No external forces such as voltages, electric fields, magnetic fields or temperature gradient are acting on the semiconductor. All properties of the semiconductor will be independent of time in this case.

The Semiconductor in Equilibrium

Equilibrium or thermal equilibrium:

Electron or hole concentrations at equilibrium:

The number of electrons per unit energy per unit volume in the conduction band is,

)()()( EfEgEn Fc=

)](1)[()( EfEgEp Fv −=For holes,

The total number of electrons per unit volume at thermal equilibrium,

dEEfEgn F

E

E cc

c

)()(0 ∫+

=χ

(A)

E

g(E)

g(E) ∝ (E-Ec)1/2

f(E)

EF

n(E) or p(E)

E E

Forelectrons

For holes

[1-f(E)]

n(E)

p(E)

Ec

EvEv

Ec

0

Ec+χ

EF

VB

CB

(a) (b) (c) (d)

Area = ∫ n(E)dE=n0

Area = p0

Since the Fermi function fF(E) decays rapidly with energy so that gc(E)fF(E) → 0 near the top of the band. Therefore, the upper limit of integration in equation (A) to be Ec+χ→ ∞. We assume that (Ec-EF)> 3kT so that

]/)(exp[)( kTEEEf FF −−≈ [Boltzman approximation]

dEkTEEEE

hmn F

CEn

C ⎥⎦⎤

⎢⎣⎡ −−

−= ∫∝ )(exp)2(4

3

2/3*

0π

This approximation is good for nondegenerate Semiconductors:

Let, ηη kTddEkTEE c =⇒

−=

ηηπ ηdekTEE

hkTmn FCn ∫

∝ −⎥⎦⎤

⎢⎣⎡ −−

=0

2/13

2/32/3*

0)(exp)()2(4

πηη η

21)2/1(

21)12/1(

0

2/1 =Γ=+Γ=∫∝ − de

)()(exp0 ccFc

c EfNkTEENn =⎥⎦

⎤⎢⎣⎡ −−

=∴

For nondegenerate Semiconductors:

232/3

2

*22 ThkTmN n

c ∝⎟⎟⎠

⎞⎜⎜⎝

⎛=

π Nc is the effective density of states in the CB

( )⎥⎦⎤

⎢⎣⎡ −−

=kT

EENp vFv exp0Similarly,

kTEvc

geNNpn /00

−=

Self-Study: Determine the expression of p0

( ) ( ) kTEvc

kTEEvc

vFFcvc

gvc eNNeNNkTEE

kTEENNpn //)(

00 expexp −−− ==⎥⎦⎤

⎢⎣⎡ −−

⎥⎦⎤

⎢⎣⎡ −−

=

I Intrinsic Semiconductors:A pair of an electron and a hole (EHP) is generated due to the thermal excitation.

Conduction electron+4

Si+4Si

+4Si

+4Si

+4Si

+4Si

+4Si

+4Si

+4Si

Eg

Ec

Ev

Ef

Valance hole

Electron and hole pairs (EHP) are generated due to the thermal excitation. no is the electron density (number of electrons / cm3), po is the hole density (number of holes / cm3) . for intrinsic materials no = po = ni . For silicon ni = 1.5 x 1010 and for GaAsni = 106 .

nn

⎥⎦⎤

⎢⎣⎡ −−

=kTEENp vF

v)(exp0

⎥⎦⎤

⎢⎣⎡ −−

=kTEENn Fc

cc)(exp

)()()( EfEgEN F=

)()()( EfEgEN F=

kTEvci

geNNn 2/−=∴

200 inpn = → Mass action law (Valid at thermal equilibrium)

Intrinsic Carrier Concentration:

n0=p0=ni; the concentration of electrons in the conduction band is equal to the concentration of holes in the valence band. The Fermi energy level for intrinsic semiconductor is called the intrinsic Fermi energy or EF=EFi, Therefore,

⎥⎦⎤

⎢⎣⎡ −−

=kTEENn Fic

ci)(exp ( )

⎥⎦⎤

⎢⎣⎡ −−

=kTEENp vFi

vi expand

00/2 pneNNnpn kTE

vciiig ===∴ −

Example 4.1: Given, EF-Ev=0.27eV, Nv=1.04×1019cm-3 at 300°K. Calculate p0 at 400°K.

⎥⎦⎤

⎢⎣⎡ −

=kTEEnn FiF

i)(exp0

⎥⎦⎤

⎢⎣⎡ −

=kTEEnp FFi

i)(exp0

And

⎥⎦⎤

⎢⎣⎡ −−

=kTEENn Fc

c)(exp0 ⎥⎦

⎤⎢⎣⎡ −+−−

=kT

EEEEN FiFFicc

)()(exp

⎥⎦⎤

⎢⎣⎡ −

⎥⎦⎤

⎢⎣⎡ −−

=kTEE

kTEEN FiFFic

c exp)(exp

Example 4.2:Given, Nc=2.8×1019cm-3, Nv=1.04×1019cm-3 at 300°K in Si. Eg=1.1eV in Si.Calculate, (i) ni at 300K

(ii) ni at 450K

Temperature dependency of nifor Si, Ge and GaAs

kTEvci

geNNn 2/−=∴

232/3

2

*22 ThkTmN n

c ∝⎟⎟⎠

⎞⎜⎜⎝

⎛=

π

232/3

2

*22 T

h

kTmN pv ∝

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

π

( ) kTEi

geTTn 2/23 −∝∴

Intrinsic Fermi-level position

For intrinsic semiconductor,

ii pn =( )

( )

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛=+−

=⎥⎦⎤

⎢⎣⎡ +−

⎥⎦⎤

⎢⎣⎡ −−

=⎥⎦⎤

⎢⎣⎡ −−

*

*

2/3

*

*

ln43

2

ln21

2

ln2

2exp

exp)(exp

n

pvcFi

n

pvcFi

c

vvcFi

c

vvcFi

vFiv

Ficc

mm

kTEEE

mm

kTEEE

NNkTEEE

NN

kTEEE

kTEEN

kTEEN

2vc EE + is the midway between Ec

and Ev

⎟⎟⎠

⎞⎜⎜⎝

⎛+=∴ *

*

ln43

n

pmidgapFi m

mkTEE

Example 4.3:

0* 08.1 mmn = 0

* 56.0 mmp =Given and in Si. Calculate EFi at 300°K

Extrinsic Semiconductor

x

As+ As+ As+ As+

EcEd

Ev Distanceinto crystal

~0.05 eV

CB

Electron Energy

N-type

If Nd >> ni, at room temperature, n0 ≈ Nd. 0

2

0 nnp i=

Ref.: S.O. Kasap, Principles of Electronic materials and devices, 3rd Edition, (McGraw-Hill, 2005)

n-type doping:

As+

e-

x

B -

Ev

Ea

Distance intocrystal

~0.05 eV

B - B - B -

h+

VB

Ec

Electron energy

If Na >> ni, at room temperature, p0 ≈ Na.

0

2

0 pnn i=

Ref.: S.O. Kasap, Principles of Electronic materials and devices, 3rd Edition, (McGraw-Hill, 2005)

B-h+

Free

B-

(a)

(b)

p-type doping:

And

Concept of donors, acceptors

Charge neutrality:

nd = Nd – Nd+ ; pa = Na – Na

-

nd & pa are the densities of electrons and holes occupying donor and acceptor levels respectively.

Statistics of donors and acceptors:

The probability function of electrons occupying the donor state is,

⎟⎠⎞

⎜⎝⎛ −

+=

kTEE

EfFd

dd

exp211

1)(

The factor ½ is due to the fact that the electron state at the donor can take an electron with spin either up or down but not both (Both electrons have a chance to occupy the state. Therefore the probability of occupancy is increased).

The density of electrons occupying the donor level Ed,

⎥⎦⎤

⎢⎣⎡ −

+==

kTEE

NEfNnFd

ddddd

exp211

)(

For (Ed-EF)>>kT ⎥⎦

⎤⎢⎣

⎡ −−=

kTEE

Nn Fddd

)(exp2

⎥⎦

⎤⎢⎣

⎡ −−=

kTEE

Nn Fddd

)(exp2

⎥⎦

⎤⎢⎣

⎡ −−=

kTEE

Nn Fcc

)(exp0 → Free electrons in CB

n0+nd → Total number of electrons

nd → No of electrons in the donor state.

( )⎥⎦

⎤⎢⎣

⎡ −−+=+=

+∴

kTEE

NN

nn

nnn dc

d

c

dd

d exp2

11 00

⎥⎦⎤

⎢⎣⎡ −−

+=

+∴

kTEE

NNnn

ndc

d

cd

d

)(exp2

1

1

0→ Determine relative ionization

⎥⎦⎤

⎢⎣⎡ −−

+=

+∴

kTEE

NNpp

pva

d

va

a

)(exp4

1

1

0Similarly for acceptor states,

Example 4.4:Given: (i) Nd = 1016 cm-3, (ii) 1017 cm-3 and (iii) 1018 cm-3 Phosphorus doping in Si at T=300°K. Ec-Ed = 0.045 eV. Calculate the fraction of the total electrons still in the donor states.

⎥⎦⎤

⎢⎣⎡ −−

+=

+∴

kTEE

NNnn

ndc

d

cd

d

)(exp2

1

1

0

A compensated semiconductor is one that contains both donor and acceptor impurity atoms in the same region. In n-type compensated semiconductor Nd> Na and p-type compensated semiconductor Na > Nd. If Na= Nd, the semiconductor is completely compensated and n0 = p0 (like intrinsic SC).

Compensated Semiconductor:

iad nNN >>− )( ad NNn −≈0ad

i

NNnp−

=2

0

ida nNN >>− )( da NNp −≈0da

i

NNnn−

=2

0

At room temperature,

1. n-type non degenerate semiconductor,

and

2. p-type non degenerate semiconductor,,

and

Charge neutrality

Charge neutrality:

In thermal equilibrium, the semiconductor crystal is electrically neutral. This charge-neutrality condition is used to determine the thermal equilibrium electron and hole concentrations as a function of the impurity doping concentration.

If doping level is comparable to ni, then, n0 and p0 can be calculated using charge neutrality:

+− +=+ da NpNn 00

For full ionization aa NN =−dd NN =+and

,

da NpNn +=+ 00

For n-type:0

2

0 nnp i=

di

a NnnNn +=+∴

0

2

0

0)( 20

20 =−−+⇒ ida nnNNn

22

0 22 iadad nNNNNn +⎟

⎠⎞

⎜⎝⎛ −

+−

=∴

dai NpNpn

+=+ 00

2

0)( 20

20 =−−+⇒ iad npNNp

22

0 22 idada nNNNNp +⎟

⎠⎞

⎜⎝⎛ −

+−

=∴

Similarly for p-type:

Example 4.5:At T =300°K in Germanium, ni=2.4×1013 cm-3. Find n0 and p0 for (i) Nd=1016

(ii) Nd=5×1013 cm-3

( ) kTEi

geTTn 2/23 −∝∴

22

0 22 idd nNNn +⎟

⎠⎞

⎜⎝⎛+=

Temperature dependence of carrier concentration:

Example 4.6:At T =550°K in Silicon, the intrinsic carrier concentration must contribute no more than 5 % of the total electron concentration. Determine the minimum donor concentration Nd required to meet this specification.

Position of Fermi energy level:

⎥⎦

⎤⎢⎣

⎡ −−=

kTEE

Nn Fcc

)(exp0 ⎟⎟

⎠

⎞⎜⎜⎝

⎛=−⇒

0lnnNkTEE c

Fc

⎟⎟⎠

⎞⎜⎜⎝

⎛=−⇒⎥⎦

⎤⎢⎣⎡ −

=i

FiFFiF

i nnkTEE

kTEEnn 0

0 lnexp

n-type:

p-type:

⎟⎟⎠

⎞⎜⎜⎝

⎛=−

⎟⎟⎠

⎞⎜⎜⎝

⎛=−

iFFi

vvF

npkTEE

pNkTEE

0

0

ln

ln

Variation of EF with doping:

As the doping levels increase, the Fermi energy level moves closer to the conduction band for the n-type material and closer to the valence band for the p-type material.

Variation of EF with temperature:

⎟⎟⎠

⎞⎜⎜⎝

⎛=−⇒

0lnnNkTEE c

Fc

Example 4.8:Given: p-type Si at 300°K doped with boron. EF-Ea=3kT. Ea-Ev=0.045eV. Calculate: Maximum doping at which the Boltzmann approximation is valid.

Example 4.7:Given: At T = 300°K, Na =1016 cm-3 and Ec-EF=0.2eV in Si.Determine: Nd

Degenerate Semiconductor:

If acceptor or donor concentration in Si is greater than approximately 1017 cm-3, then the Boltzmann approximation is less valid and the semiconductor is called degenerate semiconductor.

CB

g(E)

E

Impurities forminga band

(a) (b)

EFpEv

EcEFn

Ev

Ec

CB

VB

→ Donor electrons interact with each other.→ The single discrete donor energy level will split into a band of energies.→ As the donor concentration further increases, the band of donor states widens and may overlap the bottom of the conduction band. This overlap occurs when the donor concentration becomes comparable with the effective density of states.→ EC is therefore slightly shifted down and Eg becomes slightly narrower.→ The similar phenomena occurs in p-type degenerate semiconductor.

dEEfEgn F

E

E cc

c

)()(0 ∫+

=χ

∫∞

−+

−=

Ec F

cn

kTEEdEEE

mh

n)exp(1

)()2(4 2/1

2/3*30π

kTEE c−

=η kTEE cF

F−

=η

∫∞

−+=

0

2/1

2/1 )exp(1)(

FF

dFηη

ηηη

)(22/10 FcFNn η

π=