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Have I fried the 5V power pin on my ChipKIT Wi-FIRE?



When I move the jumper to 5V on JP1-3.3V/5.0V Shield Voltage Select J1-Shield Power Connector goes from 3.24V -> 0V

"The 5 V regulator can be completely disabled if it is not needed for a given application."

Is that done with J15(rev.D)/J16(rev.C)- 5.0V Regulator Configuration. By moving both jumpers to the right? Thats how I am getting 0V on the shield power pin.

If I move both jumpers to the left, I get 3.85Volt on the shield power pin, but that is still far from 5.0Volt! 

I can't "easily" find a description of how the many jumpers work on the Wi-Fire. 

Is it a Regulator Configuration issue that gives me 3.85V instead of 5V?

Thanks, Appreciate your help!



https://reference.digilentinc.com/reference/microprocessor/wi-fire/reference-manual is for Wi-FIRE rev.D, if possible where can I get one for rev.C?





Edited by olaf

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Hi @olaf,

If you want to be able to get 5V from the 5V pin when powering the board through USB (or a 5V external power supply), you will need to set the jumper blocks on J15 (the one with two jumper blocks and six pins) to have a jumper block going from VU to 5V0 and the second jumper block either disconnected from the board entirely or only on a single pin. I have attached an image of how this might look on the board.

The JP1 controls whether the 5V pin on the J1 header provides 3.3 V or 5V. This is in existence for shields that can be applied to the Wi-FIRE since some shield components and pins on the Wi-FIRE cannot handle having 5V directly applied to them, but you generally have to purposefully connect the 5V line to a pin (specifically analog pins) for it to become a problem. 

If you want to look, the Rev C version of the WiFIRE reference manual is available here.

Let me know if you have any questions.

Thank you,

5V selection.png

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