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Silly power loss question

I am PWMing a solar panel output or open circuit 36V to charge 2 lead acid batteries in series to 27.6V and I am trying to decide how much power is being dissipated by the switching FET. Is it relatively low because when the output is off there is no power loss in the fet and when the output is on there is only the current squared*the internal on resistance of the fet and the average power is equal to the average mark/space ratio. Or as the output is going through a low pass filter is the power equal to the averaged volt drop across the FET squared divided by the on resistance of the fet

i don't think I have explained wqhat I am asking well at all, but cannot think how better to put it?

TIA

pete d

Reply to
Pete D
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"Pete D"

** With a fixed resistance, the power dissipated is always given by the rms value of the current or voltage across that resistance. Your method of calculating the on time power and then scaling it by the duty cycle is fine.

Only time average current value comes into play is for calculating the amount of energy delivered to the battery.

........ Phil

Reply to
Phil Allison

You're also ignoring switching losses as the FET turns on and off. It spends a brief period in linear operation each time it turns on and off..

Graham

Reply to
Eeyore

thanks, that's what I hoped was the case, I can do with a relativly small heatsink in this case then, ta muchly;-)

pete d

Reply to
Pete D

Ah, right, that sounds hard to calculate;-) I'll allow for 20% extra and see how hot it gets.

Thanks both.

pete d

Reply to
Pete D

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