# converting LED lights from battery to wall wart



## drwilde (May 14, 2008)

Hey I could use some electronics help. Im sure its very simple i just dont have the electrical knowledge to figure this out. I made this spider air rig last year. the spider has 4 flashing led eyes in it that run off a AAA battery pack. it says on the pack DC 1.5 v x 2(batteries). I want to cut off the pack and splice in a wall wart so i can have them come on when the prop is activated and stop when he resets. i have a old phone charger wall wart that says output is dc 3.7 v, 350 mA. would this work or would i risk burning out the leds? i cant get to them easily to replace them if i do. I tested with a multi meter and the battery pack is actually 2.8v and the wallwart is saying 8v which confuses me. Anyone have a simple way to explain what i need to do?


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## pshort (May 6, 2008)

The batteries are reading low because they are partially discharged. The wall-wart is reading high because its output is probably 'pulsating DC', ranging from 0V to 8.2V back to 0V 120 times per second, and your voltmeter is just catching the highs. Under a full 350 mA load the RMS output (which your meter might not read) would be the 3.7V that is shown on the label.

Enough of the technical jargon...is there any reason why you can't just wire a switch between the battery pack and the prop?


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## drwilde (May 14, 2008)

thanks. the reason i need to make it a plug in is my motion sensor has 110 outlets that turn it on to jump and i just wanted the eyes to turn on and off with it instead of staying in flash mode continuously. the battery pack itself does have an on of switch.


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## drwilde (May 14, 2008)

here are pics of what it looks like

http://www.theboneyardfl.com/help_photos.html


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## Otaku (Dec 3, 2004)

Are the batteries new? That voltage seems a bit low. The 8VDC your getting from the wart just means that it's not regulated. If you know the forward voltage and current of the LEDs you can use this calculator:
http://led.linear1.org/led.wiz
to determine the resistor you need with the wart. I assume that the LEDs are pretty low voltage for 4 of them to run on a 3VDC battery pack.


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## drwilde (May 14, 2008)

yes prob old batteries. the leds are embedded in the plastic eyes and i dont want to tear open the prop to get to them so i dont have any specs on the leds themselves. anyway to guestimate?


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## drwilde (May 14, 2008)

should i not be using a non regulated adapter?


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## GhoulishGadgets (May 13, 2010)

Hi,

It does sound like an unregulated adapter if it's showing 8v unloaded.
You sure the output is given as 3.7v dc?

I assume that the LEDs flash together rather than random?
That would mean there is a small circuit in there flashing them all.

Personally, I'd go for dropping the input voltage with a diode or two...
a diode can drop around 0.6v to about 1v across itself.
start with 2 diodes, I'd go for 1n4148 or 1n4001
both have a band one end, this indicates the cathode.

so connect the -ve of the lights to the -ve of the adapter.... easy
now connect the diodes together first - connect a non-banded end of one to the banded end of the other
now of the 2 connected diodes, the non-banded end of the pair goes to the adapter +ve
the banded end of the pair goes to the +ve connection of the prop

if you test the voltage with the diodes in place, but not connected to the leds, you will see the higher, unloaded voltage reading, don't worry about this, it's normal.

if you're concerned, connect it all up, connect your meter to the prop wires (after the diodes) then very quickly switch on and off - you'll see the voltage now applied to the leds/circuit.

you may find that two diodes is too much of a drop, so you may need to remove one.

ask away as needed

I can drop a diagram if needed.

Si


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## pshort (May 6, 2008)

The cheapest way to make it work is to add a series resistor between your unregulated wall-wart and the LEDs. It will take a little investigation/testing to figure out how much current the LEDs need, and then it is easy to use the calculator that otaku suggested to find out what resistance is needed. My only concern is what effect this technique would have on the flashing function of the LEDs. You could test this out if you have resistors in the range of 100ohms to 390ohms laying around (start with the larger values, which would likely result in overly-dim LEDs and/or no flashing).

Another possibility is to add a capacitor and voltage regulator to the output of the wall-wart. The parts themselves are cheap, probably less expensive than the postage or gas needed to get them.


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## pshort (May 6, 2008)

Ghoulish -

The 3.7V is a rather common value, I have several Nokia phone chargers laying around that say 3.7V/350mA or 3.7V/300mA.


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## drwilde (May 14, 2008)

thanks for the info. yes i have some resistors. 2 types ive discovered. 336 ohm and 390 ohm im hoping that i can get these to work since i already have plenty of them.

to answer GG, yes the wall wart is 3.7v dc and yes they all flash together. after looking closer they may not even be red leds but white, the red plastic eye ball makes it look red.


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## pshort (May 6, 2008)

Also, the 8.2V doesn't surprise me. The phone charger wall-warts are often just a transformer and diodes, maybe a filter cap. The 3.7V rating would be an RMS value under full load, what he is measuring on the meter with a DC setting is probably a peak value under no load.


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## drwilde (May 14, 2008)

ok what if i throw a 336 or 390ohm resistor inbetween the wall wart and the led wires on the positive side? am i going to blow myself up? Will i burn out the lights?


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## GhoulishGadgets (May 13, 2010)

to be honest, it'll make no difference if you insert in either +ve or -ve side
if anything, those values may be too high, so you're safe - no circuit damage or to you

try the 390 first as it's highest, if the leds are dim or flash strange, connect 2 resistors in parallel to reduce the overall resistance value, post back your results if you need help

Si


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## drwilde (May 14, 2008)

thanks ill try it now. so without the resistor what would happen?


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## GhoulishGadgets (May 13, 2010)

maybe it could allow too much current into the leds.. maybe
the electronics would most likely be fine, perhaps flashing quicker.

the circuit will be built to take upto 3v or just over, (2 batteries), so applying 3.7 isn't that bad.

a quick 'flash' test should be ok, connect up - no resistors, plug-in and switch on-off quickly... see what happens.

if you do this, again, drop your meter on the leads and check the applied voltage when you 'flash' test it - check it's what you expect from the label

Si


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## drwilde (May 14, 2008)

worked fine with either resistor. thanks for the help! i knew it had to be easy, just gotta get over that fear of electricity!


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## GhoulishGadgets (May 13, 2010)

well done, good to hear

Si


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## Doc Doom (Aug 28, 2008)

drwilde said:


> thanks ill try it now. so without the resistor what would happen?


You will probably let the magic smoke out and will really be bummed out.


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## corey872 (Jan 10, 2010)

I do wish they would sell bottles of 'magic smoke' which could be put back in electronics to make them work again.

But you are right...LED's - especially the high power variety usually aren't too happy with the 'try it and see' method. In the best case, you have too much resistor, make a lot of heat and have a dim LED. Worst case, the LED is really bright...for about three seconds, then POOF!

It's always best to take a quick volt reading on your power supply and either scratch a few numbers on paper or punch up one of the many LED calculators on the internet and get close to the optimal value. That way you get a nice, bright LED which is also long lived.


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