RF workshop Alvin Tollestrup Nov /14/20101Alvin Tollestruo

E Field Emission Self Propagating electron swarm With Beam. No beam comes down Left hand side electrons in a 1 micron sphere make E=1.4 GV/m 11/14/20102Alvin Tollestruo

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. 11/14/20103Alvin Tollestruo

photons on cathode PMT gain 1700 times higher and averaged over 10 ns. Conservative limit is less than 1% of breakdown light E/P = /14/20104Alvin Tollestruo

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]=0 Solve for iR[t] R[t]= V[t]/iR[t] V[t] from 7 th order polynomial in t plus phase. See next slide 11/14/20105Alvin Tollestruo

Fitting Procedure Find start of discharge 11/14/20106Alvin Tollestruo

11/14/2010Alvin Tollestruo7

From Last two slides: I ~ 500 at pinch R ~ 0.5 microns j~ A/cm 2 So j 2 dt = = 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 11/14/20108Alvin Tollestruo

What happens at frequency shift? 1.Know C, F so calculate L 2.L= Lc Ls/(Lc + Ls) 3.Know L and Lc so solve for Ls 4.Ls =  o /2  Ln[ R 2 /R 1 ] 5.R 2 = Cavity radius so solve R 1 6.Current Ir around 600 A. 7.Calculate B =  o Ir/2  R 1 8.P = B 2 /2  o 9.P V   constant 10.Know P 2,  R 2 2, P 1 = 1000 psi 11.Find the original Radius R 1  gives the original radius about 5 microns. 11/14/20109Alvin Tollestruo

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 ] = . Total  f/f C V = electrons Rs = 4.0 K and spark current ~60 A. Too small to pinch. 11/14/201010Alvin Tollestruo

Box Cavity B=0 11/14/201011Alvin Tollestruo

11/14/2010Alvin Tollestruo12 What is wrong here?