I have a 3v DC source and a 3v light emitted diode, i want power this led with just 5 mA .

There is a formula : R = V/I

But in my case V is 0

V[source] - V[LED] = 0

So R is 0 always!!!

How can i pick a true resistor?

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    \$\begingroup\$ A resistor is not a good current source when the source and LED voltage are so close together. You may need some active circuitry to limit the current. Or add another battery and get a 4.5V source. \$\endgroup\$ – user253751 Dec 1 '17 at 1:26
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    \$\begingroup\$ Why is 0 not a valid response? There are true (almost) 0Ω resistors! \$\endgroup\$ – Curd Dec 1 '17 at 8:46

LEDs are diodes and diodes do not have a constant forward voltage. The forward voltage is actually a function of current.


enter image description here

This curve came from the datasheet of a white LED that is advertised at 3.1V. But you can see that it only achieves 3.1V at 60mA (at 25°C). The manufacturer of the LED will have chosen a nominal current that makes the LED as bright as possible without limiting the promised useful life. You certainly don't need to use that exact current value. A lower current will correspond to a lower forward voltage. You'll need to find the relevant plot in the datasheet for your LED.

For example, let's say the plot in your LED's datasheet shows 5mA corresponds to a 2.8V forward drop. Your calculation becomes:


$$\frac{3V - 2.8V}{5mA} = 40\Omega$$

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Ohm's law does have a response, it's zero ohms.

It tells you that if your actual LED has exactly 3V Vf at the rated current AND the 3V supply never goes over 3V, you do not need a resistor.

The issue here is what is the true Vf of the actual LED, or minimum Vf specified in the data-sheet, since they vary in manufacturing, and what is the maximum upper variance of your 3V supply. If you know those you can calculate the resistance required at the worst case scenario when the supply is high and Vf is low.

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