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AC Power Adapter for chipKIT WF32


HKPhysicist

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Hi @HKPhysicist,

I presume by AC power adapter you are meaning a AC to DC converter ("wall wart"); the size needed is 5.5 mm diameter with a 2.1 mm center positive pin. You can connect the device to a USB port and have it get the required power from there, presuming you set "VU Select" to "UART".

With the default jumper settings (as pictured here: https://digilent.com/reference/_media/reference/microprocessor/wf32/wf32-2.png), I have used a 12 V power supply (this one iirc, https://digilent.com/shop/12v-3a-power-supply/) as well as a 9 V power supply. If you are using external power while passing through the 5 V regulator, you'll need at least 1 Amp of current. Be aware that at high current usage the regulator will likely start power cycling due to overheating.

You can also use a 5 V external supply (which I have also done), such as this one: https://digilent.com/shop/5v-2-5a-switching-power-supply/, with VU Select set to EXT and the J17 jumper blocks set to only have VU jumpered to 5V0. You will want to by-pass the LDO if you want the 5 V rail on the board to be functional as the on-board LDO is not able to output 5 V from a 5 V source.

The onboard 3.3 V regulator which powers everything else receives it's power from VU (whether that's from the USB or UART or barrel jack) and can operate off of 4 V to 30 V, though back when Digilent sold this board I only ever tested between 5 V and 12 V.

These options are explained more in the Power Supply section of the Reference Manual here: https://digilent.com/reference/microprocessor/wf32/reference-manual#power_supply.

Thanks,
JColvin

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