I am getting an odd reading from an array of LEDs, which I could do with a little help understanding. I have a very specific project, but to abstract this question from that, let’s assume the following:
I have 24 LEDs in 4 sets, each of 6 LEDs with a 68ohm resistor. They are 1.8v LEDs using 20mA and I have a 12v supply.
simulate this circuit – Schematic created using CircuitLab
I understand how to calculate the required resistor and current for single LEDs but I used the rather excellent http://led.linear1.org/led.wiz to do the math for me on this. It concludes my array will use 80mA.
However, when I use my meter, it tells me the array is only using 8mA?!
I am planning to run this from my Arduino Micro, which can easily deliver 8mA from a pin but certainly not 80mA, so the difference is crucial to me.
Initially I thought I was reading the meter wrong (it’s new) and I was out by a factor of 10 but the only other thing of a known(ish) current I could benchmark my meter with at 12v was the Arduino itself, which is drawing 37mA, which to me sounds about right (and 370mA would seem too high for a Micro), meaning my meter is probably correct?
I know you can get variations in LEDs or whatever, but I have 3 of these arrays and they are all reading the same ~8mA. What am I missing or should this array be pulling only 8mA? How should a parallel array of in-series LEDs be calculated?
Thanks for any help.
Best Answer
How much current flows through each branch of the circuit?
All branches are in parallel with the 12V source, so there is 12V across each branch. $$12V = 6*V_{LED} + I_{BRANCH}*68\Omega$$ \$V_{LED}\$ is about 1.8V, so $$1.2V = I_{BRANCH}*68 \Omega$$ $$\frac{1.2V}{68\Omega}=I_{BRANCH}\approx18mA$$
You have four identical branches each conducting the same magnitude of current, so the total current is about 4*18mA = 72mA. The tool you used gives a reasonable estimate.
If you accidentally used \$680\Omega\$ (Blue Grey Brown) resistors instead of \$68\Omega\$ (Blue Grey Black) resistors, the current will be 10% of the calculation above, giving around 8mA.