# Wiring together LED lighting for a display

Discussion in 'LEDs and Optoelectronics' started by PotterBen, May 30, 2017.

1. ### PotterBen

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May 30, 2017
Hello,

I want some advice on wiring some small LED lights that I use for my pottery display at Art fairs to an AC/DC adapter. I have 12 sets of these LED lights wired together in parallel to run off a single AC/DC adapter. I scrapped an old AC/DC adapter I had laying around but the lights are a bit dim compared to running off the original included batteries. Something isn't right. I must have miscalculated something. The adapter I am using puts out 3V and 700mA. I did this a while ago and remember thinking then that the amps were too low. But now looking at it again, should the voltage be 6V because the original two CR2032 batteries were in series? The LED lights I am powering are these ones from amazon http://a.co/0jIfY7u I want to replace the AC/DC adapter I have now with one that is the correct size. Any advice would be greatly appreciated.

Thanks,

Ben

2. ### Harald KappModeratorModerator

9,306
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Nov 17, 2011
Welcome to EP, Ben.

Absolutely.
As for the max. current there is no information on A... You'll have to measure the current required with an ammeter/multimeter.
Or follow this estimation: A CR2032 has a capacity of ~240 mAh. The advertisement states over 24 hours of continuous operation. Let's take that number and calculate the current, then we have I = 240 mAh/24h = 10 mA per LED string. Which makes for a total of 12*10mA=120 mA for 12 LED strings in parallel. A 300 mA wall wart should really be good enough.

3. ### PotterBen

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May 30, 2017
How would I measure the current with my multimeter?

4. ### Harald KappModeratorModerator

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1,889
Nov 17, 2011
put it in series between one pole of the power supply and the corresponding pole of an LED strip. Wire the other end of teh stripü directly to the other pole of the power supply. You should measure with a power supply of 6 V. LEDs are non-linear devices. Measuring at another voltage will give you a current value that doesn't scale proportionally with voltage.
An easy way of doing this is by using the original battery case and a pair of batteries. Remove the part that connects the two batteries and connect the meter in mA range instead as shown:

5. ### Audioguru

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Sep 24, 2016
The LEDs survive when powered by the weak little battery cells because the battery cells have series resistance in them that limits the current. A CR2032 battery cell is rated at a very low current of only 0.19mA. A 6VDC power supply will quickly destroy the LEDs. You must add a separate resistor in series with each string of LEDs and the resistance must be enough to limit the current.

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6. ### BobK

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Jan 5, 2010
The internal resistance of these cells starts at about 10Ω rising to 30Ω as they deplete. At 10mA current, 20Ω (for two batteries in series) the voltage loss is only 0.2V. So it is unlikely that the LED strings have no resistor and are surviving 5.8V.

Bob

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7. ### Audioguru

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Sep 24, 2016
I agree that there is probably a series resistor inside the battery case. Their 20 LEDs string that is powered with three AA cells has an visible resistor.

8. ### PotterBen

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May 30, 2017
There was no resistor in the original battery case.

9. ### Audioguru

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Sep 24, 2016
Then the current is limited by the weak little batteries.

10. ### PotterBen

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May 30, 2017
So what size resistor should I wire before each LED light chain?
I have a 6 Volt, 1 Amp AC/DC adaptor.

11. ### Harald KappModeratorModerator

9,306
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Nov 17, 2011
Measure the current with the LED strip connected to the original batteries. Measure also the voltage across the batteries when connected to the LED strip. The measured voltage will be below 6 V!
Calculate the resistance required from R = (6 V - Vmeasured) / Imeasured.
Calcuate also the power dissipation of the resistor from P = Imeasured^2 * R
Select an approproate resistor from the nearest standard values (E12 or E24 series) with suitable power rating (probably in the range 1/2 W ... 1 W.

12. ### PotterBen

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May 30, 2017
I also have another light that runs off two AA batteries at 3V. Can I use the same 6V AC/DC adapter with larger resistors on those lights?

13. ### PotterBen

9
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May 30, 2017
This is what I measured...

CR2032 LED lights - 2.88V 0.12A

(6V - 2.88V) / 0.12A = 26
0.12A^2 * 26 = .3744

AA LED lights - 3.06V 0.03A

(6V - 3.06V)/ 0.03A = 98
0.03A^2 * 98 = .0882

So how do I use this new information to get the proper resistors? I'm new to using resistors. My electronics knowledge is pretty rudimentary.

14. ### duke37

5,231
718
Jan 9, 2011
I have not checked the calculations but
26Ω should be replaced by 27Ω, a standard value
0.5W or bigger would do.

98Ω could be replaced by 100Ω 1/8W.

15. ### PotterBen

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May 30, 2017
My Local electronics shop had the 100Ω 1/8W but not the 27Ω.
So I went with a 33Ω 1/2W. Both have a 2% tolerance.

16. ### BobK

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Jan 5, 2010
Pulling 120 mA from a CR2032 is battery abuse. Or maybe assault on battery.

What a stupid design. The batteries appear to be in series, so they are at 6V under no load. More than half the energy is then being wasted heating the batteries.

Bob

17. ### PotterBen

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May 30, 2017
Ooops I was wrong.... it was actually pulling .05A on the CR2032. I made a mistake measuring.

So then...

(6 V - 2.88V) / .05A = 62.4
.05^2 * 62.4 - .156

So I will get a 68Ω 1/4 watt resistor??

18. ### Audioguru

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Sep 24, 2016
I betcha the "6V" AC-DC adapter voltage will be much higher than 6V with the puny load of 50mA. Maybe 9V or 10V? Then the current will be 105mA and the LEDs will fail soon and the resistor will get so hot that it might glow and smoke.

50mA is too high for ordinary LEDs that are rated at 20mA. Did you replace the original weak Chinese batteries that are full of rice with much more powerful Western batteries?

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19. ### PotterBen

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May 30, 2017
I just tested the AC/DC adapter and it tested at 6.26V.
I ended up with 33Ω 1/4W resistors that I am going to wire two together to get 66Ω.
Hopefully I won't burn out all these resistors when I plug it it.

Yes, I was using brand new western CR2032 and AA batteries when performing tests.

20. ### Audioguru

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Sep 24, 2016
The original Chinese batteries were probably weak (full of rice) and could not produce more than 20mA. Calculate resistors to limit the current to 20mA, not 50mA.