I want to understand how the following circuit works. I understand OpAmps, diodes, and voltage dividers pretty well; but put this circuit together, and I don't know how it affects positive or negative voltage input.
Though not ideal it is a full wave rectifier.
For a negative input it is inverted by the op-amp thus becomes a positive output of the opposite polarity, passing through the diode. R1 and R2 set the gain at -1 so a negative input (compared to ground) becomes a positive value at the output.
For a positive input it is inverted but the diode will not let a negative value pass, thus disabling the op-amp and R1 + R2 become bypass resistors so a positive value appears at the outputs, but with 2K of resistance as they have no buffer. If the load is very light or buffered it will behave more like an ideal FW rectifier.
For a true full wave rectifier you need 2 op-amps and 2 diodes, so each polarity can be buffered.
See the 1st trace for Diode V is only < 0.25V not 0.7 because diode current with 10k is only 10uA here with 100mV input. Place with drag and drop components. Falstad is awesome to learn new things but here uses an "Ideal OpAmp" that I defined only by right mouse properties.
Rather than explain every detail, try to examine diode voltage and current and load effects and use your training and logic to figure it out for each polarity.
My simulation for you. You cannot put a cap. on this output to get an envelope voltage due to a mismatched output impedance of the diode switch in series with the output. Subtle errors arise from your choice of FET input OP Amp. So choose wisely and go for a buffered dual Op Amp FW rectifier design (search web)