this is a possible solution i understand from the text, interesting, not so difficult either, very helpful, thank you very much!
this is a possible solution i understand from the text, interesting, not so difficult either, very helpful, thank you very much!
No
see post #17 and #19
So, you believe because you drew a Nano, the Seeduino will function like one? Lost me there, chum.
this is also my first time seriously into this forum, sorry for any inconvenience, i'm a sort of novice in forums!
Not a forum problem, it's a communication problem.
That's OK
You said you wanted to learn things.
Did you read the datasheet?
downloading
It's over a 1000 pages long not something you will go through in an afternoon.
To be clear
No, this was a possible solution based on the participants here being misled into thinking you were interfacing a Nano.
Ioh, Iol = 20 mA normal, 40 mA MAX.
For your device, the current limits are much, much lower. You need a transistor driver of one form or another.
Like I recommended back in post #11
1 Like
Yes, but there appears to be an 'allergy' to such a simple solution.
This is a good place to learn electronics.
They don't oversimplify thing but they do give understandable information
As you will read in the data sheet, the SAMD21 used on that Seeeduino rates the current available to I/O based on what they call "clusters". 46 mA source and 65 mA sink per cluster. On your board D2, D3, D4, and D5 comprise a cluster so you should have the 30 mA you need available from two pins in that cluster. Therefore, this is a possible solution if you use two pins from the D2-D5 set. There may be other combinations that will work if you look into it a little deeper.
Edit:
In addition to the available cluster current, there is a per pin specification of 7mA max. In order to get approximately 30mA LED current (28mA in this case) you would have to expand on @camsysca's example and use all 4 pins from the D2-D5 cluster. Using 2 pins would not be advisable because that would violate the per pin maximum and probably damage your board.
Hopefully you have learned something from the exercise. It was a good question and stimulated a few good suggestions. I think the conclusion here is that @camsysca's suggestion in post #3 is the best solution.
If the cluster current is the limiting factor then why not 46mA from a single pin?
Nano pin can supply 30mA.
Absolute maximum is 40mA according to the atmega328 datasheet, so 30mA is comfortably under 80% of that.
Some will say 25mA or even 20mA is the safe limit, and if you want the circuit to run reliably for 20 or 30 years, maybe that's right. But how long will your LED last? LEDs get dimmer over the years as they wear out, so how long will it be before the LED is too dim to fulfil it's original purpose?
I think you skipped quite a few posts, @stirf is not using a nano
I think the limit on the SAMD21 is 7mA source if you set the drive strength to max. OP will have to use several pins to get approximately 30mA LED drive current.
So this isn't true?
I didn’t mean to limit the experiment to two pins. I’m hoping to stimulate OP to study the data sheet and come to their own conclusion. I’m sure in practice 28mA will satisfy their curiosity and I don’t think drawing an extra 0.5 mA per pin will kill the board.
Oops! Don't know how I missed that, sorry.