Electronics - drop 12V to 3.6v

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Im hoping to power one of those wind up camping lanterns from the ships battery, so i need to drop aprox 12v to 3.6v, current negligible as its only running a few LEDs. Im hoping someone can point me to a dc-dc volt dropper i can wire straight in, otherwise what resistance could i put in the circuit to achieve this. Thanks, Matt
 

johnalison

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Widely available I believe. I got mine yonks ago but someone mentioned Maplin in a similar post recently. Mine is switchable between about 2v and 12v in steps and I use it for a portable radio.
 

europe172

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I used an old mobile phone car charger, just cut off the top and wire the 12v striaght in, the output is often about 5v
 

whiteoaks7

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If they are LEDs you won't need anything more than a resistor - ohms law V = IR or R = V/I take V as the max it will see i.e. the charging voltage (say 14V) less 1.2, and divide by I (in mA) for a result in kOhms. So if it takes 30mA you get (14-1.2)/30 = 0.42 kOhms so use 470 Ohms (next nearest available size up). You also need to know the wattage for the resistor given by I*I*R, in the example (30*30*470)/1000000 = 0.423 Watts so next size up 0.5W. Watts are important if you dont want the resistor to become a fuse! Just connect the resistor in the positive line from the supply (if it's on a cig lighter plug you can lose it inside).

If you're stuck pm me with the current in mA, but I might not be able to answer for a few days.

Cheers
 

Norman_E

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Wire 4 of them in series, 4 X 3.6 volts = 14.4 volts which is the maximum voltage you should have when the batteries are on charge, so you wont blow the LEDs through having too high a voltage. I would still expect them to work when battery voltage is down to 12. If the lantern has multiple LEDs in it just put four of them in series.
 

whiteoaks7

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Sorry - made a small mistake so before anyone else pounces on me here is the correction:

i.e. the charging voltage (say 14V) less 1.2, and divide by I (in mA) for a result in kOhms. So if it takes 30mA you get (14-1.2)/30 = 0.42 kOhms

The '1.2' is correct for an LED but you will already have a resistor to allow the lamp to work at 3.6 volts so the formula should be:

i.e. the charging voltage (say 14V) less 3.6, and divide by I (in mA) for a result in kOhms. So if it takes 30mA you get (14-3.6)/30 = 0.34 kOhms so use 330 Ohms.

The resistor will need to dump 0.33 Watts or greater.


Hope that clears it up.
 
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