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I am designing a LED setup with 3 leds and a 9V battery. Nothing more I got the following specs from the seller. How much resistance do I need if I want them to shine as bright as possible without blowing up?

5mm 940nm IR Infrared Emitter LED Diameter: 5mm Wavelength: 940nm

Color: Transparent

Maximum Power: 70MW

Maximum Forward Current: 30MA

Maximum Forward Voltage: 5V

Maximum Pulse Current Peak: 75MA

Welding Temperature / Time: 240/ ≤ 5S°C / S

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    \$\begingroup\$ I doubt that the maximum output power is 70 MW. Try 70 mW. Ditto the current - milli = "m" and mega = "M". I also doubt that they are welded in place. \$\endgroup\$
    – Andy aka
    Aug 22 '20 at 17:40
  • \$\begingroup\$ Looks like a standard IR LED. I highly doubt the max forward voltage is 5V. These things are usually quite similar to red LEDs, so I'd expect max forward voltage being in the 1.5V to 2.0V area. I would believe that max reverse voltage would be 5V. \$\endgroup\$
    – Justme
    Aug 22 '20 at 18:14
  • \$\begingroup\$ Keeping in mind also that a 9 V battery has about 2 Ohms of internal series resistance and also that you likely have enough overhead voltage (as Spehro points out in a comment, your forward voltage is wrong and is probably instead the max reverse voltage) to work with for 3 LEDs in series, you might also want to consider an active current limiter approach. \$\endgroup\$
    – jonk
    Aug 22 '20 at 19:21
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IR 940nm LEDs normally have a Vf of about 1.2V at 20-30mA. Given that the specs are a bit on the 'light' side, let's design for 25mA not 30mA. If it has to survive high ambient temperature or needs to be especially reliable you probably should use a lower current (and buy the LEDs from a better source with a real datasheet).

So with the 3 LEDs in series you'll have ~3.6V. To get 25mA from your 9V battery means that there will be 5.4V across the resistor. So your resistor can be 220 ohms. Power dissipation in the resistor is 135mW so a 1/4-W resistor is fine.

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  • \$\begingroup\$ Note that at rated max V and I, 5V x 30A = 150mW, twice the rated power. That suggests that some will deliver 70mW at 2.5V and 30A, while others will require 5V and 15mA. Or the spec sheet is lying. I'd treat 2.5V and 15mA as the design limit, and step down a little from that. Or, as suggested, get some better diodes. \$\endgroup\$ Aug 22 '20 at 18:30
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    \$\begingroup\$ @GuyInchbald Forward voltage of 5V is probably an error for max reverse voltage. It’s way too high for instantaneous peak voltage of an IR LED at only 75mA pulsed. If each drops 5V chances are there won’t be much current from a 9V battery with 3 in series. \$\endgroup\$ Aug 22 '20 at 18:55

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