I’m likely to use “fairy lights”, a string of small LEDs used to light craft projects, in a theatrical prop I am building. The prop is a cake, with several working candle effects.
An Arduino Nano will create the effect.
We have a string of fairy lights. I want to use the LEDs individually driven by PWM IO ports in the Nano.
Since the original string uses the internal resistance of two CR2032 batteries, I needed to know the internal resistance to calculate dropping resistors when using the LEDs as I intended.
I found this reference. Therein is a chart from which I estimated the internal resistance to be 10-15 ohms. Two batteries would be 20-30 ohms.
I checked for typical characteristics of a white LED. I don’t know yet the actual specs, but used typical values in my calcs.
So, based on the following:
Then, I did the following calculations.
(5-2.8)=2.2v
R = 2.2/0.018 = 122 ohms
Looking at this another way, ten LEDs in parallel need 180-216 mA. That current draw over 20-30 ohms produces a voltage drop of 3.6-4.3 volts.
Something is amiss and I’m confusing myself. Is my approach correct? I don’t have the string yet. I’m tempted to wire up one LED from the string with a 100 ohm resistor and see what happens. Of course, if I can do that I will then have the string and can measure current and forward voltage directly.
Help!
An Arduino Nano will create the effect.
We have a string of fairy lights. I want to use the LEDs individually driven by PWM IO ports in the Nano.
Since the original string uses the internal resistance of two CR2032 batteries, I needed to know the internal resistance to calculate dropping resistors when using the LEDs as I intended.
I found this reference. Therein is a chart from which I estimated the internal resistance to be 10-15 ohms. Two batteries would be 20-30 ohms.
I checked for typical characteristics of a white LED. I don’t know yet the actual specs, but used typical values in my calcs.
So, based on the following:
- Supply 5V
- CR2032 internal resistance of 10-15Ω
- Vf = 2.8V
- I(led) = 0.018mA
Then, I did the following calculations.
(5-2.8)=2.2v
R = 2.2/0.018 = 122 ohms
Looking at this another way, ten LEDs in parallel need 180-216 mA. That current draw over 20-30 ohms produces a voltage drop of 3.6-4.3 volts.
Something is amiss and I’m confusing myself. Is my approach correct? I don’t have the string yet. I’m tempted to wire up one LED from the string with a 100 ohm resistor and see what happens. Of course, if I can do that I will then have the string and can measure current and forward voltage directly.
Help!