# More LED help needed



## Doc Doom (Aug 28, 2008)

Last year I bought a Spirit Grave Digger prop and want to change the lantern battery pack for a wall wart. The lantern runs on two AA batteries that powers multiple light yellow-orange LEDs in a single combined bulb configuration. The battery pack puts out just under three volts but I can't measure the current. (Apparently I have the wrong type of meter.) I want to use a 4.3 Vdc wall wart but need help to figure out what size resistor I need to add to keep the magic smoke from escaping. The wall wart puts out 4.3 Vdc with no load.

Thanks in advance for all your help.


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## Spookie_T (Jul 18, 2011)

Doc,
Do you know what the power consumption of the LED's are that you have? Or how much power the whole circuit draws?


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## Otaku (Dec 3, 2004)

Remember that a DMM measures current when wired in series with the load, and voltage in parallel. Of course, if there isn't an option on the meter for amps then you're SOL.
That said, is the wart regulated, and how many LEDs are there? Here's an easy to use calculator:
http://led.linear1.org/1led.wiz


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## Doc Doom (Aug 28, 2008)

Spookie_T said:


> Doc,
> Do you know what the power consumption of the LED's are that you have? Or how much power the whole circuit draws?


I don't know the power consumption, either of the LED or the whole circuit. My meter lets me measure voltage and resistance, put I don't see a current scale.

I don't know how to use the LED calculator because I don't know the spec on the combined LED bulb. It appears to be one single bulb, but with several LEDs. Based on typical yellow/orage LED voltages, two AA batteries don't have enough power, but somehow the thing lights up.

Not that it helps much, but here is a link to the prop
http://www.spirithalloween.com/product/sv-the-night-digger/


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## Otaku (Dec 3, 2004)

Based on what I know of this sort of prop, the manufacturer probably aimed for the least number of batteries that would work to light the LEDs. I'd say that you're safe with limiting the wart to no more than 3VDC, the highest voltage that two AAs could possibly deliver. Again, it helps to know if the wart is regulated because you don't really know what the current load is for that LED. What does the label say about the output?


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## HomeyDaClown (Oct 3, 2009)

This is pretty easy to solve actually. Figure out the current draw or load of a black box by breaking one power lead and inserting a known resistor value (in this case I'd use 1000 ohms) to complete the circuit. Power up the unit and while its running (or trying to run) measure the voltage across the resistor (one meter lead on each resistor lead). That measured voltage is the voltage drop across the resistor.

Go here and simply plug in the voltage you read and 1000 ohms for the resistor value and it will tell you the current being drawn by your unit:

http://www.onlineconversion.com/ohms_law.htm

If you measured 5 volts across the 1000 ohm resistor, the current youe unit is drawing would be .025 amp or 25 milliamps


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## Otaku (Dec 3, 2004)

Cool, learn something new every day. Thanks, Homey!


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## HomeyDaClown (Oct 3, 2009)

No Problem....

Ohm's law works every time for some strange reason

E = IR or in this case E/R = I 
Where E = Voltage Drop across the resisitor R
And R = Value of the resistor
And I = Current through the resistor (and everything else)

It's the same way a multimeter would read current, there is a shunt (resistor) inline with the load
and the meter is calibrated for voltage drop across the shunt.


If good ol Georg Ohm knew that his work in 1827 was being used today to help haunter's build crazy
Halloween props, he'd come back to haunt us all!


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## corey872 (Jan 10, 2010)

LOL, ya, but we're being taught that 5/100(0) = .025 

I'd also caution that all you're seeing is the current limiting of the resistor because the LED won't have any of it's own.

If I had to do it:

You know you have a 4.3V supply and the lantern can run on 3V. You also know there is at least one LED in there, so flowing 20mA should be safe.

Given a 4.3V supply, 3V to the lantern at 20mA, you'd need a 68 ohm resistor (1/8th watt would be plenty). So make that circuit and see what the voltage is at the lantern...probably going to be less than 3V.

The next step up would be assuming 2 LEDs @ 40mA, requiring a 33 ohm resistor, then:

3 LEDs - 60mA, 22 ohms
4 LEDs - 80mA, 18 ohms
5 LEDs - 100mA, 15 ohms
6 LEDs - 120mA, 12 ohms
...

At some point, you'll eventually build up to 3V at the lantern terminals and you then have the correct resistance.

If you notice something curious about those numbers, you'll find each step to be essentially that many 68 ohm resistors in parallel - it stands to reason if one 68 ohm resistor can pass 20mA, then two can pass 40mA, 3=60 and so on. So with a small bag of 68 ohm resistors and your volt meter, you solve the problem pretty quick.


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## Doc Doom (Aug 28, 2008)

Thanks for the help and feedback. I was stumped as to how two AAs could power that many LEDs in the first place. I'm used to wiring my LEDs in series and totally forgot about parallel hook ups. Duh!


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## HomeyDaClown (Oct 3, 2009)

corey872 said:


> LOL, ya, but we're being taught that 5/100(0) = .025


What? The numbers were 5 volts and 1000 ohms.



corey872 said:


> I'd also caution that all you're seeing is the current limiting of the resistor because the LED won't have any of it's own.


Sorry, but current in must equal current out (Kirchoff's law) for a total of zero. Besides, the technique mentioned is nothing new and has been used in electrical engineering classes for years. I have used it more times than I can count and it works period.

Everything else is pure assumption....unless you want to summon the magic smoke. If it is just a plain ol unregulated wall wart, the only thing you can assume is that it is not 4.3 volts. The output on wall warts is rated based on an expected load requirement so it rarely matches the voltage stamped on it. So now you're dealing with 2 blackboxes.

I'll put my money on a known resistor (measured) value and a known power supply (measured) voltage in any case.


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## corey872 (Jan 10, 2010)

The post originally said 1000 ohms in a couple places, then said 100 in the last sentence. You caught the 100 and corrected to 1000 after I posted, so I added the (0) to my post. Either way your formula is correct, V/R=I would hold for this case, but your example 5V/1000 ohms = .005A or 5mA, no?

I also know the shunt resistance measurement is common, though usually done with a very low value / precision resistor and a millivolt measurement across the shunt. Current in = current out holds, too - the theory is sound, it’s the application which concerns me. Shunt resistance needs to be much lower than circuit resistance for the measurement to be valid.

For sake of argument, suppose 3V on the battery and five 20mA LEDs in parallel, so the circuit runs at 3V / 100mA. You want to break that circuit and put a 1000 ohm resistor in series with the battery/LEDs? So the max current which can get through a 1K resistor at 3V is 3V / 1000 ohms or 3mA – but what does that measure? Even if you run a wire from the resistor straight back to the battery, you’d still only see 3mA because the resistor is the current limiting element.

For a shunt measurement in this range, you could use a 0.01 ohm precision resistor. Hooking that in the circuit, you’d expect to see maybe 0.001 volt across the resistor and calculate 0.001V / 0.01 ohm = 0.1A or 100mA. This would be a valid reading, the down side is precision resistors aren't much cheaper/easier to come by than a simple meter which measures current.

Doc said the wart was '4.3V with no load'. I may have erred in assuming that was an actual measured voltage? But, I’d only expect that to be lower when a load is applied. Plus he mentioned multiple LEDs but did not know how many, I recommended starting with a resistor which would be safe for the full, no load wart voltage and only pass current for 1 LED / 20mA which would be pretty safe in my book.

At any rate, wasn’t trying to ruffle feathers, just point out a few kinks in an otherwise sound application and present an alternate cheap/easy way to dial it in.


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## Doc Doom (Aug 28, 2008)

Sorry if I started something between those trying to help. My thanks to both of you. 

BTW, the 4.3 Vdc was actually measured (with no load) and just happens to also be what is printed on the back of the wall wart.


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