mechanism of electric conductance in crystals electric conducta… · the ohm’s law george ohm...
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1
Mechanism of electric conductance
in crystals
2
A
Electric current in conductors
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
3
Electric current in crystalline conductors
Electric Field +++
---
t
QI =
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
4
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
5
Conductors, Insulators and Semiconductors
Metal:
~ 1023 atoms per 1 cm3
Every atom donates 1 free electron:
n ~ 1023 cm-3
Insulator:
~ 1023 atoms per 1 cm3;
No free electrons:
n ~ 0
Semiconductor:
~ 1023 atoms per 1 cm3;
Some atoms donate free
electrons
n ~ 1010 - 1019 cm-3
6
Example problem 1
0
of
5
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?
Timed response
120120
7
Example problem 2
0
of
5
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?
Timed response
120120
8
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)
9
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.
Electron in motion in conductors
10
Electron mobility in crystals
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 = µ E
µ is called the electron mobility: µ = v/E [(m/s)/(V/m) = m2/(V.s)]
Let us ignore the
random collisions
and only monitor
the drift in the
electric field
11
A
Electric current in conductors
Electron drift velocity increases with electric field:
v = µ E
Electric field
Ele
ctri
c cu
rre
nt
Higher electron concentration
12
A
Mechanism of electric conductance
A = wire cross-section area
n= free electron concentration, i.e. the number of electrons per unit
volume
L
∆∆∆∆L = = = = v.∆∆∆∆ t
In the ∆∆∆∆t time interval, how many electrons will cross the dashed area A?
∆Ν∆Ν∆Ν∆Ν = n (v ∆∆∆∆t) A
How much charge will pass through?
∆∆∆∆Q = e ∆Ν∆Ν∆Ν∆Ν = e n (v ∆∆∆∆t) A; where e = abs (electron charge)
Avnet
QI =
∆
∆=The electric current
Gnome is
counting the
electrons
E
If the conductor volume is A×L, the total number of electrons, N = n ×A×L
13
A A = wire cross-section area
v.
Anet
QI v=
∆
∆=The electric current
The electron drift velocity, v = µ E,from this,
Mechanism of electric conductance
( )EAneI µ=
Electric current is proportional to the electric field (external parameter)
Electric current is proportional to n, µ and A (conductor parameters)
E
14
A
v.
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
VL
AneV
L
AneI
== µµ
E
Mechanism of electric conductance
( )EAneI µ=
Using E = V/L:
15
VL
AneV
L
AneI
== µµ
Mechanism of electric conductance
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:
µσ ne=
Material properties are also often characterized by
resistivity:
1 1
enρ
σ µ= =
16
VL
AI σ=
Using conductivity or resistivity the current – voltage
relationship is:
The term
or VL
AI ⋅=
ρ
1
⋅
L
A
ρ
1does not depend on the applied voltage
but only on the material properties and the sample dimensions.
is called resistance.
Using the resistance, the I-V relationship is:
VR
I1
= or RIV ⋅=
These two relationships are called the Ohm’s law
A
LR ⋅= ρ
17
The Ohm’s Law
George Ohm has established experimentally in 1827 the following
law1
I VR
=
- + I
V
V is the voltage across a conductor
I is the current thought a conductor
1/R is a proportionality factor;
R is called “resistance”
Voltage
Ele
ctri
c
curr
en
t
Less resistance
18
Ohm’s Law and electron charge transfer rate
comparison
1I V
R=
Voltage
Ele
ctri
c
curr
en
t
Less resistance
Electric field
Ele
ctri
c cu
rre
nt
Higher e
lectron concentra
tion
v = µ E
19
Experimental observations of Ohm’s Law
1I V
R=
VoltageE
lect
ric
curr
en
t
Conductor length
decreases
Current increases,
i.e. R decreases
20
Experimental observations of Ohm’s Law
1I V
R=
VoltageE
lect
ric
curr
en
t
Conductor area
increases
Current increases,
i.e. R decreases
21
Ohm’s law, resistance and resistivity - summary
V I R= × 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;
µρ
ne
1=
A
LR ρ=
ρ describes the material ability to conduct the current.
The higher ρ is, the higher R and the lower the current
VIR
=Ohm’s law in the form of I(V): the current through the
conductor = applied voltage divided by the resistance
LR
Aρ=
22
The units for resistance and resistivity
A
LR ρ=
RIV ×=I
VR =
Resistance R is measured in Ohms (Ohm, ΩΩΩΩ)
1 Ohm is the resistance of the sample that passes the current of 1A
when the voltage of 1 V is applied across it.
⋅== mOhm
m
mOhm
L
AR
2
ρ
Resistivity is measured in Ohm - meters (Ohm· m)
23
Ohms’ law using conductance and conductivity
1G R/=Conductance
L
AG σ=
I G V= ⋅
Using the conductance, the Ohm’s law can be written as
Also, from 1G R/=
1enσ µ
ρ= = σ is the conductivity of the material
(does not depend on the conductor shape)
Using conductivity, the conductance of the sample is given by
11
AG R
L/
ρ= =
LR
A,ρ=and
24
Ohm’s Law using conductance - summary
material theofty conductivi the
(wire), sample theof econductanc
,
isne
theisL
AG
whereVGI
µσ
σ
=
=
×=
The expression that relates the electric current to the applied
voltage
is called the Ohm’s Law (established experimentally in 1827)
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
25
The units for conductance and conductivity
L
AG σ=
VGI ×=V
IG =
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.
== mS
m
mS
A
LG /
2σ
Conductivity is measured in Siemens per meter (S/m)
26
Ohm’s Law in the differential (“microscopic”) form:
( )EAneI µ=
or
I A Eσ=
where A is the cross-section area,
E is the electric field in the conductor
where σ = enµ is the conductivity of
the material
or
VI A
Lσ=
where V is the voltage across the conductor
and L is the conductor length in the direction
of current flow
27
Problem 3.
What is the current in the circuit below?
1 2 3 4
0% 0%0%0%
1. 3 A
2. 4 A
3. 36 A
4. 12 A
6060
28
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
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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 2 3 4
0% 0%0%0%
180180
1. 0.043 Ohm
2. 2.17*10-2 Ohm
3. 1.39 *10-2 Ohm
4. 435 S
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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
Conclusion: wires do not change the potentials in the circuit;
Voltage across the wire can be taken as 0 in most problems
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
31
Problem 5.
Find the resistance of a carbon resistor (ρ ρ ρ ρ = 1.4 × 10-5 Ohm*m), that is
1 cm long and 0.1 mm in diameter
1 2 3 4
0% 0%0%0%
180180
1. 0.178 Ohm
2. 17.8 Ohm
3. 1.78 Ohm
4. 0.0178 Ohm
32
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 dropacross 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.
33
Problem 6. Two wires are made of the same material.
The lengths and the diameters are as follows:
1 2 3
0% 0%0%
1 mm10 cmWire 1
2 mm20 cmWire 2
DiameterLength
Which wire, #1 or #2, has higher resistance?
1. Wire #1
2. Wire #2
3. Equal resistances
120120
34
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?
1 2 3
0% 0%0% 180180
L
d
W
I
1. 5.6e-5 Ohm
2. 0.14 Ohm
3. 3.5e-4 Ohm
35
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?
1 2 3
0% 0%0% 180180
L
d
W
I
1. 5.6e-5 Ohm
2. 0.14 Ohm
3. 3.5e-4 Ohm
36
Problem 9.
What voltage is needed
to generate the current of 2A in a resistor of 100 Ohm?
1 2 3 4
0% 0%0%0%6060
1. 50 V
2. 2 V
3. 200 V
4. 25 V
37
Problem 10.
What voltage is needed
to generate the current of 2A in a resistor of 0.01 Ohm?
1 2 3 4
0% 0%0%0%6060
1. 0.2 V
2. 0.02 V
3. 20 V
4. 25 V