View
106
Download
2
Category
Tags:
Preview:
DESCRIPTION
Micro
Citation preview
MICROELECTRONICS TECHNOLOGY
Introduction
Microelectronics : Study of very small designs and componentsVLSI: Very Large Scale Integrated CircuitMicroelectronics (VLSI) technology: Study of how VLSI circuits are
made.Circuit: Discrete components soldered together
Ex: PCB Integrated Circuit: Both active and passive components housed in the
same substrateSubstrate: Slice of semiconductor that serves as the foundation for
construction of components
IntroductionClassification based on complexity
SSI :10-100 transistorsMSI : 100-1000 transistorsLSI : 1000-20,000 transistorsVLSI : 20,000-10,00,000 transistorsULSI : 10,00,000-1,00,00,000 transistorsGSI : >1,00,00,000 transistors
Elemental semiconductors: Silicon, GermaniumCompound semiconductors: InSb, InAs,GaAs,GaP,GaN,GaSb,SiCPassive components: Resistors, Capacitors etc.Active components: Transistors( BJT, MOSFET)
Material propertiesSilicon
ElementalWide Energy GapFormation of SiO2 : high-quality silicon dioxide can be grown
thermallyIndirect Energy GapEconomic Consideration
GaAsCompoundElectron mobility is higher: faster devices
TerminologiesLattice : Periodic arrangement of latticeUnit cell : representative of entire latticeBasis : same group of atomsCristal structure : lattice + basis
Crystal Structure- Simple Cubic
Ex: Polonium
structure Coordination Number=6Nearest neighbor distance: aAtoms per unit cell: (1/8)X8=1
000
x
y
z
010
011001
100110
101111
Crystal Structure- Body Centered
Ex. Molybdenum , Tentalum, Tungsten, Sodium
Structure
Coordination number=8Nearest neighbor distance:√3a/2
x
y
z
(1/2,1/2,1/2)a √3
a √2
a
Example:1 If we pack hard spheres in a bcc lattice such that the atom in the
center just touches the atoms at the corners of the cube, find the fraction of the bcc unit cell volume filled with hard sphere.
Spheres (atoms) per unit cell = (1/8) x 8 + 1= 2;Length of diagonal = √3aRadius of each sphere = √(3/4)aVolume of each sphere = 2X(4 / 3) X ∏X (a3 X 3X√3/64)= 0.67977 a3
Maximum fraction of unit cell filled = (0.67977) a3)/(a3)=0.68
a √2
aa √3
68% filled32% empty
Crystal Structure: Face centered
Copper, Gold , Nickel,Platinum, Silver
000001010011
0, 1/2, 1/2
1/2 1/2 0 (bottom)1/2 1/2 1 (top)0 1/2 1/2 (back)1 1/2 1/2 (front)1/2 0 1/2 (left)1/2 1 1/2 (right)
Nearest neighbor distance : a/√2
x
y
zCoordination number=12
Crystal Structure :Zinc blend Sphalerite Structure
Ex.: Si , GaAsCoordination number:4Distance between two neighboring atoms: (√3/4)a
000001010011
0,1/2,1/2
+1/4
Crystal Structure: Face centered 0 0 0
0 0 10 1 00 1 11 0 01 0 11 1 01 1 11/2 1/2 01/2 1/2 10 1/2 1/21 1/2 1/21/2 0 1/21/2 1 1/2
+1/4
1/4 1/4 1/41/4 1/4 5/41/4 5/4 1/41/4 5/4 5/45/4 1/4 1/45/4 1/4 5/45/4 5/4 1/45/4 5/4 5/43/4 3/4 1/43/4 3/4 5/41/4 3/4 3/45/4 3/4 3/43/4 1/4 3/43/4 5/4 3/4
Example:2A GaAs crystal is on a coordinate system such that an arsenic
atom sits at the position (000) and a gallium atom sits at (a/4,a/4,a/4) . Find the x,y,z coordinates of the other three neighbor gallium atoms to the arsenic atom at 0,0,0. What is the distance between these atoms?
1/4, 1/4, 1/4-1/4, -1/4, 1/41/4, -1/4, -1/4-1/4, 1/4, -1/4
Distance : √3a/4
Crystal planes and Miller Indices
Equation
Find the intercepts
Take reciprocalReduce(hkl)
1x y z
a b c
Miller Indices
Miller Indices :(100)Miller Indices : (110)
Miller Indices : (111)Miller Indices : (210)
3)
2)
1)
4)
Conventions
Angle between two planesCos(θ)=
Lines describing intersectionu=v1w2-v2w1v=w1u2-w2u1W=u1v2-u2v1
(hkl): negative x axis intersept
hkl: planes of equivalent symmetry
[hkl]: vector orthogonal to plane(hkl)
<hkl>: set of equivalent direction
2 2 2 2 2 2
u1u2+v1v2+w1w2
(u1 +v1 +w1 )(u2 +v2 +w2 )
The diamond Structure
•Every atom in the tetrahedral structure has four nearest neighbors•this structure that is the basic building block of the diamond lattice.
The clear diamond is pure blue contains a boron acceptor, yellow contains a nitrogen donor.
The diamond Structure
Crystal structure for silicon
Crystal structure for silicon
A p type <100> Silicon Wafer
Wafer flatsSilicon cut to 111 plane
Atomic bonding IonicCovalentMetallicVand der Waales
Metallic bond
Gallium Arsenide bonding
Ga(31) :2,8,18,3As(23) :2,8,18,5
Band DiagramThe photoelectric effect
Emission of radiation(Plank)Absorption of energy(Einstein)E=hV
Bohr Model of atomsEn=(-13.6)/n2
Lyman
Balmer
Pashen
Band Diagram : Continuous
Energy Momentum DiagramEffective mass
longitudinal mℓ =0.92 mtransverse mt = 0.19 m m the free electron mass
2
2
mvE
Metals , Insulators and semiconductors
ResistivityConductivitySheet Resistance
lR=ρ
A
*
1( )
s
n p
lR
w tl
Rt w
lR R
w
q n p
4-point probeRs = (V/I) * CF
Figure shows the top and side views of a
typical resistor with contacts at each end. If the sheet resistance of these diffused resistors were 50 ohms/square, then find the resistance of the body of the resistor.
450 Ω
Introduction: Phase DiagramsTo present properties of mixtures of materialsGold-Aluminium intermetallic
Interface between gold leads and aluminium-bonding padsWhite plague : low electric conductivityPurple plague :Voids in metal lattice
(1) Gold wire (2) Purple plague(3) Copper substrate(4) Gap eroded by wire-bond(5) Aluminum contact
IntroductionPhase diagrams: behavior of the combination of materials
fabrication process
Phase :State at which material exist
Equilibrium diagram : Phase at equilibrium conditions Phase diagrams: Phase at quasi equilibrium conditions
Composition Polarization
Unitary DiagramsPhase change in a single element
Function of temperature and pressureApply to compounds that undergo no chemical change over the range
for which the diagram is constructed
•Three line intersect at a common point
•Triple point :temperature and pressure at
which the three phases of that substance
coexist in thermodynamic equilibrium
•T= 273.16K P=0.6117P (water)
Water
Carbondioxide
Triple point:216.55 K,517PCritical point: Conditions at which no phase phase boundaries exist
Cooling from supercritical to critical
Binary DiagramsShows relationship between two components as a function of
temperaturePressure: 1atm
•The components α and β
•The possible phases are pure crystals of α, pure crystals of β, & liquid with compositions ranging between pure α and pure β.
The Lever rule
CM: initial composition(mole fraction of B in melt)
NL=number of moles of liquidNs = number of moles of solid CL,CS= composition of the liquid
and solid respectivelyNLCL and NSCS = number of moles
of B in the liquid and solid respectively
(NL+NS)CM= total number of moles of B= NLCL + NSCS
l,s are length of two lines
MS L
L S M
N C -C= =
N C -C s
l X-Y: Tie line
The Lever rule• The lever rule is used to determine the amount of each phase
exists at a given T and composition• Example: determine the amount of each in a two-phase region at
a given T and composition• The tie line at the given T is used as a lever and the composition
is used as a fulcrum• The amount of each phase is determined by the length of the
opposite arm
Recommended