Computational Plasma Physics

Kinetic modelling: Part 1 & 2

W.J. Goedheer

FOM-Instituut voor PlasmafysicaNieuwegein, www.rijnh.nl

What are kinetic methods and when do they apply

Kinetic methods retain information on the velocity distribution(hydrodynamic/fluid methods first integrate over velocity space)

Needed when distribution is non-Maxwellian

Kinetic methods are to be preferred when “mfp> L” or “coll>T”

mfp and coll depend on densities and cross-sections

But what are L and/or T??

Examples from (plasma) physics?

Kinetic models: non-Maxwellian

Collisions electrons mainly with neutral species

Low degree of ionization

Effective cooling of parts of the energy distribution function

Counteracted by Coulomb collisions at high degree of ionization (>10%)

E

Variations in space and time

L

High T Low T

x

T

T

High Pow Low Pow

t

P

Boundary layersTransition layers

Transient phenomenaSwitching onModulation

Power modulated discharges

Modulate RF voltage (50MHz)with square wave (1 - 400 kHz)

Observation in experiments UU)optimum in deposition rate

RF

Modulated discharge in SiH4

Results from a PIC/MC calculation: Cooling and high energy tail

Examples of Ion Energy DF at grounded electrode

From Th. Bisschops, Thesis TU/e, 1987

Interaction between E-field and ion motion does not result in a shifted Maxwellian

Kinetic models: strong spatial variation

Very low pressures: L = size of vessel (applies for [e,i,n])

Space charge boundary layer: L = Debye length (applies for [e,i])

Micro-structures (etched trenches): L = size of structure (applies for [e,i,n])

Shocks: L = extension of shock (applies for [e,i,n])

There may be a difference between momentum loss and energy loss

Kinetic models: strong temporal variation

Microwave discharges / high frequency RF discharges (applies for [e,(i)])

Start-up of discharges (applies for [e,i])

There may again be a difference between momentum loss and energy loss

Methods based on direct solution ofthe Boltzmann equation

Tricks to solve BE: Use symmetry if present Expand f in some small parameter

A method especially suitable for electrons

Electrons have a low mass high momentum loss in collisions energy loss in inelastic collisions

Elastic scattering redistribution over a sphere in velocity spaceSmall deviation from isotropic f in the direction of the average velocity

Therefore: expansion in Legendre polynomials Pn(cos )with the angle between average and actual velocityf = f0(v) + f1(v)P1(cos ) + f2(v)P2(cos )+…….

Note: “Amplitudes” depend on absolute value velocityand vary in space and time

An example: f0+f1cos()

-4 -2 0 2 4

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-2

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f0=v*exp(-v)

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-0.08000-0.07000-0.06000-0.05000-0.04000-0.03000-0.02000-0.0100000.010000.020000.030000.040000.050000.060000.070000.08000

f1 cos() =0.1v*exp(-v/2)cos()

f0+f1cos()

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How to calculate the amplitudes fn

V

V+dv

Transport from Neighbouring volume

Shift in velocitydue to electric field(eEt/m)

Elastic collisons move electrons from outside in

Net change in t

How to calculate the amplitudes fn

Balance of number of particles in shell between v and v+dv in dxdydzTransport in real spaceTransport in velocity spaceEffect of collisions

How to calculate the amplitudes fn

Balance of momentum in shell between v and v+dv in dxdydzTransport in real spaceTransport in velocity spaceEffect of collisions

Note that f1 is a vector, directed along the average velocity (=0)

Cutting off at f1: The Lorentz approximation

For elastic collisions with atoms/molecules, with mass M:

Special case: homogeneous, steady state

Temperature gas is zeroConstant electric field, average velocity (f1) along E

Special case: homogeneous, steady state

Solution: Backward integration, tri-diagonal system …

Special solutions:

r=-1 ; s=2 : Maxwellr= 0 ; s=4 : Druyvesteyn

Druyvesteyn has less energeticelectrons

rm vq

Reduced electric field

Inelastic collisionsCouple parts of the distribution function that are far apart

Example: Excitation-electron looses excitation energy (a few to >10 eV)-electron is set back in velocity

Source proportional to vf0(v)NMexc(v)

Same holds for ionization: Energy new electrons to be specified

Inelastic collisions: two T distribution

Noble gases have high first excitation energyFor lower energies only elastic energy losses: slow decay of f with vFor higher energies large energy losses: fast decay of f with vResulting distribution is characterised by two ”temperatures”

Eexc Eion

Inelastic collisions: ionization

Ui-du Ui+du Eion+2(Ui-du) Eion+2(Ui+du)

In ionization Eion is lostSuppose remaining energy equally dividedHow many electrons arrive between Ui-du and Ui+du

Ui-du < (U-Eion)/2 < Ui+duEion+2Ui-2du <U <Eion+2Ui+2du

So factor 2 from energy range + factor 2 from new electron:4f0(u)u1/2ion(u)

In steady state problems: new electrons neglected,Usually this has only a minor influence

An example, SiH4/H2, with inelastic collisions

EEDFs with 4eV av. Energy in SiH4/H2: non-Maxwellian

Some quantities (assuming f0 normalized)

Use of this approach in modelling

Local field approximation: Everything expressed in local E/N-fieldmobility and diffusion coefficientsreaction rates (ionization, excitation)average energy

Mean energy approximation:Use solution for various E/N-fields to construct table:(mobility, diffusion coefficient, rates)all as a function of the average energy(cf. table as function of temperature for Maxwellian f)Use fluid energy balance to obtain av.energy in simulation

Use of the mean energy approximation

Homogeneous gas of given composition, Nb1...bn

EEDF from Boltz.Eqn.Homogeneous electric field, constant in time

Mobility (e), Diffusion (De)

EEDF Average energy ( 1.5 kTe)

Reaction rates for processes Kproc (E,Nb1,Nb2,..Nbn)

Combine results in table for Kproc () , e(), De()

Modelling the electrons

+

Look-up table

One step further: time dependent E-field

Important characteristic times:

Loss of momentum: goes very fast f1 is in equilibrium with E-field

Loss of energy: Only fast in case of inelastic collision f0 can be out of equilibrium

Example: reaction of f0 in SiH4/H2

: E0cos(t): behavior depends on ratio and loss frequencies

Energy loss

Momentum loss

High frequency: smaller excursion f0

Collision frequency pressureTherefore: normalized to 1 Torr10% SiH4, 90% H2E=Emcos(t), f0 at Em, Em/2, 0, -Em/2 (1,2,3,4)

From Capitelli et al.: Pl. Chem. Plasma Proc. 8 (1988) 399-424

Time dependent, spatially inhomogeneous E field

Is possible in principle, but:More than 1 spatial dimension would take too much CPU timeReally steep gradients (sheaths) require fn with n>1

Solution: Monte Carlo methods Account in principle for all effects

Example: v*f0 in Nitrogen

E=3.6*104(x/L)5(10.8sin(t)), -L<x<L=2*80 MHz

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Position (128=0.04m)

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erg

y (1

28=64

eV

)

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Electron-electron collisions

Electrons efficiently share energy in elastic collisions

Collisions try to establish Maxwell distribution

More sophisticated operators conserve momentum and energy

Monte Carlo methods

Principle: Follow particles by - solving Newton’s equation of motion - including the effect of collisions - collision: an event that instantaneously changes the velocity

Note: The details of a collision are not modeled Only the differential cross section + effect on energy is used

Examples: Electrons in a homogeneous electric field Follow sufficient electrons for a sufficient time Obtain distribution over velocities etc. f0,f1

Positive ions in plasma boundary layer (ions have trouble loosing momentum)

Monte Carlo methods: Equation of motion

Leap-frog scheme

Alternative: Verlet schemeΔt))/2Δt(trΔt)(tr((t)v

t)O(Δt(t)ΔaΔt)(tr(t)r2Δt)(tr 42

Monte Carlo methods: B-field

Problem with Lorentz force: contains velocity, needed at time tSolution: take average

The new velocity at the right hand side can be eliminated by taking the

cross product of the equation with the vector

Monte Carlo methods: Boris for B-field

Equivalent scheme (J.P.Boris), (proof: substitution):

Monte Carlo methods: Collisions

Number of collisions: NMtot = 1/ per meter.

(x) = (0)*exp(- NMx) = (0)*exp(-x/)

dP(x)=fraction colliding in (x,x+dx)=exp(-x/)(1-exp(-dx/))=(dx/)exp(-x/)

P(x)=(1-exp (-x/))

Distance to next collision: Lcoll=-*ln(1-Rn) (Rn is random number,0<Rn<1)

Number of collisions: NMtot v= 1/ per second.

Time to next collision: Tcoll=-* ln(1-Rn)

Monte Carlo methods: CollisionsAnother approach is to work with the chanceto have a collision on vt: Pc=vt/

Ensure that vt<< to have no more than one collision per timestep Effect of collision just after advancing position or velocity introduces only small error

When there is a collision:

Determine which one: new random number

0.0 0.5 1.0 1.5 2.0 2.5 3.00.0

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fracti

on

L/

no collision colliding once colliding twice colliding> twice

Monte Carlo methods: Null Collision

Problem: Mean free path is function of velocity Velocity changes over one mean free path

Solution: Add so-called null-collision to make v*tot independent of v Null-collision does nothing with velocity

Mean free path thus based on Max (v*tot)

Is rather time-consuming when v*tot peaks strongly

Monte Carlo methods: Null Collision

v

v*

v*2

v*1

v*3

v*tot

Max

v*0

1

1+2

1+2+3

1+2+3+..N

1+2+3+..N+ 0

’s normalized to maximum:Draw random number

Monte Carlo methods: Effect of collision

Determine effect on velocity vector

Retain velocity of centre of gravity

Select by random numbers two angles of rotation for relative velocity

Subtract energy loss from relative energy

Redistribute relative velocity over collision partners

Add velocity centre of gravity

Monte Carlo methods: Effect of collision

v1,v2 velocities in lab-frame prior to collision, w1,w2 in center of mass system

Monte Carlo methods: Effect of collision

A collision changes the size of the relative velocity if it is inelastic

A collision rotates the relative velocity

Two angles of rotation: and

usually has an isotropic distribution: =Rn*

has a non-isotropic distribution

Hard spheres:

Monte Carlo methods: Rotating the relative velocity

Step 1: construct a base of three unit vectors:

Step 2: draw the two angles

Step 3: construct new relative velocity

Step 4: construct new velocities in center of mass frame

Step 5: add center of mass velocity

Monte Carlo methods: ApplicabilityExamples where MC models can be used are:

- motion of electrons in a given electric field in a gas (mixture), see practicum- motion of positive ions through a RF sheath (given E(r,t))

Monte Carlo methods: Applicability

Main deficiency of Monte Carlo: not selfconsistent

- electric field depends on generated net electric charge distribution- current density depends on average velocities- following all electrons/ions is impossible

Way out: Particle-In-Cell plus Monte Carlo approach

Particle-In-Cell plus Monte Carlo: the basics

-Interactions between particle and background gas are dealt with only in collisions

-this means that PIC/MC is not! Molecular Dynamics

-each particle followed in MC represents many others: superparticle

-Note: each “superparticle” behaves as a single electron/ion

-Electric fields/currents are computed from the superparticle densities/velocities

-But: charge density is interpolated to a grid, so no “delta functions”

Particle-In-Cell plus Monte Carlo: Bi-linear interpolation

xi=ix xi+1=(i+1)x

xs, qs=eNs

i:=i+(xi+1-xs)qs/xi+1:=i+1+(xs-xi)qs/x

zi=iz

zi+1=(i+1)z

xj=jx xj+1=(j+1)x

ij:=ij+(zi+1-zs) (xj+1-xs) qs/(x z)

zs

xs

Particle-In-Cell plus Monte Carlo:Solution of Poisson equation

Boundary conditions on electrodes, symmetry, etc.

Electric field needed for acceleration of particle:(bi)linear interpolation, field known in between grid points

2

Particle-In-Cell plus Monte Carlo:Full cycle, one time step

Collisionnew v

Interpolatecharge to grid

Solve Poissonequation

Interpolate fieldto particle

Check lossat the walls

Move particlesF v x

Particle-In-Cell plus Monte Carlo:Problems

Main source of problems: Statistical fluctuations

Fluctuations in charge distribution: fluctuations in Eaverage is zero but average E2 is not numerical heating

Sheath regions contains only few electrons

Tail of energy distribution contains only few electronslarge fluctuations in ionization rate can occur

Particle-In-Cell plus Monte Carlo:Problems

Solutions:

-Take more particles (NB error as N-1/2 ) , parallel processing!

-Average over a long time

-Split superparticles in smaller particles when neededrequires a lot of bookkeeping, different weights!

Particle-In-Cell plus Monte Carlo:Stability

Plasmas have a natural frequency for charge fluctuations:

The (angular) Plasma Frequency:

And a natural length for shielding of charges:

The Debye Length:

Stability of PIC/MC requires:

Power modulated discharges

Modulate RF voltage (50MHz)with square wave (1 - 400 kHz)

Observation in experiments UU)optimum in deposition rate

Modulated discharges

Results from a PIC/MC calculation: Cooling and high energy tail

RF

Void

Crystal (21010 m-3)7.5 m radius

1-D Particle-In-Cell plus Monte Carlo Simulationof a dusty argon plasma

Capture cross section

Scattering:Coulomb, truncated at d

L/4L/8

w is energy electron/ion

Charging of the dust upon capture of ion/electron

The total charge is monitored on the gridpointsCharge of superparticle is added to nearest gridpointsDivision according to linear interpolationLocal dust charge is total charge divided by nr. of dust particlesThis number is density*dz*a2

For Monte Carlo the maximum v is computedNull-collision is used

Note the difference with the collisions with the uniformbackground gas: here we construct a grid-related probabilityof an event

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.00

2.50x1015

5.00x1015

7.50x1015

1.00x1016

1.25x1016

1.50x1016

Ne

N+

Den

sity

(m

-3)

x/L

Simulation for Argon, 50MHz, 100mTorr, 70V, L=3cm

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00

1x1015

2x1015

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d/e

Ne

N+

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sity

(m

-3)

x/L

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EE

DF

(ar

b.u

n.)

Energy (eV)

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EE

DF

(ar

b.u

n.)

Energy (eV)

dustfree with dust

Vd6V

Simulation for Argon, 50MHz, 100mTorr, 70V, L=3cm

dustfree with dust

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IED

F (

arb

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.)

Energy (eV)

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F (

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.)

Energy (eV)

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Av. El. Energy

Ion.Rate

3kT

e/2 (

eV),

Ion

.Rat

e (a

rb.u

n.)

x/L

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0.5

1.0

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Av. El. Energy Ion.Rate

3kT

e/2 (

eV),

Ion

.Rat

e (a

rb.u

n.)

x/L

Simulation for Argon, 50MHz, 100mTorr, 70V, L=3cmGeneration of internal space charge layers

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0-1x1014

0

1x1014

2x1014

3x1014

Net

ch

arg

e / e

x/L

0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000-20000

-15000

-10000

-5000

0

5000

10000

15000

20000

Av.

Ele

ctri

c F

ield

(V

/m)

x/L

An internal sheath is formedinside the crystal

Ions are accelerated beforethey enter the crystal

This has consequences forthe charging + shielding

Particle-In-Cell plus Monte Carlo:What if superparticles collide?

Example: recombination between positive and negative ions

Procedure: number of recombinations in t: N+N-Krec t

corresponds to removal of corresponding superparticles randomly remove negative ion and nearest positive ion but: be careful if distribution is not homogeneous Again a grid-based probability can be used

A more sophisticated approach: Direct Simulation Monte Carlo

DSMC: Basics

Divide the geometry in cells

Each cell should contain enough testparticles

Newton’s equation: as before, but keep track of cell number

Collisions: choose pairs (in same cell!) and make them collide

Essential: the velocity distribution function is sum of -functions

DSMC: Choosing the pairs

Add null collision

Chance of collision of particle i with j is Pc=(Npp/Vcell)*Max(v)t

Average number of colliding pairs: n(n-1)* Pc/2

Select randomly n(n-1)* Pc/2 particle pairs (make sure no double selection)

See if there is no null collision, again with random number, comparingthe real chance for this collision (vr) with the maximum Max(v)

Perform the collision

DSMC: An example

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# p

art

icle

s

Energy (arb.un.)

25 50 75 100

Relaxation of a mono-energetic distribution to equilibrium20000 particles, hard sphere collisions, one cell contains all particles

Fast when possible, kinetic when needed:Hybrid models

Examples of hybrid models: Hydrodynamic ions and cold electrons Monte Carlo for fast electrons (tail EEDF)

Boltzmann electrons, Monte Carlo for ions

MHD model for plasma, Monte Carlo neutrals

Problems are due to coupling: iterations needed

B2-EIRENE for Magnum-psi

5 slm H2Th= 2eVTe=12 eV10^24 m^-2s^-1

T profile at inlet

Recycling at the target