Avisos
Vaciar todo

Adding LEDs to 45° fan nozzle (MK3)  

  RSS
michael.p77
(@michael-p77)
Eminent Member
Adding LEDs to 45° fan nozzle (MK3)

I am investigating how easy it is to take the new 45° fan nozzle on my MK3 and modify it to hold two super bright bright LEDs, found here https://www.adafruit.com/product/754 and locate them on either side of the shroud angled toward the print nozzle.

The LEDs are 3v and consume a maximum of 20 mA.

Does the power supply have a 3v output, does anyone have any specifications on the MK3 power supply or better yet a reference manual?

Respondido : 07/01/2019 5:15 am
toaf
 toaf
(@toaf)
Noble Member
Re: Adding LEDs to 45° fan nozzle (MK3)

the Pi connector has a 3.3 pin1 and 17

I have a Prusa,therefore I research.

Respondido : 07/01/2019 5:33 am
florian.c2
(@florian-c2)
New Member
Re: Adding LEDs to 45° fan nozzle (MK3)

There should be 5V outputs on the board that are meant for the cooling fans. 2x20mA is not much so it should be fine.
However you will need to add resistors in series to your LEDs to limit the current to 20mA per LED or you WILL fry the led and/or the board.
Here's how to do it:

You find a power output that is 4.5V or more (3V won't do it because it will be tricky to fit the right resistor). If that power output is more than 7V or so, I recommend connecting your LEDs in series.
You will need to limit current to 20mA (or less) to avoid frying the LEDs.
Follow this formula:

Your voltage is going to be the PSU voltage minus the LED voltage drop (3V here, find the datasheet for more accuracy!). I'm going to assume we are using a 12V output here (that I am not sure is available on the board btw, but you'll get the idea)
We will put the LEDs in series with the resistor:
----|>-----|>------R--------
The LEDs drop 2*3V so we will have:
U = 12V - 2*3V = 6V of voltage accross our resistor.
U = R*I so to get I=20mA or less:
R = U / I = 6 / 0.020 = 300 ohms or more. You can use a 330ohm resistor, that should be the next highest available value.
Now for safety let's check how much power that resistor will have to dissipate so that the LEDs won't fry themselves and let the magic smoke out:
P = R * I^2 = 330 * 0.02^2 = 0.12 Watt

Be sure that your resistor is rated for this much power, and be careful, they can get quite hot since they are small. You can either use one that is rated for a higher power (they will have the same resistance but be physically bigger, so they don't get that hot) or take two resistors of double the value (660ohm) soldered in parallel.

For this reason I do not recommend powering your LEDs off the power supply's 24V output, you will need to dissipate too much power in your resistor (unless you are able to power more than two LEDs in series. If you could add something like 7 of them, which would be 21V, you would only need a resistor for 3V/20mA, or even less depending on the actual resistance of the LEDs. You will need to test this).

Respondido : 13/01/2019 9:15 am
Compartir: