Just to be clear a 7805 is not a converter like Perry is suggesting but a regulator, which drops voltage leaving current the same and dissipating the extra energy as heat, not very energy-efficient of course.
yes, thanks for confirming this - it's what i had in mind with my rudimentary knowledge of electronics.
You may use a resistor to drop the excess voltage. Or a few diodes. Use the linear regulator as the last step.
ahh, good tips - thanks !!
obviously they should be properly rated, my stock of resistors are just the 1/4W ones though, but i do have a few rectifier diodes to line up in series.
Yes, if you use a boost converter then the output current will be lower than the input current.
ok, good - thanks !!
(I'd better hide in case it goes BANG!)
i think... :-P
That's using a PCB as heatsink or nothing as heatsink, using a block of copper that 19 becomes 3, and the power handling is 30W for
a 90 deg rise above ambient.
ahh, i see - that's very handy to know - the datasheet wasn't very informative on that, it didn't even mention heatsinks at all which was strange - although i can't say what's normal, i think i've seen a datasheet which gave specs with heatsink usage advice but i don't know if that's the norm.
Perhaps cut up a couple of aluminium cans, open them out and arrange three or four layers together with a hole through them, use a washer to hold them together where you screw them to the tab of the 7805. The separate layers can fan out as fins on the heatsink.
Thank you, very handy tip !!
Stop fiddling with linear regulators, and buy a switching (buck) converter.
yes sir - just purchased and on the way !
Don't you have a car cigarette lighter style USB charger (same thing).
oh - you mean like phone chargers that plug into the mains AC ?
yeah - those supply 5V don't they (usually at 1A) - now, there's a thought...
Thanks again everyone for such informative tips.