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Engineering 45. Electrical Properties-1. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Learning Goals – Elect. Props. How Are Electrical Conductance And Resistance Characterized?

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Engineering 45

ElectricalProperties-1

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]


Learning goals elect props
Learning Goals – Elect. Props

  • How Are Electrical Conductance And Resistance Characterized?

  • What Are The Physical Phenomena That Distinguish Conductors, Semiconductors, and 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

()

Volt Meter

Bulk Matl

AmpMeter

I

Battery

Electrical Conduction

  • The Constant of Proportionality, R, is

    • Called the Electrical RESISTANCE

    • Has units of Volts/Amps, a.k.a, Ohms (Ω)

  • This Expression is known as Ohm’s Law


Electrical conduction cont

Fluid↔Current Electrical Current (I), and Electrical Potential (V) in Many Bulk MaterialsFlow Analogs

Electrical Conduction cont.

  • 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

    • Think of

      • Voltage as the “Electrical Pressure”


    Electrical resistivity

    Consider a Section of Physical Material, and Measure its Electrical Current (I), and Electrical Potential (V) in Many Bulk Materials

    Resistance

    Geometry

    Length

    X-Section Area

    Resistance, R

    Matl Prop → “”

    Area, A

    Length, L

    Electrical Resistivity

    • R↑ as L↑

      • R  L

    • R↑ as A↓

      • R  1/A

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


    Electrical resistivity cont

    Thus Expect Electrical Current (I), and Electrical Potential (V) in Many Bulk Materials

    Resistance, R

    Matl Prop → “”

    Area, A

    Length, L

    Electrical Resistivity cont.

    • 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 → [Ω-m2]/m

      • ρ → Ω-m


    Electrical conductivity

    conductANCE is the inverse of resistANCE Electrical Current (I), and Electrical Potential (V) in Many Bulk Materials

    Conductance, G

    Matl Prop → “”

    Area, A

    Length, L

    Electrical Conductivity

    • Similarly, conductIVITY is the inverse of resistIVITY

    • Units for 

      •  = 1/ρ → 1/ Ω-m

    • Now Ω−1 is Called a Siemens, S

      • σ → S/m


    Ohm related issues

    Recall Ohm’s Law Electrical Current (I), and Electrical Potential (V) in Many Bulk Materials

    L

    V

    Ohm Related Issues

    • J  Current Density in A/m2

    • E  Electric Field in V/m

      • In the General Case

    • E =J is the NORMALIZED, Resistive, Version of Ohm’s Law


    Normalized conductive ohm

    Recall Ohm’s Law Electrical Current (I), and Electrical Potential (V) in Many Bulk Materials

    L

    V

    Normalized, Conductive Ohm

    • Rearranging

    • G is Conductance

    • Recall also

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


    Some conductivities in s m

    Metals Electrical Current (I), and Electrical Potential (V) in Many Bulk Materials

    Conductivity (107 S/m)

    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  10-17


    Conductivity example
    Conductivity Example Electrical Current (I), and Electrical Potential (V) in Many Bulk Materials

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

    100m

    -

    e

    I = 2.5A

    +

    -

    Cu wire

    DV

    • Recall

    • Also G by σ & Geometry

    • For Cu: σ = 6.07x107 S/m


    Conductivity example cont
    Conductivity Example cont Electrical Current (I), and Electrical Potential (V) in Many Bulk Materials

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

    100m

    -

    e

    I = 2.5A

    +

    -

    Cu wire

    DV

    • Solve for D

    • Sub for Values


    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

    Electronic Conduction

    • 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

    Solid State Energy Band Theory

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


    Band theory cont

    The N atoms per m SOLID STATE with EQUILIBRIUM SPACING r3 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

    Band Theory, cont.

    • 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 SOLID STATE with EQUILIBRIUM SPACING r

    ΔE 15 eV

    N  5 x 1028 at/cu-m

    Then the difference between allowed Energy Levels within the Band, E

    Energy Band Calc

    • The Thermal Energy at Rm Temp is 25 meV/at, or about 1026 times E

      • Thus if bands are Not Completely Filled, e- can move easily between allowed levels


    Electronic conduction metals

    In Metals The Electronic Energy Bands Take One of Two Configurations

    Partially Filled Bands

    e- can Easily move Up to Adjacent Levels, Which Frees Them from the Atomic Core

    Overlapping Bands

    e- can Easily move into the Adjacent Band, Which also Frees Them from the Atomic Core

    Energy

    Energy

    empty

    band

    empty

    GAP

    band

    partly

    filled

    filled

    valence

    valence

    band

    band

    filled states

    filled states

    filled

    filled

    band

    band

    Electronic Conduction - Metals


    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

    V+

    E-Field

    V-

    Net e- Flow

    Current Flow

    Metal Conduction, Cont.

    • Metallic Conduction Model → Electron-Gas or Electron-Sea

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


    Insulators semiconductors

    Energy State”, and are Not Free to Leave the Atom Core

    Energy

    empty

    ConductionBand

    ConductionBand

    empty

    band

    band

    ?

    GAP

    GAP

    filled

    filled

    Valence

    valence

    band

    band

    filled states

    filled states

    filled

    filled

    band

    band

    7

    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


    Metals vs t vs impurities

    The Two Basic Components of Solid-St Electronic Conduction State”, and are Not Free to Leave the Atom Core

    The Number of FREE Electrons, n

    The Ease with Which the Free e-’s move Thru the Solid

    i.e. the electron Mobility, µe

    6

    Cu + 3.32 at%Ni

    5

    -m)

    Cu + 2.16 at%Ni

    4

    Resistivity, 

    deformed Cu + 1.12 at%Ni

    3

    -8

    Cu + 1.12 at%Ni

    (10

    2

    “Pure” Cu

    1

    0

    -200

    -100

    0

    T (°C)

    Metals -  vs T,  vs Impurities

    • Consider The ρ Characteristics for Cu Metals


    Metals vs t impurities cont

    Since “Double Ionization” of Atom Cores is difficult State”, and are Not Free to Leave the Atom Core

    n(Hi-T)  n(Lo-T)

    Thus T, Impurities and Defects must Cause Reduced µe

    These are all e- Scattering Sites

    Vacancies

    Grain Boundaries

    6

    Cu + 3.32 at%Ni

    5

    -m)

    Cu + 2.16 at%Ni

    4

    Resistivity, 

    deformed Cu + 1.12 at%Ni

    3

    -8

    Cu + 1.12 at%Ni

    (10

    2

    “Pure” Cu

    1

    0

    -200

    -100

    0

    T (°C)

    Metals -  vs T, Impurities cont

    • Impurities; e.g., Ni above

    • Dislocations; e.g., deformed


    Metal mathiessen s rule

    The Data Shows The Factors that Reduce State”, and are Not Free to Leave the Atom Coreσ

    Temperature

    Impurities

    Defects

    These Affects are PARALLEL Processes

    i.e., They Act Largely independently of each other

    Metal  - Mathiessen’s Rule

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


    Resistivity relations for metals

    Temperature Affects may be approximated with a Linear Expression

    Resistivity Relations for Metals

    • For A Single Impurity That Forms a Solid-Solution

    • 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%)

    • Where

      • 0 is the Resisitivity at the Baseline Temperature, Ω-m

      • a is the Slope of ρvs T Curve, Ω-m/K


    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

    ρRelations for Metals cont

    • 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

    Est. but in the Formation of a 2σfor a Cu-Ni alloy with yield strength of 125 MPa

    From Fig 7.19 Find Composition for Sy = 125 MPa

    180

    160

    140

    strength (MPa)

    120

    10

    0

    21 wt%Ni

    8

    0

    Yield

    60

    0

    10

    2

    0

    3

    0

    4

    0

    5

    0

    wt. %Ni, (Concentration C)

    r

    5

    0

    Ohm-m)

    40

    3

    0

    Resistivity,

    -8

    2

    0

    (10

    1

    0

    0

    0

    10

    2

    0

    3

    0

    4

    0

    5

    0

    wt. %Ni, (Concentration C)

    Example  Estimate σ

    • So need 21 wt% Ni

      • Find ρfrom Fig 18.9

    •   30x10-8Ω-m

      • And σ= 1/ρ

    •  σ= 3.3x106 S/m


    All done for today
    All Done for Today but in the Formation of a 2

    UsingBandGaps

    To Make

    LEDs


    Chabot Engineering but in the Formation of a 2

    Appendix

    Bruce Mayer, PE

    Licensed Electrical & Mechanical [email protected]


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


    Whiteboard work
    WhiteBoard Workhttp://www.chemistry.wustl.edu/~edudev/LabTutorials/PeriodicProperties/MetalBonding/MetalBonding.html

    • 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


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