Brian, I am very familiar with 61000-4-6, and I think this is where the confusion is coming from. 61000-4-6 specifies the 6dB adjustment not because of loss in the fixture, but because the test level is specified as an open-circuit voltage. However, the calibration must be done in a terminated system, which will reduce the measured voltage by half. There is a 9.6 dB associated with the voltage divider network formed by the 150-50 ohm conversion. GR1089 references 61000-4-6 for some reason. I think it should only reference the mil std test, IMHO. The test level is specified as a current level, and the cal fixture is a simple loop wtih only one current path. So, I don't see the justification of applying a 6dB reduction in the calibration level. Side note: I gave a presentation recently on 61000-4-6, and in researching the standard I was never able to find a good reason why the test level is specified as an open circuit voltage. Thanks, Bob Richards, NCT.
"Kunde, Brian" <[email protected]> wrote: Bob, I'm not familiar with GR1089 but somewhat with EN61000-4-6 which is a conducted immunity test standard. When calibrating the CDNs or Injection clamps, you have to compensate for the losses in the test fixture. For the CDN, it comprises of a 150 to 50 ohm adaptor at the input and output of the CDN. The fixture acts as a voltage divider, the the measured power levels during calibration would be much lower than what is expected on the CDN output. So you have to account for this loss. In EN61000-4-6 section 6.4.1 "Setting of the output level at the EUT port of the coupling device" it has an equation for calculating the measured power level when you perform the calibration test: Umr = Uo / 6 (in linear quantities) or Umr = Uo - 15.6db (in logarithmic quantities where Uo is the test voltage from table 1. There is a note at the bottom that says, "The factor 6 (15.6db) arises from the e.m.f. value specified for the test level. The matched load level is half the e.m.f. level and the further 3:1 voltage division is caused by the 150 ohm to 50 ohm adaptor terminated by the 50 ohm measuring equipment". For Current Clamps the equation is Umr = Uo / 2 (in linear quantities) or Umr = Uo - 6db (in logarithmic quantities). I am assuming the test fixture acts like a voltage divider so you will only be measuring a fraction of the true voltage at the CDN in the the calibration setup. So yes, if you are not factoring in the fixture losses when you calibrate, you would be testing at too high of level. I hope this is what you are referring to. The Other Brian _____ - ---------------------------------------------------------------- This message is from the IEEE Product Safety Engineering Society emc-pstc discussion list. Website: http://www.ieee-pses.org/ To post a message to the list, send your e-mail to [email protected] Instructions: http://listserv.ieee.org/request/user-guide.html List rules: http://www.ieee-pses.org/listrules.html For help, send mail to the list administrators: Scott Douglas [email protected] Mike Cantwell [email protected] For policy questions, send mail to: Jim Bacher: [email protected] David Heald: [email protected] All emc-pstc postings are archived and searchable on the web at: http://www.ieeecommunities.org/emc-pstc ______________________________________________________________________ This email has been scanned by the MessageLabs Email Security System. For more information please visit http://www.messagelabs.com/email ______________________________________________________________________
