You need to know the forward voltage drop of the LED, and the current rating. A typical red LED will have a forward voltage drop of ~1.4 VDC, which is essentially independent of current. Other colors have slightly higher voltage drops, with blue being the highest at ~3 VDC
Basically, you subtract the LED voltage drop from the supply voltage, and then use ohm's law to select a resistor that will drop the remainder of the voltage at the desired current. An example:
Powering a green LED with a voltage drop of 1.6 V and a current rating of 15 mA. Supply voltage 24 VDC.
24VDC (supply voltage) - 1.6 V (LED drop) = 22.4 V (drop across resistor)
Resistance = 22.4V / 0.015A = 1493 ohms
Closest standard value would 1.5K ohms.
Power dissipated in the resistor would be:
22.4V * 0.015A = .336 Watts
Next higher standard value would be 1/2 W
[This message has been edited by NJwirenut (edited 05-22-2004).]
#38421 - 05/22/0409:08 PMRe: very basic electronics!!!!!!
No, not really because the LED draws so little current. I normally just ignore that the LED is in the circuit and select the series resistor to limit the current to the required value when placed across the supply voltage.
I normally just ignore that the LED is in the circuit and select the series resistor to limit the current to the required value when placed across the supply voltage.
Ignoring the LED voltage drop doesn't much matter when the supply voltage is much higher than the LED drop. But if the supply voltage is 2V you definately need to consider the LED voltage drop. For a red LED (1.4V) you'd need a 40 ohm resistor to pass 15ma thru the LED. Ignore the LED drop and select a resistor based on 2V/15ma = 133 ohms, and use it with the red LED and you'll only get 4.5ma thru the LED. The LED will be much dimmer than desired.
#38424 - 05/23/0401:43 AMRe: very basic electronics!!!!!!
I think everyone else has beaten me to all the calculations, but one other area where you might want to adjust the resistor values (and thus the LED's forward current) more accurately is where you're using a dual red/green LED.
By careful adjustment of the red and green currents you can obtain varying shades of yellow and orange for some applications.
#38426 - 05/26/0412:49 AMRe: very basic electronics!!!!!!
I usually use this method: (derived from R=E/I) R=(V-Vf)/I R=Resistor Needed V=Voltage applied to the resistor and LED Vf is the forward voltage of the LED. I is the current used by the LED (typically 20mA for red.) Remember, I is in Amps, so you must convert the LED's req'd mA into A. An LED is a current device, not a voltage device. It will always drop the voltage by Vf. Adding more current will make them brighter.