power electronics - How to create a zero crossing detector using a full bridge wave rectified circuit

Currently

1) I have a full bridge wave rectified ac waveform of 10Vmax.

2) I also have a potentiometer adjustable DC waveform.

3) My rail voltages are +12V and -12V.

4) I currently supply my full bridge wave rectified wave in to the non inverting terminal and the DC voltage to the inverting terminal.

5) Using an opamp or a comparator, I want to compare the two signals and produce a square wave which varies from 0 to a positive voltage at the output.

6) At every zero crossing , my square wave should go to zero. In all the other times it must stay positive.

I have created an incomplete schematic below and have also tried many different methods but I am having difficulties in achieving my result.

Thank You

^{simulate this circuit – Schematic created using CircuitLab}

EDIT

New schematic

Removed the Buffer, Positive Feedback resistor, Ac source floating and load resistor added.

^{simulate this circuit}

Answer

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Modified OP circuit.

Your second circuit has no AC return path. This modification fixes that.

10 V RMS will peak at 14.1 volts which will overload the input to the comparitor. Rearrangement of the two 1k resistors will drop this to 7 V.

Figure 2. The LM311 has the emitter of the open collector output available on a separate pin.

Using the LM311 you can tie the emitter of the output transistor to ground (or anything else) to prevent negative output excursions.

When I simulate this circuit in the circuit lab , it didn't give me any square wave at the output of the comparator? Any ideas on that?

Figure 2. Running the simulation on Figure 1 with the potentiometer K factor set at 0.15 results in the above simulation. (Settings: 0 - 0.1 s, 0.001 s steps.)

You probably had the pot setting too high in your simulation.

Don't forget that most comparitors have open-collector outputs so I've added R3.