Created Mon, 10 Jun 2013 15:45:40 +0000 by alnath
Mon, 10 Jun 2013 15:45:40 +0000
Hi u all :)
I'm using a GSM shield, and it works fine, but it is hard to debug my sketch, because I can't use the software serial library as I could with arduino. The GSM shield has jumpers that allow to switch the serial RxTRx pins from pin0/pin1 to pin7/pin8 (pins that can be defined for softwareserial functions) .... unfortunately, Chipkit Uno32 2nd UART port uses pin39 and pin40 !! Then, my question is : would it work if I placed the jumpers of the shield to use pin7 and pin8 instead of pin0 and pin1 , and, if I wired togetherpin39 -> pin7 and pin40->pin8 on the Uno32 ? If yes, in which state should I configure pin7 and pin8 in the sketch ??
Another good point if it worked : it would allow me to upload the sketch without removing the shield, as I must do now, because the upload stops if I don't :?
Thanks for any answer :D
Tue, 11 Jun 2013 09:52:08 +0000
I answer to myself...in case somebody else has the same question, for this or another shield : I tried, and it works ! I didn't declare anything for pin7 and pin8 in the sketch. Put a wire between 7 and 39 , another one for 8 and 40 , and use Serial1
Sure it's only for debugging time, because it uses 6 pins where only 2 are required ;) , but it's helpfull and, as pins 0 and 1 are not used by the shield, no need to remove the shield each time I upload the mdified sketch.
hope this will help some of you :)
Tue, 11 Jun 2013 10:14:22 +0000
All pins default to input mode (which is a "high impedance" mode) so shouldn't affect the serial signal adversely. It would even be possible to "snoop" on the serial signal using those pins should you so wish - could be handy for debugging ;)
Be careful that you never set those pins as output though, or you could fry the chip. You might be better off doing the link with some low value resistors (100Î© or so) just to limit the amount of current that could flow between the pins should a sketch set the pins to output at some point in the future (when you've forgotten what you did ;) )
Tue, 11 Jun 2013 13:51:15 +0000
Thank you Majenko, especially for the wise advice :) I didn't think I could fry the chip if I set the pins as output, I thought all pins had limiting resistors, but only analog ones actually have a 200Î© limiting resistor. Fortunately, I first tried with the default mode for the pins, and it worked ! but I'll try with 100 Î© resistors right now, just in case.... ;)
Tue, 11 Jun 2013 14:13:59 +0000
If you connect two pins together, and set them both as output with one high and one low you are effectively creating a direct connection between +3.3V and GND with two very small resistors. That would cause quite a large current to flow, which would cause the MOSFETs in the IO pins to go into meltdown breaking those two IO ports.
Tue, 11 Jun 2013 14:56:18 +0000
sure, you're right !