Sinking 5v on a 3.3v board

I feel like this is a silly question, but I'm a bit uneasy about it so I figured I'd ask anyways...

I have some intricate LED devices (eg an RGB LED pen, for long-exposure light drawing) with built in resistors chosen for a 5v source. I'd like to be able to use these with a 3.3v Arduino clone. Assuming I have a 5v source available, is there any problem with sinking the LEDs into a 3.3v device?

Feeling a bit rusty, thanks for any advice.

Do I understand correctly: You are going to power your LEDs from a separate 5V supply. However, the Arduino will be controlling them?

Are you trying to do something like this:

5V->LED Strip-> Arduino (HIGH turns leds off, LOW turns LEDs On?)

I'm not sure this will work for your purpose and depending on the voltage drop of your leds, it might burn your arduino. The Atmega `68 IO pins are rated for Vcc + 0.5 = 3.3+.5=3.8

You would be better off using an NPN transistor switching circuit with you arduino switching the base.


The I/O pins have diodes to ground and VCC. So you can’t connect anything higher than VCC + or about 0.5v.

You can use an array of transistors (ULN200x series are popular if overkill) or just discrete NPNs or N-Channel FETs.

Which clone-board are you using that you can’t run it at 5V?

Assuming I have a 5v source available, is there any problem with sinking the LEDs into a 3.3v device?

To answer the question directly, yes there is a problem. It is that the output pin when off will be exposed to 5V. This is greater than the 3V3 of the supply and so will cause damage.