I had a bit of a play last night. Putting two of these things in series reduced the current to about 45 mA!
That produced a very noticeable decline in light output.
I wasn't looking to tame the light output down too much.
Really I was just looking to drop the current back a bit, more toward what I was expecting.
(My son said he thought they were 1W, which would have been about 80mA per module.)
Being the owner of enormous bundles of brand new resistors and other bits, (there's a story for another day, give me a shout if you are in need of a few hundred schottky diodes) I suppose I could play around with leaving the modules in parallel, and just insert a resistor.
I'm a tad rusty, I left high school 45 years ago, but lessee:
V=IR
so
V/I=R
I have V=13, and I=0.360 in my circuit.
Therefore:
R = V/I = 13/0.36=36 ohms is the theoretical resistance of the modules when the lights are on.
Now, I want to limit the current to, say, 250mA
So I need a total resistance of R = V/I = 13/.25 = 52 ohms.
I've already got 36 ohms in the modules, so my current limiting series resistor needs to be 52-36 = 16 ohms
Now, if I have 16 ohms and 250mA then I have a voltage drop across the resistor of:
V=IR=0.25 x 16 = 4 volts
P = VI = 4 x 0.25 = 1 watt power handling capacity for the resistor.
So right now, I'm thinking I have an LED module that flows 360mA at 13V, and if I wish to reduce the current to 250mA, then I need a series resistor of 16 ohms and minimum 1 watt.
Also, the voltage across the LED module is now 13-4 = 9
Using the previously calculated 36 ohms, back calculating gives me V/R=I=9/36=0.25Amps
It all adds up as it should, I think.
I can't think in watts, apparently.
In LED's it's current that determines light output, so they say.
So current reduced from 360 to 250mA = 30% reduction.
If it
is watts, then I have 9x0.25 = 2.25W going through the module.
That's a reduction of more than 50%. Too much, and simpler to just use one module rather than two in parallel with a resistor.
It will be interesting to see which way it works
To me, the logic is sound, it remains only to download an app for the phone to measure lumens (there must be losses - there are always losses), and give it a try.
Anyone wanna check my logic?