Formule correcte pour la résistance de limitation de courant LED?

J'essaie de déterminer quelle résistance de valeur utiliser dans un circuit à LED. L'équation que j'utiliserais pour faire ceci est:

Seems logical, and makes complete sense. The answers to the question How do I calculate the resistor value for a simple LED circuit? confirm this too.

J'ai les LED suivantes:

• $V_f = 3.3V$
• $I_{f_{typ}} = 20mA$

En utilisant une alimentation 5V:

• $V_{cc} = 5V$

En les connectant à l'équation ci-dessus, on obtient:

All good so far.

However, if I use the calculator at http://led.linear1.org/1led.wiz, that gives me 100Ω. If I use the ElectroDroid app on my phone, that gives me 85Ω.

So, I assume that the linear1 calculator is using a different method of calculating this resistor value; is there some better way of doing this?

Your calculation is correct. linear1 rounds up to the next E12 value, which happens to be 100$\Omega$. The nearest E12 value would have been 82$\Omega$, and that would still be safe, because, even if the current will be higher, the difference will be small, within the 10% tolerance of the E12 series.

edit
Purists may say I'm cutting corners here. Russell has a long answer about iterating the solution, and others whine (hey, no offense!) about rounding up being more safe. My answer is meant to be pragmatic; no professional design engineer can afford to spend 15 minutes to calculate the resistor for a classical color LED. If you stay well below the maximum allowed current you'll have enough headroom to allow some rounding, and the rounded value won't be noticeable in brightness. For most LEDs perceived brightness doesn't increase much above a value of typically 20mA, anyway.

Your formula is correct BUT to do it properly you need to iterate the result. This is because LED forward voltage drop is non linear with current (or current is non linear with forward voltage drop. In many cases this effect is not significant, but in some cases it can lead to results which are 2:1 or more in error.

Where there is plenty of "headroom" voltage for the series resistor - the difference between Vcc and Vf - the original result is liable to be close enough to correct so as to not matter. But if headroom voltage is small with respect to Vf, changes in LED Vf with current will change headroom which will change current which will change Vf which will ... . This really does happen in real world situations.

For white LEDs Vf is typically in the 2.9V to 4V range with more typical values 3.3 - 3.8V until quite recently and say 3.0 - 3.3V in more modern higher efficiency LEDs. In serious production applications Vf will be available in "bins" so can be guaranteed within about +/- 0.1V at a given current. In retail sales you may get samples from every bin going and Vf may be eg 3.3V for one LED and 3.6V for another nominally identical one.

If operating from 5V the headroom will be 1.7V and 1.4V respectively for a current variation of about (1.7-1.4)/1.7 =~18%. Add to that slight shifts in Vf with current as above and 20% variations in If may result between "identical" LEDs. In most cases this is not going to make the slightest practical difference. Light output is approximately proportional to current - 20% variation in light output is not detectable by eye by all but the most skilled or experienced of viewers.

If this was a say 5 Watt power LED the difference in LED dissipation may be 1 Watt and this MAY make a difference in operating temperatures and lifetime.

All of which leads to the advice that in "serious" applications LEDs should be driven from a constant current source if you care about the true operating current. In "indicator" roles or low level illumination applications this may not matter. In high power applications or where LED lifetime matters then constant current drive is essential.