I have been trying to get a feel for measuring (accurately +-3%) the true rms current going to a motor that is fed from a PWM or PAM type drive(say up to 500HP, 480V). I do not have access to a drive or motor right now to do any testing. Wouldn't the motor act as an integrater, thereby smoothing out the current signal enough to allow the CT to get an accurate coupling? This has been bugging me for a while and I would like to know what others think about this, thanks to all buzzp
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standard CT's and VFD's 1
- Thread starter buzzp
- Start date
electricpete
Electrical
An interesting case Mr Ants.
Tough to draw any conclusions from a distance. It would be interesting to know the drive input voltage and number of phases, and do you know the actual horsepower drawn from the load (which would help narrow down which is correct... note there are likely harmonics on both input and output).
If I were to take a guess, I would say that the output is reading high due to presence of harmonics. We have just finished saying how a ct should provide output current which accurately reflects a fixed fraction of input current, regarldless of harmonics. Why would a clamp-on be different?
There's may be a good reason. A ct depends on the principle of canceling the amp-turns of output with amp-turns of the input, which gives good reproduction even in the presence of harmonics.
But a clamp-on ammeter may simply develop a voltage v=dPhi/dt which is related to the rate of change of flux and rate of change of input current. Higher harmonics would be weighted proportionately higher than the fundamental due to higher rate of change. I've never studied the principle of operation of ammeter probes but I have noticed that some are rated in voltage output (20A in=2v out) and some are rated in current output (20A in =200mA out). If you have the kind rated with a voltage output this theory seems plausible.
But it leads to another question... what kind of current sensing device is used in your switchboard? A current to voltage transducer? or a ct driving a standard switchboard ammeter?
I'm just thinking out loud on this one... probably way off base.
Tough to draw any conclusions from a distance. It would be interesting to know the drive input voltage and number of phases, and do you know the actual horsepower drawn from the load (which would help narrow down which is correct... note there are likely harmonics on both input and output).
If I were to take a guess, I would say that the output is reading high due to presence of harmonics. We have just finished saying how a ct should provide output current which accurately reflects a fixed fraction of input current, regarldless of harmonics. Why would a clamp-on be different?
There's may be a good reason. A ct depends on the principle of canceling the amp-turns of output with amp-turns of the input, which gives good reproduction even in the presence of harmonics.
But a clamp-on ammeter may simply develop a voltage v=dPhi/dt which is related to the rate of change of flux and rate of change of input current. Higher harmonics would be weighted proportionately higher than the fundamental due to higher rate of change. I've never studied the principle of operation of ammeter probes but I have noticed that some are rated in voltage output (20A in=2v out) and some are rated in current output (20A in =200mA out). If you have the kind rated with a voltage output this theory seems plausible.
But it leads to another question... what kind of current sensing device is used in your switchboard? A current to voltage transducer? or a ct driving a standard switchboard ammeter?
I'm just thinking out loud on this one... probably way off base.
Thanks Electricpete, valid remarks. I omitted to say that these measurements were taken at full speed(50 Hz, 1490rpm) and full voltage applied to the motor terminals (525 volt on both drive input and output, 3 phase).
I believe the load measured on the output side of the VFD is the correct one (14.3Amps)because of DOL measurements taken previously, and I also believe the switchboard device to be measuring accurately.
The clamp-on CT is 1A = 1mA device.
The current sensing device on the VFD switchboard is manufactured by LEM. It is based on the Hall effect principle and allows for the measurement of dc and ac pulsed and complex currents with galvanic isolation. In this case the output signal is voltage. For more info you can visit LEM at
Regards
Ants
I believe the load measured on the output side of the VFD is the correct one (14.3Amps)because of DOL measurements taken previously, and I also believe the switchboard device to be measuring accurately.
The clamp-on CT is 1A = 1mA device.
The current sensing device on the VFD switchboard is manufactured by LEM. It is based on the Hall effect principle and allows for the measurement of dc and ac pulsed and complex currents with galvanic isolation. In this case the output signal is voltage. For more info you can visit LEM at
Regards
Ants
- Thread starter
- #23
I also believe it is the filtering. Generally, the converter is not filtered (the input) but the output is often filtered within the drive design. The filtering to prevent trash on the utility is often an option but may become mandated depending on where you live (i think IEEE 519 specifys allowable harmonic distortion from the converter section of the drive). But this brings up an interesting question: if the CT's are accurate regardless of harmonics then why would the clamp on meter be off this much? The design of the converter and inverter are obviously different. I have tested a small 3HP 230 volt single phase drive with standard CT's, a shunt with a scope, and the current the drive says the motor is pulling. I found these three data sources do NOT indicate the same current draw. I tend to believe the shunt with the scope connected across it (yeah I know the shunt has inductive properties). I plan to due more testing (did this test 5 years ago when I was even dumber than I am now, hehe) with the CT's within the next six months. I will post my findings. Thanks
Buzzp:
The motor acts as a low pass filter for the current.
The current XFORMER -- which acts as a bandpass and not as highpass filter -- may transmit enough harmonics but
the phase shift may vary within the band.
I would suggest using an interface based on analog optocoupler.
<nbucska@pcperipherals.com>
The motor acts as a low pass filter for the current.
The current XFORMER -- which acts as a bandpass and not as highpass filter -- may transmit enough harmonics but
the phase shift may vary within the band.
I would suggest using an interface based on analog optocoupler.
<nbucska@pcperipherals.com>
The only way to measure the current will be to use a hall effect CT...I know this from experience. I have installed two different systems of 250kW and 315kW - current measurement was originally class 1 CT feeding a DCS (3% Z reactor was also installed). The results were not linear. Changed to Hall effect CT and checked with a flukescope - very accurate. Don't ask me why!!!
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestions:
1. Reference
Gordon R. Slemon "Magnetoelectric Devices, Transducers, Transformers and Machines," John Wiley and Sons, Inc., New York, 1966, Section 3.2.2 Variable Frequency Operation (under Operating Characteristics of Transformers) generally analyzes the transformer dependency on frequency. A graph of |Eload/Esource| versus frequency is provided. The distributed capacitances between the primary and secondary windings becomes influential at higher frequencies.
2. Traditionally, the analysis of a transfer function is in relationship with the input type, i.e.
Iout(s) = H(s) x Iin(s)
For all practicall purposes, the Iin(s) is different from 1 in Laplace s-domain or Dirac function delta(t) in the time domain. Usually, Iin(s)= w/(s**2 + w**2) ~ sin(wt) or Iin(s)=s/(s**2 + w**2) ~ cos(wt) for the fundamental waveform (no harmonics involved). The harmonics somewhat complicate the Laplace analysis approach; however, it can be used with the nonlinear approach to the modeling via the multidimensional Laplace transform.
1. Reference
Gordon R. Slemon "Magnetoelectric Devices, Transducers, Transformers and Machines," John Wiley and Sons, Inc., New York, 1966, Section 3.2.2 Variable Frequency Operation (under Operating Characteristics of Transformers) generally analyzes the transformer dependency on frequency. A graph of |Eload/Esource| versus frequency is provided. The distributed capacitances between the primary and secondary windings becomes influential at higher frequencies.
2. Traditionally, the analysis of a transfer function is in relationship with the input type, i.e.
Iout(s) = H(s) x Iin(s)
For all practicall purposes, the Iin(s) is different from 1 in Laplace s-domain or Dirac function delta(t) in the time domain. Usually, Iin(s)= w/(s**2 + w**2) ~ sin(wt) or Iin(s)=s/(s**2 + w**2) ~ cos(wt) for the fundamental waveform (no harmonics involved). The harmonics somewhat complicate the Laplace analysis approach; however, it can be used with the nonlinear approach to the modeling via the multidimensional Laplace transform.
-
1
- #27
What a lot of good information to this post. To keep it simple and add my two cents worth, The current flowing to a motor on the output of a PWM inverter is very nearly sinusiodal for frequencies below the line frequency, but begins to square up for frequencies above line frequency. As the frequency of the current passed through a CT is reduced, the effective VA rating of the CT reduces. Provided that the CT is loaded with a burden that is well below the maximum VA rating of the CT, it will maintain reasonable accuracy at the lower frequencies. For higher frequencies, the waveform distortion will impact on the accuracy and this is dependent on the phase shift through the CT and the responce of the measuring circuit to the waveform.
Mark Empson
Mark Empson
electricpete
Electrical
Ants - Your question now boils down to: Why would the input current to your drive read lower than actual.
I can think of one reason:
Imagine a single-phase full-wave rectifier feeding a large capacitor. There will be no current during the majority of the cycle when Vac<Vcap. But for a very short period of time near the peak of the ac voltage waveform Vac>Vcap and a very large current will flow. Thus the waveform is zero most of the time and very high for short bursts twice per cycle. Three phase will behave similarly. Series input inductances will tend to spread the current spikes out somewhat, but they will still be very high and narrow.
Given that your current input waveform is likely close to zero most of each cycle very high for small fraction of each cycle, there are two possible reasons why this current might be sensed low:
#1 - (more likely)- Saturation of your current probe. If the current probe is acting like a ct, then a high peak primary current can cause saturation (even in the presence of moderately low burden..your multimeter). It may sound a little paradoxical given that the reading is far less then the maximum possible output reading of your probe, but in fact it is very possible.
#2 (less likely) - Your multimeter samples the waveform periodically. If it is sampling at only 1khz then it would only get 8 samples per half cycle. That might significantly miss the bulk of the large peak.
Once again just thinking out loud. Any thoughts?
I can think of one reason:
Imagine a single-phase full-wave rectifier feeding a large capacitor. There will be no current during the majority of the cycle when Vac<Vcap. But for a very short period of time near the peak of the ac voltage waveform Vac>Vcap and a very large current will flow. Thus the waveform is zero most of the time and very high for short bursts twice per cycle. Three phase will behave similarly. Series input inductances will tend to spread the current spikes out somewhat, but they will still be very high and narrow.
Given that your current input waveform is likely close to zero most of each cycle very high for small fraction of each cycle, there are two possible reasons why this current might be sensed low:
#1 - (more likely)- Saturation of your current probe. If the current probe is acting like a ct, then a high peak primary current can cause saturation (even in the presence of moderately low burden..your multimeter). It may sound a little paradoxical given that the reading is far less then the maximum possible output reading of your probe, but in fact it is very possible.
#2 (less likely) - Your multimeter samples the waveform periodically. If it is sampling at only 1khz then it would only get 8 samples per half cycle. That might significantly miss the bulk of the large peak.
Once again just thinking out loud. Any thoughts?
- Thread starter
- #29
I think, a factor in reading low, is also related to the high crest factors of the waveform. If the measurement technic is just measuring the peaks and then scaling back then this would be the reason it reads lower (this is usually the case). Unless they have made there own algorithm, hardware off the shelf (trms converters) has problems above crest factors of 5 (was case few years ago, may have changed now).
It seems there are some theorhetical arguments that say standard CT's will work on SOME vfd applications, depending on range of frequency, carrier frequency, acceleration and deceleration time, type of motor, etc.. The real world arguments all point to it not working except in a few applications. I believe, if I were a designing a product to use on vfd's, I would definately use hall effects or similar type of sensor to avoid confusion to the end user. Would all agree with this? And thanks for all the posts!
It seems there are some theorhetical arguments that say standard CT's will work on SOME vfd applications, depending on range of frequency, carrier frequency, acceleration and deceleration time, type of motor, etc.. The real world arguments all point to it not working except in a few applications. I believe, if I were a designing a product to use on vfd's, I would definately use hall effects or similar type of sensor to avoid confusion to the end user. Would all agree with this? And thanks for all the posts!
Do not forget that the inverter acts a little bit like a transformer giving an output that is Variable Voltage Variable frequency. As the output frequency is reduced, so is the voltage. Just like a transformer, the VA into the drive equals the VA out of the drive plus the losses. If the frequency is under line frequncy, then the voltage will be below line frequency and the output current will be higher than the line current.
At line frequency, the voltage will still be a little lower and so the current on the output will be a little higher.
Mark Empson
At line frequency, the voltage will still be a little lower and so the current on the output will be a little higher.
Mark Empson
electricpete
Electrical
I don't have the practical experience with VFD's to agree with your statement.
I don't see the theoretical basis for why CT's would read low. Although I'll admit now that doesn't mean much.
The discussion of crest factors seems to imply you think the CT will saturate due to high magnitude current spike for brief time. But it saturates based on high flux which is related to the integral of primary current over time.... so a high (amplitude)narrow (time width) spike is no worse than a low wide spike.
There is a link below which discusses performance of Hall transducers is VFD's. Apparently they are used so there must be a reason why we use them. You've got my curiosity up now.
I don't see the theoretical basis for why CT's would read low. Although I'll admit now that doesn't mean much.
The discussion of crest factors seems to imply you think the CT will saturate due to high magnitude current spike for brief time. But it saturates based on high flux which is related to the integral of primary current over time.... so a high (amplitude)narrow (time width) spike is no worse than a low wide spike.
There is a link below which discusses performance of Hall transducers is VFD's. Apparently they are used so there must be a reason why we use them. You've got my curiosity up now.
TO ALL
WE ARE GOING TO INSTALL COMPRESSED AIR SYSTEM WHERE THE MOTOR OF 45 KW WILL BE DRIVEN BY VFD & IN CASE OF FAILURE OF VFD THE SAME MOTOR WILL BE RUNNING ON DOL. THE CURRENT MEASUREMENT WILL BE TALLIED IN THIS CASE DURING COMMISSSIONING. MY DOUBT IS WHETHER THERE IS ANY RESIDUAL HARMONICS WHICH WILL GIVE ERROR WHEN THE MOTOR WILL RUN ON DOL AFTER RUNNING ON VFD. IF NOT THEN SOME GROUND IS THERE TO ANALYSE THE DIFFERENCE OF CURRENT MEASUREMENT ON VFD.
WE ARE GOING TO INSTALL COMPRESSED AIR SYSTEM WHERE THE MOTOR OF 45 KW WILL BE DRIVEN BY VFD & IN CASE OF FAILURE OF VFD THE SAME MOTOR WILL BE RUNNING ON DOL. THE CURRENT MEASUREMENT WILL BE TALLIED IN THIS CASE DURING COMMISSSIONING. MY DOUBT IS WHETHER THERE IS ANY RESIDUAL HARMONICS WHICH WILL GIVE ERROR WHEN THE MOTOR WILL RUN ON DOL AFTER RUNNING ON VFD. IF NOT THEN SOME GROUND IS THERE TO ANALYSE THE DIFFERENCE OF CURRENT MEASUREMENT ON VFD.
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