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Old 27th February 2007, 04:28 PM   #1
zilog is offline zilog  Sweden
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Join Date: Oct 2005
Default input capacitor ripple current rating for push-pull

I want my smps with input voltage range 10-16V, 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 under-rated with respect to rms rating, there is usually only a few caps on the input while my calculations tell me that I need 12-13 pieces of 2200uF MV-WX caps @ 16V. I know that these caps can withstand 2-3 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.
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Old 27th February 2007, 07:58 PM   #2
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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.
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Old 27th February 2007, 10:31 PM   #3
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Join Date: Apr 2006
Location: Minnesota

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.

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.

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Old 28th February 2007, 10:08 AM   #4
zilog is offline zilog  Sweden
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 push-pull converter with total duty-cycle of 2*D (D being per switch leg) and I_{fp} being the equivalent flap-topped 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 MV-WX (1.82Arms per piece) will suit my needs then.
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