Wagner Lipnharski says:
I have a circuit that works from 3.8 to 5.2Vdc, some tests were done to try to feed some extra micro or milliamps to the VDD via an input pin, connected to another machine that supplied its data lines with steady TTL 5V or zeroV. The idea was to "steal" this small current back into my circuit VDD, when its battery was low. Primarily we used Schoktty diodes from the data lines to VDD, then we thought why not allow the protective input diodes (inside of the chip) to do this job... wrong action. What happened was just as you described. When the input pin received the external 5Vdc data line, and the local VDD was 4Vdc, the local circuit sucked more current from the weak battery instead to relief it by using the input data line deviated current.... it took a while to understand why...If the input current is caused by a voltage above the chip VDD, remember that the input protective diodes in real are not "diodes" but silicon junctions to do the job. Some chips use a technique to keep the VDD steady when an overvoltage enters a pin, they increase the supply current in a proportional way. This is done to make sure the protection does not interfere into the other parts of the chip, otherwise the deviated current to VDD would be poured all over the other parts of the chip. If the protection would be just a diode connecting the pin to VDD, and if you apply 6 Volts to that pin, your VDD "would" increase to 5.4 Volts, or 6 minus the internal diode voltage drop, right?, it means all the chip would be supplied with 5.4V... it could be a problem.