I am little confused by the requirements of GR1089, if I choose to use 1.2/50 surge generator instead of 2/10 generator. For the 800V 1.2/50 pulse, the standard specifie that a 6 ohm resistor should be added to the generator output. However, there is a 2 ohm built-in resistor in the generator that I have (KeyTek), and I am not sure whether I must have the total output impedance of 6 ohm or 8 ohm. Following the current-limiting logic described in connection with using a 2/10 surge-generator, I tend to believe I can limit the current to 100 A, which would mean that for 800V surge I need 6 ohm in addition to the built-in 2 ohm. But the standard only mentions 6 ohm, so maybe that's the total output R I can have. If I use 6 ohm, I not only test more severely by the nature of the surge waveform (1.2/50 has faster rising edge and longer duration than 2/10), but also by allowing more current through my DUT. Is there anyone on the list who knows the answer? Thanks, Neven - ---------------------------------------------------------------- This message is from the IEEE Product Safety Engineering Society emc-pstc discussion list. Website: http://www.ieee-pses.org/
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