Simple project needs help.

Thread Starter


Joined Mar 17, 2009
I'm so ashamed I forgot this important information after going to school for electronics many years ago, but here I am asking for some help. This is a very simple circuit, but I forgot OHMs Law and I'm not good with math.

Supply : 12v 300mA

LED 1 : 3.5/4v 20mA
LED 2 : 3.5/4v 20mA
LED 3 : 3.5/4v 20mA

Series Circuit

I need to step the current down from 300mA to 20mA to keep the LEDs from burning out. Please remind this old man the formula so I can perform future projects w/o feeling stupid.



Joined Jul 17, 2007
I=E/R, or Current = Voltage / Resistance.
Conversely, R=E/I, or Resistance = Voltage/Current
Also, E=IR.
So, if you're going to wire the LEDs in series, you add up their total Vf, and subtract that from your supply voltage.
Then you figure out what size resistor you need to limit the current across the remaining voltage from the previous step. Since R=E/I, Rlimit = RemainingVoltage/DesiredCurrent

For example, let's say your LEDs have an average Vf of 3.8v @ 20mA.
Rlimit = (Vsupply - TotalVfLED)/20mA = (12 - (3*3.8))/0.02A = (12-11.4)/0.02 = 0.6/.02= 30 Ohms.

If you're using a regulated 12v supply, that should work OK.
However, if you're thinking of using it in a motor vehicle, it won't work so well.
The electrical system in a vehicle may vary from a low of 11.4v (battery discharged) to over 14v when running. You'll need a different plan of attack in that case.