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Here’s my analysis on your circuit based on the simulation I did. Please refer on the circuit below.
During the positive cycle of the input, D1 is open and D2 is shorted. This makes U1 an inverting amplifier with a gain of -2. With this, the output of U1 will be inverted (180degrees out-of-phase with the input). The positive current will flow through R3 and the negative current will flow through R4. Since the voltage at R3 is positive and the voltage at R4 is twice R3 (output of inverting amp) but opposite in polarity, then the voltage at the output of U2 (being an averaging circuit) is equal to the voltage at R3 so it's positive (please see equation below). Note that the gain accuracy of U1 will not be affected since D2 is inside the feedback loop.
During the negative cycle of the input, D1 is shorted and D2 is open. This makes the output of U1 clamped to 0V or 0.7V for the diode drop and disconnected to U2 by D2. Then U2 acts as a unity-gain inverter with R5 as the feedback and R3 as the input resistor. The output of U2 will then be the same as the input but will be inverted. The positive current will flow through R4 and the negative current will flow through R3.
The diode, resistor and capacitor at the output will determine the response time of your output. Without these, the output waveform will be a full-wave rectified signal. With these, the output will be nearly pure DC as long as the value of C (uF) and R are both higher (Mohms).
So with this, I can see that your circuit can be used for bidirectional current sense applications.
Hope I helped you on this. Apology for the very late response.
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