Chapter 1 
ECE 4339-4119 - U Houston
Han Le - Copyrighted

1. Review of resistivity and conductivity

1.1 Resistance & conductance

Suppose we have a copper wire. We apply 1 V, and obtain 0.142 A.

1.1.1 Resistance

What is the wire resistance?

R=(V/Ic) /. V-> 1 /. Ic-> 0.142

1.1.2 Conductance

What is the wire conductance? Conductance is: G=(1/R):

(1/R)

1.2 Scaling the wire dimensions

1.2.1 Scaling the length

Suppose we double the length of the wire, what will be the resistance?

1.2.2 Scaling the wire cross section

Suppose we obtain a new wire with diameter twice the diameter of the current wire, what will be the resistance?

1.3 Resistivity and conductivity

Current= number of charges passing per unit time. Analogy: traffic and the road size. What happens when 4-lane traffic is converted into 2-lane traffic? or vice versa?
   

Resistance inversely decreases as a function of cross section area
                                                  R∝(1/A)

If we double the length, each half has only 1/2 voltage, hence, 1/2 the current.
The total resistance is thus doubled.
Resistance increases linearly as a function of length:
                                                  R∝(l/A)
There is a coefficient to make the two side equal:
                                                R=ρ(l/A)
ρ is called the resistivity. It is a property of the material, and has nothing to do with the length or cross section area of any particular piece of a material.
A related definition is conductivity, defined as:
                                              σ=(1/ρ)

Discussion: fundamentally, what determines the conductivity of a material? (we’ll discuss more in Chapter 3).

From periodic table (example: http://www.ptable.com/) 

({{Element, Conducting
electrons, Atomic W, Density 
(g/cm3), Conductivity
(1/Ohm-cm), Cond 
elec density ((cm^(-3)))}, {Au, 1, 196.966569, 19.30, 4.51671×(10^5), 5.900866179326097`*^22}, {Cu, 1, 63.546, 8.94, 5.959×(10^5), 8.472279545935227`*^22}, {Al, 3, 26.9815386, 2.70, 3.546×(10^5), 1.8078786840940202`*^23}, {Ni, 1, 58.6934, 8.908, 1.443×(10^5), 9.139909949895559`*^22}})

Source code: Review of resistance and conductance

Block[{MatpatmL,matlistname
,mpos,resistivity
,matlist,metalhue,INIT,RUN,L,A,r,R,G
,wiregr
}
,Manipulate[
If[INIT== 1,

MatpatmL=1./{ 5.959×(10^5),4.51671×(10^5),3.546×(10^5),1.443×(10^5) } ; 

matlistname={"Cu","Au","Al","Ni"};
metalhue={Hue[0.05,1,1],Hue[0.15,1,1]
,GrayLevel[0.8],GrayLevel[0.8]};
matlist=Table[Button[matlistname[[m]],BaseStyle->{14,FontFamily-> "Calibri"}
,Background->Hue[(m-1)*0.15+0.1,1.,1.]
],{m,1,Length[matlistname]}];
INIT=0 
,TextCell["Not ready"
,Black,20,Italic,FontFamily->"Calibri",Background->GrayLevel[0.5]]];
Manipulate[
If[RUN== 1,
Manipulate[
mpos=Position[matlist,mat][[1,1]];
resistivity=MatpatmL[[mpos]] ;
r=(Sqrt[A/π]) ;
wiregr=Graphics3D[{metalhue[[mpos]],Cylinder[{{0,0,0},{L,0,0}},5.r]}];
R= resistivity*L*100./A;
G=1./R  ;
Column[{
TextCell["wire transverse scale = 50 x longitudinal scale"
,Black,20,Italic,FontFamily->"Calibri",Background->Hue[0.5,1,1,0.5]],Grid[{{Show[wiregr, PlotRange-> {{0,10.},5.5*{-(Sqrt[0.1/π]),(Sqrt[0.1/π])}
,5.5*{-(Sqrt[0.1/π]),(Sqrt[0.1/π])}}
,Boxed-> False,ImageSize-> {300,200}]
,BarChart[{R},ChartStyle->Red ,PlotRange-> {0,0.75}
,AspectRatio->10., AxesLabel->{"","Resistance (Ω)"}]
,BarChart[{G},ChartStyle->Hue[0.38,1,1],PlotRange-> {0,600.}
,AspectRatio->10., AxesLabel->{"","Conductance (S)"}]}}]
}]
,{{mat,matlist[[1]],Button["Choose material ->",BaseStyle->{14, Bold,Black,FontFamily-> "Calibri"},Background->Hue[0.3,1,1,1]
]},matlist,ControlType->SetterBar},{{L,1.,Button["Length (cm)",BaseStyle->{14, Bold,White,FontFamily-> "Calibri"},Background->Hue[0.8,1,0.7]
]},1,10.}
,{{A,0.01,Button["Cross section (mm^2)",BaseStyle->{14, Bold,White,FontFamily-> "Calibri"},Background->Hue[0.17,1,0.4]
]},0.01,0.1}
]

,TextCell["Not running"
,Black,20,Italic,FontFamily->"Calibri",Background->GrayLevel[0.5]]]
,Row[{Button["Toggle box to start, and turn off when not running",BaseStyle->{16,White, Bold,FontFamily-> "Calibri"}
,Background->Black],Control[{{RUN,0,"->"},{0,1}}]
}]
]
,Button["REVIEW OF RESISTANCE/CONDUCTANCE",BaseStyle->{24,Black, Bold,FontFamily-> "Calibri"}
,Background->Hue[0.4,1,1]]
,Row[{Button["Click once to initialize",BaseStyle->{16,White, Bold,FontFamily-> "Calibri"}
,Background->Red],Control[{{INIT,0,"->"},{0,1}}]
}]
]
]

Demo: Review of resistance and conductance

Question: what do you think is the reason for semiconductors to have such a wide range of conductivity (but being neither good conductors nor good insulators)?

2. Crystal graphics

2.1 Basic crystallography

Demo

http://demonstrations.wolfram.com/SomeRepresentativeCrystalStructures/

http://demonstrations.wolfram.com/CubicCrystalLattices/

http://demonstrations.wolfram.com/DiamondLattice/

http://demonstrations.wolfram.com/TheStructureOfDiamond/

http://demonstrations.wolfram.com/SpherePacking/

2.2 Crystallographic planes

http://demonstrations.wolfram.com/CrystallographicPlanesForCubicLattices/

http://demonstrations.wolfram.com/MillerIndicesForASimpleCubicLattice/

http://demonstrations.wolfram.com/ReciprocalLattice2D/

http://demonstrations.wolfram.com/GrapheneBrillouinZoneAndElectronicEnergyDispersion/

Homework

Source code: 2D crystal structure

Block[{off,on,b,atomA, atomB ,U,V,lg1,lg2,pU,pV,pg},
Manipulate[
If[INIT== 1,
{off,on}={Button["OFF",BaseStyle->{16, Bold,White,FontFamily-> "Arial"}
,Background->Red]
,Button["ON",BaseStyle->{16, Bold,White,FontFamily-> "Arial"}
,Background->Hue[0.36,1,0.4]]} ;
atomA=Graphics[{Blue,Disk[{0,0},0.11] },ImageSize->{27.5,27.5}];
atomB=Graphics[{Red,Disk[{0,0},0.08]} ,ImageSize->{20,20}];
INIT=0 
,TextCell["Not ready"
,Black,20,Italic,FontFamily->"Calibri",Background->GrayLevel[0.5]]];
Manipulate[
If[RUN== 1,

Column[{
TextCell["Vary various parameters below to create a 2D crystal\nHomework: create lattice for graphene"
,Black,20,Italic,FontFamily->"Calibri",Background->Hue[0.15,1,1]]
,
Manipulate[
b= bL*{Cos[bθ*Pi/180.],Sin[bθ*Pi/180.]} ;
lg1=Table[Line[{j *b,j *b+{na*a,0}}],{j,0,nb}];
pU= rU[[1]]*{a,0}+rU[[2]]*b;
pV= rV[[1]]*{a,0}+rV[[2]]*b;
lg2=Table[Line[{{i*a,0},{i*a,0}+nb *b}],{i,0,na}];
pg=Table[Point[(i-1)*{a,0}+(j-1)*b],{i,1,na+1},{j,1,nb+1}] ;
range={{-0.5,1.1*(na*a+nb*b[[1]])},{-0.5,1.1*(nb*b[[2]])} };
U=Table[Disk[(i-1)*{a,0}+(j-1)*b+pU,0.11],{i,1,na+1},{j,1,nb+1}] ;
If[TrueQ[aV=="None"]==False,
V={If[TrueQ[aV==atomA],Blue,Red],Table[Disk[(i-1)*{a,0}+(j-1)*b+pV,
If[TrueQ[aV==atomA],0.11,0.08]],{i,1,na+1},{j,1,nb+1}]}
,V={PointSize[0.0001],Point[{0,0}] }] ; 
Graphics[{Thickness[0.005],Hue[0.35,1,0.5],Arrow[{{0,0},{a,0}}],Hue[0.1,1,0.8],Arrow[{{0,0},b}]
,If[TrueQ[grid==on],{Thickness[0.001],{Orange,lg2},{Green,lg1},pg},]
,If[TrueQ[atoms==on],{Blue,U,V}, ]}
,PlotRange-> range]
,Grid[{{Button["atom U",BaseStyle->{14,Black Bold,FontFamily-> "Calibri"}
,Background->Hue[0.05,0.4,1]]
,atomA,Spacer[20]
,Button["atom V",BaseStyle->{14,Black Bold,FontFamily-> "Calibri"}
,Background->Hue[0.7,0.4,1]]
,Control[{{aV,atomB,""},{atomB,atomA,"None"}
}]}}]
,Grid[{{Control[{{a,1,"Basis vector a"},0.5,1.5}],Control[{{bL,1,"Basis vector b length"},0.5,1.5}]}
,{
,Control[{{bθ,90.,"Basis vector b angle"},45.,90.}]}
,{Control[{{na,4,"Number of unit cell along a"},{2,3,4,5,6,7},ControlType->SetterBar}],Control[{{nb,4,"Number of unit cell along b"},{2,3,4,5,6,7},ControlType->SetterBar}]}
,{Control[{{rU,{0.5,0.5}},{0,0},{1,1}}],Control[{{rV,{1,1}},{0.25,0.},{1,1}}]}
,{Control[{{grid,off},{on,off}}],Control[{{atoms,off},{on,off}}]}},Spacings->{2,1}]
]
}]

,TextCell["Not running"
,Black,20,Italic,FontFamily->"Calibri",Background->GrayLevel[0.5]]]
,Row[{Button["Toggle box to start, and turn off when not running",BaseStyle->{16,White, Bold,FontFamily-> "Calibri"}
,Background->Black],Control[{{RUN,0,"->"},{0,1}}]
}]
]
,Button["Ch. 1: Homework: Illustration of a 2D crystal",BaseStyle->{24,Black, Bold,FontFamily-> "Calibri"}
,Background->Hue[0.4,1,1]]
,Row[{Button["Click once to initialize",BaseStyle->{16,White, Bold,FontFamily-> "Calibri"}
,Background->Red],Control[{{INIT,0,"->"},{0,1}}]
}]
]
]

Demo 2D crystal - HW graphene

3. Defect: edge dislocation

http://demonstrations.wolfram.com/MovementOfAnEdgeDislocation/