I have a couple answers. In IEC950 2.1.10 the standard calls for a time constant of 1 second. In IEC1010, the time allowed is 5 seconds to get to a 'safe voltage'. For the same circuit, after 5 time constants of decay, the voltage has decayed to less than 1 percent of the original value. Even at the highest input voltage, say 1,500 Volts, the terminal is below 15 volts.
On the more practical side, I like the AC test method. Although I haven't ever confirmed this to be a problem, some circuits may act differently under a DC input. Connect a switch & oscilloscope to the line. Triggering is a little tricky. If you were to say a 10 percent error is acceptable, it is not hard to get a 'good' measurement. Get your calculator or trig book out... Since the sine waveform has large flat areas at the high voltage points, you will find that it is above 90 percent of peak voltage almost 30 percent of the time. (28.7percent) So one out of three measurements should be good. arcsin (.9) = 64 degrees so it is above 90 percent from 64 to 90 degrees of the wave. 26/90 = 0.29 I find that quite often the first or second measurement is good. The older standard said to try 10 times. Of course Murphy visits occasionally and I do remember one time trying 12 times to get a measurement I liked. Standard disclaimers apply; opinions expressed are personal, not corporate. Ray Kato, Product Safety Engineer Tektronix Inc., Beaverton, OR USA [email protected] Phone(503)627-2993 FAX(503)627-3838 ------------- Original Text From: <[email protected]>, on 10/24/97 2:24 PM: Just a quick testing tip: You don't have to disconnect the AC right at the peak. If the circuitry downstream from the line-to-line cap can handle DC, then use a DC power supply to charge the capacitor up to the same voltage as the peak of your AC supply voltage (e.g. 1.414 x 120Vac = 170Vdc). This avoids the tricky business of managing to disconnect right at the peak of an AC waveform. Regards, Jim Eichner Statpower Technologies Corp. Burnaby, B.C., Canada [email protected] Any opinions expressed are those of my invisible friend, who really exists. Honest. -----Original Message----- From: jppena@anetMHS (Juan Pedro Pe±a){MHS:[email protected]} Sent: Thursday, October 23, 1997 5:07 AM To: emc-pstc@anetMHS (EMC-PSTC){MHS:[email protected]}; JEichner; bceresne Subject: RE: Capacitor discharge: IEC950 2.1.10 Many other standards state that voltage shall decay under a safety level (i.e., 60 V). However, to measure it is necessary to cut the power just in the top of the wave, and it is difficult. If we speak about percentage, that problem won't exist: the decay time until 37% of it initial value depends only on the time constant of the circuit, independently of the initial voltage. But you are right and that definition may force an excessive requirement. Maybe, the best option could be to mix both methods. Juan P. Pena / Electrical Safety Area Company: CETECOM, S.A. (http://www.cetecom.es) e-mail: [email protected] ---------- De: Munechika Gomi[SMTP:[email protected]] Enviado el: viernes 3 de octubre de 1997 8:35 Para: EMC-PSTC Asunto: Capacitor discharge: IEC950 2.1.10 Would anyone tell me why IEC950 allows a capacitor having a discharge means in a time constant not exceeding: ---, i.e. during one time-constant the voltage decays to 37 % of its original value? The reason of my question is: If the original voltage is 120 V ac, 37 % or its original will exceed 60 V dc which I think is hazardous in terms of voltage. Is there any implied consideration in IEC950 for energy level other than voltage? Thank you for your help in advance. Mike Gomi _
