I am using a CT current sensor to measure the current consumption of an AC device. Is there a relatively accurate way to get the RMS signal from the sensor? Because the sensor signal I put into the oscilloscope, I see that it is not a perfect sine wave, I make a circuit to take the peak amplitude and divide it by the root of 2, it is no longer accurate. I say it's not accurate because I measured the current with a clamp meter for comparison. The measurement results were about 50% different than what I measured with the clamp meter. I find the RMS value measured by the oscilloscope quite accurate and I think its calculation is more complicated. Thank you everyone.

Here is an image of the waveform I measured.

 

1) Sample at above 1600Hz. Square the value each sample. Do a running average with an IIR filter. Take the square root of the output.
or
2) Use an RMS-to-DC converter
https://uk.farnell.com/c/semiconductors-ics/data-signal-conversion/rms-to-dc-converters

 

1) Sample at above 1600Hz. Square the value each sample. Do a running average with an IIR filter. Take the square root of the output.
or
2) Use an RMS-to-DC converter
https://uk.farnell.com/c/semiconductors-ics/data-signal-conversion/rms-to-dc-converters

Thank you for your helpful answer. What I'm aiming for is signal processing at the hardware level. Is there an opamp circuit that can calculate RMS signals or do I have to trade off by using a dedicated RMS calculator IC at a higher cost?

 

hi frank,
If you are considering additional circuitry, these two PDF's may offer a solution.

E

 

Thank you for your helpful answer. What I'm aiming for is signal processing at the hardware level. Is there an opamp circuit that can calculate RMS signals or do I have to trade off by using a dedicated RMS calculator IC at a higher cost?

It’s not easy to accomplish the “square” function with op-amps with any degree of accuracy. It generally involves a lot of very closely matched bipolar transistor, similar to the logarithmic and exponential amplifiers. Matched transistors are a lot easier to fabricate when they are all on the same chip.
https://patentimages.storage.googleapis.com/08/ab/84/99f50a6fbae9d0/US7081787.pdf

 

hi frank,
If you are considering additional circuitry, these two PDF's may offer a solution.

E

Thanks for your supporting documentation <3

 

It’s not easy to accomplish the “square” function with op-amps with any degree of accuracy. It generally involves a lot of very closely matched bipolar transistor, similar to the logarithmic and exponential amplifiers. Matched transistors are a lot easier to fabricate when they are all on the same chip.
https://patentimages.storage.googleapis.com/08/ab/84/99f50a6fbae9d0/US7081787.pdf

Thank you for giving me more useful information. Maybe I have to accept the high price of my product in exchange for more accuracy in measuring current.

 

A small micro-controller could easily do the sampling required by post #2 and would be very cheap to implement.

 

Examine the correct meaning of RMS, root mean square.

The RMS value of a signal represents the DC value that would have the same power as the signal in question.

This is determined by computing the square, mean, root. In other words, one computes the square, followed by the mean, followed by the square root.

The calculation of dividing by the square root of 2 applies to sinewaves. It does not apply to other random waveforms.

Before the advent of modern electronics, RMS could be determined accurately using a hot wire ammeter.

With MCU software solutions, the proper method of determining RMS is to do exactly as stated above. Take continuous samples, and compute the square and take the mean over a given period. Then compute the square root of the mean.

Sorry, there is no easy hardware solution.

 

Examine the correct meaning of RMS, root mean square.

The RMS value of a signal represents the DC value that would have the same power as the signal in question.

This is determined by computing the square, mean, root. In other words, one computes the square, followed by the mean, followed by the square root.

The calculation of dividing by the square root of 2 applies to sinewaves. It does not apply to other random waveforms.

Before the event of modern electronics, RMS could be determined accurately using a hot wire ammeter.

With MCU software solutions, the proper method of determining RMS is to do exactly as stated above. Take continuous samples, and compute the square and take the mean over a given period. Then compute the square root of the mean.

Sorry, there is no easy hardware solution.

I got it, thank you very much :3

 

You could also use a True RMS measuring IC such as this.

 

True RMS measuring IC

AD737JRZ-RL
LTC1966CMS8#TRPBF
I like the 8 and 10 pin versions.

It looks like your load is a PC or TV or something like that. Use a heater or some resistive load to see a sine wave. You are getting current pulses at the peak of the sine wave. Very typical of a power supply with a full wave bridge as a load.
It also could be you don't have the burden resistor right.