I have to admit that I haven't read all the posts, but I noticed one thing awhile "up". PLEASE don't drive a LED directly from a PIC pin! LEDs are not 'lamps'. LEDs ARE DIODES! It is CURRENT, not voltage that lights them. Since they are a diode, if you put a slowly rising voltage across them, they (at first) draw no current whatsoever. Then, when they get to the diode forward voltage (which is determined by the technology of the LED), all of the sudden, they light. So, you can have no light at all at 2.2V, and burn them up at 2.3V. If a LED with a forward voltage of 2.3V was connected to a 3V source, it wouldn't last long. So why do they work when you connect them directly to PIC pins? It is because the PIC pins have very small FETs driving them, and they have an "effective" resistance. That is, they cannot supply a huge amount of current. This current-limiting allows both the PIC and the LED to survive.
You have seen the LED flashlights that have nothing but white LEDs and 3 - AAA batteries. No resistor. The forward voltage of most white LEDs is between 3.1 and 4V (Wikipedia). The 3 batteries in series produce about 4.5V. It all works because batteries are not perfect power sources, they have an Equivalent Series Resistance (ESR) that acts like a resistor in series with the battery. This ESR allows things to work - kind of. But have you ever noticed that while you expected the LED flashlights to work forever - they don't? They fail because the makers don't put in a little extra resistance in the leads of each LED to balance the LEDs and to control the current.

Back to PICs: The PIC pins outputs are not really resistors. They may act somewhat like resistors, but they are not. The "resistance" of the pin can vary quite a bit over Vcc variations and from device to device and even pin-to-pin. As a result, the results can be unpredictable. Please put a resistor (at least) in series with your LED.