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Calculating LED resistor

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cobra1

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I want to build an LED array using red and blue leds.

i have found a good calculator on the net that works out the required resistor, it even draws out the circuit for the array.

the problem i have is that it only bases this on using 1 type led.

the leds im using are:

Red:
2.4v forward voltage @ 20ma

Blue:
3.6v forward voltage @ 20ma


how would i work out the limiting resistor for say 8 RED and 2 BLUE leds??

source voltage is 24v

if there is a formula (im guessing there is) could someone post it, it would help a lot



EDIT: just realised this isnt possible, i have also solved my problem
 
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You cannot run it with just one string, so split it in two identical strings of 5 LEDs each:

Vf(4*reds + blue) = 4*2.4 + 3.6 = 13.2V.

Starting from 24V, the drop across the resistor will be 24-13.2 = 10.8V

Resistor to drop 10.8V @ 20mA = E/I = 10.8/0.02 = 540 (round up to std. value of 560Ω)

Power dissipation in resistor = E^2/R = 0.2, so a 1/4 or 1/2W resistor will work.
 
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hi mike, i am using series parallel arrangement, i just though of doing 2 seperate strings before you put your reply. should have thought of this earlier on really
 
You cannot run it with just one string, so split it in two identical strings of 5 LEDs each:

Vf(4*reds + blue) = 4*2.4 + 3.6 = 13.2V.

Starting from 24V, the drop across the resistor will be 24-13.2 = 10.8V

Resistor to drop 10.8V @ 20mA = E/I = 10.8/0.02 = 540 (round up to std. value of 560Ω)

Power dissipation in resistor = E^2/R = 0.2, so a 1/4 or 1/2W resistor will work.

But isn't it 10.8V @ 200mA because each string is 100mA and there are 2 strings (10 LEDs at 20mA each) so 10.8/0.2 = 54 (56Ω Resistor)
 
But isn't it 10.8V @ 200mA because each string is 100mA and there are 2 strings (10 LEDs at 20mA each) so 10.8/0.2 = 54 (56Ω Resistor)

Voltz,

First, the OP said that his LEDs are rated at only 20mA.

Second, if you think that it is ok to share one resistor between two parallel strings of LEDs, then you need to read this thread.
 
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I thought it would be ok seeing as there are so few LEDs, each branch is 100mA anyway isn't it (5 LEDs at 20mA) each
10.8/0.1 = 108 (110)
 
I thought it would be ok seeing as there are so few LEDs, each branch is 100mA anyway isn't it (5 LEDs at 20mA) each
10.8/0.1 = 108 (110)

No, that is where you are getting confused. If I have 5 LEDs in series and those LEDs are 20 mA the current for the string is 20 mA and not 100 mA. Forget resistors for a moment. Let's say I have 5 each LEDs rated as 1.2 volts @ 20 mA. If I place 5 of them in series I will need a voltage source of 1.2 * 5 = 6 and a current of 20 mA. Does that make sense to you now?

Ron
 
i have another question, if the numbers were to add up exactly, i.e in reloadrons example above, 1.2v per led x 5 = 6v @ 20ma

if you were to supply a stable 6v to this circuit is a resistor required, im guessing a 1ohm resistor should still be used here??
 
i have another question, if the numbers were to add up exactly, i.e in reloadrons example above, 1.2v per led x 5 = 6v @ 20ma

if you were to supply a stable 6v to this circuit is a resistor required, im guessing a 1ohm resistor should still be used here??

Things are seldom perfect but nice round numbers work well for examples. :)

Actually in the real world you would not want to run a LED at its maximum current. Running a little below is the way to go for a nice long life. Additionally when choosing a resistor to limit the current to a LED or LED string you likely won't find a resistor of the exact value the math delivers. Short of special purpose or precision resistors you would use a generic value one size up from what the numbers call for. For example a number like 40 Ohms isn't common but 47 Ohms is so you would go with a 47 Ohm resistor rated for the proper wattage.

Ron
 
No LEDs have a characteristic voltage drop of 1.2v. The minimum is 1.7v for red LEDs. About 1.9 to 2.2v for green and other colours. White is about 3.4v to 3.6v.
 
i know there isnt an led with a drop of 1.2v, the voltage was used as an example earlier on as it was easy to explain
 
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