Determining the correct resistor for an LED

I am new to circuitry, and I recently learned that to avoid breaking the LED, I would need to get a resistor with a certain resistance; the formula is Resistance = (Voltage of Battery - Desired Voltage for LED)/(Desired current for LED). However, I am confused about the characteristic of this formula, that different combinations of the "Desired Voltage for LED" and "Desired current for LED" can result in the same "Resistance". Where is the determining factor that actually allows the current and voltage to be the "Desired value" of the LED? This formula tells me that a resistor with certain resistance can result in multiple combinations of voltage and currents, which may be different from the actual desired value for the LED.

by totoro711
February 13, 2021

About the prefered current, that depends on the illumination of the area where you use the LED. Never exceed 30 mA to not shorten dramatically the LED life. 20mA is probably appropriate if you use it in daylight. 12 mA if you are in a somehow dark room, but 7 mA is a very dimmed shinning.

If you don't have the LED specs, neither a ohm-meter giving you the value of the voltage threshold of diode (the voltage allowing 1mA to flow through, in general), you can assume 1.6 volt for a red LED, 1.9 for a green and 2.5 for a blue. It CAN be larger than that, but if so, less current would flow through, and without problem for the LED. A simple test with a breadboard could also be a "non destructive" test where you could experiment the optimal resistor for your use.

It could be easier to target a constant current with a resistor, from R = V / I, no LED. Then use a current mirror, and on the parallel path, then the voltage drop of the specific diode won't matter much since you would impose the (constant) current through the LED (or many LED in series). You could even add some of them, keeping the same lightning intensity, since you forced then a (constant) current through them, not a voltage. You could also use any constant current source circuit instead of the curent mirror. That constant intensity of light emitted would be impossible if you were adding the LED in series with a single resistor (and no constant current).

Note that if you have to use many LED, you can also use a sequence of parallel paths, each path having its own LED and its own resistor. You can so add as many paths as you want, as long as the constant voltage source can supply the total current required.

by vanderghast
February 14, 2021

See Circuit with two different LED, same current. I estimated the single resistor like this: ( 3.3 volt - 0.8 volt) / (0.020 A) where 0.8 volt is the price I have to pay to be in the active region of the transistors and I targetted around 20mA as current. Sure, in the end, having to pay less than 0.8 volt, I got a little bit more than 20mA. On the other hand, you can't use a LED with a threshold much larger than 3.3-0.8 = 2.5, which means that we are border line if we have to use blue DEL, as example. You would have to increase the voltage of the source. Also note that you can dim all the LED at the same time by increasing the value of the single resistor (a potentiometer). You may also note that, in theory, we only loose 20mA * 3.3V in heat, the power dissipated by the single resistor (if we don't take into account the transistors who dissipate too some heat) and so, large part of the energy from the battery ( total of 3.3 V * (3 * 20mA)) is use for the lightining, and not into heat.

by vanderghast
February 14, 2021
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1 Answer

Answer by Foxx

Here is a graphical method for sizing an LED series resistor.

+1 vote
by Foxx
February 15, 2021
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