standard CT's and VFD's
standard CT's and VFD's
(OP)
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





RE: standard CT's and VFD's
RE: standard CT's and VFD's
In a very basic sense, CTs are considered better at passing higher frequencies than PTs because of fewer turns of heavier wire, minimizing reactance. Contrast this to a PT with many turns of smaller wire having comparatively higher reactance. It seems generally accepted that for CTs, up to the 50th harmonic should not be a significant problem. Of course, high-crest-factor waveforms will more easily saturate CTs than traditional sine waves.
RE: standard CT's and VFD's
RE: standard CT's and VFD's
RE: standard CT's and VFD's
My traditional model will include:
a linear magnetizing branch (L), purely resistive burdren R. And higher frequency actually gives the core less time to go into saturation so the linear magnetizing branch assumption will not be violated by high frequency content... ie high frequency doesn't push core into saturation... low frequency does).
With all quantities referenced to the secondary:
I1 = Im + I2 = V2/(L*s) + I2 = (R*I2)/(L*s) + I2 = I2* [1 + R/(Ls)] I2/I1 = 1 / [1 + R/(Ls)] = s / [ s + R/L]
This transfer function has a zero at the origin and a pole at s = -R/L. For typical installation L/R corresponds to a large time constant and the pole -R/L is very close to the origin.... perhaps at 0.001 sec^-1. The pole and zero being so close together means that their effects will cancel for all (complex) frequencies except those very close to the pole zero pair.
So we have unity gain for high frequency and approaches zero for low frequencies (high pass filter). The cutoff perhaps being on the order of R/L~0.001hz which is very slowly varying.
H(s) = s / [ s + R/L] transforms into a quirky "impulse response" h(t) which itself includes an impulse (delta). It's a little easier to look at the step response which is (1/s) * H(s) = 1 / [ s + R/L] or step_response(t) = exp(-Rt/L) So the system responds perfectly to the high-frequency components in the rising edge of the step, but decays away as the steady-state (dc) response starts to kick in.
The low pass-filter concept fits well with our intuitive idea that a transformer cannot pass dc. But anyone who has viewed an oscillograph of a fault knows that the decaying dc offset does show up. The explanation: as shown above, that dc component would decay away from the output only if it lasted perhaps a few hundred seconds (not typical).
But I'll admit that the traditional model is after all geared toward power frequency analysis. I wonder whether there might be other components of the model which are unimportant at power frequency but would show up for high frequency. For instance, a parasitic parallel capacitance on the output would limit the ability to produce high frequency output. I don't believe that the series reactances will play any significant role with increasing frequency. Even though their impedance increases with frequency, the magnetizing branch impedance increases proprotionately, combine this with the fact that the primary side is acting like a curent source and it's hard to develop a scenario where increased magnetizing current would develop.
I know that plenty of well-educated people have written papers on capacitive switching oscillographs with very high-frequency content, and I have never seen any mention of concern for lack of fidelity of the ct at these high frequencies. Although I have seen a discussion that the mechancial inertia of old-style recording devices limits high frequency content. (I guess the modern day equivalent is sampling rate).
Sorry for a little bit of rambling. In summary, I'm pretty sure that busbar and Gordon are correct. And likewise you are correct in your analsysis that the inductive nature of the motor limits the very high frequency harmonics in the current... and at these lower frequencies below a few khz the existing model of a ct (which won't have any significant shunt capacitance and will therefore act as high-pass filter) should remain accurate.
RE: standard CT's and VFD's
I showed that the impulse response h(t) = invlapalce ((s/(s + R/L) = delta(t)-exp(-R*t/L)*R/L where delta(t) is the unit impulse. Therefore our model predicts that an impulse (which includes all frequencies low and high) is passed immediately right through the system... and the -exp(-R*t/L)*R/L represents a low-magnitude error term which persists for a very long time.
(that error term will only become signficant when input is very slowly varying... in which case the error term tends to cancel out that low frequency input).
Once again, all of the above is predicated on the correctness of "the model".
RE: standard CT's and VFD's
RE: standard CT's and VFD's
Busbar dont take offense that I questioned your comment about the 50th harmonic. I am sure you know what your talking about and I am not a new kid on the block either (7years). This is what these forums are for, discussions. Enough said.
Based on what I am hearing(and maybe I havent clearly defined what I am getting at), I could take a signal generator (sin wave) and assuming I could source 40 amps with this and put this conductor through a CT (50:5 2.5VA rating) with a frequency of 3KHz and the CT would couple the signal perfectly and be linear throughout the operating range specified by manufacturer(think they usually say 60-80% of CT primary is where the accuracy spec applies)? I don't have the means to come up with a source to do something like this as most people won't but it would be an interesting experiment. Your mention of crest factor brings an interesting card to the table. If I have a waveform with a crest factor of 5 or greater (some drives are this bad). The CT is not going to see the crest of this waveform. With that said the frequencies that are causing this crest factor are likely the ones containing the largest distribution of power in the frequency domain 5th and 7th usually. So why doesnt the CT couple this higher frequency signal? I better read electricpetes message real close before I rattle on to much. Thanks Buzzp
RE: standard CT's and VFD's
Using same model as before:
Im = 1/Lm* Integral(v(t))dt ~ 1/Lm*Integral(i(t))dt if we assume resistive burden. Im and Lm are quantities associated with the magnetizing branch. v is voltage accross the magnetizing branch.
Phi = Im*Lm/N = (1/N) * Integral(i(t))dt
Saturation occurs when Phi reaches a limiting value PhiMax. Whether or not Phi reaches PhiMax depends upon the area under the curve Integral(i(t)dt. The maximum area under that curve occurs when we integrate over a half of a cycle. And for a given magnitude of I, the low frequency will have a much higher area-under-the-half-cycle-curve than the high frequency (because the time interval is much longer). Saturation will occur at much lower current magnitudes for low frequency (we aready know that dc is effective at pushing a ct into saturation). Current magnitude for causing saturation increases as the frequency of the current increases.
With all that said, it's really the combined effect of all frequencies that is important in determining whether we will reach saturation, not the effect on each individual frequency acting alone.. i.e. how high will flux be when we apply the total non-sinusoidal current. That can be assessed by looking at the max area under the curve of what you see on your trace, and determining an equivalent sinusoid which has same area under the curve. Even though your calc is a little skeewed by the fact that your beginning waveform is suspect of being distorted, it should give you a ballpark idea.
RE: standard CT's and VFD's
Epete's observation of: "...how high will flux be when we apply the total non-sinusoidal current..." is just about to the edge of my understanding and practicality with respect to field measurements.
Sort off topic, but there is an IEEE tutorial collection of papers that deals with measurements in “nonsinusoidal situations” that gives an idea of the complexity of such a task, and hardly broaches the area of instrument transformers. FIFW, I can’t imagine any significant/interesting power levels above the 25th harmonic. (The effects of beer on the accuracy of revenue metering is even covered in the tutorial.) I can post the subject IEEE publication number if anyone’s interested.
RE: standard CT's and VFD's
H(s) = s / [ s + R/L], which can be rewritten as
H(s) = 1 - (R/L) / [ s + R/L].
The "1" term is a perfect all-pass filter. The
"- (R/L) / [ s + R/L]" is the low-magnitude slowly-decaying error term which acts somewhat similar to an integrator. It only comes into play for slowly varying inputs. If you convolve this error term with a rapidly varying sinusoid over several cycles of input you will get zero (similar to convolving ac input with dc system response... yields average value of ac which is zero). You can only get a large error response over a long period of time with a slowly-varying input.
busbar - I would be interested in the IEEE document number if you have it available.
RE: standard CT's and VFD's
RE: standard CT's and VFD's
Thanks in advance
Pitat
RE: standard CT's and VFD's
(I'll keep my fingers crossed that it looks the same as it does on my preview)
I1 V
==>=======
| |
| Im | I2
Lm R
| |
| |
__ __
- -
I1 is primary current (on secondary basis).
I2 is secondary current
Im is magnetizing current.
Lm is magnetizing inductance
R is resistive burdern
Circuit: the primary current divides into two parts: the secondary current I2 through the relay/meter R and the magnetizing current Im through Lm. If Im is zero, there is no error and I2=I1. As we increase the burden or increase the current, we increase the voltage V=I2*R and increase the resulting current Im = (1/L)*Integral(v).
By KCL:
I1 = Im + I2 = (1/L)*Integral(v)dt+I2
substitute v=I2*R
I1 = (1/Lm)*Integral(I2*R)dt + I2
Take laplace transform of both sides
I1 = (R*I2)/(Lm*s) +I2 = I2*[R/(Lm*s)+1]
Look at the transfer fucntion H(s) which will be defined as ratio of output I2 to inptu I1
H(s) = I2/I1 = 1 / [1 + R/(Lm*s)] = s / [ s + R/Lm]
This transfer function has been analysed above in a number of ways. One more way (maybe simpler?) is to substitute s=jw to give the transfer function as function of radian frequency w=2pif (where j=sqrt(-1)).
H(s) = I2/I1 = = jw / [ jw + R/Lm]
For high frequency = large w, jw dominates the denominator and the ratio approaches 1. The system transmits high frequency with gain of 1.
For small w->0, the denominator approaches constant R/Lm and the numerator approaches zero and the ratio approaches zero. The system does not transmit low frequency.
I hope this has made it easier. If not, let me know once again and I'll try to do a better job perhaps on paper.
RE: standard CT's and VFD's
http://www.geindustrial.com/industrialsystems/publications/dcpub.jsp
Go to protective relays or to transformers for more info about ct's.
I took a look at the art and science of protective relaying chapter 7
http://www.geindustrial.com/industrialsystems/pm/notes/artsci/art07.pdf
On page 5 of 19 of that file, they describe a procedure for calculation of ct accuracy using secondary excitation curve. I believe that what I have done is similar except I have assumed linear magnetizing branch. We have previously discussed that the effect of considering saturation would be more significant on the low frequency end of the transfer function then on the high end. (it would chop/distort and reduce rms of low frequency signals more than high frequency signals, assuming single-frequency signals of same amplitude). Also if we try the linear model to solve a problem, we can use the results to check if our linearity assumption was correct.
RE: standard CT's and VFD's
RE: standard CT's and VFD's
Analyse the above circuit remembering that an inductor acts like a short-circuit to dc-like low frequencies and acts like an open circuit to very high frequencies (impedance Z=2pi*l*F is proportional to F).
If we apply a dc primary current, then all the current will flow through the inductor (since it acts liks a short circuit) and none will flow through the resistor (meter or relay). Behavior of low-frequency is similar to dc.
If we apply a high frequency, no current will pass through the inductor (high-impedance to high frequency), and all current will flow through the resistor (meter or relay).
Above predicated on the model. I read recently that power transformer eddy current losses increase drastically with frequency. Those eddy current losses would be represented by another resistor in parallel with Lm. My guess is that to model the losses properly the value of the resistance would have to decrease (draw more resistive SHUNT current) as frequency increases. It is not signficant enough to incorporate into the normal power frequency model (like the one shown in Art and Science of Protective Relaying), but it may be an effect that becomes important at higher frequencies. (?)
RE: standard CT's and VFD's
RE: standard CT's and VFD's
Now although we have become quite good at it, some things are still lacking. Example: I noticed that when measuring on the input side of the drive (no line filter, small load) with a true RMS meter (Fluke 87III and a Fluke i200 AC current clamp) I get about 1.4 Amps, but when measuring on the output((no filter) I get 14Amps, which corresponds with the drive's indication.
All the drives are PWM, makes use of 6 pulse Rectifiers and utilizes LEM (Linear Electromagnetic Modules) on the output for current sensing.
Is this difference in value because of signal noise imposed by the different switching frequencies between the input and output power equipment? Or am I just using the wrong equipment?
Thanks in advance. I don't think buzzp expected so many responses. Obviously this is a subject of large debate!
RE: standard CT's and VFD's
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.
RE: standard CT's and VFD's
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 www.lem.com
Regards
Ants
RE: standard CT's and VFD's
RE: standard CT's and VFD's
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>
RE: standard CT's and VFD's
RE: standard CT's and VFD's
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.
RE: standard CT's and VFD's
Mark Empson
http://www.lmphotonics.com
RE: standard CT's and VFD's
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?
RE: standard CT's and VFD's
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!
RE: standard CT's and VFD's
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
http://www.lmphotonics.com
RE: standard CT's and VFD's
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.
http://www.ab.com/drives/techpapers/ieee/APEC99%20current%20sensors.pdf
RE: standard CT's and VFD's
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.