If my calculations are right, when sinking current, my 16f1828 is working out at about 52 Ohms of series resistance per pin - now whether the resistance comes by way of a resistor, or the resistive properties of the silicon inside the PIC IO driving circuitry...resistance is resistance?
V Supply = 3.268V
V drop measured across blue LED = 2.931V
remianing V drop across PIC pin = 3.268V-2.931V = 0.337V
Current through LED (inferred by measuring current into whole circuit - 11 LEDS - then taking a little off (3mA for the PIC) & dividing by 11 = 6.4mA
Resistance of PIC pin R = V/I =....or 0.337V/0.0064A = 52.65 Ohms.
So why not simply treat the PIC's pin as a series 52 ohm resistor for the purposes of LED current calculations? (ie what am I missing?)
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