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I am trying to design a precision rectifier, to detect AC current through a current transformer. With a 22 Ω burden resistor and a test bulb, the CT I have gives me a pretty ugly 20 mV sinusoid with a 5% duty factor, which I have modeled as pulses in the simulation below:

sinusoid with 5% duty factor

I am trying to use a first non-inverting amplifier stage, followed by a precision half-wave rectifier. Each circuit taken separately in a simulator works fine, but as soon as I combine the two everything breaks down.

This is a snapshot of the amplifier simulation (5 V voltage source on the right, LM324 op-amps):

enter image description here

This is a snapshot of the half-rectifier simulation:

enter image description here

And the complete circuit that doesn't work at all:

enter image description here

For starters the negative rail goes up to 0.8 V and the positive one to 5.8 V :( What am I missing?

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  • \$\begingroup\$ What is the purpose of the right side buffer op amp? \$\endgroup\$ – Navaro Apr 18 at 16:21
  • \$\begingroup\$ Simply a voltage follower to try to keep the virtual ground at 0V \$\endgroup\$ – Franck Apr 18 at 16:46
  • \$\begingroup\$ What op-amp and what battery supply? \$\endgroup\$ – Andy aka Apr 18 at 16:59
  • \$\begingroup\$ The op-amp model is of an LM324, the battery is a 5V voltage source in the simulator (switching DC power supply eventually in real life) \$\endgroup\$ – Franck Apr 18 at 17:26
  • \$\begingroup\$ BTW, changing the battery to 15V makes the simulator work for a couple iteration, and it then crashes... so my guess is that the problem comes from getting too close to the rails. \$\endgroup\$ – Franck Apr 19 at 11:40
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I've gone down that rabbit hole a few years ago (designing precision rectifiers for current transformers). The results were unsatisfactory.

Here's a better suggestion: try to use a diode bridge directly, instead of a precision rectifier. Simply insert the diode bridge between the C+/C- terminals of the CT and the 22 Ω load, and do away with the precision rectifier. Because it's a current transformer, not a voltage source, it "compensates" for the diode drop. You could try simulating this circuit with a sinusoidal current source in place of the current transformer to see what I mean.

When I did this in my circuit, the result was cheaper and better, with no downsides, as I recall.

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  • \$\begingroup\$ To be honest I am doing this mostly to get a a better understanding of op-amps. \$\endgroup\$ – Franck Apr 18 at 16:48
  • \$\begingroup\$ OK, there is certainly lots of room for learning here (though I wouldn’t recommend this kind of circuit as a beginner lesson in op amps). Just be aware of not falling into the trap of thinking that, since the only tool you have is a hammer, everything should be treated as a nail. Maybe this problem is a screw and you need a screwdriver. \$\endgroup\$ – swineone Apr 18 at 16:59
  • \$\begingroup\$ +1 for lateral thinking .I have done this many times but have difficulty in explaining that this can and does work . \$\endgroup\$ – Autistic Apr 19 at 0:00
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Taking your suggestion I came up with the circuit below. I only need to detect the A/C current, not measure it, I beefed up the burden resistor and I clamped the voltage using two zeners.

The circuit works in the simulator, the thing that bothers me is that I don't have a good discharge path through the capacitor. With no series resistor it takes 1nF to keep the mosFet decently on (Vgs drops to 3V at the end of the discharge), higher values take forever to shutdown when the CT current is removed: enter image description here

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  • \$\begingroup\$ Franck - Hi, It's not clear whether you are trying to imply questions and get replies here, by writing things like "the thing that bothers me is [...]", "higher values take forever to shutdown [...]. (a) Questions are not allowed in answers (especially self-answers) as you have written, and: (b) Since you have marked an answer as "accepted" then many people who answer questions will skip reading this one now, as the question has apparently been solved. For those reasons, I doubt you will get many, if any, responses to the points in your answer. If you want to continue this topic, let me know \$\endgroup\$ – SamGibson Apr 19 at 17:37
  • \$\begingroup\$ To discharge the capacitor, why not use a resistor in parallel? If you only need a pulse, select one such that the RC time constant formed with the capacitor is considerably faster than the signal frequency (I would start at 1 ms and tweak it from there). By the way, the other comment makes sense -- maybe you should edit the question with the new circuit, and I can amend my answer with this comment. \$\endgroup\$ – swineone Apr 19 at 21:43
  • \$\begingroup\$ @Sam: I guess I couldn’t make up my mind, I can’t say I am too comfortable yet with the single question/single answer format :) I did want to post my best attempt at following the suggestion to try a completely different approach, I dont have any specific question at this point. \$\endgroup\$ – Franck Apr 20 at 17:33

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