On Jan 22, 2008 7:06 PM, Tamas Rudnai wrote: > Hi All, > > Looking at the PicKit2 schematics, there is one very interesting to see. > When PicKit2 programs a 3.3V part, it reduces the Vdd to 3.3V, that's fine, > that's done by a PWM driven PN dual mosfet. But was wondering what happens > with the PGC/PGD/PGM lines. It can be seen that each line has a 10R resistor > and a bipolar that puts the exceeding voltage to the ground. As 10R is far > too small, too much current would have been taken out from the pin I was > wondering how it works. > > It turned out, that the output pin of a pic drops the voltage down by > withdrawing too much current from it. Therefore as the current increases the > voltage decreases, so less and less current will be flown through on that > 10R resistor. > > The question is that is this a sufficient way to limit voltage on output > pin? It seems so weird for me but maybe it was the easiest way to do that. > Any thoughts? > This has been disucssed in pickit-devel (http://groups.google.com/group/pickit-devel) mailing list. http://groups.google.com/group/pickit-devel/browse_thread/thread/2dd07bed3164d97c/7c076ab1e6e79583 http://forum.microchip.com/tm.aspx?m=283929 Regards, Xiaofan http://mcuee.blogspot.com -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist