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need to understand!

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watzmann

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Hi All ,

my pic circuit is working well , but i noticed something and need to know what does it mean ,

when i finished testing my circuit i removed it's positive terminal from the battery that mean my circuit will not run because there are no supply .

but

my circuit LED indicator still light up and i found out that's because i still connecting my photosensor to one of the input ports and it supply 1 to this port.

my question is :

how this positive signal goes from this port to supply all the circuit? , is it normal with pic circuits ?......or i have short in my circuit ?
 
is the battery completely removed? Or is the battery still connected to the photo sensor but not to Vdd? I would just use one connector to the battery for your power.

Also, the PIC documentation shows how the internals of the ports are wired, so you can trace the currents path from that input and see where it goes. But I always assumed the ports would be tri-stated on power off.
 
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Ambient said:
is the battery completely removed? Or is the battery still connected to the photo sensor but not to Vdd? I would just use one connector to the battery for your power.

Also, the PIC documentation shows how the internals of the ports are wired, so you can trace the currents path from that input and see where it goes. But I always assumed the ports would be tri-stated on power off.

Was I completely wasting my time with my previous post?.

I'll try again! - there are protection diodes from most PIC I/O pins to Vss and Vdd. If the external sensor circuit is still powered, and feeding power to the I/O pin, then this will feed through the (now forward biased) diode to Vdd, and most likely cause the PIC to continue running.

I't a very well known phenomenon, and not just for PIC's, but for any extremely low consumption micro-controller with protection diodes on the inputs (most of them?).
 
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