Alvin Tollestruo 1

RF workshop

Alvin TollestrupNov.14 2010

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EField Emission

Self Propagating electron swarm

With Beam. No beam comes down Left hand side.

106 electrons in a 1 micron sphere make E=1.4 GV/m

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Comparison of VacRF and HPRF

•1. Same asperities for both cavities.•2. Breakdown process very different. In the Paschen region the breakdown is determined completely by Townsend avalanche process. Physics is well known and documented.•3. After break down, the major part of the cavity energy goes into the gas, not the electrodes. As a result, training can be very different. •4. The gas collision frequency if very high compared to the cyclotron frequency so there is not a big effect from B. The effect of B with beam needs to be checked as there could be effects on the diffusion of the plasma from the beam transit and hence on the recovery time.•5. Vacuum cavities have dark current. This has taught us about the cavity surface in its equilibrium condition. We haven’t found a similar tool for HPRF.We have tried to see light when there is not break down but without success.6. The rest of this talk will be about the breakdown physics. This is not really what we want, but as we have no beam we hope it will teach us about some of the pertinent physics processes taking place.

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3.9 108 photons on cathode

PMT gain 1700 times higher and averaged over 10 ns. Conservative limit is less than 1% of breakdown light

E/P =10.2

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Details of discharge

Cavity L C

Spark L R ic[t[= V’[t]iL[t]= iL[0] +1/L Integral[V[t]iR[t]+ic[t]+iL[t]=0Solve for iR[t]R[t]= V[t]/iR[t]V[t] from 7th order polynomialin t plus phase. See next slide

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Fitting Procedure

Find start of discharge

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From Last two slides:I ~ 500 at pinch R ~ 0.5 microns j~ 8 1010 A/cm2

So j2 dt = 6 1021 10-9 = 6 1012 Much greater than limit given below.Arc melts electrode but field emission can’t unless asperity is very well insulated so can use many cycles

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What happens at frequency shift?

1. Know C, F so calculate L2. L= Lc Ls/(Lc + Ls)3. Know L and Lc so solve for Ls4. Ls = mo /2p Ln[ R2 /R1 ]5. R2 = Cavity radius so solve R1 6. Current Ir around 600 A.7. Calculate B = mo Ir/2p R1 8. P = B2 /2 mo 9. P Vg = constant 10. Know P2 , p R2

2 , P1 = 1000 psi11. Find the original Radius R1 12. = 5/3 g gives the

original radius about 5 microns.

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Alvin Tollestruo 10

Discharge of the Second Kind

E__

+

•1. Plasma has - & + charges that can be separated by the field.•2. Similar to having a losey dielectric. The charges move back and forth making inelastic collisions. This generates the R. But it is also non uniform. Div[ P ] = r.•Total df/f C V = 5 1012 electrons •Rs = 4.0 K and spark current ~60 A. Too small to pinch.

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Box Cavity B=0

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What is wrong here?