ICD-U and 3.3V PIC connection

bwhiten · Joined: 26 Nov 2003 Posts: 151 Location: Grayson, GA

I am using a +3.3V PIC for the first time. The connection diagrams for for the ICD-U all show +5V on connector pin 5 for the ICD to use for pulling up the PIC signals. For a 3.3V device is 3.3V here OK or do I need to provide 5V? I have the 5V rail available, just confused by the documentation........ again Embarassed

newguy · Joined: 24 Jun 2004 Posts: 1909

The "5V" line on the ICD-U should really be labelled "Vdd". You connect that line to the target's Vdd. There is a jumper inside the ICD that controls whether the ICD powers the target or not. I can't remember off the top of my head what the jumper's state is when you buy it. I think that it has to be removed (ie don't allow ICD to power target) if you're programming a 3.3V board.

Come to think of it, I'm 99.9% sure it has to be removed for low voltage targets.

Dimmu · Joined: 01 Jul 2007 Posts: 37

Hi,

You have to remove the jumper from the ICD. Then you program your PIC as if it was a 5V device.

If you have a 'j' series PIC, you must also select the 'J' option from the ICD software.

Dimmu