Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • ENGR-45_Lec-08_ElectProp-Metals.ppt1

Bruce Mayer, PE Engineering-45: Materials of Engineering

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 45

ElectricalProperties

-1



Learning Goals – Elect. Props How Are Electrical Conductance And

Resistance Characterized? What Are The Physical Phenomena That

Distinguish Conductors, SemiConductors, and NonConductors (i.e., Insulators)?

For Metals, How Is Conductivity Affected By Imperfections, Temp, and Deformation?

For Semiconductors, How Is Conductivity Affected By Impurities (Doping) And Temp?



Electrical Conduction Georg Simon

Ohm (1789-1854) First Stated a Relation for Electrical Current (I), and Electrical Potential (V) in Many Bulk Materials

The Constant of Proportionality, R, is• Called the Electrical

RESISTANCE• Has units of Volts/Amps,

a.k.a, Ohms (Ω)

RIV This Expression is

known as Ohm’s LawBattery

Bulk Matl

Volt Meter

AmpMeter

I

()

http://www.physicslessons.com/ohmslaw-equation.gif








http://images.google.com/imgres?imgurl=www.mines.edu/fs_home/tboyd/GP311/MODULES/RES/NOTES/ohm.gif&imgrefurl=http://www.mines.edu/fs_home/tboyd/GP311/MODULES/RES/NOTES/ohm.html&h=226&w=143&prev=/images?q=ohm's+law&start=80&svnum=10&hl=en&lr=&ie=UTF-8&oe=UTF-8&sa=N



Electrical Conduction cont. Fluid↔Current

Flow Analogs

Think of • Voltage as the

“Electrical Pressure”

• Current as the “Electrical Fluid”

• Wire as the “ Electrical Pipe”

Just as a Small Pipe “Resists” Fluid Flow, A Small Wire “Resists” current Flow• Thus Resistance is a

Function of GEOMETRY and MATERIAL PROPERTIES– Next Discern the

Resistance PROPERTY

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgele/ohmpoi.gif



Electrical Resistivity Consider a

Section of Physical Material, and Measure its• Resistance• Geometry

– Length– X-Section Area

Thinking Physically, Since R is the Resistance to Current Flow, expect

R↑ as L↑• R L

R↑ as A↓• R 1/A

Area, A

Length, L

Matl Prop → “”

Resistance, R



Electrical Resistivity cont. Thus Expect

This is, in fact, found to be true for many Bulk Materials

Convert the Proportionality () to an Equality with the Proportionality Constant, ρ

Units for ρ• ρ → [Ω-m2]/m• ρ → Ω-m

Area, A

Length, L

Matl Prop → “”

Resistance, R

AL

R

ALR



Electrical Conductivity conductANCE is

the inverse of resistANCE

Similarly, conductIVITY is the inverse of resistIVITY

Units for σ• σ = 1/ρ → 1/ Ω-m

Now Ω−1 is Called a Siemens, S• σ → S/m

Area, A

Length, L

Matl Prop → “σ”

Conductance, G

RG 1

1

LA

LAG

1R1



Ohm Related Issues Recall Ohm’s Law

E = ρJ is the NORMALIZED, Resistive, Version of Ohm’s Law

J Current Density in A/m2

E Electric Field in V/m• In the General Case

JEAI

I

LV

orALRIV L

V

dLdVE



Normalized, Conductive Ohm Recall Ohm’s Law

G is Conductance Recall also

ReArranging

VGV1RV

orALRIV

RI

IL

V

VVGI

then

LA

LAG

EJ

LA

VI

• J = σE is the Normalized, Conductive Version of Ohm’s Law



Some Conductivities in S/m Metals 107

SemiConductors• Si (intrinsic) 10−4

• Ge 100 = 1• GaAs 10− 6 • InSb 104

Insulators• SodaLime Glass 10− 11

• Alumina 10− 13

• Nylon 10−13

• Polyethylene 10−16

• PTFE 10−170

1

2

3

4

5

6

Cu Al Brass SS

Con

duct

ivity

(107 S

/m)

Metals



Conductivity Example

Recall

What is the minimum diameter (D) of a 100m wire so that ΔV < 1.5 V while carrying 2.5A?

100mCu wire I = 2.5A- +e-

V

VI G orVG

I

Also G by σ & Geometry

VI

LD

LAG

42

For Cu: σ = 6.07x107 S/m



Conductivity Example cont

Solve for D

What is the minimum diameter (D) of a 100m wire so that ΔV < 1.5 V while carrying 2.5A?

100mCu wire I = 2.5A- +e-

V

Sub for Values VILD

VI

LD

4

4

2

mm86981m

mm1000

V51mV

A10076

m100A5247

...

.

D



Electronic Conduction As noted In Chp2

Electrons in a FREE atom Can Reside in Quantized Energy Levels• The Energy

Levels Tend to be Widely Separated, Requiring significant Outside Energy To move an Electron to the next higher level

In The SOLID STATE, Nearby Atoms Distort the Energy LEVELS into Energy BANDS• Each Band Contains

MANY, CLOSELY Spaced Levels



Solid State Energy Band Theory Consider the 3s Energy Level, or Shell,

of an Atom in the SOLID STATE with EQUILIBRIUM SPACING r0

By the Pauli Exclusion Principle Only ONE e− Can occupy a Given Energy Level



Band Theory, cont. The N atoms per

m3 with Spacing r0 produces an Allowed-Energy BAND of Width ΔE

Most Solids have N = 1028-1029 at/m3

Thus the ΔE wide Band Splits into 1029/m3 Allowed E-Levels

Leads to a band of energies for each initial atomic energy level • e.g., 1s energy band

for 1s energy level



Energy Band Calc Given

• ΔE 15 eV• N 5 x 1028

at/cu-m Then the

difference between allowed Energy Levels within the Band, δE

The Thermal Energy at Rm Temp is 25 meV/at, or about 1026 times δE• Thus if bands are

Not Completely Filled, e− can move easily between allowed levelsatmeVE

mateVNEE

328

328

103

10515



Electronic Conduction Metals In Metals The

Electronic Energy Bands Take One of Two Configurations

1. Partially Filled Bands• e− can Easily move Up

to Adjacent Levels, Which Frees Them from the Atomic Core

2. Overlapping Bands• e− can Easily move into

the Adjacent Band, Which also Frees Them from the Atomic Core

Energy

filled band

filled valence band

empty band

fille

d st

ates

filled band

Energy

partly filled valence band

empty band

GAP

fille

d st

ates



Metal Conduction, Cont. Atoms at Their

Lowest Energy Condition are in the “Ground State”, and are Not Free to Leave the Atom Core

In Metals, the Energy Supplied by Rm Temp Can move the e− to a Higher Level, making them Available for Conduction

Metallic Conduction Model → Electron-Gas or Electron-Sea

Net e- FlowCurrent Flow

E-Field V-V+

• Note: e−’s Flowing “UPhill” constitutes Current Flowing DOWNhill

http://www.bcpl.net/~kdrews/periodic%20chart/e_sea.gif



Insulators & Semiconductors Insulators:

• Higher energy states not accessible due to lg gap– Eg > ~3.5 eV

Semiconductors:• Higher energy states

separated by smaller gap– Eg < ~3.5 eV

Energy

filled band

filled Valence band

empty band

fille

d st

ates

GAPConduction

Band

7

Energy

filled band

filled valence band

empty band

fille

d st

ates

GAP?Conduction

Band



Metals: ρ vs T, ρ vs Impurities The Two Basic

Components of Solid-St Electronic Conduction• The Number of

FREE Electrons, n

• The Ease with Which the Free e−’s move Thru the Solid– i.e. the electron

Mobility, µe

Consider The ρ Characteristics for Cu Metal & Alloys

T (°C)-200 -100 0

Cu + 3.32 at%Ni

Cu + 2.16 at%Ni

deformed Cu + 1.12 at%Ni

123456

Resis

tivity

, ρ

(1

0-8

-m

)0

Cu + 1.12 at%Ni

“Pure” Cu

charge electronic theis q Where

1 ee nqnq



Metals - vs T, Impurities cont Since “Double

Ionization” of Atom Cores is difficult• n(Hi-T) n(Lo-T)

Thus T, Impurities and Defects must Cause Reduced µe

• These are all e- Scattering Sites– Vacancies– Grain Boundaries

T (°C)-200 -100 0

Cu + 3.32 at%Ni

Cu + 2.16 at%Ni

deformed Cu + 1.12 at%Ni

123456

Resis

tivity

, ρ

(1

0-8

-m

)

0

Cu + 1.12 at%Ni

“Pure” Cu

– Impurities; e.g., Ni above– Dislocations; e.g.,

deformed



Metal - Mathiessen’s Rule The Data Shows

The Factors that Reduce σ• Higher

Temperature• Impurities• Defects

These Affects are PARALLEL Processes• i.e., They Act

Largely independently of each other

The Cumulative Effect of ||-Processes is Calculated by Mathiessen’s Rule of Reciprocal Addition

diTtotal

diTtotal

or

1111



Resistivity Relations for Metals Temperature

Affects may be approximated with a Linear Expression aTT 0• Where

– 0 is the Resisitivity at the Baseline Temperature, Ω-m

– a is the Slope of ρ vs T Curve, Ω-m/K

For A Single Impurity That Forms a Solid-Solution

iii cAc 1• Where

– A is an Alloy-Specific Constant, Ω-m/at-frac

– ci is the impurity Concentration in in the atomic-fraction Format At-frac = at%x(1/100%)



ρ Relations for Metals cont In alloys where the impurity results, not

in Solid-Solution, but in the Formation of a 2nd Xtal Structure, or Phase, Use a Rule-of-Mixtures Relation for ρi

• Use Vol-Fractions as the Weighting Factor

2211 VVi • Where

– ρk is the Resistivity of phase-k

– Vk is the Volume-Fraction of phase-k

Plastic Deformation• There is no Simple

Relation for This– Consult individual metal

or alloy data



Example Estimate σ Est. σ for a Cu-Ni

alloy with yield strength of 125 MPa• From Fig 7.16

Find Composition for Sy = 125 MPa

So need 21 wt% Ni• Find ρ from Fig 18.9

Yiel

d stre

ngth

(MPa

)

wt. %Ni, (Concentration C)0 10 20 30 40 5060

80100120140160180

21 wt%Niwt. %Ni, (Concentration C)

Resis

tivity

,

(10-

8 Ohm

-m)

10 20 30 40 500

1020304050

0

ρ 30x10-8 Ω-m• And σ = 1/ρ

σ = 3.3x106 S/m



All Done for Today

UsingBandGapsTo Make

LEDs

http://www.chemistry.wustl.edu/~edudev/LabTutorials/PeriodicProperties/MetalBonding/diode_movie.html




Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Engineering

Appendix



http://www.chemistry.wustl.edu/~edudev/LabTutorials/PeriodicProperties/MetalBonding/MetalBonding.html





WhiteBoard Work Derive Relation for e- Drift Velocity, vd

Calculate the Drift Velocity in a 20 foot Gold Wire Connected to a 9Vdc Batt• Assume Au Atoms in the Solid Are Singly

Ionized, contributing 1 conduction-e- per atom (monovalent)

Compare (random) THERMAL Velocity

)000260(/1173003 2

1

mphskmKvmkTv

Te

eTe

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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