Electronic – Can a phase margin become bigger than 180° and is this bad

Some comments from my side:

1.) The stability check in the BODE diagram concerns the LOOP GAIN response only (because once you did mention "closed-loop system" in your text.)

2.) The shown system is "conditionally stable". That means: It is stable - regardless the properties at the frequency A. However, if you REDUCE the gain within the loop until the gain crosses the point A (the phase remains unchanged) the closed-loop system will be unstable.

Such conditional stable system should be avoided because a gain reduction can happen due to aging or other damping effects. Remember: Classical feedback systems with a continuos decreasing loop phase will become unstable (under closed-loop conditions) for rising gain values (beyond a certain limit) only.

As to your next question - the input signal Vi does not influence stability properties at all. Stability is determined by the loop components only. That is the reason, we investigate the loop gain only.

EDIT: Here is an explanation why the closed loop (your example) will be stable: If a closed-loop system is unstable, this point of instability also must be "stable". That means - either we will have "stable" and continuous oscillations or the output is latched at one of the supply voltage rails. In both cases, this point of instability is fixed.

Now - what happens at the point A in your example? Here we have a rising phase which is identical to a NEGATIVE group delay at this point (group delay is defined as the negative phase slope). This is an indication for the unability of the closed-loop system to let the amplitudes rise (oscillations or latching at the supply rail). Rather, the system returns to a stable operating point.

A final information: The stability check investigates either (a) the -180deg line or (b) the -360 deg line. This depends on what you are investigating: (a) Either the simple product GH or (b) the loop gain LG which is LG=-GH.

The diagram can be rearranged so that \$H_2G_2\$ is an inner loop and $$\frac{G_1H_1G_2}{1+G_2H_2}$$ is the outer loop.

It is required for the outer loop to be stable. The inner loop may be unstable with the full loop being stabilised by controller \$G_1\$. This is generally only done when absolutely necessary to stabilise an unstable plant.

Practically speaking, both loops should be required to be stable in a system such as this with standard op-amps.

As a rule of thumb for design simplicity, the \$G_2H_2\$ loop is designed to be stable with significantly higher bandwidth than the final loop bandwidth so that its phase delay can be neglected in the design of the outer loop; i.e. its phase/magnitude contribution to the overall loop is negligible at crossover.

To the OP's Update: The point of which loop to analyse: A straightforward analysis of the OP's loop (also the Updated equivalent arrangement) shows the stability analysis to be: $$G_2(G_1H_1+H_2)+1=0$$

for the rearranged circuit with the inner loop we get: $$\frac{G_1G_2H_1}{1+G_2H_2}+1=0$$

Applying simple algebra we see that it's the same result, as expected: $$G_2(G_1H_1+H_2)+1=0$$

Therefore you can analyse the loop several ways and get the same result. Considering the loop to consist of an inner and outer loop, is however, very convenient.

UPDATE: Here's the analytic justification for the loop stability equation to be $$G_2(G_1H_1+H_2)+1=0$$ Apologies for quality. I hope to enter as native format when I get time: