Semiconductors 3

8/14/2019 Semiconductors 3

1/16

Semiconductors & EM Theory 1 Lecture 3

Section 3: Electric Field E in a Semiconductor

Aims of this section

To introduce the concept of electric field E, and to find the

response of charge carriers within a semiconductor device to anapplied electric field.

Objectives of this section

After completing this section, you should

understand what is meant by the electric fieldEbe able to findEoutside a point charge and between the plates

of a parallel plate capacitor using Gauss law

understand what controls the motion of charge carriers withina semiconductor

be able to calculate their average velocity


2/16

3.1 Definition of the electric fieldE

Basic experimental observation:

Like charges repel, unlike charges attract

Force on charge q defined as:

F qE= (3.1)

e.g. for point charges



3/16

3.2 The flux ofE

FluxofE EA=

Note that while E is a vector, its flux is a scalar (i.e. it has

magnitude only).



4/16

3.3 Gauss Law for the flux ofE

Draw an imaginary 3-D closed surface that fully encloses the

charge (or distribution of charges), so that at all points on the

surface, the lines ofEproduced by the enclosed charge cut the

surface at right angles and E is the same everywhere on thesurface. The flux ofEthrough the surface is

Fluxof areaof surfaceE E=

Gauss law states that:

Fluxof through closed surface charge enclosed bysurfacer

E = 0

(Remember - examples should clarify)



5/16

0128 854 10= . F/ m, the permittivity of free space

r is a number called the relative permittivity of the material


Material Relative

Permittivity r

Si 11.9

Ge 16.0GaAs 13.1

SiO2 (silicon dioxide) 3.9


6/16

E for a point charge :

Mag. ofE equal on surface as spheres are concentric Lines of E are all perpendicular to sphereGauss states :

Flux ofE =E x area of surface =E x 4r2 =0r

Q

Re-arranging : 204 r

QE

r

=


Imaginary Sphere

of radius r

Lines ofE

+


7/16

3.4 Use of Gauss Law to find E between the plates of a

parallel plate capacitor



8/16

To findEusing Gauss law:

For the beancan Gaussian surface shown:

Fluxof area of end plateE E EA= =

Using Gauss law

Fluxofr

E EAQ

= =

0

=EQ

A r 0



9/16

Worked Example

A square piece of SiO2 of area 1 mm2 and thickness 0.1 mm is

metallised and connected across a battery. If a charge of 0.1 nC

is stored on the plates, find the magnitude of the electric field in

the SiO2.

Neglect edge effects, in which case

EQ

A= =

=

r . .V/ m . V/ m

0

10

12 6

610

3 9 8 854 10 102 9 10

Note the SI units ofE, which are volts per metre (V/m).



10/16

3.4 Electric field gradient - differential form of Gauss Law

e.g. very flat beancan of thickness x, with ends of areaA

( ) ( )Fluxof E E x x A E x AdE

dx

x A= +



11/16

Total charge enclosed in bean can is x .

Application of Gauss law then tells us that

Fluxofcharge enclosed

r r

EdE

dxx A

A x = =

0 0

Cancelling the volume x on both sides then gives

dE

dx =

r 0

(3.4)


This is what we mean by the differential form of Gauss law.


12/16

3.5 Motion of charges in a vacuum (i.e. free space)

e.g. a particle of mass m carrying a charge q placed in a

uniform electric fieldEin free space.

Force on this charged particle is F qE= .

Acceleration of particle:

ad F

m

qE

m= = =

v

dt

Velocity at time t(starting from rest):

v = =atqEt

m

Distance moved:

xqEt

m=

2

2



13/16

3.6 Motion of charges within a conductor

In free space, a charge will accelerate indefinitely to veryhigh velocities in response to a uniformE.In a conductor, collisions either with other electrons or with

the surrounding crystal lattice, which keeps its velocity to alimiting value.

Average velocity of electron is

v =qE

m



14/16

The scattering time controls the electrical conductivity of the

conductor.

We usually write the average velocity of charges within a

conductor

v = =qEm

E

where is called the mobility of the charges, and equals

=q

m

In a metal, all of the mobile charge carriers are electrons withthe same scattering time.



15/16

In a pure or doped semiconductor, the mobile charge carriers

are electrons and/or holes, which have different scattering

times!Material Electron mobility n Hole mobility p

Si 1400 2 1 1cm V s 500 2 1 1cm V s

Ge 3900 2 1 1cm V s 1900 2 1 1cm V s

GaAs 8500 2 1 1cm V s 400 2 1 1cm V s

The unit of mobility is 2 1 1cm V s

We always define mobility to be a positive quantitywhether it applies to electrons or holes.

A closing thought:

Electrons move opposite toE, but holes alongE. However, thetotal current due to the two sorts of carriers adds together!

Think about it!



16/16


Summary

You should now be familiar with the following terms:

Electric fieldFlux of electric fieldGauss lawPermittivity of free space and relative permittivityCarrier mobilityThe visualisation of electric field lines is very important.

You will not be asked to perform detailed derivations based on

Gauss law, but you should be familiar with its application tothe two examples that weve outlined. Having found E, you

should be able to find the velocity of each type of charge carrier

in a semiconductor in response toE.

Semiconductors 3

Documents

Transcript of Semiconductors 3