Let's think for a minute.
It helps to think of diodes as voltage variable resistors. Try drawing the circuit with the diodes replaced with 1-ohm resistors with a switch in series. When the voltage across the diode is greater than the diode drop, the switch closes and it looks like a 1 ohm resistor. Then evaluate what the opamp does each way.
I can't quite tell from your description whether you're using a back to back clipper to ground from the output end of the feedback loop or the inverting input end of the feedback loop.
Diodes to ground from the output end can be connected up either directly to ground or through a capacitor. In a single supply setup, directly to ground clamps the output of the opamp to one diode above ground, which may be hard on the opamp if it's being fed a reference voltage of 4.5V. It sits in current limit until the signal pulls it below the diode drop.
If you're using bipolar supplies, or tie the diodes to Vref or through a capacitor, then the diodes act as a hard current limit on the output. The voltage can't go above the diode's voltage drop at whatever current limit the opamp has. So the opamp is running dead linear until you trigger the diode switch on, and then the diode can eat as much current as the opamp can pour out.
This is only varied a bit by the fact that a resistor/resistor/capacitor Vref or a capacitor to ground is not a perfect short. These setups will let the capacitors charge up and unclamp the thing at some point. Diodes to Vref may cause low frequency motorboating because they give you mixed positive and negative feedback, and in some cases could cause oscillation. Not often, but could.
Diodes from the inverting input to Vref will have different effect depending on whether you drive signal into the noninverting input or the inverting input. If you use the inverting input for signal in, the diodes have no effect whatsoever. The inverting input in this case sits at virtual ground, moves no more than millivolts, and never turns the diodes on.
If you drive it from the noninverting input, the diodes do their one-ohm dance because signal on the non-inverting input moves the virtual ground around, and will eventually cause a diode to switch over to low resistance.
When that happens, feedback is made inoperative and the opamp blams against the power supply in whichever direction the signal is pushing it, as it cannot supply enough current through the feedback resistor to balance.
So signals under a diode drop are normal, and peaks over a diode drop at the + input will cause a sudden wham to the power supply.
Anyway, use the switched resistor model to think about what happens.
And *do* think. Good practice.