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Silly power loss question

P

Pete D

Jan 1, 1970
0
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
 
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Phil Allison

Jan 1, 1970
0
"Pete D"
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


** 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
 
E

Eeyore

Jan 1, 1970
0
Pete said:
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?

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
 
P

Pete D

Jan 1, 1970
0
Phil said:
"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
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
 
P

Pete D

Jan 1, 1970
0
Eeyore said:
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
Ah, right, that sounds hard to calculate;-)
I'll allow for 20% extra and see how hot it gets.

Thanks both.


pete d
 
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