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Old 11th February 2008, 09:11 PM   (permalink)
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If your alternator over-charges the battery at 14.4V and the 78M05 draws 1.2A then it must dissipate 11.3W which is not cool.
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Old 11th February 2008, 09:16 PM   (permalink)
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Uhmm.

Well, what that meant was, a USB port on a PC can only supply a max of 500mA. I know the wall charger charges the device a heck of a lot faster, so it stands to reason the device draws more.

I dont really know what I meant with that.

You've succeeded in confusing me there, with my own words :P

Anyway, from the looks of that link you just pin the 7805 in and put capacitors on the in/out (for smoothing im guessing?).
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Old 11th February 2008, 09:27 PM   (permalink)
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Quote:
Originally Posted by wetlegs6
Uhmm.

Well, what that meant was, a USB port on a PC can only supply a max of 500mA. I know the wall charger charges the device a heck of a lot faster, so it stands to reason the device draws more.

I dont really know what I meant with that.

You've succeeded in confusing me there, with my own words :P

Anyway, from the looks of that link you just pin the 7805 in and put capacitors on the in/out (for smoothing im guessing?).
I was just talking about the word "push"--to oversimplify, voltage is "pushed", and current is "pulled". But yeah, if the charger can draw more than 500mA when available, it will, and it would presumably use that to speed up the charging.

The input-side cap is sometimes called a "bypass" capacitor and you use it to help stabilize the voltage if the power supply is not very close to the regulator. The output-side cap helps keep the regulator from oscillating and also helps transient response.


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Old 11th February 2008, 09:28 PM   (permalink)
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The capacitors must be very close to the pins of the regulator. They keep it from oscillating.

You never said how much current is needed at 1.75V. If it is low then two resistors can make it from the 5V supply.
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