pinkmouse said:Didn't you ask this question a while back, or am I going mad in my old age?
Dissipated power is a product of the voltage dropped and the current used. For that much voltage drop, even at low current, I would heatsink the regs. Better safe than sorry.
IOW, if the input is 35V, and the output is 15V, dropped V is 20V. With a load of say 25mA (probably less) P = 20 x 25 mW = half a watt. Maybe just a small sink would be wise, as Pinkmouse said.
Jan Didden
No problem Wynand, I'm sure I've done the same thing before!
Jan's post gives you the numbers, though I was rather hoping you'd do the maths and get a proper feel for it, then you can use it again in the future. I've found that working out simple questions like this helps a lot to get a feel for a circuit and leads to greater understanding in the long term, and as a relative newbie, I certainly need all the understanding I can get.
Jan's post gives you the numbers, though I was rather hoping you'd do the maths and get a proper feel for it, then you can use it again in the future. I've found that working out simple questions like this helps a lot to get a feel for a circuit and leads to greater understanding in the long term, and as a relative newbie, I certainly need all the understanding I can get.
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