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27th February 2007, 04:28 PM  #1 
diyAudio Member
Join Date: Oct 2005

input capacitor ripple current rating for pushpull
I want my smps with input voltage range 1016V, maximum D=0.8 and assumed Eff=0.8 to withstand 150W continuous output power. Calculations from Pressman give the rms current through each transformer leg is 15*1.56*sqrt(0.4) = 14.8Arms giving a total Irms of 2*14.8 ~30A through the center tap leg. Since I will be feeding this smps from a car battery through inductive and resistive cables, am I correct to assue that the input capacitor bank will see the full 30Arms ripple current? If so, why do every commercial car amp seem so underrated with respect to rms rating, there is usually only a few caps on the input while my calculations tell me that I need 1213 pieces of 2200uF MVWX caps @ 16V. I know that these caps can withstand 23 times that rms current at 25 degC as opposed to the specifiec 105 degC, but still..
Any opinions on this? Its not only that I need high ripple endurance, I also need low ESR and a couple mF to reduce voltage droop. 
27th February 2007, 07:58 PM  #2 
diyAudio Member
Join Date: Jan 2007

Needing such large capacitor banks implies that the input to your transformer is experiencing voltage droop, which you do not want anyway. Normally, the cables going to the battery are thick enough to permit the battery to supply most of the peak current.

27th February 2007, 10:31 PM  #3 
diyAudio Member
Join Date: Apr 2006
Location: Minnesota

zilog,
It's a good idea to put some high quality capacitance across the bus near the power switches and transformer. You could calculate the impedance of the cable, but at high frequencies it will be large. Assuming that most of the high frequency current is flowing through the caps you can use this to find the cap current. http://www.geocities.com/CapeCanaveral/Lab/9643/rms.htm Enter the data for you waveform and then click on one of the calculation boxes. Obviously, only the AC component flows through the caps  not the DC component. Rick 
28th February 2007, 10:08 AM  #4 
diyAudio Member
Join Date: Oct 2005

I have plated a bit with the maths now, and have come up with much better numbers. For a pushpull converter with total dutycycle of 2*D (D being per switch leg) and I_{fp} being the equivalent flaptopped current of the trapezoidal currents in the switcher legs,
I_{rms} = I_{fp}^2\sqrt{2D} I_{DC} = 2DI_{fp} I_{AC} = \sqrt{I_{rms}^2  I_{DC}^2} This gives a DC current of 18.72A and an AC current of 9.3A if I operate the smps with Eff = 0.8, 2*D = 0.8, Vin = 10V and Pout = 150W. Guess 5x1000uF MVWX (1.82Arms per piece) will suit my needs then. 
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