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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

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT

New schematic

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

schematic

simulate this circuit

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  • \$\begingroup\$ There's too much going wrong there to begin to put it right point by point. So make a new circuit with (a) without the buffer (b) with the AC source floating with respect to your power supplies (c) without feedback round the comparator (d) load the output of the bridge with a resistor to get clean zeros. If you can't 'b', then you need a different configuration (perhaps your DC power also comes from your AC source?). You may be better off not rectifying your AC, and using two open collector comparators (393 are dual anyway), and gating the outputs of a 'higher than' and a 'lower than'. \$\endgroup\$ – Neil_UK Jun 29 '17 at 4:37
  • \$\begingroup\$ To avoid the xy problem: do you have to use a bridge rectifier (e.g. as part of an exercise specification) or would you be happy with a simpler zero crossing detection solution? \$\endgroup\$ – replete Jun 29 '17 at 4:50
  • \$\begingroup\$ @Neil_UK Did all the changes you mentioned \$\endgroup\$ – Amy Jun 29 '17 at 4:55
  • \$\begingroup\$ @replete As long as I am getting a square wave which ranges from 0 to a positive value, I am fine with it. \$\endgroup\$ – Amy Jun 29 '17 at 4:56
  • \$\begingroup\$ @replete When the ac waveform crosses the zero, my square waveform must also go to a zero and rise to a positive value once again. Well I don't mind comparing the ac waveform with even 1V DC. \$\endgroup\$ – Amy Jun 29 '17 at 5:00
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schematic

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.

enter image description here

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?

enter image description here

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.

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  • \$\begingroup\$ Thanks for modifying my circuit. But 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? \$\endgroup\$ – Amy Jun 29 '17 at 8:32
  • \$\begingroup\$ <<Using the LM311 you can tie the emitter of the output transistor to ground (or anything else) to prevent negative output excursions>> I grounded the emmiter of the output to get rid of the negative portions of the square wave, but then it flattens out the whole waveform. Any idea on this? \$\endgroup\$ – Amy Jun 29 '17 at 11:46
  • \$\begingroup\$ Sorry, @Amy, I've just seen this. Did you figure out the problem? What voltage did it flatten out at? \$\endgroup\$ – Transistor Jul 2 '17 at 21:40
  • \$\begingroup\$ I implemented your circuit and I have posted a thread regarding the problems I faced . electronics.stackexchange.com/questions/314350/… \$\endgroup\$ – Amy Jul 4 '17 at 6:22
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Source link:

The circuit in this Design Idea generates a zero-crossing pulse off the ac mains and provides galvanic isolation. The falling edge of the output pulse happens at approximately 200 μsec before the zero crossing. You can use the circuit to safely stop the triggering of a thyristor gate, giving it time to properly turn off. The circuit generates short pulses only when the mains voltage is approximately 0V, thereby dissipating only 200 mW at 230V and a 50-Hz input.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

Rearranging the circuit you can get the same for your XFORMER input:

schematic

simulate this circuit

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Zero crossing detector

This is an old Elektor design. First of all the zero detection is realized on the mains side. The value of the capacitor C1 is important to obtain the desired output pulse. I would strongly discourage the use of transformer since the use of it creates a zero crossing not in sync with mains.

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  • \$\begingroup\$ That's got to be one of the worst schematic layouts for understanding it I've seen in some time. I'm sure it comes from Elektor. But it is UGLY! For example, how easily can you tell that \$D_3\$ serves a function of protecting the BE junction of \$T_1\$? Sure, if you know to look for it, you can find that fact. But cripes it is hard to see in there. That said, interesting circuit. I'd modify it with an additional resistor inserted just before the base of \$T_1\$. Would be nice to see an analysis of this circuit and a better layout. Plus I'd seal the unit using Cotronix ceramics. \$\endgroup\$ – jonk Jun 29 '17 at 16:37

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