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CWRF 2012 SHORT CIRCUIT ENERGY CONSIDERATIONS USING THE EXAMPLE OF A PULSE STEP MODULATOR BASED HVPS CWRF 8. 11. May 2012 Andr Spichiger, Michael Bader, Marcel Frei SHORT CIRCUIT ENERGY Content Motivation: Why this topic? Tube


  1. CWRF 2012 SHORT CIRCUIT ENERGY CONSIDERATIONS USING THE EXAMPLE OF A PULSE STEP MODULATOR BASED HVPS CWRF 8. – 11. May 2012 André Spichiger, Michael Bader, Marcel Frei

  2. SHORT CIRCUIT ENERGY Content Motivation: Why this topic? Tube Specifications How can the arc energy be measured or calculated HVPS Stored Energy Energy dissipation / arc energy Reduction of arc energy Inverse Voltage- System Summary 2 5/11/2012

  3. MOTIVATION Why this topic? Questions during bids and acceptance tests Increasing filter / ripple requirements  what are the limitations? Frequently discussed issue: Cable length between the supply and the tube Important for a safe tube operation To give an idea / feeling about the figures 3 5/11/2012

  4. SPECIFICATIONS What is the energy dissipated in a load (tube) arc? Arc energy specifications Usually in the region of 10..20 Joule 15 Joule corresponds to 15 bars of chocolate lifted 1m. 4 5/11/2012

  5. VERIFICATION 5 5/11/2012

  6. VERIFICATION Measuring the arc energy Simple test with a wire as proposed in tube Datasheets Using a real fuse I2t value Accurate Current measurement is possible (e.g. pearson coil, coaxial shunt) Accurate Arc Voltage measurement is not easy to do We proposal: Measuring the current and post-calculating the energy dissipation with a defined arc Model Calculating the arc energy Arc Model Proposal for vacuum tubes: Series connection of DC Voltage source 100V and Resistor 200mOhm. Simulations provide good basis for later verification System optimization can be easily done in simulations 6 5/11/2012

  7. BASIC MODEL 7 5/11/2012

  8. STORED ENERGY Basic model with a 2 – pole RLC filter network System parameter 35kV / 3.5A, a rather small system. ARC energy specifications 15 Joule Stored energy is much higher Parameter Value Stored Energy Filter Inductor 16 mH 0.1 Joule / 6 Joule 5 mH 0.03 Joule / 15 Joule Filter Capacitor 200nF 123 Joule High Voltage Power 60 modules 3420 Joule Supply 57 Joule / module Cable 100 pF / meter 0.06 Joule / meter 8 5/11/2012

  9. ENERGY DISSIPATION Where is the energy dissipated in case of a short? What happens with the stored energy in the filter QL QC? Simulation model parameters: Turn off delay 10 us Voltage drop in the diodes: 2V / Diode Short circuit model: 100V / 0.2 Ohm 9 5/11/2012

  10. SIMULATION MODEL FILTER LFilter 1 2 out 5m HV-Supply LOAD RFilter 2 200 ARC U5 D2 TOPEN = 100u Dbreak R10 R_ARC 1 10k 0.2 V6 V8 35000Vdc V_ARC 120Vdc 100Vdc R6 1 200n 0.72 IC = -35k ARC TCLOSE = 90u TTRAN = 0.1u 2 RCLOSED = 0.001 ROPEN = 1Meg 0 10 5/11/2012

  11. SIMULATION: 5 MH FILTER INDUCTANCE Discharge of filter cap 200A Inductance current 100A SEL>> 2A I(RFilter) I(ARC:1) I(LFilter) 4.0 2.0 0 51us 100us 150us 200us 250us 300us 350us 400us S(W(R_ARC)+W(V_ARC)) Time 11 5/11/2012

  12. SIMULATION: 5 MH FILTER INDUCTANCE 200A 100A SEL>> 0A I(RFilter) I(ARC:1) I(LFilter) 8.0 about 7 Joule 4.0 0 0.05ms 0.20ms 0.40ms 0.60ms 0.80ms 1.00ms 1.20ms 1.40ms 1.60ms 1.80ms S(W(R_ARC)+W(V_ARC)) Time 12 5/11/2012

  13. SIMULATION: 16 MH FILTER INDUCTANCE 150A 100A 50A SEL>> 0A I(RFilter) I(ARC:1) I(LFilter) 2.0 1.0 0 55.3us 80.0us 120.0us 160.0us 200.0us 240.0us 280.0us 320.0us 353.2us S(W(R_ARC)+W(V_ARC)) Time 13 5/11/2012

  14. SIMULATION: 16 MH FILTER INDUCTANCE 150A 100A 50A SEL>> 0A I(RFilter) I(ARC:1) I(LFilter) 4.0 about 3.3 Joule 2.0 0 0.053ms 0.200ms 0.400ms 0.600ms 0.800ms 1.000ms 1.200ms 1.400ms 1.600ms 1.800ms 1.996ms S(W(R_ARC)+W(V_ARC)) Time 14 5/11/2012

  15. SIMULATION: WITH CABLE 100 METER CABLE ADDED TO THE SIMULATION 705A big oscillation 400A 0A SEL>> -225A I(RFilter) I(ARC_Cab:1) I(LFilt_Cab) 2.0 1.0 0 80us 100us 120us 140us 160us 180us 200us 220us 240us 260us 280us 300us S(W(R_ARC_Cab)+W(V_ARC_cab)) Time 15 5/11/2012

  16. SIMULATION: WITH CABLE AND SNUBBER FILTER SNUBBER RSnubber LFilt_Cab T3 50 1 2 out_cab LOSSY LSnubber 16m 1 2 LEN = 100 100uH L = 250n 0 0 HV-Supply C = 100p G = 10p R = 0.1m 2 RFilt_Cab 200 LOAD ARC U4 D1 TOPEN = 100u Dbreak R_ARC_Cab 1 R8 0.2 10k V5 V4 35000Vdc V_ARC_Cab 120Vdc 100Vdc R4 1 200n 0.72 IC = -35k ARC_Cab TCLOSE = 90u TTRAN = 0.1u 2 RCLOSED = 0.001 ROPEN = 1Meg 0 16 5/11/2012

  17. SIMULATION: WITH CABLE AND SNUBBER 375A smaller peak, no oscillation 250A 125A SEL>> 0A I(RFilter) I(ARC_Cab:1) I(LFilt_Cab) 1.5 1.0 0.5 0 80.0us 100.0us 120.0us 140.0us 160.0us 180.0us 200.0us 220.0us 240.0us 260.0us S(W(R_ARC_Cab)+W(V_ARC_cab)) Time 17 5/11/2012

  18. ENERGY DISSIPATION SUMMARY Where is the energy dissipated in case of a short? Filter Capacitor: Almost everything is dissipated in the filter resistor Filter Inductance: Before shutdown of the PSM the L is charged  Energy After PSM shutdown the energy is dissipated in the short and the diodes of the PSM Cable snubber are reducing oscillations and Arc Energy 18 5/11/2012

  19. OPTIMIZATION How can the energy be reduced, measures? Filter can be optimized according operating point. More complex filters instead a simple RLC network. Tradeoff: Filter vs. Energy Using Cable Snubbers for long cables Thomson patented Inverse voltage Operation Mode 19 5/11/2012

  20. KGP5: INVERSE VOLTAGE OPERATION Application example: KGP5 Test stand for Tubes 160 kV / 3.2 MW CW Higher power in pulsed mode 5 us Rise time Arc energy can be set between 2 Joule and 20 Joule This specifications required new solution: Inverse Voltage Operation Mode 20 5/11/2012

  21. KGP5: INVERSE VOLTAGE OPERATION Principle 21 5/11/2012

  22. KGP5: INVERSE VOLTAGE OPERATION Short Circuit Current Simulations Green: output current without inverse voltage operation Red: output current with inverse voltage operation 1.0KA 1.0KA 0.8KA 0.6KA 0.5KA 0.4KA 0.2KA 0A 0A -0.2KA -0.5KA 0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 0s 10us 20us 30us 40us 50us 60us 70us 80us 90us 100us i(v2) i(v2) Time Time 22 5/11/2012

  23. KGP5: INVERSE VOLTAGE OPERATION Short circuit Test on Tube teststand Adjustable arc energy 2 – 20J Main Issue with IVO was the 20J not the 2J  2Point Regulation implemented Top Trace: Vout = 102kV; Energy Setting = 2.5J Charge Measured 29mAs (2.9J @100V) Bottom Trace: Two point regulation until energy is reached 23 5/11/2012

  24. KGP5: INVERSE VOLTAGE OPERATION Two point regulation for a defined Q 24 5/11/2012

  25. SUMMARY Summary Bigger capacitors or inductors des not necessarily mean more energy in the arc. Considering the whole system is important! Degrees of freedom allow optimization  therefore it is important to know the real requirements. May be an iterative process together with the customer. Thomson’s Inverse Voltage Supplies can reduce the energy to a minimum. 25 5/11/2012

  26. THANK YOU

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