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

Question

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}

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'. — Neil_UK, Jun 29 '17 at 04:37
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? — , Jun 29 '17 at 04:50
@replete As long as I am getting a square wave which ranges from 0 to a positive value, I am fine with it. — Amy, Jun 29 '17 at 04:56
@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. — Amy, Jun 29 '17 at 05:00
@Amy, OK, now we can see what we're doing, connect the bottom end of R1 to the mid point of V2 and V3 so we have a reference for the bridge output voltage. Try simulating the resulting circuit. Note that many comparators have an open collector output, so you may need a pullup resistor from positive rail to the output. Due to the forward drop of the diodes, the output will have a minimum low time, which may or may not be OK for you. That can be fixed, but with a different configuration. — Neil_UK, Jun 29 '17 at 05:11
@Amy I've for the moment neglected the fact that the AC peak voltage is higher than the comparator supply. You'd handle that by making the undesignated 1k resistor larger, and either potting it down with a resistor to our reference point, or (better) diode clamping it to the reference, so it doesn't exceed the comparator's +ve rail voltage. — Neil_UK, Jun 29 '17 at 05:21
Apart from making the circuit of OP function properly one important situation is overlooked. OP is using 10VAC expected to come from a transformer. Apart from many other issues the output from the transformer is never in sync with the mains. The aproach should be find the zero crossing on mains level and isolate afterwards with an optodectector. — Decapod, Jun 29 '17 at 06:42

Marko Buršič · Answer 1 · 2017-06-29T07:06:26.850

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}

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

schematic

^{simulate this circuit}

Transistor · Accepted Answer · 2017-06-29T09:20:22.957

1

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.

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.

edited Jun 29 '17 at 09:20

answered Jun 29 '17 at 07:03

Transistor

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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? – Amy Jun 29 '17 at 08:32
<> 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? – Amy Jun 29 '17 at 11:46
Sorry, @Amy, I've just seen this. Did you figure out the problem? What voltage did it flatten out at? – Transistor Jul 02 '17 at 21:40
I implemented your circuit and I have posted a thread regarding the problems I faced . https://electronics.stackexchange.com/questions/314350/a-problem-regarding-connecting-the-grounds-within-two-seperate-circuits – Amy Jul 04 '17 at 06:22

score 0 · Answer 3 · answered Jun 29 '17 at 08:22

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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.

answered Jun 29 '17 at 08:22

Decapod

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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. – jonk Jun 29 '17 at 16:37

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

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