Pin on PIC drive

Christophe · Joined: 10 May 2005 Posts: 323 Location: Belgium

Hi,

this is the scheme:
http://expand.xs4all.nl/uploadarchief/download.do?file=schma.jpg

I need to drive ENABLE_CHARGING LOW to disable charging. This means that the pin on the PIC (I'm using PIC16lF877A Vdd = 3V) is on Vadap via a pull-up resistor. Vadap = 8V ; can that harm the pic?

thx

newguy · Joined: 24 Jun 2004 Posts: 1907

That will definitely harm the PIC. Use a transistor to pull that line low instead.

PIC --- R --- base of BJT (or gate of FET), emitter (or source) to GND

The collector/drain of the transistor then goes to your ENABLE CHARGING line. The only difference is that now a logic high on the PIC line will bring ENABLE CHARGING low.

Guest · Posted: Thu May 12, 2005 9:43 am

There is another way without requiring additional components and without harming the PIC. Use RA5 of the PIC to drive this PIC. RA5 is open drain.

newguy · Joined: 24 Jun 2004 Posts: 1907

Good point, only it's A4 that's open drain, not A5.

The only thing is that A4 can only tolerate a maximum of 8.5V (first page of chapter 17 of the 877A's data sheet). If the line that it is connected to is a nominal 8V, it doesn't leave much room for error.

Guest · Posted: Mon May 16, 2005 1:43 pm

can I put a diode in series with the resistor? then I have 8V-0.7V = 7.3 op RA4

newguy · Joined: 24 Jun 2004 Posts: 1907

I guess so. I'm just overly cautious - experience has taught me to be that way. As I said before, I'd do things differently.