You can use pin 13 as an input just like any other. The fact that it has an LED and 1K across it shouldn't matter for 99.99% of the inputs you will be using.
If it is a problem then unsoldering either the LED or the resistor will disconnect them.
If you really want to disconnect the pin 13 LED (I'd be really certain before you commit yourself) , I'd just take a small pair of pliers to the 'L' LED and twist it off.......
i`m setting up a "comm" line from pin 13 of one arduino to the pin 13 of another.
That should be no problem.
getting 1 volt readings when the pin is "HIGH"....booooo
That is not being caused by the internal LED you have something else wrong. Grounds not joined or pin mode not correct (one has to be an input and the other an output).
The location of the resistor depends what board you have but you should be able to trace it back from the LED.
my 10k to ground, and the two leds are munching a considerable amount of that MEGA power!!
No do the sums.
1K resistor (assume no LED) = 5mA
10K resistor = 0.5mA
So for two connected together you have 5 + 5 + 0.5 = 10.5mA.
An arduino pin can comfortably supply twice that. And four times that at an absolute maximum.
how sure are you it is not a problem with the two 1k resistors and the two leds? like from 1-100%?
I am 100% sure, maths don't lie, and I have been doing this stuff for over 40 years. You have something else wrong.
signal path is (as i understand it)
arduino A
diode to ground
1k resistor to ground
my 10k resistor to ground
This is wrong, the output pin goes to one end of the 1K the other end goes to one end of the LED and the other end of the LED then goes to ground.
Why do you want a 10K resistor in place as well? It will do no harm but no good either.
What vintage of arduino do you have?
or perhaps it would be a parallel resistance equation...something akin to 1k resistance to ground. i`m not sure what to do....
what should i try?
Just have a wire between both pin 13 from both boards. Also have a wire between both boards ground pins. Be sure you are setting pinMode() statements correctly for both boards, one as output (the sender) and one as input (the listener). Both on board leds will operate showing the state of the digital out to digital in connection.
It really is just that simple.
Lefty
that would give something along the order of 4k resistance wouldn`t it?
No, it gives what I said before 1K with 5V across it gives 5mA so two of them gives 10mA.
Or resistors in parallel two 1Ks give 500R and 500R across 5V gives 10mA, it is the same because it is two ways of calculating the same thing.
If you don't believe me then believe retrolefty (we are both saying the same thing)
it must be the code.
**Yes **
should i move the posting to the "code room" from here...or perhaps you guys have some ideas and are saavy with the standard firmata...`?
Keep it here. I have no experiance with firmata, sorry.
Lefty
i`m using the firmata that comes with the arduino
So now you tell us !!!!!
A bit vital that, why are you using this, is it you can't write your own code?
Just like lefty I can write my own code and so have never used it. It could be that firmata doesn't allow that pin to be an input because mostly it is used as an output.
Can you try it just writing your own sketch.
ya i could grumpy_mike...but "i gots to have good dialogues with the puter". myz programming skillz for byte transmission on a serial interface between max/msp and arduino are none too keen...
i think i am better off using the standardfirmata and modulating it with the hopes that it can be trimmed of the fat that i need it not for. #
i hope this doesnt offend...but i
m terminating this sarah conner or er uhm...ending this post and following the thread on the programming portion of this forum. not entirely sure how to terminate...but i`ll see if i can do it.
sorry folks...butchered that. had thought it would be easy to terminate a thread that i made. guess it`s around forever and ever and ever and ever. no delete found.
posted issue in software and programming portion of the forum as i found out it is most likely a problem with this area.