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I'm just trying out Arduino Uno for the first time with 2 blinking LEDs on a breadboard. All the tutorials on the Internet seem to use a resistor. I do know the function of resistors, but does it really matter here? These LEDs are working just fine without a resistor.

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(if you would ever return here): don't hit and run. You asked a question, got an answer, and you were gone. You haven't upvoted the answer (you didn't have enough reputation) and you haven't accepted it. Since you don't seem to come back I suppose your question has been answered. Give credit where credit's due. – Federico Russo Jun 3 '12 at 9:38
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The good part is that person cared to ask. So future generations are now able to find the question and answer – Rocket Surgeon Jun 4 '12 at 2:48
Ok. Sorry about that, I'm new here. – 40Plot Jun 14 '12 at 13:34

1 Answer

up vote 21 down vote accepted

Naughty! :-). If they say to use a resistor there's a good reason for that! Switch it off, NOW!

The resistor is there to limit the LED's current. If you omit it the current limiting has to come from the Arduino's output, and it will not like it. How do you find out what the resistor needs to be? You do know Ohm's Law? If you don't, write it down in big letters:

\$ V = I \cdot R \$

Voltage equals current times resistance. Or you could say

\$ R = \dfrac{V}{I} \$

It's the same thing. The voltage you know: Arduino runs at 5V. But not all that will go over the resistor. The LED also has a voltage drop, typically around 2V for a red LED. So there remains 3V for the resistor. A typical indicator LED will have a nominal current of 20mA, then

\$ R = \dfrac{5V - 2V}{20mA} = 150\Omega \$

The Arduino Uno uses the ATmega328 microcontroller. The datasheet says that the current for any I/O pin shouldn't exceed 40mA, what's commonly known as Absolute Maximum Ratings. Since you don't have anything to limit the current there's only the (low!) resistance of the output transistor. The current may so well be higher than 40mA, and your microcontroller will suffer damage.

edit
The following graph from the ATmega's datasheet shows what will happen if you drive the LED without current limiting resistor:

enter image description here

Without load the output voltage is 5V as expected. But the higher the current drawn the lower that output voltage will be, it will drop about 100mV for every extra 4mA load. That's an internal resistance of 25\$\Omega\$. Then

\$ I = \dfrac{5V - 2V}{25\Omega} = 120mA \$

The graph doesn't go that far, the resistance will rise with temperature, but the current will remain very high. Remember that the datasheet gave 40mA as Absolute Maximum Rating. You have three times that. This will definitely damage the I/O port if you do this for a long time. And probably the LED as well. A 20mA indicator LED will often have 30mA as Absolute Maximum Rating.

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I don't have any resistor at the moment. For the time being, can I just hook up a bunch of LEDs in series to mimic the resistor? – 40Plot May 31 '12 at 13:26
Not really. 3 and you won't get any light, because 3 x 2V > 5V, and 2 you'll have the same problem, only 1V drop instead of 3V. I added to my answer to switch it off now! – stevenvh May 31 '12 at 13:28
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I lost count on how many times I have written that calculation in answers here. I should have a script for it :-) – stevenvh May 31 '12 at 13:49
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@JohnR.Strohm I wouldn't call it an industry standard. – m.Alin May 31 '12 at 14:29
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@JohnR.Strohm - Like m.Alin says that's not an industry standard. Digikey lists 10mA nominal LEDs ranging from 0.4mcd to 1000 mcd, and 20mA LEDs ranging from 0.1 mcd to 54000 mcd. There's no line in it, and no guarantee that your 10mA LED will be good visible. Red LED voltage typically ranges from 1.8V to 2.2V. For a red LED 1.6V is exceptionally low. – stevenvh May 31 '12 at 15:37
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