Mechanism of electric conductance
in crystals
1
Electric current in conductors
A
When the conductor is in electric field, the field accelerates free
electrons
Electrons moving (drifting) in electric field transfer the electric
charge.
Moving electric charges create electric current through the
conductor
2
Electric current in crystalline conductors
-
Electric Field
Q
I=
t
+
+
+
Factors determining the electric current in crystalline conductors:
1. How many free electrons the crystal has
2. How fast the charge carriers (i.e. electrons)
can move in the electric field
3
How many free electrons are there in the conductor?
Metals have ~ 1023 atoms per 1 cm3
Every atom donates 1 free electron:
Therefore, every 1 cm3 of the metal contains ~ 1023 free electrons
n ~ 1023 cm-3 = 1023 (1/cm3)
n is the electron concentration
The concentration shows the number of particles per unit volume
4
Conductors, Insulators and Semiconductors
Metal:
~ 1023 atoms per 1 cm3
Insulator:
~ 1023 atoms per 1 cm3;
Semiconductor:
~ 1023 atoms per 1 cm3;
Every atom donates 1 free
electron:
No free electrons:
n ~ 1023 cm-3
Some atoms donate free
electrons
n~0
n ~ 1010 - 1019 cm-3
5
Example problem 1
The metal bar dimensions are 2mm x 3 mm x 0.5 mm;
Free electron concentration in the metal is 2x1023 cm-3.
How many free electrons the bar contains?
0
of
5
120
Timed response
6
Example problem 2
Thin-film wire is 2 µm thick, 5 mm long and 20 µm wide.
Free electron concentration in the metal is 5x1022 cm-3.
How many free electrons the wire contains?
0
of
5
120
Timed response
7
How fast the electrons can move:
electron mobility in crystals
Equilibrium condition, no electric field (voltage) applied
Free electron experiences very frequent collisions with atoms in the
metal.
As a result it moves randomly (with the velocity of around 105 m/s).
On average, the electron does not go anywhere!
Average electric current is equal to zero
(There is a flicker charge transfer, or the NOISE current though)
8
Electron in motion in conductors
Electric field applied:
There is an electric force, F = e E exerting on any free electron.
Electron still experiences very frequent random collisions.
However, after each collision the electron’s velocity has a component toward
the positive electrode (against the field direction)
On average, the electron drifts from negative electrode toward positive.
There is a current flowing through the metal.
9
Electron mobility in crystals
Let us ignore the
random collisions
and only monitor
the drift in the
electric field
Ignoring the collisions, which are completely random, we can say that
average electron velocity (drift velocity) is proportional to the electric
field applied:
v~E
v=µE
µ is called the electron mobility: µ = v/E [(m/s)/(V/m) = m2/(V.s)]
10
Electric current in conductors
A
Electron drift velocity increases with
electric field:
Electric current
v=µE
er
h
Hig
en
c
on
c
n
o
r
ct
e
l
e
n
o
i
t
tra
Electric field
11
Mechanism of electric conductance
∆L = v.∆ t
A
E
Gnome is
counting the
electrons
A = wire cross-section area
L
n= free electron concentration, i.e. the number of electrons per unit
volume
If the conductor volume is A×L, the total number of electrons, N = n ×A×L
In the ∆t time interval, how many electrons will cross the dashed area A?
∆Ν = n (v ∆t) A
How much charge will pass through?
∆Q = e ∆Ν = e n (v ∆t) A;
The electric current
I=
where e = abs (electron charge)
∆Q
=env A
∆t
12
Mechanism of electric conductance
v.
E
A
A = wire cross-section area
∆Q
The electric current I =
=env A
∆t
The electron drift velocity, v = µ E,
from this,
I = (e n µ A) E
Electric current is proportional to the electric field (external parameter)
Electric current is proportional to n, µ and A (conductor parameters)
13
Mechanism of electric conductance
E
v.
A
1. The external power source (the battery) creates a potential
difference along the conductor: V;
2. The potential difference (the voltage) creates an electric field
across the conductor
E = V/L;
3. Electric field accelerates free electrons and produces the
current
I = (e n µ A) E
Using E = V/L:
A
A

I = e n µ V =  e n µ V
L
L

14
Mechanism of electric conductance
A
A

I = e n µ V =  e n µ V
L
L

This product contains only the material parameters,
does not depend on applied voltage or on the geometrical
shape of the conductor.
It is called conductivity:
σ =enµ
Material properties are also often characterized by
resistivity:
1
1
ρ= =
σ enµ
15
Using conductivity or resistivity the current – voltage
relationship is:
A
I =σ V
L
The term
 1 A
 ⋅ 
ρ L
or
1 A
I= ⋅ V
ρ L
does not depend on the applied voltage
but only on the material properties and the sample dimensions.
L
is called resistance.
R= ρ⋅
A
Using the resistance, the I-V relationship is:
1
I= V
R
or
V =I ⋅R
These two relationships are called the Ohm’s law
16
The Ohm’s Law
George Ohm has established experimentally in 1827 the following
law
V is the voltage across a conductor
I is the current thought a conductor
1/R is a proportionality factor;
R is called “resistance”
+ I
V
Electric
current
1
I= V
R
r
s
s
Le
e
ce
n
ta
s
i
s
Voltage
17
Ohm’s Law and electron charge transfer rate
comparison
1
I= V
R
r
e
gh
i
H
ele
ra
t
en
c
n
o
c
n
o
r
t
c
it on
Electric field
Electric
current
Electric current
v=µE
r
s
s
Le
e
ce
n
ta
s
i
s
Voltage
18
Experimental observations of Ohm’s Law
Electric
current
1
I= V
R
Voltage
Conductor length
decreases
Current increases,
i.e. R decreases
19
Experimental observations of Ohm’s Law
Electric
current
1
I= V
R
Voltage
Conductor area
increases
Current increases,
i.e. R decreases
20
Ohm’s law, resistance and resistivity - summary
I =V
R
V = I ×R
R= ρ
L
A
Ohm’s law in the form of I(V): the current through the
conductor = applied voltage divided by the resistance
Ohm’s law in the form of V(I): the voltage needed to
maintain the current I = current times resistance.
L is the length of the conductor
in the direction of the current flow;
A is the cross-section area;
L
R=ρ
A
ρ describes the material ability to conduct the current.
The higher ρ is, the higher R and the lower the current
1
ρ=
enµ
21
The units for resistance and resistivity
V
R=
I
V = I ×R
Resistance R is measured in Ohms
(Ohm, Ω)
1 Ohm is the resistance of the sample that passes the current of 1A
when the voltage of 1 V is applied across it.
L
R=ρ
A

A 
m2
ρ =R
Ohm
= Ohm ⋅ m

L 
m

Resistivity is measured in Ohm
- meters (Ohm· m)
22
Ohms’ law using conductance and conductivity
Conductance
G =1 / R
Using the conductance, the Ohm’s law can be written as
I = G ⋅V
Also, from G =1 / R
σ=
1
ρ
=enµ
L
and R = ρ ,
A
1 A
G =1 / R =
ρ L
σ is the conductivity of the material
(does not depend on the conductor shape)
Using conductivity, the conductance of
the sample is given by
A
G =σ
L
23
Ohm’s Law using conductance - summary
The expression that relates the electric current to the applied
voltage
is called the Ohm’s Law (established experimentally in 1827)
I = G ×V , where
A
G =σ
L
σ =enµ
is the conductance of the sample (wire),
is the conductivity of the material
A is the conductor cross-section area;
L is the conductor length along the current direction
n is the concentration of free electrons
µ is the electron mobility
24
The units for conductance and conductivity
I = G ×V
I
G=
V
Conductance is measured in Siemens
(S)
1 S is the conductance of the sample that passes the current of 1A
when the voltage of 1 V is applied across it.
G =σ
A
L
L  m

σ =G
S
=
S
/
m

A  m2
Conductivity is measured in Siemens
per meter (S/m)
25
Ohm’s Law in the differential (“microscopic”) form:
I = (e n µ A) E
where A is the cross-section area,
E is the electric field in the conductor
or
I = σ AE
where σ = enµ is the conductivity of
the material
or
V
I=σ A
L
where V is the voltage across the conductor
and L is the conductor length in the direction
of current flow
26
Problem 3.
What is the current in the circuit below?
1.
2.
3.
4.
3A
4A
36 A
12 A
60
0%
1
0%
0%
2
3
0%
4
27
Resistivity of different materials
Wires are used to connect different components in the network;
Wires have very low resistance
Resistors are used to dissipate the power and to change the voltage
(potential).
Material
Electric Resistivity (×10-9 Ohm·m)
Aluminum [Al]
27
Aluminum Alloy
50
Brass
20 - 61
Carbon [C]
1.4 × 104
Copper [Cu]
17
Copper Alloy
17 - 490
Gold [Au]
24
Iron [Fe]
97
28
Problem 4.
Find the resistance of a wire made of copper [Cu] (ρ = 17*10-9 Ohm-m).
The wire is 1 m long and is 1 mm in diameter.
1. 0.043 Ohm
2.
2.17*10-2 Ohm
3. 1.39 *10-2 Ohm
4. 435 S
180
0%
1
0%
0%
2
3
0%
4
29
Solution
Find the resistance of a wire made of copper [Cu] (ρ = 17*10-9 Ohm-m).
The wire is 1 m long and is 1 mm in diameter.
R = ρ *L/A
L = 1 m; D = 1mm = 10-3 m;
The area, A = π*D2/4 = 3.14*(10-3)2/4 = 7.85*10-7 m2
The resistance
R = 17*10-9 Ohm*m*1m/ 7.85*10-7 m2 = 2.17*10-2 Ohm = 21.7 mOhm
Find the voltage across this wire, in the current through it is 1 A
V = I ×R = 1 A × 21.7 m Ohm = 21.7 mV
Conclusion: wires do not change the potentials in the circuit;
Voltage across the wire can be taken as 0 in most problems
30
Problem 5.
Find the resistance of a carbon resistor (ρ
ρ = 1.4 × 10-5 Ohm*m), that is
1 cm long and 0.1 mm in diameter
1.
2.
3.
4.
0.178 Ohm
17.8 Ohm
1.78 Ohm
0.0178 Ohm
180
0%
1
0%
0%
2
3
0%
4
31
Solution
Find the resistance of a carbon resistor, which is 1 cm long and 0.1 mm
in diameter
R = ρ *L/A; ρ = 1.4 × 104 * 10-9 Ohm*m = 1.4 × 10-5 Ohm*m ;
L = 1 cm = 10-2m; A = 7.85*10-9 m2
R = 1.4*10-5 Ohm*m*10-2 m/ 7.85*10-9 m2 = 17.8 Ohm
Compare: Cu – wire, R1 = 21.7 mOhm ≈ 0.02 Ohm;
Carbon-resistor, R2 = 17.8 Ohm ≈ 18 Ohm;
Given electric circuit with the current of 1 A, the voltage drop
across the Cu- wire, V1 = I*R1 = 0.02 V;
The voltage drop across the Carbon-resistor, V2 = I*R2 = 18 V;
Resistors significantly change the potentials in the circuit.
32
Problem 6. Two wires are made of the same material.
The lengths and the diameters are as follows:
Length
Diameter
Wire 1
10 cm
1 mm
Wire 2
20 cm
2 mm
Which wire, #1 or #2, has higher resistance?
1. Wire #1
2. Wire #2
3. Equal resistances
0%
1
0%
2
0%
3
120
33
Problem 7. The bar made of carbon (ρ = 1.4 × 10-5 Ohm*m)
The current flows through the bar in the direction shown by a blue arrow.
L = 10 cm; W=5 mm; d = 2 mm. What is the resistance of the bar?
d
L
I
W
1. 5.6e-5 Ohm
2. 0.14 Ohm
3. 3.5e-4 Ohm
0%
1
0%
2
0%
3
180
34
Problem 8. The bar made of carbon (ρ = 1.4 × 10-5 Ohm*m)
The current flows through the bar in the direction shown by a blue arrow.
L = 10 cm; W=5 mm; d = 2 mm. What is the resistance of the bar?
I
d
L
W
1. 5.6e-5 Ohm
2. 0.14 Ohm
3. 3.5e-4 Ohm
0%
1
0%
2
0%
3
180
35
Problem 9.
What voltage is needed
to generate the current of 2A in a resistor of 100 Ohm?
1.
2.
3.
4.
50 V
2V
200 V
25 V
60
0%
1
0%
0%
2
3
0%
4
36
Problem 10.
What voltage is needed
to generate the current of 2A in a resistor of 0.01 Ohm?
1.
2.
3.
4.
0.2 V
0.02 V
20 V
25 V
60
0%
1
0%
0%
2
3
0%
4
37