34

The p-n junction diode

We are now in a position to explain the rectifying action of a p-n junction diode. We

shall also derive an equation for the current-voltage characteristic. In order to analyse the

device in as simple a manner as possible we shall assume that the p-n junction is made by

bringing pieces of p-type and n-type semiconductor together. We note, however, that this is

not how such diodes are made. The lattice must be perfectly continuous across the 'junction'

- only the doping density may differ on either side - so junctions are most commonly made by

diffusion of impurities. If you want to learn more about how devices are actually made then

Introduction to Semiconductor Microtechnology by D.V. Morgan and K. Board is a good

reference.

The junction in equilibrium (no bias)

We begin by reminding ourselves of the typical values of electron and hole densities in

n- and p-type semiconductors.

If we assume the n-region is doped with ND=1021m-3 then

€

n = ND =1021m−3; and p =ni2

ND

~ 2x1011m−3

where we have taken ni=1.4x1016 m3 for silicon.

If the p-region is doped with an acceptor density, NA=1020 m-3, then

€

p = NA =1020m−3 and n =ni2

NA

=1012m−3

This suggests that there is a huge density gradient across the junction.

35

The first figure below shows the distribution of charges when the two materials are brought

into contact. The mobile current carriers are shown together with their parent donors and

acceptors which are, of course, immobile.

The mobile current carriers will tend to diffuse down the huge concentration gradients.

Electrons in the n-region will diffuse toward the p-region and likewise holes in the p-region

will diffuse toward the n-region. Initially interdiffusion of carriers will inevitably lead to

recombination. Hence near the junction the number of mobile current carriers will be

dramatically reduced. However in removing the negatively charged electrons from the

originally neutral n-side the latter becomes positively charged in the thin layer where the

electron concentration is depressed. The positive charges are clearly the single excess fixed

positive charges on the nuclei of the immobile donor atoms. An entirely analogous situation

occurs on the p-side where a thin negatively charged layer appears. The fixed negative

charges in this case are due to the ionised immobile acceptor atoms. This leads to a

'transition' region between the p- and n-side which is more or less depleted of current

36

carriers. The region is often called the depletion region which, as we shall see later, extends

to a fairly well defined distance on either side of the junction.

The fixed charges in the depletion region inevitably give rise to an electric field in the

junction region which is directed from the positive charges toward the negative ones.

This field is so directed as to cause holes to drift back toward the p-side from where

they are diffusing in such a way as to counteract the diffusion since there can be no net hole

current across the junction in equilibrium. A similar argument applies to electrons.

Once the balance of drift and diffusion has been achieved we can write

€

Jn = neµnE + Dn edndx

= 0

and

€

Jp = peµpE −Dp edpdx

= 0

Since the balance of drift and diffusion applies separately for electrons and holes this is

referred to as the principle of detailed balancing.

It is straightforward to re-arrange these two equations as

€

E =Dp

µp

. 1pdpdx

= −Dn

µn

1ndndx

37

We recall that D and µ are related via the Einstein relationships,

€

µn

Dn

=µp

Dp

=ekT

which permits us to rewrite the previous equation as

€

ekT

E =1pdpdx

= −1ndndx

Now since the electric field is, by definition, the negative potential gradient

€

E = −dVdx

we can write, using separation of variables

€

−ekT Vn

Vp

∫ dV =pn

Pp

∫ 1pdp = −

nn

n p

∫ 1ndn

where the integration takes place between any two planes,

€

x1 and

€

x2, outside the depletion

region and the subscripts on the carrier densities specify the n or p side of the junction, i.e. np

indicates the electron density in the p-type material. Performing the integrals gives

€

ekT

Vn −Vp( ) = npppn

= n

nnnp

and hence there is a potential difference, Vn-Vp, between the two regions which is referred to

as the built in potential, Vo. Thus

€

Vo =kTen nn

np

=

kTen

pppn

38

or, alternatively, in a form that will be useful later

€

nnnp

=pppn

= exp eVo

kT

Now since nn=ND and pp=NA we can write

€

np = ni2 / pp = ni

2 /NA and hence

€

Vo =kTen nn

np

=

kTen ND NA

ni2

and hence, depending on the doping densities the built-in voltage may be calculated. If

ND=1024 m-3 and NA=1022 m-3 then in silicon at 300K

€

Vo ~ 0.8V. At slightly lower doping

densities, NA=1020 m-3 a slightly lower value of

€

Vo ~ 0.5V is found.

We note that this built-in potential difference, like all contact potentials, cannot be

measured directly with an ordinary voltmeter, because there is no source of energy to drive

even the smallest current through the instrument.

We can now see the sequence of events leading up to the built-in field as

39

We could have obtained the same result by an easier (?) method, but without the

physical insight, by considering the energy level diagram for the junction. As we have seen

previously equilibrium requires that the Fermi level be the same throughout the material.

The potential energy difference between free electrons in the conduction band on either

side of the junction, Ecp-Ecn, is equal to the 'built-in' potential energy, eVo. We recall that we

can write the electron concentrations in the conduction band as

nn=Nc exp-(Ecn-EF)/kT

for electrons on the n-side and similarly as

np=Nc exp-(Ecp-EF)/kT

for the electrons on the p-side. Thus

40

€

nnnp

= expEcp − Ecn( )

kT= exp eVo

kT

and hence

€

Vo =kTdn nn

np

=

kTen

NA ND

ni2

as before.

The depletion region

As we have seen there are very few carriers in the transition region and so to a good

approximation we can consider the space charge within the depletion region to be due

entirely to the fixed donor and acceptor atoms. Thus the space charge density on the n-side

is simply eND since there are ND ionised donor atoms there whereas on the p-side the charge

density is -eNA due to the ionised acceptors.

We now derive an expression for the width of the depletion region and start by assuming that

the transition/depletion region extends xp into the p-side and xn into the n side. Charge

neutrality requires

eNA xpA = e ND xnA

41

where A is the cross sectioned area of the diode. Thus

xp NA = xn ND

We start with the p-side and notice that the total charge in the shaded region a distance y from

the transition region edge is given by

Q= - NA Ae y

The flux density, D, is given by D=Q/A= - NAA ey and hence the electric field, E=D/ε may

be written as

€

E = −NA eε

y

where ε is the dielectric constant. If we now change the variable to x, measured from the

metallurgical junction, as in the first diagram i.e. let x=y-xp so that

€

Ep = −NA eε

x + xp( )

an entirely analogous argument gives the electronic field on the n-side as

€

En =NDeε

x − xn( )

42

Thus the electric field varies linearly in both regions and reaches a maximum at x=0 given by

€

Emax = −NAeε

xp = −NDeε

xn

Now since

€

E = −dVdx

the potential difference, the built in voltage, Vo may be calculated

as

€

Vo = − E dx−x p

xn

∫

However since E is discontinuous at x=0 we must split this integral up into two parts

€

Vo = −−x p

o

∫ Ep dx −o

xn

∫ En dx

The integrals are routine and yield

€

Vo = e NA xp2 + ND xn

2( ) /2ε

Vo = - ∫ Edx 'area'

43

The total width of the depletion region may now be worked out in terms of Vo by using this

equation together with the space charge neutrality condition, xpNA=xnND. These reveal

€

w = xn + xp =2εe

NA + ND

NAND

Vo

and

€

xn =NA

NA + ND

w

together with

€

xp =ND

NA + ND

w

We notice that the depletion region extends more deeply into the region of lowest

doping. Indeed in the case of a p+-n diode where the p side is much more heavily doped than

the n, i.e. NA»ND , we find

€

w ~ 2εVo

eND

and xn»xp so the depletion region is almost entirely confined to the relatively lightly doped n

side.

In order to get an idea of the distances and field involved let's look at a silicon p-n

junction doped with NA=5x1021m-3 and ND=1021m-3. The built in voltage is given by

€

Vo =kTen NAND

ni2

= 25n

5x1042

2x1032

mV = 0.6V

The relative permittivity of silicon is 11.9 and hence

44

€

w =2 11.9x8.85x10−12( )

1.6x10−196x1021

5x142

0.6 = 0.97µm

and

xn= 0.81µm; xp= 0.16µm.

Finally the maximum electric field may be found from the previous formulae on page 42

together with the expressions for xn or xp or, more easily by remembering that the built in

voltage is given by the "area" under the electric field triangle,

€

Vo = − Edx∫ , i.e.

€

Vo =12Emax w

hence

€

Emax =2Vo

w

which, in our case, is 1.2x106 V/m.