LIsta Aplicações de Derivadas.tex

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Notes on calculating the Fermi level

M.P. Vaughan

April 18, 2014

1 Calculating the Fermi level

1.1 Free carrier concentration

1.1.1 Non-degenerate semiconductor

The free electron concentration n can be given by an integration over energy

n= 2

f()D() d, (1)

wheref() is the occupation probability andD() is the density of single-spinstates. The factor of two incorporates electron spin. For a non-degenerate

semiconductor we can use Boltzmann statistics, so that, relative to = C

f() = exp

C kBT

, (2)

where kBT is Boltzmanns constant and T is the temperature. Assuming aspherical, parabolic conduction band, the density of states is

D() =(2me)

3/2

42h3 ( C)1/2 (3)

for Cand zero otherwise, where me is the electron effective mass andC is the energy at the conduction band edge. Hence,

n=(2me)

3/2

22h3

C

exp

C kBT

( C)1/2 d. (4)

Putting

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x=

C

kBT (5)

and using

0

x1/2ex dx=

2 , (6)

we have

n= 2

me

2h2kBT

NC, (7)

whereNCis known as the effective density of states. A similar result can be

derived for holes

p= 2

mh

2h2kBT

NV, (8)

wheremh is the hole effective mass.

1.1.2 Degenerate semiconductor

For a degenerate semiconductor, we must use Fermi-Dirac statistics. Theoccupation probability is then given by the Fermi factor

f() = 11 + exp ([ F] /kBT) , (9)

whereF is the Fermi energy. Putting

=F C

kBT , (10)

we can write

n = (2me)

3/2

22

h3

0

x1/2

1 +ex

dx,

= NC2

0

x1/2

1 +ex dx,

= NCF1/2(), (11)

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where

F1/2() = 1

(j+ 1)

0

xj

1 +ex dx (12)

is the Fermi-Dirac integral and is the gamma function. A similar relationholds for holes.

Fj() does not have an exact analytical solution. However, we may usethe approximation

Fj() 1e +Cj()

(13)

where, for j = 1/2

C1/2() = 3(/2)1/2

+ 2.13 +| 2.13|12/5 + 9.6

5/123/2 . (14)

1.1.3 Non-parabolic energy bands

When the dispersion relations are non-parabolic, the density of states forspherical energy surfaces is

D() =(2m

e

)3/2

42h3 1/2

()d()

d (15)

where() is defined by

() =h2k2

2me(16)

In this case, the integral Eq. (??) must usually be solved numerically.

1.2 Donors and acceptors

The free carrier concentration is greatly affected by doping with donorsand/or acceptors. If ND and NA are the doping concentrations of donorsand acceptors respectively, then the ionised concentrations are given by

N+D =ND

1 1

1 + (1/2)exp([D F] /kBT)

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and

NA = NA

1 + 4 exp ([A F] /kBT) (17)

whereD and A are the respective ionisation energies. For an n-type semi-conductor, we usually haveF A, and we can take NA NA. D is foundfrom the hydrogenic model

D =me

h2

e2

4

(18)

where is the permittivity of the material.

1.3 Calculating the Fermi energy

The Fermi level is calculated by imposing the condition of charge neutrality

n+NAp N+D = 0 (19)For an n-type semiconductor, we can take p 0 andNA NA, as describedpreviously. Equation (??) must then be solved forF numerically. However,we may draw a few qualitative conclusions before doing so. With the as-sumptions above, we have n N+D , so we require these two quantities tovary at the same rate with both F and T.

Clearlyn will increase with both F and T, since increases in both willincrease the occupation probability at higher electron energies. InspectingEq. (??), we see that an increase in F will always be accompanied by adecrease inN+D . The dependency with Tdepends on the value ofF relativeto D.

Case (1) : F < DIn this case, N+D decreases with T, tending towards a limiting value ofND/3. IfFwere to increase withT, thenN

+

D must necessarily decreasewhilstnmust necessarily increase. This cannot be the case and henceFmust decrease with T.

Case (2) : F =DHere, N+D is fixed at ND/3, independent of T. Since n will increasewithT,Fmust change. Now any increase in N

+

D means that we have

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N+D > ND/3 but this can only be the case for F < D. Therefore F

must decrease.Case (3) : F > D

This case is not quite so straightforward, since N+D increases monotoni-cally withT. Therefore an increase ofN+D withTmight reasonably beaccompanied by either an increase or decrease in F especially wheredN+D/dT was large. However, as T , N+D ND/3, for fixed F.IfFwas to always increase withT,N

+

D could never exceed this value.At the same time, n would increase inexorably. So, for large Tat least,Fmust decrease with temperature.

In conclusion, we can say that most of the time, at least, F decreaseswith temperature. Exceptions may arise when F > D (in which case thesemiconductor is likely to be degenerate). However, at high temperatures,dF/dT should always be negative.

For the numerical solution of Eq. (??) we use a bisection method, inwhich two values of F are 1 and 2 are chosen so that the net charge isnegative (or positive) for 1 and positive (or negative) for 2. The averageM = (1+ 2)/2 is then found for which the net charge is calculated. Inthe next iteration, M is then substituted for either 1 or 2 depending onthe sign of the calculated charge. In this way, the interval [1, 2] becomesprogressively smaller until the routine exits when

|1

2

|< , for a chosen

tolerance.

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