Overview
Topics:
•Resistivity
Resistivity, Sheet Resistance and
•Sheet resistance
Mobility
•Examples
•Conductivity and Electric Field
•Current Density
•Carrier concentration
•Mobility
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Resistivity
Resistivity
•Resistivity is the property of a material that
•Calculate resistance between front and back faces :
determines the electrical resistance of a specific
sample.
R= ρ
l
l
w× h
= ρ
l
A
•A is the area through which the current must pass
and can be used where the cross-section is not
h
rectangular, e.g. for a wire with circular crosssection.
•If length increases, resistance increases and vice
w
versa. Opposite for area.
•The block of material has resistance (R) and
•The S.I. unit of resistivity is the Ω m, but the
resistivity (ρ).
practical unit is the Ω cm.
• ρ remains constant regardless of size or shape.
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Sheet resistance
Sheet Resistance
R = 4 Rs
•In IC technology we often have conducting layers
with constant depth (h in figure).
R =
•Resistance therefore depends upon the aspect ratio
as viewed from the surface (l/w).
•Define sheet resistance (R ) so that resistance is:
s
•R
s
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The relation between ρ and R
s
R=R
Rs
l
R
w
s
But since l = w:
is given in Ω /square where l = w.
l
w
s
is:


l
= ρ

w×h
ρ =R × h
s
Using sheet resistance, define resistance of an area
(e.g. in a silicon IC) by number of squares.
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Example 2
Example 1
An aluminium rod 30cm long with a cross-section of
An area of undoped (high resistivity) silicon contains
0.5cm x 0.25cm has a voltage of 100mV applied across
a region of doped silicon (so much lower resitivity) for
it and a current of 13.9A is observed to flow. Calculate
fabricating a 20 kΩ resistor, as shown in the figure. If
the resistivity, ρ, of the aluminium.
the resistivity, ρ, of the doped silicon is 10
-3
Ω m,
calculate the sheet resistance, R , the number of
s
•R = V/I = 0.1/13.9A = 7.2 x 10
-4
squares of doped silicon required for the resistor, and
Ω
hence the length of the resistor.
•R = ρl/wh or ρ = Rwh/l
so ρ = 7.2 x 10
-4
x 0.005 x 0.0025/0.3 = 3 x 10
-8
Ω m.
2µm
0.8µm
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Example 2 Solution
Range of Resistivity Values
• We can assume that the doped silicon is a conducting area
surrounded by insulator.
• Rs = ρ/h = 10
-3
-6
/0.8 x 10
Aluminium
3
= 1.25 x 10 Ω/square = 1.25kΩ/square.
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Intrinsic Si
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Doped Si
• No of squares = aspect ratio l/w since all squares must have
sides of length w (in this case 2µm). R = Rsl/w or l/w = R/Rs.
-10
0
3.10 -8 Ω m
• No of squares = Value of resistor/Rs = 20k/1.25k = 16 squares.
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Insulators
3.10 3 Ω m
10
Ω cm
Log Scale
20
Ω m
• Length of resistor = No of squares x length of side of a square
=16 x 2µm = 32µm.
Values cover 30 orders of magnitude!
• Alternatively to find the length:
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R = Rsl/w or l = Rw/Rs = 20 x 10
-6
x 2 x 10
/1.25 x 10
3
= 32µm.
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Conductivity and Electric Field
Conductivity
Electric Field
• The conductivity is the
reciprocal of resistivity, i.e.
the conductivity σ is given
by:
σ =
-1
-1
m
the current per unit
1
• The electric field, E
-1
) is the
Siemen (S), so conductivity
J =
.
I
i.e.
-2
.
• Also
E= −
dV
E
J
A
•J has units of A m
-1
actually has unit S m
I
= R
density J is:
l
-1
J
V
=
area A, the current
V
• E has units of V m
E
flowing through an
potential difference V is:
, but the unit of
current density to get:
•If I is the total current
distance l apart with
E=
•Divide electric field by
area.
potential.
between two points
ρ
conductance (Ω
•The current density is
• The gradient of the
• The S.I. unit is obviously
Ω
Current Density
.
l
A
A
l
= ρ
= ρ
•This is just a slightly
different form of
dx
Ohm’s Law.
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Electric Current
Carrier Concentration
•Current is a flow of charge.
•Electric current carried by charged particles
(positive or negative).
•We can measure current by measuring the charge
that flows by in unit time (I = Q/t).
•Called carrier concentration n or p (depending
whether the charge is negative or positive).
•Current is carried by charged particles (carriers)
each carrying a known charge.
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•S.I. unit for n and p is carriers/m
•Current is therefore determined by three factors:
(or simply
-3
m
– The amount of charge on each carrier.
).
•Practical unit used is cm
-3
rather than the m
-3
.
– 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
Mobility
•Another important factor contributing to current.
•Each carrier has charge.
•Carrier mobility (µ).
•Negative charge carriers are electrons and
charge is electronic charge (q = 1.602 x 10
-19
C) .
•Carriers accelerated by electric field, but
decelerated again by collisions with atoms.
•Positive charge carriers have positive charge of
•Leads to a mean carrier velocity, v , per applied
same amount.
electric field.
•Sometimes charge carriers have multiples of q,
but not in normal conduction, and we will not
•The formula for mobility is:
consider these more unusual situations.
µ =
2
•Units are m
v
-1
s
velocity
electric field
-1
=
v
E
. Practical unit is cm
2
v
-1
s
-1
.
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