Lec 20

# Lec 20 - ENGR ENGR 4250 Advanced Materials Engineering...

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ENGR 4250 ENGR 4250 – Advanced Materials Engineering Lecture 20 Electrical Properties

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Chapter 18: Electrical Properties SUES TO ADDRESS ISSUES TO ADDRESS. .. How are electrical conductance and resistance haracterized characterized ? What are the physical phenomena that distinguish conductors, semiconductors, and insulators? For metals, how is conductivity affected by imperfections, T , and deformation?
Electrical Conduction Ohm's Law: Δ V = I R voltage drop (volts = J/C) resistance (Ohms) C = Coulomb current (amps = C/s) I e - A (cross sect. area) Δ V L

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Electrical Properties • Which will conduct more electricity? D A A 2 D l l I VA RA = = ρ Analogous to flow of water in a pipe • So resistance depends on sample eometry etc geometry, etc.
Electrical Conduction Ohm's Law: Δ V = I R voltage drop (volts = J/C) resistance (Ohms) C = Coulomb current (amps = C/s) I e - A (cross sect. Resistivity, ρ and Conductivity, σ : -- eometry-independent forms of Ohm's Law area) Δ V L geo e y depe de o s o O s a E : electric resistivity hm ) -- Resistivity is a material property & is independent of sample Δ I V field intensity (Ohm-m) J : current density ρ = A L Resistance: conductivity σ = 1 ρ σ = ρ = A L A L R

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Definitions Further definitions J = σ ε <= another way to state Ohm’s law J current density flux a like current I = = ε electric field potential = V / l or ( Δ V / Δ l ) area surface A J = σ ( Δ V / Δ l ) Current carriers Electron flux conductivity voltage gradient • electrons in most solids • ions can also carry (particularly in liquid solutions)

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Conductivity: Comparison
Conductivity: Comparison Room T values (Ohm-m) -1 METALS conductors = ( Ω -m) -1 CERAMICS 0 1 Silver 6.8 x 10 7 Copper 6.0 x 10 7 Soda-lime glass 10 Concrete 10 -9 3 -10 -10 -11 Iron 1.0 x 10 7 Aluminum oxide <10 -13 Silicon 4 x 10 -4 SEMICONDUCTORS Polystyrene <10 -14 5 POLYMERS Germanium 2 x 10 0 GaAs 10 -6 Polyethylene 10 -15 -10 -17 sulators Selected values from Tables 18.1, 18.3, and 18.4, Callister 7e . semiconductors insulators

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Example: Conductivity Problem What is the minimum diameter ( D ) of the wire so that Δ V < 1.5 V? 100m Cu wire I = 2.5A - + e - Δ V < 1.5V 2.5A 100m V L R Δ = = olve to get 187mm 6.07 x 10 (Ohm-m) 7- 1 I A σ 4 2 D π Solve to get D > 1.87 mm
Conduction Electronic: Solids Ionic: Liquids

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Electron Energy States

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Electronic Band Structures Adapted from Fig. 18.2, Callister 7e .

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Band Structure Valence band – filled – highest occupied energy levels Conduction band – empty – lowest unoccupied energy levels alence band Conduction band valence band Adapted from Fig. 18.3, Callister 7e .
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Lec 20 - ENGR ENGR 4250 Advanced Materials Engineering...

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