I'm designing a circuit with an opamp buffer at the input, which runs of batteries.
As the input does what it likes, the input of the opamp can see voltage even when the opamp is powered down. As the opamp inputs are rated at +0.5V V+ (rail) and -0.5V V-(rail), it is common practice to bolt diodes across the inputs of the opamps to the rails.
My question is, where does this current go? Say the positive rail is at 0V and 5V is applied to the input, so the diode (seen in the diagram below) clamps at 0.7V. Well, this makes sense but it's a positive rail so it's not grounded, its just sitting there floating, so where does this current flow? Surely this will just cause the positive rail to flow to Vin-Vdrop = 4.3V. Same goes for the negative rail just vice versa. Can anyone explain how this works?
I have attached supporting works below.