Rikard, I'm late to the party as usual, but... On Sat, 31 Mar 2007 11:08:13 +0200, Rikard Bosnjakovic wrote: > On 3/31/07, wouter van ooijen wrote: > > > - an unloaded small 9V transformer can give much more than 9V (measure > > or find the specs), calculate with at least 12V, maybe even 18V > > Using my Fluke 179 (RMS-DMM), it said 9.6VAC for the unloaded > transformer (no bridge). Right, so if this is a "True RMS" meter, it would be reading 9.6V RMS - so the *peak* voltage would be 9.6 * sqrt(2), so way above your capacitors' 10V maximum. Nobody has emphasised this: the peak voltage is what matters for a capacitor's survival. > Measuring over "+" and "-" on the bridge, it > said 9.5VDC. Adding a load (100k) over the bridge yielded no change > for the voltage (9.5VDC still). But it isn't DC. Without any capacitance it's full-wave-rectified AC, so a meter set to a DC range won't give you an accurate result, although it may be close to the RMS voltage depending on the design of the meter, so even then you're a factor of sqrt(2) too low for calculating the capacitor's rated voltage. > Could it be the transformer is marked with its effective voltage? Else > I should have seen sqrt(2)*9 on my DMM, no? No, unless it's a peak-reading meter (very rare) it will read RMS or something like it. The transformer markings show its output as an RMS voltage, and the actual output will vary with the input voltage and the load, but you need to rate for the peak, not the RMS voltage. So even if the transformer is spot-on in its output, the peak voltage will be 9 * 1.414, about 12.73V so you need at least a 16V rated capacitor even before considering tolerances on input voltage and so on. Cheers, Howard Winter St.Albans, England -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist