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5 Volt Regulator question (7805). Answered

My students are building a shake cell phone charger and I know there's going to be a lot of current (hopefully) and voltage making it's way to the 7805.  I don't want anything to be lost as heat, so does the 5 volt regulator need a heat sink??  Thank you so much!!!


With what you'll be producing with a shake generator, there'll be no problem with over current. The problem will be to get enough current!
To that end, you'd be better off using a low dropout regulator to reduce the losses.

Hi Andy!
Thanks for the info. Can you tell me anything about the low dropout regulator? Do I just put it in place of the 7805 or do I need to hook up the circuit differently? Also, does it output a particular voltage? I don't want my students to fry their cell phones. Thank you!!


The one I've linked comes in various voltages but the 5V version one is the one you need and is a drop-in replacement for the 7805. There are others available too if you search for 5V LDO regulator.  Anything 100mA or upwards would be fine for your application.
As Frollard says, you'll also need a bridge rectifier and a capacitor - The rectifier to convert to rough DC and the capacitor to smooth it.

Here's your problem:

Anything over 5 volts going into a regulator will get burned off as heat. period.
Thats how it regulates.

Your shake weight will probably produce intermittent AC at ridiculo-voltage, depending on the speed of shaking, n factor on the magnets, and number of coils.

so...build your coils so that their RMS voltage (root mean squared), i.e. average while shaking + including the downtime at the end of each stroke will be around 6.5 volts.

Why 6.5? because you will run the power through a full wave bridge rectifier, and each of those diodes will drop just over half a volt.

THEN you feed the output into a big capacitor to smooth out the output. Then it will connect to a low dropout (read higher efficiency) regulator.

Even more efficient would be a 5v boost/buck regulator.