I have 15VAC and want to make a power indicator light.
I noticed that different color LEDs seem to pull more power then others. I can hook up two green off the same resistor but when I try a red and green just the red comes on and the green goes out. I also have some weird one I want to try like pink and white. The pink seems to be a UV led lighting up a phosphorous? One cool thing about my eye condition is I see the spectrum of things. With most LEDs its monochromatic, but on this pink one I see the pink color and a deep blue band thats at much less intensity. Also white LEDs don't look like RGB, but don't look like when I look at a incandescent light bulb either (I will see a solid rainbow smear with bright yellow and very little after green). Some white LED's look like the spectrum of HID's where I will see sharp bands of color, but not what you would expect if you held a prism up to it.
Anyways. I know the E = 15V- 0.6V for the 1/2 wave rectifying diode but this s not enough to figure out ohms law. Do you just put in some arbitrary resistor then figure out the mA for each LED? MY meter doesn't do current even with a new fuse.
Hi,
We can still approximate using Ohm's Law but we have to subtract the LED voltage first. That means we have to know or can measure what is called the "characteristic" voltage of the LED.
For example, a small 5mm white LED might have a characteristic voltage of 3.5 volts. If we want to know the current draw with a 5v power source and a 100 ohm resistor we first subtract the 3.5 from the 5 and then use Ohm's Law.
5-3.5 equals 1.5, then 1.5/100=0.015 which is 15ma.
For a red LED it might have a characteristic voltage of 2v. Now we do the same thing but use 2v instead:
5-2 equals 3, and then 3/100=0.030 which is 30ma which may be too high for this LED so we might increase the resistor to 200 ohms and now we have:
5-2 equals 3, and then 3/200=0.015 which is 15ma and now we are more in the right range of current.
In a formula this would look like this:
i=(Vcc-Vd)/R
where
i is the current through the LED,
Vd is the characteristic voltage of the LED,
R is the resistor value we chose.
Now many times we want to know the required resistor (R) value, so we turn that around and get:
R=(Vcc-Vd)/i
and this directly calculates the resistor value.
Note that this is always an approximation because the LED voltage is not exactly what we use in the formula, but it's close. To know for sure it is a good idea to measure the current.
Lastly, when using higher power LED's it is required to calculate the power dissipation in the resistor. The power dissipation is:
Power=R*i*i
which is simply the resistor value times the current times the current a second time, or just R*i^2.
The power rating of the resistor should be as a min about two times that value, so the resistor power rating should be at least:
ResistorPowerRating=2*R*i*i
and that is a min. If the resistor will be confined in a closed container with no air flow the power rating must be increased more. The final project must be tested to make sure the resistor does not overheat.