I want to wire up an IR LED on a battery. I've seen many tutorials telling me that I should add a resistor, but none says what resistor to use to maximize battery life. So a 680ohm resistor will do the job of powering up the LED, but will a 47k resistor do the job too and keep the current still running (which will be more beneficial for the battery)?

I wonder if the LED (and all the diodes) need a minimum current, or a minimum voltage to operate, because I want to know how will adding one more diode in series or in parallel affect my choice of a resistor.

Also is my thinking correct? Say I have a 9V battery and a 0.7V drop LED. This means that if I place a resistor in series, the voltage drop through the resistor will always be 8.3V. So the current through both the resistor and the LED will be determined by the current through the resistor R which will be 8.3V / R.

  • \$\begingroup\$ What does the datasheet say? \$\endgroup\$ – Ignacio Vazquez-Abrams Apr 24 '14 at 17:14
  • \$\begingroup\$ I haven't bought any yet... \$\endgroup\$ – nass Apr 24 '14 at 17:19
  • \$\begingroup\$ possible duplicate of How can I efficiently drive an LED? \$\endgroup\$ – Ricardo Apr 24 '14 at 17:19
  • \$\begingroup\$ I think this question How can I efficiently drive an LED? will tell you what you want to know about LEDs. \$\endgroup\$ – Ricardo Apr 24 '14 at 17:20
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    \$\begingroup\$ What @Ignacio meant is that datasheets are publicly available on the Internet. Just Google your IR LED part number or code and you'll get the datasheet. It will tell you everything about the IR LED. It may take a while for you to learn how to read a datasheet, but once you do it, it will become your primary source of info about the parts you will be working with. \$\endgroup\$ – Ricardo Apr 24 '14 at 17:23

Taking a look at a datasheet for a popular IR LED, the Lumex OED-EL-1L2, we can see the IR light output vs. current curve:

enter image description here

So if you only give it 1mA, it gives you 0.2mW/sr radiant intensity. At 20mA, you get 10mW/sr radiant intensity.

So, to maximize battery life, you should use the minimum current that will produce a light intensity that your circuit can (with an appropriate safety margin accounting for all the things that can change, including LED aging and temperature) work with.

If the sr unit used for the radiant intensity is unfamiliar- that's the SI unit of solid angle, so it's power per unit of solid angle.

enter image description here

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    \$\begingroup\$ Worth noting from Spehro's graph and numbers : 20x the current = 50x the brightness, so low currents are relatively inefficient. You get more IR power from your battery by driving at higher current (say 20ma) and saving power by shortening the pulses instead. \$\endgroup\$ – Brian Drummond Apr 24 '14 at 18:56

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