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I'm new to electronics and Arduino. I bought the Arduino starter pack and I did the first exercise: connect a LED and a resistor.

But after I tried to connect more LEDs, and when I connect 3 LEDs they light up little (they seem off if I don't look carefully).

enter image description here

Why is this happening?

I know that when I use only one LED, it has a potential difference of 2V and need 5 - 20mA or it will burn, so:

R=V/I => 5-2/0.015 = 200ohm <210ohm

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    \$\begingroup\$ Do you know anything about Ohm and Kirchhoff laws? \$\endgroup\$
    – Eugene Sh.
    Commented Aug 3, 2017 at 18:54
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    \$\begingroup\$ If one LED has a drop of 2V.. how do you expect 5V to light three in series? \$\endgroup\$
    – Trevor_G
    Commented Aug 3, 2017 at 18:54
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    \$\begingroup\$ So you recommend starting with Ohm law and Kirchhoff? \$\endgroup\$
    – Albert Lazaro de Lara
    Commented Aug 3, 2017 at 18:58
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    \$\begingroup\$ I recommend a book such as The Art of Electronics \$\endgroup\$
    – Eugene Sh.
    Commented Aug 3, 2017 at 18:59
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    \$\begingroup\$ LEDs are like battery voltages that add up so 2+2+2=6V ythen you match the supply voltage as close as possible to drop 1~2V across the current limiting R. from 7~8V , which may be inconvenient so choose a better array of voltages that matches your available sources. But like coin cells LED's also have internal resistance determined by Power rating \$\endgroup\$
    – D.A.S.
    Commented Aug 4, 2017 at 0:46

2 Answers 2

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I know that when I use only one LED, it has a potential difference of 2V and need 5 - 20mA or it will burn, so:

Good. Then you should know that if you have three of those LEDs in series that you will need 6 V. Since you only have 5 V that explains the dim LEDs.

R=V/I => 5-2/0.015 = 200ohm <210ohm

Not quite. It should be \$ R = \frac {V}{I} = \frac {5 - 3 \times 2}{0.015} = \frac {-1}{0.015} = -66.6 \; \Omega \$. Since negative resistors don't exist this is a sign that something is wrong.

The most you can light on 5 V is two LEDs but even that depends on the colour (which in turn depends on the doping added to the LED material).

enter image description here

Figure 1. Forward current vs voltage for a range of typical LEDs. Note the variation between colours. Source: LEDnique.com.

Be careful with your micro-controller output pin. There will be a maximum current that it can safely provide and the voltage will droop a bit - maybe 0.5 V or so at that current. There is also a maximum that all the outputs can provide at one time because the current has to be sourced or sunk from the Vss or Vdd pin. The datasheets will contain all this information.

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  • \$\begingroup\$ My micro-controller is an Arduino Uno and I use the 5V pin, but I'm only doing the tutorials. \$\endgroup\$
    – Albert Lazaro de Lara
    Commented Aug 3, 2017 at 19:11
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    \$\begingroup\$ @Albert: Thanks for accepting my answer but I recommend that you un-accept for a day or two to encourage other answers. Start to read through the Uno datasheet to figure out some of the stuff I mentioned above. At first you won't understand much but it will start to make sense the more widely you read. \$\endgroup\$
    – Transistor
    Commented Aug 3, 2017 at 19:13
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The way i understand diodes, LED's are doped(for various wavelengths) semiconductors, they create a potential difference barrier through diffusion. enter image description here

For current to flow through it we break this potential difference barrier by applying potential difference(this is where the term 'forward bias' comes in) now if you do KVL in your case 2V difference for each of 3 diodes would result in total 6V difference that is required to break the barriers in series for flow of current and that is why 5V supply wont work here.

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