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Accuracy current probe 1

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hhhansen

Electrical
Jan 14, 2004
61
Hi
How do I calculate the overall accuracy of an AC current probe (In= 1000 A) when measuring the motor current for a frequency drive having 2 kHz switching frequency. The measured current is approx. 400 A.
The accuracy of the current probe is 0,5 % of 1000 A (sinusoidal current), the bandwith is 10 kHz.

B.R.
hhhansen
 
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Thats not something you can necessarily calculate without including the manufacturer to understand the construction of the probe. The probe will smooth out the signal as the motor will as well. They do make true RMS probes as well which is what you should have. In any case, your best bet is to compare the drives output current to your probe reading or better yet, use a shunt with a good scope. I have seen drive (single phase 5HP) readings be off by quit a bit as well (I used a shunt with scope to verify a current that the drive was saying and the drive was off about 15%). So it will depend on what the drive mfg is doing to measure the current as well.
To add to your question: Does anyone have any experimental data on probe accuracy on the output of drives?
 
You are viewing the waveform or measuring the current? If you are measuring the waveform in its output of VSI, you will get the sinus waveform with harmonics. For measuring the current you should have true RMS meter.
 
Suggestion: When it comes to measuring harmonics and associated errors of their measurements, the worst case ± error may be considered.
Irms,total=sqrt[(Irms1±0.5%)^2 + (Irms2±0.5%)^2 + (Irms3±0.5%)^2 + ... + (Irmsn±0.5%)^2]
Now, if +0.5 upper error is considered, then it can be extracted, leading to:
Irms,total,+0.5%=1.005 x sqrt[(Irms1)^2 + (Irms2)^2 + (Irms3)^2 + ... + (Irmsn)^2]
Irms,total,-0.5%=0.995 x sqrt[(Irms1)^2 + (Irms2)^2 + (Irms3)^2 + ... + (Irmsn)^2]
The expression may be normalized per 1000A.
 
Following up on Jbartos answer, how would you suggest that I calculate Irms(n)normalized to 1000 A, knowing that the switching frequency is 2 kHz?
 
Suggestion: By knowing the 400A current, it may be assumed that the 0.5% error holds true for all currents less than 1000A. This may or may not be correct. Check with the current probe manufacturer regarding a character of "The accuracy of the current probe is 0,5 % of 1000 A (sinusoidal current), the bandwidth is 10 kHz."
Specifically, whether or not 0.5% accuracy is constant from 0A to 1000A.

 
I would try installing the same probe at the incoming of the inverter(Pure sine wave) and compare with standard meter/ VFD display. It should be reading the current reading accurate enough ignoring the VFD losses.
I searched a lot but could not find a way to accurately measure the PWM current. I wonder how drive measures/ controls, is it on incoming AC side ? may someone enlighten us.

Best regards,
 
PWM signals are generally measured using shunts with software RMS measuring or they may use hall-effect sensors to accurately measure the current (aftermarket in my experience like Load Controls Inc.). (think of methods to measure DC)

I did compare a drive reading with a shunt with an accurate rms meter and found the drive readings were a little low. This told me that the circuitry in this 5HP drive was using some generic hardware/software to come up with this number. I do not know for sure how drive mfg's do this but have learned not to trust their readings (at least the inexpensive single phase drives).

Clamp-ons are not that accurate to begin with depending on the conductor size. As an experiment, place your clamp-on on some current carrying conductor and move it slowly around the conductor and watch the reading change more than it should based on the accuracy they give you at that current level. I have found that most exceed their accuracy spec by simply moving the meter window in relation to the conductor.
 
Suggestion: A true RMS meter is required, e.g. Simpson
 
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