Running PIC18F2580 at +3.3v .. Are there any consequences?

[TokTok12]

Hi all,

I’m looking to change my design in terms of powering my PIC18F2580 (IC3 on my schematic below) from my MCP1703 +3.3v voltage regulator (IC2 on my schematic below) - **broken link removed**.

See my schematic screenshot below:



Currently I’m supplying power to my PIC18F2580 from my +5V LM7805 (IC1 on my schematic above). Based on my understanding of the “Absolute Maximum Ratings” PIC18F2580 datasheet it is possible to run the PIC at +3.3V i.e. **Voltage on VDD with respect to VSS ......................................................................................................... -0.3V to +7.5V** - see page 416 of the datasheet.

The reason for the change is due to the feedback received from my board house whereby they mention by having two different supply voltages on my board I could have “unknown long term consequences”.

What could these be? Could having two different supply voltages cause a problem with my Tx Rx interfacing between my ETRX357HR Zigbee module(IC7 on my schematic above) and my PIC18F2580.

As for the MCP1703 if I were to change my design this would mean directly providing the MCP1703 with incoming +12v supply voltage from my PSU which would be OK based on the MCP1703 “Absolute Maximum Ratings”.

Any help is appreciated.

Thanks,
TokTok.

 

[milan.rajik]

You can replace PIC18F2580 with PIC18LF2580. 18LF series run at 3.3V. The pinout will be same.

 

[bluelasers]

Hello,

You have two options.
1) You can switch to the PIC18LF2580 which allows 3.3v operation. (Max operating speed is 25 MHz)
2) Use logic level translators to convert between the 3.3v and 5v TTL. (Single direction channel will work since this is UART. Make sure they can keep up with the baud rate.)

Make sure that you check the power rating of MCP1703. Converting from 12v to 3.3v is a lot of heat.

I hope this helps