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Joined: May 2003
Posts: 107
J
james S Offline OP
Member
when using resistors to limit the current for say an L E D have you not got to take into account the voltage drop the resistor will create if the led is dependent on a particular value of voltage?

i know silly questions time!!!

can't see wood for the trees maybe?

Joined: Sep 2001
Posts: 806
N
Member
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).]

Joined: Dec 2003
Posts: 886
H
Member
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.

-Hal

Joined: Apr 2002
Posts: 2,527
B
Moderator
See also www.ledmuseum.org/ledlinks.htm for several “resistor calculators” and a lot of LED-application data.

www.ledmuseum.org is itself an admirable and enormous collection of practical LED information and history, and well known for many merciless LED-flashlight reviews in “The Punishment Zone.”




[This message has been edited by Bjarney (edited 05-22-2004).]

Joined: Nov 2002
Posts: 790
W
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Quote
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.

Joined: Dec 2003
Posts: 886
H
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Agreed.

Joined: Aug 2001
Posts: 7,520
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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.

Joined: May 2004
Posts: 13
4
Member
not to confuse anyone...make sure that your not dealing with an OLED also...it is organic in nature but has a different set of rules...

when in doubt...put 480 on it !!!

that was a joke...don't play with 480 and led's...they can curl your toes !


...Despite all my RAGE, I am STILL just a rat in a cage...
Joined: Jun 2004
Posts: 176
P
Member
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.


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