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I've built a circuit on a breadboard using a 10K potentiometer, a 220 ohm resistor and a LED. They are connected in series. Using Ohm's law, you find that:

Total resistance \$R = 10000 + 220 = 10220\Omega\$

Battery: 9V.

\$I = \frac{V}{R} = \frac{9}{10220} \approx 0.88mA\$

Isn't that current too tiny for a LED? But even tough, it lights up at this resistance. Can anyone explain?

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  • \$\begingroup\$ The blue led in youtube.com/watch?v=yxoKb4d3gFA is powered at 0.3mA, 300 microamps. It's clearly visible in full daylight, when the white leds are on, and in the dark. I have other leds that are reasonably still on at 100 microamps and below. \$\endgroup\$
    – Passerby
    Mar 25, 2016 at 3:52
  • \$\begingroup\$ I messed with an LED once that you could see glow by attaching one side to ground and holding the other side. Basically it was acting as a diode detector for all the line voltage noise that my body was picking up from the mains. \$\endgroup\$
    – Daniel
    Mar 25, 2016 at 3:54

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Since the pot is in series you should be able to vary the current- assuming the LED drops 2V you would have from (9-2)/220 = 32mA to (9-2)/10.2K = 0.7mA.

Note that there is no need to have more than a couple significant figures for the current calculations- the battery voltage , LED drop and resistances are not that accurately defined.

Modern LEDs can be fairly bright with 0.7mA DC so if you have a good LED it could well appear reasonably well illuminated. On the other hand, 32mA may be too much current for a small LED.

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