

The n0 and p0 Equations

The thermal-equilibrium concentration of electrons

8

The lower limit of integration is Ec

The upper limit of integration should be the top of the allowed

conduction band energy.

However, since the Fermi probability function rapidly

approaches zero with increasing energy we can take the upper

limit of integration to be infinity.

0 ( ) ( )c Fn g E f E dE

9

If (Ec- EF) >> k T , then (E - EF) >> kT

Let

* 3/2

3

4 (2 )( ) n

c cm

g E E Eh

[ ( )]1( ) exp

( )1 exp

FF

F

E Ef E

E E kT

kT

* 3/2

3

4 (2 ) ( )exp[ ]

C

Fno c

E

m E En E E dE

h kT

cE E

kT

* 3/21/2

03

0

4 (2 ) ( )exp[ ] exp( )

c Fnm E En d

h kT

1/2

0

1exp( )

2d

*

3/2

0 2

2 ( )2( ) exp[ ]

c Fnm kT E En

h kT

Boltzmann

approximation

10

The thermal-equilibrium electron concentration in the

conduction band

Nc is called the effective density states function in the

conduction band.

The thermal-equilibrium concentration of holes in the valence

band is

11 ( )

( )1 exp

FF

f EE E

kT

* 3/2

3

4 (2 )( )

p

v v

mg E E E

h

0 ( )[1 ( )]v Fp g E f E dE

*3/2

2

22( )n

c

m kTN

h

0

( )exp[ ]c F

c

E En N

kT

11

The lower limit of integration is taken as minus infinity instead

of the bottom of the valence band since the probability function

of holes approach zero when energy is minus infinite.

1 ( )1 ( ) exp[ ]

( )1 exp

FF

F

E Ef E

E E kT

kT

* 3/2

0 3

4 (2 ) ( )exp[ ]

vE

p Fv

m E Ep E E dE

h kT

( )F vIf E E kT

* 3/2 0

' 1/2 ' '

0 3

4 (2 ) ( )exp[ ] ( ) exp( )

p F vm kT E E

p dh kT

3/2*

v0 3

2 ( )2 exp

p Fm kT E E

ph kT

' ( )vE E

kT

12

The effective density of states function in the valence band is

The thermal-equilibrium concentration of holes in the valence band

The magnitude of Nv is also on the order of 1019 cm-3at T = 300 K

for most semiconductors.

0

( )exp[ ]F v

v

E Ep N

kT

3/2*

2

22

p

v

m kTN

h

13

The effective density of states functions, Nc and Nv, are constant

for a given semiconductor material at a fixed temperature.

Table

Effective density of states function and effective mass values

3/2*

2

22

p

v

m kTN

h

3/2*

2

22 n

c

m kTN

h

14

Example:Calculate the probability that a state in the conduction band is occupied by an

electron and calculate the thermal equilibrium electron concentration in silicon

at T= 100 K.

Assume the Fermi energy is 0.25 eV below the conduction band. The value of

Nc for silicon at T = 100 K is Nc = 2.8 x 1019 cm-3.

The probability that an energy state at E = Ec is occupied by an

electron is given by

The probability of a state being occupied can be quite small, but the fact that

there are a large number of states means that the electron concentration is a

reasonable value.

Solution:

5( )1 0.25( ) exp[ ] exp( ) 6.43 10

0.02591 exp( )

F

c Fc

c F

E Ef E

E E kT

kT

19 15 3

0

( ) 0.25exp[ ] (2.8 10 )exp( ) 1.8 10

0.0259

c Fc

E En N cm

kT

15

Example

Solution

The hole concentration is

Calculate the thermal equilibrium hole concentration in silicon at

T= 400 K.

Assume that the Fermi energy is 0.27 eV above the valence band

energy. The value of Nv for silicon at T = 300 K is Nv = 1.04 ×1019 cm-3.

19 19 3400(1.04 10 )( ) 1.60 10

300vN cm

19

0

15 3

( ) 0.27exp[ ] (1.60 10 )exp( )

0.03453

6.43 10

F vv

E Ep N

kT

cm

The Intrinsic Carrier ConcentrationFor an intrinsic semiconductor,

The concentration of electrons in the conduction band ni is

equal to the concentration of holes in the valence band pi .

The Fermi energy level is called the intrinsic Fermi energy, or

EF = EFi .

16

Eg is the bandgap energy

0 exp[ ]Fi ci c

E En n N

kT

0 exp[ ]v Fi

i i v

E Ep p n N

kT

2 exp[ ]exp[ ]Fi c v Fii c v

E E E En N N

kT kT

2 exp[ ] exp[ ]gv c

i c v c v

EE En N N N N

kT kT

17

Fig The intrinsic carrier concentration of Ge,

Si, and GaAs as a function of temperature.

For Eg = 1.12 eV, ni = 6.95 x 109cm-3 from

the equation for silicon at T = 300 K .

The commonly accepted value of ni for silicon

at T = 300 K is approximately 1.5 × 1010cm-3

This theoretical function does not

agree exactly with experiment.

The Intrinsic Fermi-Level Position

Since the electron and hole concentrations are equal

18

If we take the natural log of both sides of this equation

The midgap energy1

( )2

c v midgapE E E

( ) ( )exp[ ] exp[ ]c Fi Fi v

c v

E E E EN N

kT kT

*

*

1 3( ) ln( )

2 4

p

Fi c v

n

mE E E kT

m

1 1( ) ln( )

2 2

vFi c v

c

NE E E kT

N

3/2*

2

22

p

v

m kTN

h

3/2*

2

22 n

c

m kTN

h

19

The intrinsic Fermi level is

below the center of the bandgap.

exactly in the center of the bandgap

above the center of the bandgap.

*

*

3ln( )

4

p

Fi midgap

n

mE E kT

m

* *

p nm m

* *

p nm m

* *

p nm m

20

Example

To calculate the position of the intrinsic Fermi level with respect

to the center of the bandgap in silicon at T = 300 K.

The density of states effective carrier masses in silicon are

Solution

The intrinsic Fermi level with respect to the center of the

bandgap is

If we compare 12.8 meV to 560 meV, which is one-half of the

bandgap energy of silicon, we can, in many applications,

simply approximate the intrinsic Fermi level to be in the

center of the bandgap.

* *

0 01.08 0.56n pm m m m

*

*

3 3 0.56ln( ) (0.0259) ln( )

4 4 1.08

0.0128 12.8

p

Fi midgap

n

Fi midgap

mE E kT

m

E E eV meV

DOPANT ATOMS AND ENERGY LEVELSThe real power of semiconductors is realized by adding small,

controlled amounts of specific dopant, or impurity, atoms.

21

The doped semiconductor, called an extrinsic material

The phosphorus atom without the donor electron is positively

charged. At very low temperatures, the donor electron is bound

to the phosphorus atom.

adding a group V element,

such as phosphorus

22

Fig The energy-band diagram showing (a) the discrete donor energy

state and (b) the effect of a donor state being ionized.

The electron in the conduction band can now move through the

crystal generating a current, while the positively charged ion is

fixed in the crystal.

The donor impurity atoms add electrons to the conduction band

without creating holes in the valence band. The resulting

material is referred to as an n-type semiconductor.

The donor

electrons jump to

the conduction

band with thermal

energy

23

Adding a group III element, such as boron, as a substitution

impurity purity to silicon.

One covalent bonding

position appears to be

empty

Fig Valence electrons may gain a small amount of

thermal energy and move about in the crystal.

The "empty" position associated with the boron atom becomes occupied, and

other valence electron positions become vacated. These other vacated electron

positions can he thought of as holes in the semiconductor material.

24

Fig The energy-band diagram showing (a) the discrete acceptor

energy state and (b) the effect of a acceptor state being ionized.

If an electron were to occupy this "empty" position, its energy

would have to be greater than that of the valence electrons.

The acceptor atom can generate holes in the valence hand

without generating electrons in the conduction band. This type

of semiconductor material is referred to as a p-type material

Acceptor atom

gets electrons

from the

valence band

with thermal

energy.

Ionization Energy

Energy required to elevate the donor electron into the

conduction band.

Bohr theory

The most probable distance of an electron in a hydrogen atom

from the nucleus from quantum mechanics is the same as

Bohr radius.

The coulomb force of attraction between the electron and ion

equal to the centripetal force of the orbiting electron. This

condition give a steady orbit.

25

2 * 2

24 n n

e m v

r r

26

If we assume the angular momentum is also quantized, then we can

write n is a positive integer

The assumption of the angular momentum being quantized

leads to the radius being quantized

The Bohr radius is

normalize the radius of the donor orbital to that of the Bohr radius

the relative dielectric constant of the

semiconductor material

rest mass of an electron

effective mass of the electron

in the semiconductor.

Substitute *

n

nhv

m r into

2 * 2

24 n n

e m v

r r

*

nm r v n

2 2

* 2

4n

nr

m e

2

00 2

0

40.53Aa

m e

2 0

*

0

( )nr

r mn

a m

27

If we consider the lowest energy state in which n = 1 , and if we

consider silicon in which = 11.7 and the conductivity effective

mass is m*/m0 = 0.26. then we have thatr

r1 = 23.9 A. This radius corresponds to approximately four lattice

constants silicon. Recall that one unit cell in silicon effectively

contains eight atoms, so the radius of the orbiting donor

electron encompasses many silicon atoms.

=> The donor electron is not tightly bound to the donor atom.

1

0

45r

a

28

The total energy of the orbiting electron is given by

The kinetic energy is

Since

The potential energy is

The total energy is

E T V

* 21

2T m v

*

nm r v n2 2

* 2

4n

nr

m e

* 4

2 22( ) (4 )

m eT

n

2 * 4

2 24 ( ) (4 )n

e m eV

r n

* 4

2 22( ) (4 )

m eE T V

n

29

Table Impurity ionization energies

in silicon and germanium

In silicon the ionization energy is E = -25.8 meV, much less

than the bandgap energy of silicon. This energy is the

approximate ionization energy of the donor atom.

Germanium and silicon have different relative dielectric constants and

effective masses, resulting in different ionization energy.

THE EXTRINSIC SEMICONDUCTOR

A material has impurity atoms.

One type of carrier will predominate.

The Fermi energy will change as dopant atoms are added.

n type: the density of electrons is greater than the density of holes

n0>p0 majority carrier: electrons; minority carrier: holes

the Fermi energy is above the intrinsic Fermi energy

p type: the density of holes is greater than the density of electrons

n0<p0 majority carrier: holes; minority carrier: electrons

the Fermi energy is below the intrinsic Fermi energy

30

31

Fig. Density of states functions.

Fermi-Dirac probability function, and

areas representing electron and hole

concentrations for the case when EF

is above the intrinsic Fermi energy.

n type

32

Fig Density of states functions,

Fermi-Dirac probability function, and

areas representing electron and hole

concentrations for the case when EF

is below the intrinsic Fermi energy.

p type

33

The electron concentration in extrinsic semiconductor

The intrinsic carrier concentration

0 exp[ ]F Fii

E En n

kT

0

( )exp[ ]F Fi

i

E Ep n

kT

( )exp[ ]c Fi

i c

E En N

kT

0

( ) ( )exp[ ]c Fi F Fi

c

E E E En N

kT

The n0 and p0 Product

34

• The product of n0 and p0 is always a constant for a given

semiconductor material at a given temperature.

• The equation is invalid if the Boltzmann approximation is

not valid since it is derived by the Boltzmann approximation

and c F F vE E kT E E kT

2

0 0 in p n

0 0 exp[ ]g

c v

En p N N

kT

0 0

( ) ( )exp[ ]exp[ ]c F F v

c v

E E E En p N N

kT kT

The Fermi-Dirac Integral

35

If the Boltzmann approximation (E-EF>>kT) does not hold.

If we again make a change of variable and let

The integral is defined as

Fermi-Dirac integral𝐹 Τ1 2(𝜂𝐹) =

0

∞

𝜂 Τ1 2𝑑𝜂

)1 + exp(𝜂 − 𝜂𝐹

1/2* 3/2

0 3

( )4(2 )

1 exp( )c

cn

FE

E E dEn m

E Eh

kT

cE E

kT

F c

F

E E

kT

* 1/23/2

0 2

0

24 ( )

1 exp( )

p

F

m kT dn

h

36

Fig The Fermi-Dirac integral as a function

of the Fermi energy

The Fermi-Dirac integral,

is a tabulated function.

The Fermi energy is

actually in the conduction

band.

0Fif

/F C FE E kT

1

2

1/2 ( )0 1 FF

dF

e

37

the Fermi level is in the valence band.

* ' 1/2 '3/2

0 2 ' '

0

2 ( )4 ( )

1 exp( )F

pm kT dp

h

' vE E

kT

' v F

F

E E

kT

' 0Fif

38

To calculate the electron concentration using the Fermi-Dirac

integral. Let = 2 so that the Fermi energy is above the

conduction band by approximately 52 meV at T = 300 K.

Example

F

Solution

The Femi-Dirac integral has a value of F1/2 (2) = 2.3

With the Boltzmann approximation

no = 2.08 × 1020 cm-30

( )exp( )c F

c

E En N

kT

For silicon at 300K, Nc = 2.8 × 1019 cm-3

19 19 3

0

2(2.8 10 )(2.3) 7.27 10n cm

* 1/23/2

0 0 1/22

0

2 24 ( ) ( )

1 exp( )

nt F

F

m kT dn n N F

h

STATISTICS OF DONORS AND ACCEPTORS

The probability function of electrons occupying the donor state is

39

nd is the density of electrons occupying the donor level

Ed is the energy of the donor level

The factor 1/2 in this equation is a direct result of the spin factor.

Each donor level has two possible quantum states (spin

orientations). The insertion of an electron into one quantum state,

however, precludes putting an electron into the second quantum

state.

Nd is the donor concentration

11 exp( )

2

dd

d F

Nn

E E

kT

40

The density of electrons occupying the donor level is equal to

the donor concentration minus the concentration of ionized donors

The concentration of holes in the acceptor states

Na is the concentration of acceptors

Na- is the concentration of ionized acceptors

g is a degeneracy factor（简并因子） g=4 in silicon and gallium

arsenide.

11 exp( )

aa a a

F a

Np N N

E E

g kT

d ddn N N

Complete Ionization and Freeze-Out

41

The Boltzmann approximation is also valid

The ratio of electron in the donor state to the total number

of electrons in the conduction band plus donor state.

0

( )2 exp( )

( ) ( )2 exp( ) exp( )

d Fd

d

d F c ddd c

E EN

n kTE E E En n

N NkT kT

2 exp( )1

exp( )2

d d Fd d

d F

N E En N

E E kT

kT

0

( )exp( )c F

c

E En N

kT

d FIf E E kT

42

The factor (Ec - Ed) is just the ionization energy of the

donor electrons.

At room temperature, the donor states are essentially

completely ionized.

=> All donor impurity atoms have donated an electron

to the conduction band.

Small value

At room temperature, there is also essentially complete

ionization of the acceptor atoms.

=>each acceptor atom has accepted an electron from the

valence band.

0

1

( )1 exp( )

d

c c dd

d

n

N E En n

N kT

43

Fig Energy-band diagrams showing complete ionization of

(a) donor states and (b) acceptor states.

Complete Ionization at T=300 K

Partial ionization of donor or acceptor atoms when 0K<T<300K

44

Conduction band Conduction band

Valence band Valence band

Fig Energy-band diagram at T = 0 K for

(a) n-type and (b) p-type semiconductors.

Nd+=0

=>Each donor state must contain an

electron.

=> EF>Ed

Freeze-Out at T=0K

Na- =0

=>Each acceptor state does not

contain electron.

=> EF>Ev

CHARGE NEUTRALITY

In thermal equilibrium, the semiconductor crystal is

electrically neutral. =>The electrons are distributed among the

various energy states, creating negative and positive charges,

but the net charge density is zero.

A compensated semiconductor is one that contains both donor

and acceptor impurity atoms in the same region.

An n-type compensated semiconductor occurs when Nd > Na,

A p-type compensated semiconductor occurs when Na> Nd.

If Nd = Na, it is an intrinsic material.

45

46

Equilibrium Electron and Hole Concentrations

Fig Energy-band diagram of a

compensated semiconductor

showing ionized and un-ionized

donors and acceptors

The charge neutrality condition is

n0: total electron concentration

= thermal electrons + donor electrons

p0: total hole concentration

= thermal holes + acceptor holes

pa: the hole concentration in acceptor states

nd: the electron concentration in donor states

0 0( ) ( )a a d dn N p p N n

0 0a dn N p N

47

If we assume complete ionization, nd and pa are both zero

Two factor to impact n0

--- the concentration of impurity atoms

--- the intrinsic carrier concentration

2

0

0

ia d

nn N N

n

0 0a dn N p N

2 2

0 ( )2 2

d a d ai

N N N Nn n

48

Fig Energy-band diagram showing the redistribution

of electrons when donors are added

49

Fig Electron concentration versus temperature showing the

three regions: partial ionization, extrinsic, and intrinsic.

As the temperature increases, additional electron-hole pairs are

thermally generated so that the ni term may begin to dominate.

Complete

ionization

( )exp[ ]c Fi

i c

E En N

kT

50

Similarly

If Na-Nd>>ni , then

0 a dp N N

2 2

0

0 ( )

i i

a d

n nn

p N N

2 2

0 ( )2 2

a d a di

N N N Np n

2

0

0

ia d

nN p N

p

POSITION OF FERMI ENERGY LEVEL

51

The position of the Fermi energy level is a function of the

doping concentrations and as a function of temperature

For an n-type semiconductor, Nd>>ni then0 dn N

As the donor concentration increases, the Fermi

level moves closer to the conduction band.

where

0

( )exp[ ]c F

c

E En N

kT

0

ln cc F

NE E kT

n

2 2

0 ( )2 2

d a d ai

N N N Nn n

ln cc F

d

NE E kT

N

0lnF Fi

i

nE E kT

n

0 exp[ ]F Fii

E En n

kT

52

For a p-type semiconductor

0lnFi F

i

pE E kT

n

The difference between the intrinsic Fermi level and the

Fermi energy in terms of the acceptor concentration

As the acceptor concentration increases, the Fermi level

moves closer to the valence band.

ln vF v

a

NE E kT

N

0

ln vF v

NE E kT

p

If we assume that Na>>Ni

53

Fig Position of Fermi level for an (a) n-type and (b) p-type semiconductor.

Fig Position of Fermi level as a

function of donor

concentration (n type) and

acceptor concentration (p type).

54

0lnFi F

i

pE E kT

n

0lnF Fi

i

nE E kT

n

Fig Position of Fermi level as a function of

temperature for various doping concentrations.

As T increases, ni will increase F FiE E

( )exp[ ]c Fi

i c

E En N

kT

Summary

The concentration of electrons and holes

55

• Using the Maxwell-Boltzmann approximation

• The intrinsic carrier concentration is

0 ( ) ( )c Fn g E f E dE 0 ( )[1 ( )]v Fp g E f E dE

2 exp[ ]g

i c v

En N N

kT

0

( )exp[ ]c F

c

E En N

kT

0

( )exp[ ]F v

v

E Ep N

kT

56

• The fundamental relationship of the hole and electron

concentration is 2

0 0 in p n

• The concept of doping the semiconductor with donor atoms and

acceptor atoms to form n-type or p-type material.

• The electron and hole concentrations is a function of impurity

doping concentrations

• The position of the Fermi energy level is a function of impurity

doping concentration

0lnFi F

i

pE E kT

n

0lnF Fi

i

nE E kT

n

2 2

0 ( )2 2

d a d ai

N N N Nn n

2 2

0 ( )2 2

a d a di

N N N Np n

Homework 5Consider germanium with an acceptor concentration of Na =

1015 cm-3 and a donor concentration of Nd = 0. Plot the

position of the Fermi energy with respect to the intrinsic

Fermi level as a function of temperature over the range

200K T 500K