pn junction ex 2008

8/3/2019 Pn Junction Ex 2008

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PHYS208p-n junction

January 15, 2010

1


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List of topics

(1)

Density of states

Fermi-Dirac distribution

Law of mass action

Doped semiconductors

Dopinglevel

p-n-junctions


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1 Intrinsic semiconductors


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List of topics

Density of states

The number of conduction electrons is equal to the number of holes.

The density of states for the free electron in the conduction band is

gc(E) =V

2

2meh2

3/2(E Ec)

1/2. (2)

me is the electron effective mass.

Ec is the energy at the conduction band edge.


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For the free holes in the valence band, the density of states is

gv(E) =V

22m

h

h2

3/2

(Ev E)1/2. (3)

mh is the hole effective mass.

Ec is the energy at the valence band edge.


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In conduction band there are very few electrons. This means that

fe(E) =1

eE

kT + 1

> kT (7)


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The concentration of electrons in the conduction band is

ne =1

V

Ec

gc(E)f(E)dE = 2m

ekT

2h2

3/2

eEckT (8)

gc(E) =V

2

2meh2

3/2(E Ec)

1/2. (9)

Go to the chemical potential


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For holes in valence band

fh(E) = 1 fe(E) eEkT (10)

fh is the probability that a valence orbital is occupied.

gv(E) =V

2

2mh

h2

3/2(Ev E)

1/2. (11)

The concentration of holes in the valence band is

nh =1

V

Ev

gv(E)fh(E)dE = 2

mhkT

2h2

3/2eEvkT (12)


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List of topics

Law of mass action

Since the number of holes is the same as the number of electrons in an intrinsicsemiconductor

ne = nh = ni (13)

ne = 2

mekT

2h2

3/2

eEckT (14)

nh = 2

mhkT

2h2

3/2eEvkT (15)

ne(T)nh(T) = 4

kT

2h2

3(mem

h)3/2e

EgkT = n2i , (16)


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List of topics

ni(T) = 2 kT2h2

3/2

(mem

h)3/4e

Eg2kT (17)

ni is the intrinsic carrier density.

Eg = Ec Ev

Eg is the energy gap.

The result is independent of and it drops exponentially at the band gapincreases.

The expression (16) is the law of mass action.

Since ne(T) = ni(T), we obtain by equating (17) and (8)

= Ec 1

2Eg +

3

4kT ln

mhme

(18)


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List of topics

Doped semiconductors

(19)

Impurity atoms with more valence electrons than the rest of the atoms in thecrystall are called Donors and it forms n-type semiconductors.

Impurity atoms with less valence electrons than the rest of the atoms in thecrystall are called Acceptors and it forms p-type semiconductors.

If the material to be doped is silicon (4 valence electrons), one can add phos-phorus (5 valence electrons) to get a n-type semiconductor and one can addboron (3 valence electrons) to get a p-type semiconductor.


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Si

+4

Si

+4

Si

+4

Si

+4

Si

+4

Si

+4

+4

Si

Si

+4 +5

P

CONDUCTION

(a)

ELECTRON

-q

Si

+3

B

+4

Si

+4

Si

+4

Si

+4

Si

HOLE

+4

Si

+4

Si

+4

Si

+4

+q

(b)


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List of topics

Doping level

We want to show that the energy level for the extra electrons will be rightbelow the conduction band for a n-doped semiconductor.

As a model we use the Hydrogen Hamiltonian, where the electronic mass isreplaced by the corresponding effective mass.

H = h2

2me2

e2

4r0r(20)

r is the relative dielectric constant since we are not in vacuum.

We know that the energies in hydrogen are given by

En = me

2h2

e2

40

21

n2=

13.6

n2eV (21)


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me = Cme (22)

Then the energy of the groundstate (n = 1) in our model is

Emodel = me2h2

e2

40r

2=

me

2h2

e2

40

2 C2r

= EhydrogenC

2r(23)

Example

For phosphorus doping of silicon r = 11.7 and m

e 0.2me

The binding energy in our model is then

E 0.02 eV (24)

The energy of the doping states are

Ed

= Ec

+ E (25)

Since E < 0 the states are right below the edge of the conduction band.

If we look at the holes in the same way as the electrons it can be shown thatthe energy level for the extra holes will be right above the edge of the valenceband.


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n-doped

p-doped


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Ev

Ec

f( , ,T)

k T 1+e

1


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Ev

Ec

f( , ,T)

k T 1+e

1

Ev

Ec


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Ev

Ec

f( , ,T)

k T 1+e

1

Ec

Ev


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Ev

Ec

f( , ,T)f( , ,T)1f( , ,T)

k T 1+e

1

f( , ,T)

Ec

Ev


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List of topics

2 p-n junctions(26)

In the p-doped and the n-doped materials the fermi levels will be at differentenergies.


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From thermodynamics we know that the fermi levels must be the same if theyare joined together.


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When p-doped and n-doped materials are joined together, there is an inbalancebetween electrons and holes on each side.

Electrons and holes diffuse diffustion current.

Positive charge on n-side and negative on p-side potential driftcurrent.

At equilibrium 0 = jdrift +jdiff


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The diffusion current is given by Ficks law,

J = Dn (27)

where D is the diffusion constant and n is the density of carriers. Regardingthe electrons we find that the charge current caused by diffusion is

jdiff = eJ = eDn (28)

The current caused by the electric field

jdrift = E = enmobE, (29)

where again n is the carrier density and mob is the mobility. At equilibrium

0 = jdrift +jdiff = en(r)mobE + eDn(r) (30)

Assuming that there only is change in the x-direction, the equation becomes

n(x)mobEx(x) = D

dn(x)

dx , (31)which gives us the electrical field in terms of the electron distribution:

E(x) = D

mob

1

n

dn

dx. (32)


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Interface between the n-doped andthe p-doped semiconductor is very

depleted of charge carriers.

The resistance there will be muchlarger than in the rest of the semicon-ductor.

Practically all of the potential fallwill be in this region. Thus

Vc = () () =

Exdx

=D

mob

n()n()

1

n(x)

dn

dxdx =

D

mob

n()n()

dn

n=

D

mobln

n()

n().


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n(x) eeExx/kBT. (35)

n(x) emobE

xx/D

. (36)

Since these two exponentials must be the same to ensure consistency, we findthat

D

mob=

kBT

e, (37)

which is the Einstein-Nernst equation. Upon substitution

Vc =kBT

eln

NDNAn2i (T)

. (38)

Example

Si crystal: NA = ND = 1016

cm3

.kTe 0.025 V. For Si when T = 300 K, ni 10

10 cm3.

ThenVc = 0.72 V (39)


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Assume (x) has a constant value in a region of thickness dp and anotherconstant value in a region of thickness dn.

Everywhere else we assume (x) = 0.

Assume all electrons from the region dn fill all the holes in the region dp.

NAdp = NDdn.


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From Poissons equation we know that dEx/dx = 4(x). Thus fordp < x < 0

Ex(x) = 4eNA(dp + x), (40)

and for 0 < x < dn Ex(x) = 4eND(dn x). (41)

Outside the depletion zone we assume there is no electric field.

S E d /d h b l d b


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Since Ex = d/dx this can be realised by

(x) =

0 x < dp2eNA(dp + x)

2 dp < x < 0

Vc 2eND(dn x)2 0 < x < dnVc x > dn

(42)

C i i 0 i


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Continuity at x = 0 gives

2eNAd2p = Vc 2eNDd

2n.

Defining d = dp + dn, and using dnND = dpNA we can write

dn =NA

NA + NDd, dp =

NDNA + ND

d.

Inserting this for dn and dp we find

Vc = 2e

NAN

2D + NDN

2A

(NA + ND)2

d2 = 2ed2

NANDNA + ND

,

or solving for d,

d = Vc2e

NA + ND

NAND. (43)

It i diffi lt t i fl th d ift t i th l t i fi ld i


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It is very difficult to influence the drift current since the electric field is verylarge at the junction.

It is possible to change the diffusion current by changing the density of electrons

and holes.At Vexternal = 0 the current over the p-n junction is

j = jdiffjdrift = j0 j0 = 0.

IfVexternal = 0 jdrift will not be much changed, but in the diffution current there

will be an extra Boltzmann factor eeV/kBT. Thus the current with an externalfield is

j = (eeV/kBT 1)j0. (44)

At V > 0 there is a current growing exponentially as we increase V, and atlarge negative values of V we are left with only the drift current which is very

small.


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pn junction ex 2008

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