Roger, 20kA at 10kV probably cannot be done effectively -- this implies a source impedance of the generator of only 0.5 ohms. So far no problem, but what happens when you add any kind of load ? The available current into a 0.5 ohm load is only 10kA; into a 0.1 ohm load it's only 16.7kA.... But that's not the only problem: with the 8/20us waveform, the di/dt at 20kA says that the inductance of wire will have an effect: V drop (across a piece of wire) = Ldi/dt: the available current is then 10kV/(0.5 ohms + the effective impedance of the inductor with the 8us rise time) --- if you assume 0.1uH/10cm of #20 wire (use your own numbers if you think I'm wrong) the current is then 19kA if the resistance is 0 ohms for 10cm of wire........ smaller wire is worse, bigger wire helps, but someone told me once that building steel is considered to be about 0.1uH/foot, so maybe that's worst case.... All of that only says that to get 20kA into a cable of any length at all will require a generator with a much higher compliance voltage --- at least 40kV and a 2 ohm source...... Certainly out of our ballpark at Thermo...
Best Regards, Michael Hopkins Manager, EMC Technologies Control Technology Division Compliance Test Solutions Thermo Electron Corporation One Lowell Research Center Lowell, MA 01852 Tel: +1 978 275 0800 ext. 334 Mobile: +1 603 765 3736 [email protected] One Thermo, committed to integrity, intensity, innovation & involvement From: Roger Hsu [mailto:[email protected]] Sent: Tuesday, March 23, 2004 11:02 PM To: [email protected]; [email protected] Subject: 10kV/20kA 8/20us surge All, Does any one know a test lab that can perform the 10kV/20kA 8/20us-current surge for the DC power supply. Is this requirement same as GR-1089:2002 R4-45 requirement? (GR-1089:2002 R4-45 did not mention the test voltage) I also need to look for the manufacturer for the above surge generator. Thanks and regards, Roger
