Hi, everybody!!!

I am trying to measure small mains current. My loads are between 6-12 W up to about 600 W. I tried with Hall effect based ICs but they appeared to be quite noisy and not suitable for such small loads. Then decided to go with CT. Now I took CT 500:1, with burden 330 Ohm and op amp (so called precision rectifier) for buffering and rectifying my signal. With this setup I can achieve almost linear output 70mA -> ~70mV and 440mA -> ~500mV - measured with non RMS multimeter.
My intend is that this output voltage would be measured by MCU.
Because the shape of the rectified sin, didn't appear to be good? , I placed 4,7uF cap at the output. Is it worth filtering the input of the op amp (the secondary of CT)? Or is it possible to use bridge rectifier instead of op amp precision rectifier (having in mind the voltage drops over each diode)?

Thanks very much for help!!!!

 

I am trying to measure small mains current. My loads are between 6-12 W up to about 600 W. I tried with Hall effect based ICs but they appeared to be quite noisy and not suitable for such small loads. Then decided to go with CT. Now I took CT 500:1, with burden 330 Ohm and op amp (so called precision rectifier) for buffering and rectifying my signal. With this setup I can achieve almost linear output 70mA -> ~70mV and 440mA -> ~500mV - measured with non RMS multimeter.
My intend is that this output voltage would be measured by MCU.
Because the shape of the rectified sin, didn't appear to be good? , I placed 4,7uF cap at the output. Is it worth filtering the input of the op amp (the secondary of CT)? Or is it possible to use bridge rectifier instead of op amp precision rectifier (having in mind the voltage drops over each diode)?

It would help if you could post a complete schematic diagram of the opamp precision rectifier circuit you are trying to use, then we could help you better. If you are using the "classic" 2-opamp, 2-diode precision full-wave rectifier circuit, filtering the output should be easy; take a look at the diagram on page 6 of this document,

https://www.site.uottawa.ca/~rhabash/ELG4135L8.pdf

and note that the capacitor C provides output filtering.

 

Thank you for the provided document. As I see on p 6 it is a half wave rectifier and the second op amp is a filter
I am using 2 op amp, 1 diode schematic. Here is the circ I am trying.
The graph posted above shows its input - red, and output - blue

Can I use such filtering over each op amp on my circ?

 

Oh, my... you are right, that is a half-wave rectifier. I gave you the wrong circuit. This is the one I should have given you:

http://m.circuitdiagram-schematic.com/full-wave-rectifier-circuit-with-averaging-filter/

Putting a capacitor from an opamp output to ground does absolutely nothing to filter the opamp output; all it can do is make the opamp unstable and cause it to oscillate. Don't do that.

 

Why you not just move the C1(4.7uF) to the positive input?
And does C1(4.7uF) enough to do the filter job?

 

You should be able to use a diode bridge directly at the transformer output with the burden resistor at the bridge output since a CT is a current device and the forward drop of the diodes should not significantly affect the voltage across the resistor.
You can then add an RC filter at the burden resistor output to filter the rectified voltage and give a DC value equal to the average (not RMS) value of the current.

Note that the current output may not look sinusoidal since many loads look nonlinear, such as rectifier DC supplies, and thus draw currents that are non-sinusoidal.

 

OBW0549, thank you for the suggested circ, I can try it out!

ScottWang, thanks for the ideas! If C1 is raised for let's say 22uF and more, then the output rectified sin is more smoothed, but then time constant rises making the circ slower.
The idea of moving the cap C1 to +5V seems interesting - I must try it

 

ScottWang, thanks for the ideas! If C1 is raised for let's say 22uF and more, then the output rectified sin is more smoothed, but then time constant rises making the circ slower.
The idea of moving the cap C1 to +5V seems interesting - I must try it

I mean that moving the C1 to the (+) positive or non-invert input of the op amp, not the +Vcc.
But if you want to try what you think then it's ok, anyway it just a test.

 

You should be able to use a diode bridge directly at the transformer output with the burden resistor at the bridge output since a CT is a current device and the forward drop of the diodes should not significantly affect the voltage across the resistor.
You can then add an RC filter at the burden resistor output to filter the rectified voltage and give a DC value equal to the average (not RMS) value of the current.

Note that the current output may not look sinusoidal since many loads look nonlinear, such as rectifier DC supplies, and thus draw currents that are non-sinusoidal.

I tried sth like that before but with burden before the bridge and another after, with another ct 2500:1 and Schotky diodes. Then there was no need for filtering, since the graph seemed like normal rectified sin wave (just a small cap.). I remember that, due to very low sec voltage at my low primary currents, instead of using 50-100Ohm burdens as specified, I had to place 4,7kOhm before and after the bridge rectifier.
Is it safe/reasonabole using such high resistance burden?
What is better to use only a burden after the bridge rec., before it or like me two?

 

I mean that moving the C1 to the (+) positive or non-invert input of the op amp, not the +Vcc.
But if you want to try what you think then it's ok, anyway it just a test.

Ok, now I got it

 

Can I omit the load resistor R4 just after the last op amp? I mean can I put that signal to ADC without load directly from the op amp output?

 

Can I omit the load resistor R4 just after the last op amp? I mean can I put that signal to ADC without load directly from the op amp output?

If your ADC provide the +5V to Vcc then it's OK, otherwsie you may need a voltage divider using two resistors.

The R4 you used just for the load current not the voltage.