0

I have a few MakerBot Mightyboards running off ATmega1280 chips that can no longer use their standard USB port to connect to a computer (to make a long story short, there's a little inductor that blew up between the USB port and the 8u2 chip and it is essentially impossible to replace). I have noticed, however, that they have headers for the RXD1 (PD2) and TXD1 (PD3) pins on the 1280 as seen in this picture:

PD0-PD3 pins on Mightyboard Rev H

Would it be possible to hook up an external 8u2 (or similar) chip programmed as a USB to serial converter (something like this) and use it as if it were the normal USB port (e.g. for uploading firmware)?

(I wanted to add atmega1280 as a tag, but it doesn't seem to exist)

  • 1
    How on earth did you manage to blow up a resistor?!?!?! – Majenko Apr 2 '18 at 10:37
  • Just noticed I actually got mixed up and the parts I'm talking about are inductors. They're tiny surface-mounted ones, and they're placed right next to the USB port. The boards are quite old (about 5 years) so I'm guessing a slight power surge over USB caused them to literally blow up (there's burn marks on the boards that have this). – Nicolas Gnyra Apr 2 '18 at 16:34
1

Not directly, no.

The USB interface connects to TX0 and RX0 - you have access to TX1 and RX1. The bootloader is set to look at TX0 and RX0.

You would have to compile and install a new version of the bootloader that is set to use TX1 and RX1.

| improve this answer | |
  • Thanks! I highly doubted it would work out-of-the-box like that. Do you happen to know what changes I would have to make to the bootloader for this to work (and where I could get said bootloader)? – Nicolas Gnyra Apr 2 '18 at 17:36
  • No idea, I have never compiled the bootloader. I think there's something included with the IDE somewhere. Or Google optiboot as an option. – Majenko Apr 2 '18 at 17:54
  • Alright, I'll try that out. It might take a while though since I'm waiting on parts. I'll make sure to post any new info I might have. – Nicolas Gnyra Apr 2 '18 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.