7d6cb5ab   
   From: mynamespacedbydots@xs4all.nl   
      
   >"Patrick Turner"  wrote in messageThe usual saw   
   >tooth wave seen at the top of caps in PSUs usually shows   
   >   
   [...]   
   >   
   >the rise due to charge time being less than the fall due to DC flow to   
   >the load. The rectifier must always provide the DC flow out plus the   
   >excess during charge time to get the Vdc to remain steady. Usually its   
   >possible to experiment with a small series R to measure peak charge   
   >currents and usually its not more than 6 times DC to the amp.   
   >   
   >In my case with +/- 70V rails and 800 watts of draw, Idc = 5.7A dc and   
   >ripple at 100,000uF = 0.125Vrms, or about 0.38Vpk to pk.   
   >   
   >I leave the better minds amoung you to tell me what my peak charge   
   >currents may be but the 35Amp bridges have never failed.   
      
      
   Assume the ripple described is an ideal, asymmetrical, sawtooth.   
   On the negative(trailing) slope discharge of the cap takes place.   
      
   Voltage across a cap is proportional to the current.time product   
   devided by the cap's capacity, or U= I*t/C.   
   Derivative from that is t= C*U/I, or for any slope: dt= C*dU/I   
      
   For your example case, time for the negative(trailing) slope,   
   must have been dt= 0.1F*0.38V/5.7A= 6.67mS.   
   (assuming C= 0.1F but if not sure apply C= I*dt/dU, under the   
   condition that Udc >> Uripple, to gain the actual capacitance).   
      
   So charge time left during a 10mS .au mains half-cycle is 3.33mS   
   From above derive I= C*U/t or for the positive slope: I= C*+dU/dt   
   Again assuming an ideal sawtooth, the (mean) charge current, but   
   not the rectifier current, must be I= 0.1F*0.38V/3.33mS= 11.4A   
   The resulting (mean) rectifier current would be 11.4A+5.7A= 17.1A   
      
   In practice, of course, you will not see an ideal shaped sawtooth.   
   Yet, peak charge current is determined easily if dU/dt is taken   
   from the steepest part of the (positive) slope, and if the cap's   
   capacity is known you can calculate the peak charge current, and,   
   summed with the load current (Idc), the peak rectifier current.   
      
   In some cases where I checked peak charge vs. mean charge currents   
   I found factors varying between 2-3, say 2.5 for the time being.   
   For the example a 2.5*11.4A +5.7A = 34.2A peak rectifier current.   
      
   Higher factors, maybe up to 6 as you mention, are possible if low   
   ESR cap's are applied together with a low winding resistance etc.   
   So, checking with a low ohm series R is a good idea provided that   
   R << ESR + winding R + other parasitics.   
      
   Regards,   
   Gio   
      
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