Electronic – Problems amplifying mV potential differences

As you say, instantaneous power is the product of the instantaneous voltage and current. Average power is simply that averaged over time. Note that the instantaneous power can be both positive and negative.

To get a reasonable average, you want eliminate the 50 or 60 Hz components. One way to do this is to keep a running average over whole line cycles. This will be easier if you look at the voltage zero crossings and average all the values between one and the next. You could then additionally low pass filter those readings a bit to reduce random noise and to make the result more smooth.

Another way that doesn't rely on detecting individual power line cycles is to heavily low pass filter the instantaneous power values so that 50 Hz is attenuated to the point you don't care. That of course will work at least as well with 60 Hz then too. Let's say you are sampling the voltage and current at 10 kHz rate, which is 200 readings per 50 Hz line cycle. 3 poles of low pass filtering with a filter fraction of 1/1024 should do it. That assumes each pole follows the algorithm:

FILT <-- FILT + FF(NEW - FILT)

where FF is the filter fraction. Note that the multiply by FF is then really accomplished by a right shift of 10 bits, which is easy to do in a microcontroller. The step response of that filter is .1% after 200 iterations, or one 50 Hz line cycle. The step response after 5000 iterations, or 1/2 second, is 87%. That is about right for displaying to a human observer.

At first, the principle of "virtual ground" can be applied during DESIGN of opamp-based amplifiers. This simplifies calculations - and the error is in most cases acceptable. Error? Yes - because there is always a differential voltage between both opamp inputs, which is exactly Vdiff=Vout/Aol. (Aol=open-loop gain of the opamp). Because of the large values for Aol (1E4...1E6 for lower frequencies) this diff. voltage Vdiff is in the µV range.

However, because this is not true for larger frequencies, the closed-loop gain will deviate from the calculated value for rising frequencies.

Regarding your last sentence: Yes - introducing additional delay in the feedback path will cause additional phase shift - and this can lead to instability/oscillations.

EDIT: "...until the virtual ground is re-established and the cycle repeats."

I suppose, with the above cited sentence you are asking for something like a "sequence" which leads to the steady-state conditions after applying an input signal, correct? This is, indeed, a question which deserves some explanations.

Example: Inverting opamp-based amplifier with a gain of "-2". Input: +1V step (t=0).

At the very beginning (t>0), the feedback is not yet active and the output will jump to the maximum negative voltage (supply rail). Now the feedback network causes the inverting terminal to become negative - and the output starts to go to positive voltages. However, this will not continue again and again because the opamp has internal delay elements (causing bandwidth limitations and phase shift). That means: The output does not "jump" to other values but it takes some time to reach the upper rail. But, in reality, the output will NOT reach the upper rail because on the way to the maximum positive output the output voltage crosses some finite negative values - and for an output value of app. Vout=-1.999V there will be an equilibrium between input and output. Explanation:

Vout=-1.999V and Vin=+1V cause a very small voltage between both resistors (at the inv. input terminal) which - when multiplied with Aol - is exactly the assumed output voltage (in the example: Vout=-1.999V.) This equilibrium state is stable.