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I'm attempting to build a homemade head tacking unit for use in computer games using infrared LEDs for my kid. I followed one tutorial where they were wired in parallel. After doing more research, I read posts on here and on Nuts & Volts that wiring LEDs in parallel is not ideal because one or more LEDs will hog the current. This appears to be true because the one LED with the longest distance from the power source always seems dimmer.

I figured out the resistance needed to wire the LEDs in parallel. There are three LEDs with a forward voltage of 1.2V and a continuous forward current of 100mA. They are powered by a 3V source (two AA batteries), requiring a resistance of 6 ohms.

If I wire them in series, the LEDs take more voltage (3.6) than the batteries can supply. I don't want to move to a bigger battery box as this is head mounted. Can the LEDs still work with a bit less voltage. I did a quick test and they did not seem to light up. Maybe moving to a 9V battery is an option, but those things are far more expensive than AAs :)

If I am stuck with wiring in parallel I read it's better to put a resistor on each LED diode. How do I calculate the resistor value I on each LED. I know wiring resistors in parallel halves their resistance (thanks Element14!), so if there are three paths, do I need 18 ohm resistors at each LED to get to the 6 overall ohms the circuit needs?

Yes, I am very new to all of this.

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  • \$\begingroup\$ at 1.2 volts per LED, at 3.3 volts for the battery (supplying 0.3 amps total), you can have a resistor in series with each LED, of value (3.3 - 1.2)/0.1 == 2.1/0.1 = 21 ohms. A 22 ohm resistor (5% or 10% tolerance) can be used. The power is 2.1volts * 0.1 == 0.21 watts. So a big resistor. Or use 39 ohms or 43 ohms or 47 ohms, for half the current, twice the battery life, half the light output. Your resistor has heat removal challenge; use a wide bit of copper braid (or solder wick) to attach to the +3.3 volt end of the three resistors and remove the heat. \$\endgroup\$ – analogsystemsrf Sep 14 '19 at 9:09
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When each LED has its own resistor you only consider one LED when calculating it.

Effectively you have three independent circuits.

The presence of the others will cause the battery to sag a little, but the operating voltage will vary through their life anyway.

Note that LEDs to be detected should be modulated, this seems to be a fairly crude approach using power instead of sophistication to make them distinct.

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  • \$\begingroup\$ So I tried it with your method and it worked. I don't know what the modulation comment means. Like I said in my opening post, I barely know what I'm doing. :). \$\endgroup\$ – Ferkner Sep 21 '19 at 3:08
  • \$\begingroup\$ I have components to make a second headset. If I wanted to use one resistor at the start of the circuit containing the three IR LEDs, can I safely use one of those calculators.With a 9V battery, 3 IR LEDs that are 1.5V each with a current of 100mA each in series require 45 ohms of resistance. So one 47ohm resistor should work with no problems? \$\endgroup\$ – Ferkner Sep 21 '19 at 4:00
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    \$\begingroup\$ @Ferkner: 9 volt batteries are pretty feeble - 100 mA is a very heavy load for them. \$\endgroup\$ – Peter Bennett Sep 25 at 23:12
  • \$\begingroup\$ I'd probably go with 2xAA or 2xAAA and 3 parallel circuits each being the series combination of an LED and a resistor chosen for its current alone. \$\endgroup\$ – Chris Stratton Sep 25 at 23:19
  • \$\begingroup\$ I ended up using another tutorial to build one that is powered via USB. It ended up working great, \$\endgroup\$ – Ferkner Sep 27 at 3:38

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