# Help with calculating how to operate LED”s correctly

#### Icanmakeit67

Joined Sep 23, 2018
12
I’m trying to use LED’s in a Halloween project. I have several fairy light strings that each contain 20 tiny LEDs. When purchased they were provided with twin button cell power modules. Each contains two 2032 button cells for a 6 V power supply. I have a lead acid 12 V battery I want to use to run the entire project. I have several 12 V programmable light strips and want to power the fairy lights also at 12 V. I do not have specs for the Individual LEDs in the light strings. I do know the entire string runs at 6 V as I said and I know the current the entire string draws. There are two colors one string green one string purple they draw current differently. How can I calculate the current limiting resistor so I can replace the 6 V or two button cell power supply with a 12 V power supply. Any help would be gratefully appreciated.

#### ericgibbs

Joined Jan 29, 2010
17,454
hi 67,
Welcome to AAC.
E

#### BobTPH

Joined Jun 5, 2013
6,535
Be very careful. Those lights are most likely relying on the high internal resistance of the CR2032s. Just providing a stiff 6V will most likely burn them out.

You need to measure the voltage and current going into the string of lights.

Also, using a 12V battery will be wasting at least half the power if you limit the current with just a resistor. To utilize the full capability of the battery you would need a switch mode constant current supply.

Bob

#### WBahn

Joined Mar 31, 2012
28,510
I’m trying to use LED’s in a Halloween project. I have several fairy light strings that each contain 20 tiny LEDs. When purchased they were provided with twin button cell power modules. Each contains two 2032 button cells for a 6 V power supply. I have a lead acid 12 V battery I want to use to run the entire project. I have several 12 V programmable light strips and want to power the fairy lights also at 12 V. I do not have specs for the Individual LEDs in the light strings. I do know the entire string runs at 6 V as I said and I know the current the entire string draws. There are two colors one string green one string purple they draw current differently. How can I calculate the current limiting resistor so I can replace the 6 V or two button cell power supply with a 12 V power supply. Any help would be gratefully appreciated.
Are the two strings always one or always off together? Or do you need to have one on when the other is off some of the time?

How much current does each string draw? Since they are powered by button cells I'm going to guess that it's not very much.

Let's say that one of the string draws 100 mA from the 6 V cells. You would need to drop the other ~6 V across your resistor when there is 100 mA flowing through it. Ohm's Law then gives us R = V/I = 6 V / 100 mA = 60 Ω.

Another way to figure it out is to start with a fairly high valued resistor and measure the voltage across the place where the button cells were originally. Keep trying lower-valued resistors until you get close to 6 V.

#### WBahn

Joined Mar 31, 2012
28,510
Be very careful. Those lights are most likely relying on the high internal resistance of the CR2032s. Just providing a stiff 6V will most likely burn them out.
Using a voltage dropping resistor will not provide a stiff 6 V supply. But this is a situation in I agree that the usual recommendation of using a voltage regulator to generate the 6 V rail is risky.

Normally using a resistor to drop the voltage is not a good idea because of the poor regulation. But since this is a steady load, that shouldn't be too much of a problem.

Wasting 50% of the energy is annoying, but probably not a deal stopper since we're talking about currents that coin cells are expected to provide for a reasonable amount of time.

#### BobTPH

Joined Jun 5, 2013
6,535
Using a voltage dropping resistor will not provide a stiff 6 V supply. But this is a situation in I agree that the usual recommendation of using a voltage regulator to generate the 6 V rail is risky.
Of course it wouldn’t. I was warning the OP that using, say, a 6V motorcycle battery to replace the two coin cells could be a disaster.

Bob