I just received this part and am trying to figure out the power wiring. The breakout pins called “VCC” and “5V” I think are tied together and map to what Atmel calls “VCC” and “AVCC”?
The breakout pin called “3.3V” is the output of the onboard 3.3V regulator?
The PDF schematic (http://dlnmh9ip6v2uc.cloudfront.net/dat … ut-v11.pdf) shows an SJ1 “Power Select”. I don’t see that physically on this board. Where is this?
I just will provide 5V to the “5V” and “VCC” pins.
Thank you.
Are we talking about this product?
https://www.sparkfun.com/products/11117
harrybstoner:
I just received this part and am trying to figure out the power wiring. The breakout pins called “VCC” and “5V” I think are tied together and map to what Atmel calls “VCC” and “AVCC”?
Correct, on the AVR chip “VCC” and “AVCC” are seperate pins, but are often connected together on the pcb.
The breakout pin called “3.3V” is the output of the onboard 3.3V regulator?
It’s connected to it, yes. It’s not really a breakout pad intended to connect wire to. But it is intended as an easy access voltage selection option. A solderjoint is easy to remove/reconfigure, yet also static and more resiliant than a switch or jumperpins which can be accidentally pulled/flipped.
The PDF schematic (http://dlnmh9ip6v2uc.cloudfront.net/dat … ut-v11.pdf) shows an SJ1 “Power Select”. I don’t see that physically on this board. Where is this?
If your board looks different from the pictures on the product page, then I suggest you make a picture and show what’s on yours. On the product page I clearly see solderjumper pad in the forward corner showing the 5 volt side containing a solderbridge. If you would remove that, and solder the middle to the 3.3 volt side then the microcontroller would operate on 3.3 volt basis. It still gets the 5 volt from the USB connector though, or whatever you connected to the 5 volt input.
I just will provide 5V to the “5V” and “VCC” pins.
Thank you.
No need to additionally connect VCC, as the solderjumper is placed already.
The USB connector is followed by a PTC-fuse before it is labeled as 5 volt in the schematic. So you won’t have that protection if you power it at the 5 volt point with your own powersupply. Also, you loose the serial port connection if you don’t use the USB connection. Quite a drawback in the testing and developing phase.
Valen:
Are we talking about this product?
Yes, that is the board I am referring to.
If your board looks different from the pictures on the product page, then I suggest you make a picture and show what’s on yours. On the product page I clearly see solderjumper pad in the forward corner showing the 5 volt side containing a solderbridge. If you would remove that, and solder the middle to the 3.3 volt side then the microcontroller would operate on 3.3 volt basis. It still gets the 5 volt from the USB connector though, or whatever you connected to the 5 volt input.
Thank you - I see it more clearly now. There are 3 pads and the solder blob “jumpers” 2 of them. On my actual piece the solder blob obscured the fact that it was covering two pads and I didn’t clue in.
The USB connector is followed by a PTC-fuse before it is labeled as 5 volt in the schematic. So you won’t have that protection if you power it at the 5 volt point with your own powersupply. Also, you loose the serial port connection if you don’t use the USB connection. Quite a drawback in the testing and developing phase.
I see what you mean about the fuse. However I don’t quite understand your last remark regarding the serial connection?
Thanks for your help.
Well, when developing and testing it can be useful to send text strings and data from your program to your pc to check if everything is working as it should. Or else you can only hope the program is doing what you think it should be doing. If you only supply it with 5 volt power and run it on it’s own then you don’t have that feedback.