P-N JUNCTION Lecture 2

EBB 215

(Dr Zainovia Lockman)

2015

The Depletion Region

The Basic Structure of a pn Junction

p n

So far we have been considering the properties of semiconductor materials: electron and hole concentration, band diagram, Fermi level determination, conductivity and current.

We are now considering what would happen when p type and n type semiconductor are brought into contact with one another.

A pn junction is a single crystal semiconductor material which one region is doped with acceptor impurity atoms to form p region and the adjacent region is doped with donor atoms to from the n region. The interface separating them is the junction.

Diffusion of Carriers

Once a junction is created, diffusion of carriers will happen across the junction. To explain this, let us consider a step junction.

A step junction is a junction with uniform doping concentration on both the p and n regions and there is an abrupt change in doping at the junction.

Since there is a concentration gradient across the junction, the carriers from n will diffuse to p and likewise p to n: electrons are diffusing to the p region from the n region and holes are diffusing to the n region from the p region.

The carriers are the majority carriers.

If there is no external connections to the semiconductor, then the diffusion process cannot continue indefinitely.

The donor and acceptor atoms

Recall that the electrons from n type material come from dopant atoms. Once the atoms have been ionised to produce free electrons, the dopant atoms become positively charged. And once the electrons have all diffuse away to the p region, the positively charged donor atoms will be left behind.

Donor atoms CANNOT move.

Similarly, holes from acceptor atoms will diffuse away to the n region leaving behind negatively charged acceptro atoms. These atoms cannot move either.

The diffused electrons and holes will annihilate each other when they meet each other.

P N

Holes diffuse to the n region

- +

+ +

-

+

+

+

Electrons diffuse to the n region

-

- Ionised acceptor atom (-ve charged)

+ Ionised donor atom (+ve charged)

e

Ionised acceptor atom (-ve charged)

e e

e

e e

h

h

- h

-

h

-

This will create

P N

- -

-

-

-

-

- -

-

-

-

-

- -

+ +

+

+

+

+ + + + +

+ +

+

+

A region with lack of mobile charge carrier:

Depletion region

The Space Charge Region

P N

-

-

-

-

-

- -

+

+ + +

+ +

+

+

Space

charge

region

Na negative

charge

Nd positive

charge

The Electric Field and Forces acting on Charged Carriers

P N

-

-

-

-

-

- -

+ + +

+ +

+

+

Diffusion

force on

holes

Depletion

region

Diffusion force

on electrons E field

E-field force on electrons E-field force on holes

The Force Acting on Charged Particles

The diffusion force acts on the charged particles.

In the same time, electric field in the space charged region produces another force on the electrons and holes which is in opposite direction

In thermal equilibrium, the diffusion force and the E-field exactly balance each other

No charged carriers can move in the depletion region as the net effect

Zero Applied Bias

If there is no bias (no voltage) applied to the junction, the junction is in thermal equilibrium

If the junction is in thermal equilibrium, what happened to the Fermi level?

What do you think? What happened to the Fermi Level when the two materials

are at thermal equilibrium?

The Fermi Level is CONSTANT throughout.

Sketch the band energy

diagram for the n and p type to show the invariance of the

Fermi Level a equilibrium!

Isolated Energy Band Diagrams

WHEN THE 2 SEMICONDUCTOR MATERIALS OF

DIFFERENT TYPES ARE IN CONTACT .

Upon contract between the two materials (p and n) electrons flow from the n to the p because there are more free electrons in the n than in the p.

As the electrons move towards the p region. They leave behind the ionised donor atoms (positively charged) that are locked to the semiconductor lattice

At the same time holes flow from the p region to the n region and leave negatively charged atoms. This separation of changes set up and electric field.

An equilibrium condition is reached whereby the Fermi Level will be continuous across the sample

Energy Band Diagram at Equilibrium:

Built in Potential Barrier

A potential barrier is created at the space charge region (depletion region)

Electrons from n will see this as an obstruction for it to move to p region

The barrier is called a built in potential barrier, Vb

The built in potential maintains equilibrium between majority carrier electrons in n and minority carrier electrons in p.

And the majority carrier holes in p and minority carrier holes in p.

Built in Potential Barrier Vb

The built in field at the junction alters the band.

It cannot be measured.

The Vb only maintains equilibrium because of this voltage formed, no current can flow across the depletion region. Vb can be expressed by the following equation:

i

dab

n

NN

q

kTV

2ln

C A L C U L AT I O N

Given a p-n junction with n-type Si consisted of dopants, ND=10

14 cm-3. If ni = 9.65 x 109cm-3, calculate.

po which will be assumed to be the same as NA)

Then determine the built in potential of this junction.

Solution:

Use

i

dab

n

NN

q

kTV

2ln

Built in potential

Constants: k and T

Dopants concentration:

P-type: NA N-type: ND

Intrinsic carrier concentration: ni

= 9.65 x 109cm-3

The Electric Field

Since there is a separation of positive and negative space charge densities, an electric field is built up in the depletion region

P N

xp xn W

Assume abrupt

junction.

Assume space

charge region

ends in the n

region at xn and at

xp in the p region

The Space Charge Density, (C/cm3)

-

+

+eNd

-eNa

(C/cm3)

x

Depletion region

xp xn

xn

-xp

P N

Three properties of Junction: (charge distribution/density at junction), E (electric

field) and (potential) can be determined by Poissons equation

Volume charge density

Permittivity of semiconductor

Electric

Field

dx

xdEx

dx

d )()(2

2

Potential

Rewrite the Poisson Equation:

The distribution of charges at the

junction

-

+

+eNd

-eNa

-xp

xn

(C/cm3)

-xp

The integration E

Use Poisson's equation to get electric field:

Integrate the charge density to get E: for p region (e = q)

And integrate charge density at the n region

We can set E = 0 at x =-xp and at x=xn

dx

xdEx

dx

d )()(2

2

1

)(Cx

eNdx

eNdx

xE aa

2

)(Cx

eNdx

eNdx

xE dd

In the p region

0)(

xxxxeN

E ppa

nnd xxxx

eNE

0)(

In the n region

The Plot of Electric Field

E

Emax

-xp xn

Maximum E value can be produced from this plot. And E can be seen reducing as we go from the junction to xpor xn

The Second Integration Consider an n region. The potential or voltage can be obtained by integrating the field, E:

22

.

2'

2

'

2

2

.

2)

2()(

2

)2

()(

)()()(

p

s

an

d

p

s

a

nd

nd

xeNx

xxeN

x

xeN

C

Given

Cx

xxeN

x

dxxxeN

dxxEx

The Plot of Electric Potential

-xp xn

x

Vb

The potential in the n region is seen to be quadratic i.e. a curve as shown in the next slide would be expected. Apparently, when x = xn then V = built in potential.

The potential through the junction shows the quadratic dependence on distance both in the p region and the n regions, depending on the dopant concentration the potential in the p region can recedes to zero or a very small value.

As a Summary

P-n junction made

Depletion region formed

Built in potential defined

Set boundaries and charge distribution plot

Poisson's equation to define the depletion region

From the equation: First integration gives electric field

Second integration gives potential

From the second integration we can have another equation for built in potential

Self Assessment A wafer of n-type Si is given to you. The wafer is then subjected to p-type doping in an ion implantation chamber with very high concentration of p-type doping.

How does the depletion region of this junction will look like?

Describe the characteristics of the space charge region.

State the Poission equation and use it to explain these characteristics.

One Sided Junction