Electronic – Why does an op amp need to be powered before applying the input signal

Here is a link to an article I wrote a while back about this class of problems.

A device such as the LM7705 or an inverting charge pump such as the LM2661 will generate the negative voltage needed for the amplifier to output 0 volts with 0 volts input

These devices will generate the negative voltage with reference to your circuit common.

Without seeing the input signals, I'd say the circuit is working just fine. That is, it's working as it should given the topology. Unfortunately, it won't work as an instrumentation amplifier, or even as a differential amplifier. And I'm pretty sure a differential amplifier with gain is what you intended.

First, of course, is the fact that you're feeding an AC-coupled signal to a single-supply op amp, and this is guaranteed to clip all inputs which are less than the average. Are you sure you want to do that? Granted, it's a poor man's rectifier, but it's bad for the op amps.

Each amplifier has a nominal gain of 63. Let's assume the each has an input of 10 mV. Then the output of the lower amp will be 630 mV. If the upper amp were operating from a (very large) split supply, it would put out 630 mV - (63 * 620 mV), or -38.4 volts. This, obviously, is not going to happen. The problem you have is that your topology only works as a differential amplifier when the gain is 1.

There is a variant on this which supports gains >1

^{simulate this circuit – Schematic created using CircuitLab}

Common mode range is limited.

Another problem you have is that your input AC coupling networks are not obviously matched. Identical AC signals can produce non-zero outputs due to differential phase shift. Combined with the clipping produced by your AC coupling circuit, I'd expect some very odd results.