bestpicks411 Posts: 3, Reputation: 1 New Member #1 Jan 27, 2014, 10:45 AM
String of LEDs
Have a string of LEDs in series and I used a resistor calculator to find the resistor needed for it to work properly with my 14 volt input. However each led has forward current of 100ma and I'm not sure if I have to worry about the input amperage. I'm using a ac/12vdc adapter that states it puts out 150ma and I don't know if that divides equally over the LEDs like voltage. Will this work?
 ebaines Posts: 12,129, Reputation: 1307 Expert #2 Jan 27, 2014, 11:10 AM
If the LEDs are in series then the same current flows through all of them, so its not an issue. Current divides only if the circuit divides - such as wiring the LEDs in parallel - which is not what you're doing.

How many LEDs are you wiring, and what resistor value did you come up with? I think it should be around 150 ohms, right?
 bestpicks411 Posts: 3, Reputation: 1 New Member #3 Jan 27, 2014, 12:37 PM
Thanks, the input voltage is actually 15.7 and I have 10 LEDS
Wavelength:
890 nm

Angle of Half Intensity:

±22 °

Intensity:

70 mW/cm2

Forward (Drive) Current:
100 mA

Forward Voltage:

1.4 V

So I'm using a 18ohm 1/2 watt resistor, correct? Do I have to worry about what amperage the input is because I have some other adapters as well. One of which is 300ma. I just want to be sure I don't burn them out. The amps is the part I don't have a full understanding on.
 ebaines Posts: 12,129, Reputation: 1307 Expert #4 Jan 27, 2014, 02:18 PM
I think 18 ohms is too low: 15 volts/18 ohms = 0.887 A, which will blow out the power supply. Given that you want 100mA the resistor should be 15.7 V/0.1A = 157 ohms. The power rating of the resistor is i^2R = 0.1^2 x 157 = 1.57 W.
 bestpicks411 Posts: 3, Reputation: 1 New Member #5 Jan 27, 2014, 05:25 PM
my understanding of that formula is to divide the leftover voltage by the .1A as in 15.7 V- (10x1.4 V) / .1A = 17 ohms. I got that from 2 sites, one of which is How to make LEDs glow not blow! | Let's Make Robots!
are they wrong?
 ebaines Posts: 12,129, Reputation: 1307 Expert #6 Mar 31, 2014, 05:25 AM
Yes, that will work. I didn't realize you had the forward voltage rating for the LED - my bad.

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