Just trying to figure out what the voltage should be on pin 2 (vdd) of the ICD connector for my application using a 3.3v powered PIC18F26K80.
Should it be 3.3 or 5v?
My understanding is, that this pin powers the programming device (which also monitors the circuit voltage.
Also, my MCLR pin is pulled up to 3.3v.
Is there a possibility that the programming voltage could damage the 3.3v system?
The reason I'm using 3.3v is because I want to use an SPI display that is only 3.3v.
I could power the PIC at 5v and use open collector outputs and pullups but I thought making it all 3.3v might be a simpler solution.
newguy
Joined: 24 Jun 2004 Posts: 1903
Posted: Tue Oct 14, 2014 7:31 am
Pin 2 is whatever your processor's Vdd is. What programmer are you using?
Years ago the CCS programmers (ICD-S and I think the first generation ICD-U) came with an internal jumper pre-wired so that the programmer supplied a 5V Vdd. With the jumper removed, connection of this 5V supply was disabled. For quite some time now their programmers default to NOT supplying Vdd.
Ttelmah
Joined: 11 Mar 2010 Posts: 19372
Posted: Tue Oct 14, 2014 7:47 am
Key thing is the connections on MCLR. This pin has to go up to the Vpp voltage when programming. This is why the pull-up should be about 10K, so that excess current will not flow into the supply when this happens. From the data sheet:
"A suggested starting value is 10 k. Ensure that the
MCLR pin VIH and VIL specifications are met."
jmaudley
Joined: 01 Feb 2011 Posts: 32 Location: 53
Thanks
Posted: Tue Oct 14, 2014 8:26 am
Thanks for the info. That confirms my assumptions.
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