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