Not the right question. An LED is a current controlled device. You can put
any positive voltage you like across it so long as the maximum current rating
is not exceeded. You want to limit current, not stop it.
The cheap calculation is simply to divide the expected voltage by the desired
current, then choose a resistor that matches the value of the result. So for
example if you had a 30V source and you wanted 15 mA of current across the
LED you would divide 30V/0.015A giving 2000 ohms. So put a 2K resistor
in serial with the LED.
Now there's actually a bit more to it. First off the LED will drop some of
the voltage. That's what's the maximum voltage rating. Say that the LED drops
1.7V. Well that's 1.7V that the resistor won't see, so the resistor will
drop the other 28.3V. So the actual current is 28.3V/2000 -> 14.1 mA. Less
than the 15 mA you wanted. If you really want 15 mA you have to subtract
the LED voltage from the source, then divide by the current. So use
(30V - 1.7V) / 0.015 mA -> 1886 ohms.
The second thing is the resistor power which is the voltage dropped times the
current across it. So the power is 28.3 * 15 mA -> 0.4245 W which means you'd
need a 1/2 W resistor.
-In article <[email protected]>,
->I am making an LED light.
-> How do I stop current from getting to the
->LED if the voltage is greater than it's max rating?
-Not the right question. An LED is a current controlled device. You can put
-any POSITIVE voltage you like across it...
I hate it when I forget something. All diodes have a Inverse voltage limit.
LEDs have a particularly low one (like negative 5-10V) that will cause them
to go up in smoke. So if there is any chance that the LED will see negative
voltage, be sure that you limit it. This is usually done by putting a second
diode (possibly an LED) in anti-parallel to the first.
So an example that pulls it all together. A while ago I experimented with using
cheap motion sensor floodlights for my homebuilt alarm system. I didn't want
to take the time to strip out the high voltage electronics, plus I liked the
fact that you could use a standard junction box. So I removed the light sockets
and was left with a wire pair that developed 120V AC when activated. Since I
didn't want that type of voltage floating around my system, I used an opto-
isolator to capture the signal. The opto has an LED in it. Using the
calculations outlined before I choose a 2W 12K resistor to get 14 mA at the
peak 170V and wired a second LED in anti-parallel. A clean 60 Hz signal
resulted on the other side when activated.
-> In article <[email protected]>,
-> -In article <[email protected]>,
-> ->I am making an LED light.
-> -> How do I stop current from getting to the
-> ->LED if the voltage is greater than it's max rating?
-> I hate it when I forget something.
-You forgot something else, subract the led forward voltage from the supply
-voltage before dividing by the current required.
-So with a 2.1v led on a 9v supply, for 15mA current you'll need series
-resistor of (9-2.1)/.015 = 460 ohms, nearest common value is 470R which will
-give you 14.7mA.
Re-read my original post. Right after the "Now there's a bit more to it than
that..." both the forward LED drop and resistor power rating are discussed.