Miller op-amp resistive feedback

When a Miller-opamp is compensated, it is assumed that the opamp is unit-gain mode with capacitive loading. My question is:

If the opamp is unity-gain mode using resistive feedback (Rf = R1) with a capacitive load, then:
1) The gain of second stage is now gm*(Resistive feedback) which reduces the miller-compensation? Is it such a way that when second-stage gain reduces, the second pole-moves away and somehow balances itself?
2) The resistive load seen by the second stage: Is it Rf + R1 or Rf || R1?

3) If a miller-opamp is loadded with an oscilloscope's 1MOhm resistance, then the total gain of the OTA would be gain of diff-pair which is usually a few 100 mutliplied by second stage, which is gm*1MegaOhm? In which case the total gain may be severely effected?



PS: These are not homework questions or anything, just trying to full understand the miller-amp.

Miller-opamp isn't a clear technical term in my view, rather a sloppy phrase. I assume, that you mean a miller compensated two-stage OP?

1) The feedback doesn't affect the compensation directly. For the analysis, you should distinguish between amplifier open loop gain and closed loop gain.

The output load, e.g. the feedback resistor however affects the open loop gain, and to some extent the miller pole. Although a resistive load changes the open loop gain and also the pole frequency, it keeps the gain-bandwidth product in a first order.

2) I leave this simple network calculation problem to your analysis.

3) I don't understand the involved calculation. The load effect of the oscilloscope probe has to be calculated like every other load. I would mainly expect the capacitive part of the probe load affecting the OP gain. Or are you referring to DC gain? For an OP with real transistors, the output resistance is most likely below 1 Mohm.

P.S.: Rf = R1 is not unity gain mode, it's an inverting amplifier (G=-1). Both differ regarding feedback factor, unity gain (G=+1) is the more critical case for stable compensation.