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Resistivity is the property of a material thatdetermines the electrical resistance of a specificsample.

The block of material has resistance (R) andresistivity ().

remains constant regardless of size or shape.

Resistivity

l

h

w

4

4

R= l

w h=

l

A

Resistivity

Calculate resistance between front and back faces :

Ais the area through which the current must passand can be used where the cross-section is notrectangular, e.g. for a wire with circular cross-section.

If length increases, resistance increases and viceversa. Opposite for area.

The S.I. unit of resistivity is the m, but thepractical unit is the cm.

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Sheet resistance

In IC technology we often have conducting layerswith constant depth (hin figure).

Resistance therefore depends upon the aspect ratioas viewed from the surface (l/w).

Define sheet resistance (Rs) so that resistance is:

R=Rs

l

w

Rsis given in /square where l = w.

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The relation between and Rsis:

But since l = w:

Using sheet resistance, define resistance of an area(e.g. in a silicon IC) by number of squares.

R =

Rs

4

Sheet Resistance

R= 4 Rs

Rs

l

w

= l

wh

=Rs h

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Example 1

An aluminium rod 30cm long with a cross-section of0.5cm x 0.25cm has a voltage of 100mV applied acrossit and a current of 13.9A is observed to flow. Calculatethe resistivity, , of the aluminium.

R= V/I= 0.1/13.9A = 7.2 x 10-4

R= l/whor = Rwh/lso = 7.2 x 10-4x 0.005 x 0.0025/0.3 = 3 x 10-8m.

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0.8m

2m

Example 2

An area of undoped (high resistivity) silicon containsa region of doped silicon (so much lower resitivity) forfabricating a 20 kresistor, as shown in the figure. Ifthe resistivity, , of the doped silicon is 10-3m,calculate the sheet resistance,R

s, the number of

squares of doped silicon required for the resistor, andhence the length of the resistor.

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Example 2 Solution

We can assume that the doped silicon is a conducting areasurrounded by insulator.

Rs= /h= 10-3/0.8 x 10-6= 1.25 x 103/square = 1.25k/square.

No of squares = aspect ratio l/wsince all squares must havesides of length w(in this case 2m). R= R

sl/wor l/w= R/R

s.

No of squares = Value of resistor/Rs

= 20k/1.25k = 16 squares. Length of resistor = No of squares x length of side of a square

=16 x 2m = 32m.

Alternatively to find the length:

R= Rsl/wor l= Rw/R

s= 20 x 103x 2 x 10-6/1.25 x 103= 32m.

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Range of Resistivity Values

Values cover 30 orders of magnitude!

Log Scale

22

20

12

10

2

0

-8

-10 m

cm

3.103m

Intrinsic Si

3.10-8m

Aluminium

Doped Si Insulators

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Electric Field

The gradient of thepotential.

The electric field, Ebetween two pointsdistance lapart withpotential difference Vis:

Ehas units of V m-1.

Also

Conductivity and Electric Field

Conductivity

The conductivity is thereciprocal of resistivity, i.e.the conductivity is givenby:

The S.I. unit is obviously -1 m-1, but the unit ofconductance (-1 ) is theSiemen (S), so conductivityactually has unit S m-1.

=1

E =

V

l

E=

dV

dx

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Current Density

The current density isthe current per unitarea.

IfIis the total currentflowing through anareaA, the currentdensity Jis:

Jhas units of A m-2.

J = IA

Divide electric field bycurrent density to get:

This is just a slightlydifferent form ofOhms Law.

E

J=

Vl

IA

= RA

l=

i.e. EJ=

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Electric Current

Current is a flow of charge.

We can measure current by measuring the chargethat flows by in unit time (I = Q/t).

Current is carried by charged particles (carriers)each carrying a known charge.

Current is therefore determined by three factors:

The amount of charge on each carrier. The number of charge carriers in unit volume(known as the carrier concentration or density).

The velocity of the charge carriers.

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Carrier Concentration

Electric current carried by charged particles(positive or negative).

Called carrier concentration nor p(dependingwhether the charge is negative or positive).

S.I. unit for nand pis carriers/m3(or simplym-3).

Practical unit used is cm-3

rather than the m-3

.

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Carrier Concentration

Each carrier has charge.

Negative charge carriers are electrons andcharge is electronic charge (q= 1.602 x 10-19C) .

Positive charge carriers have positive charge ofsame amount.

Sometimes charge carriers have multiples of q,

but not in normal conduction, and we will notconsider these more unusual situations.

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Mobility

Another important factor contributing to current.

Carrier mobility ().

Carriers accelerated by electric field, butdecelerated again by collisions with atoms.

Leads to a mean carrier velocity, , per appliedelectric field.

=

velocity

electricfield=

v

E

Units are m2v-1s-1. Practical unit is cm2v-1s-1.

v

The formula for mobility is: