I have a Brinkman LED flashlite I want to convert to an IR led. A schematic of the Brinkman driver is here:
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I have a Brinkman LED flashlite I want to convert to an IR led. A schematic of the Brinkman driver is here:
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Fine the power supply rail-to-rail voltage with the LEDs in use operating. (Just for reference, so many of the "usual" / "regular sizes" white LEDs have a rated maximum continuous current of 30 mA and peak of 100 mA.)
Infrared LEDs that only tolerate 50 mA at max. current drop approx.
1.3-1.5 volts at max. current. Subtract 1.3 volts from the rail-to-rail DC supply voltage (unlikely to exceed the battery voltage if the battery voltage is normally near or over 4.5 volts). Divide this difference by .05 amp and you get a moderately alarmist required resistor in ohms. Use the next higher value among those available at your favorite parts supplier.- Don Klipstein ( snipped-for-privacy@misty.com)
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It looks like there's a mistake in that circular wiring diagram.
You don't need the circuit if you use IR LEDs because IR LEDs need only
1.5V or so to conduct that much current, compared to well over 3V for white LEDs. So you can just remove the circuit and replace it with a single resistor. The resistor should be 1.5V / .05A or 30 ohms.ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.