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I've been for quite some time now, trying to create a precise full-wave rectification of a signal coming from a CT sensor SCT-013-030.

Here's the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Now, actually the 7805 voltage regulator is also providing regulated power supply to the remaining of the circuit, which is basically a XBee that's activating a given relay. If you feel it's required I'll update the post.

I first started by doing only a half-wave rectification. So I hadn't even the second op-amp in place, and the following was being displayed in the scope:

Yellow waveform is Vin, Blue waveform is V1

Basically the yellow waveform is Vin, blue waveform is V1. I must say that, considering the resistors (the same 10K at 1%), I can't actually understand that voltage boost.

When I try to apply the Full-wave rectification I get this:

precise Full-wave rectification

Here's a zoomed screenshot of the last scope image:

Full-wave Zoomed

I would probably say that I must first understand why am I getting such a boosted voltage on the output, while doing only a half-wave rectification.

The input signal is around 50hz. Another thing is that the bridge rectifier is actually a 4 diodes set up.


So, after compiling all the great comments and answers I tried to implement all the suggestions. And I must say that I'm rather confused.

Here's the current circuit. Note that I left out the AC/DC conversion to simplify:

schematic

simulate this circuit

As a first approach I decided to do the full rectification, but without the dual power supply. For visibility purposes I've increased the power being used on the power cord being measured:

SIngle power supply

Afterward I've implemented the dual power supply making use of a inverting regulator ICL7660CPAZ, I got the following:

dual power supply

And last by not least I've disconnected whatever was attached to the power cord, and got this funky wave form:

At lower power

I'm completely lost here...


So, I have to conclude that the problem is with my implementation of the precise full-wave rectifier. Even if the schematics that I'm presenting here, make sense, I'm problably not implementing it correctly.

So, bare it with me... but I feel compelled to post my breadboard image, because sincerely I'm out of ideas... the next step is to buy a different op-amp...

enter image description here

Let me know if you can see anything terribly wrong... needless to say that I've rebuilt the same circuit 3 different times...

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  • \$\begingroup\$ A couple things catch my attention. First, are your op amp voltage rails connected properly? Typically, the + rail is on the same side of the symbol as the + input, which isn't how yours appears to be. Second, do you have both positive and negative rails on your op amps, or just + and common? You don't show a negative rail. I'm reasonably confident your circuit won't work without a negative rail, though your waveform doesn't look like I'd expect if that was the problem... \$\endgroup\$ – Stephen Collings Oct 2 '13 at 1:00
  • \$\begingroup\$ Just + and common... I'll do some modifications and post the results later on... \$\endgroup\$ – cvicente Oct 2 '13 at 13:36
  • \$\begingroup\$ Look at your full wave rectification. On the peaks the output is double the input. Did you flip two resistors around? What does this mysterious "power chord" have to do with anything? Also, the half wave should not be that tiny. Use the rectifier from my circuit. \$\endgroup\$ – Matt Young Oct 3 '13 at 2:15
  • \$\begingroup\$ @MattYoung I'll have another look to what I have in the breadboard this night. Nevertheless, although your circuit it's easier to read and more organized, and with the exception of the bias on the non-inverting pin (since I'm using a inverting voltage regulator), it seems the same as the last one I posted. Another thing is that, all last 3 screen-shots of the scope are from, supposedly, a full-wave rectification. I had no intention on these last three screenshots to have half-wave rectification. \$\endgroup\$ – cvicente Oct 3 '13 at 8:55
  • \$\begingroup\$ And the last screen-shot, why do I get no rectification at low voltages, when using the inverting voltage regulator? I'm almost inclined to post a picture of the breadboard... :-P \$\endgroup\$ – cvicente Oct 3 '13 at 8:56
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Below is a schematic, from ESP, of a full-wave precision rectifier:

enter image description here

First, note the orientation of the diodes which differ from your circuit. In this circuit, the output of the 1st op-amp is negative during the input positive half-cycle and zero during the negative half-cycle.

Second, and once again, note that the 1st op-amp produces a negative voltage. But, in your circuit, that's impossible. The op-amp output can only be positive.

If you insist on using a single supply, you could bias the input signal and non-inverting inputs at one-half the supply voltage and then remove the output voltage offset via capacitive coupling.

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  • \$\begingroup\$ Thank you for the answer Alfred. Yes I was aware that the op-amps were in fact in the opposite direction. I did that, since I was trying to do a step-by-step kind of thing. What I didn't remember was that, by leaving it that way and keep using a single supply, my wave form would never go negative when implementing the full-wave approach. I'll modify the circuit this night and post the results. \$\endgroup\$ – cvicente Oct 2 '13 at 8:36
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You need to modify the circuit. What you have will not work.

schematic

simulate this circuit – Schematic created using CircuitLab

You need to put half the power rail bias on the noninverting pin to allow the signal to go "negative", and the RC filter at the output is to remove that DC offset.

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  • \$\begingroup\$ Thank you for the answer Matt. I've an inverting voltage regulator. Could I use a dual power supply instead? The RC filter is a great idea. I'll try it as soon as I get home this night. \$\endgroup\$ – cvicente Oct 2 '13 at 8:42
  • \$\begingroup\$ Using a dual supply would make your life easier for sure. You could eliminate the need to have a bias voltage on the noninverting pins and the RC filter on the output. \$\endgroup\$ – Matt Young Oct 2 '13 at 14:47
  • \$\begingroup\$ I'll try to use the ICL7660CPAZ and let you know of the result. \$\endgroup\$ – cvicente Oct 2 '13 at 22:27

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