In the attached schematic, I want a .850V reference on GPIO.0, which I get when the PICkit2 is plugged in. But when I unplug the PICkit2, the voltage jumps to 1.182V. Why? What have I done wrong?
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You are correct on both counts, but why does the PICkit2 load the divider? Vcc is unchanged. ETA: This is a superfluous question; just ignore it. I learned something useful today; watch out for loading during in-circuit programming.You have made a voltage divider: V out = V in * R3/(R3 +R4)...so 1.176 ~= 1.182 is what is expected...
The pickit may very well have been loading the divider....
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