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
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.
- Thread starter
- #4
Ct's will pass some of the harmonic content of the motor current(depends on the severity of the distortion). Usually, if the drive has any type of load side reactor (filter), the harmonic content is negligible and difficult to see on a scope. I know the CT is not going to spit out a TRMS signal. I want to know how much of the signal is coupled to the CT. In otherwords, how accurate would the waveform out of the CT be as compared to the actual wave the motor sees? The standard CT obviously can not do a good job coupling the high frequency content of the signal from the drive. So if we took a shunt and a standard CT(with burden resistor) and placed both of these voltage signals into the identical circuitry to determine the TRMS current, what would the difference in readings be? I would assume the shunt would do a better job of representing the real waveform the motor is seeing or maybe the CT signal is doing a good job of representing the real signal. So if I designed a something to measure this current and wanted to use standard CT's for the sensor part of this device and ignoring the error of the signal conditioning circuitry, would there be additional error in measurement (other than the standard error of the CT selected)? I know standard CT's are not designed to couple 1kHz signals, but is this necessary since most of the power distribution will be in the selected frequency of the drive (say 30Hz-120Hz)? Thanks, Buzzp
- Thread starter
- #5
Busbar, i just missed your post before I sent mine. Your saying a standard CT (50-60Hz) variety would couple up to the 50th harmonic? Ummm. I dont believe it is really coupling that high frequency it just cant respond quick enough to the changes and basically is an integrater and simply shows a almost linear output of the energy content of the waveform going to the motor. I don't know, it will be interesting to see what others say. Thanks Buzzp
electricpete
Electrical
I believe that Gordon and busbar are right that the traditional model of a CT predicts that it will act like a high pass filter.
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.
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.
electricpete
Electrical
If the discussion of why the model predicts a hi-pass filter was not clear, I should add one more point....
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".
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".
- Thread starter
- #9
Your assuming the drive is producing a 60 Hz waveform so 50th harmonic is 50 times 60 which is 3kHz. I havent had time to read all of electricpetes comments (printed out and read when have more time) and I am sure they are good ones. I will likely comment on them nest week. What assumptions were made in these theorhetical calculations as to the core of the CT? This obviously has something to do with the saturation level of the CT. Typical power CTs are made with tape wound steel of various variety, this is not the case with PT's. That is why standard PTs will not couple high frequency signals.
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
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
electricpete
Electrical
As far as the impact of core saturation on CT's abaility to transmit high-frequency, I believe there is no significant impact unless the magnitude of the high frequecy far exceeds the magnitude of the power frequency (saturation affects the ability to transmit low frequency much more than it affects the ability to transmit high frequency... for the same current magnitude, assuming single-frequency sinusoidal input).
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.
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.
Folks-- what started as an off-the-cuff shot-from-the-hip remark sorta got me in trouble, and I didn't expect that. I just sorta figured that 3kHz/50th and 3% were in the ballpark.
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.
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.
electricpete
Electrical
One more comment (more is better, right? ;-)... you said the ct acts like an integrator and I don't totally agree with that. My model above predicts
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.
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.
sparky1976
Electrical
Electricpete could you send me your post with simple drawings to my email ledig18@yahoo.com, I have similar problem on big motors that use VFD.I'm a slow learner I just can't understand by reading words but need drawings to explain it.
Thanks in advance
Pitat
Thanks in advance
Pitat
electricpete
Electrical
I can appreciate that the explanation I provided was not particularly easy to follow (a lot of rambling). Let me try again here starting with a circuit diagram for my model:
(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.
(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.
electricpete
Electrical
You can find a lot of useful info from the GE page
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
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.
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
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.
- Thread starter
- #17
electricpete
Electrical
One more explanation might make more sense (last one?):
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. (?)
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. (?)
- Thread starter
- #19
Yamin, Please keep us posted on the accuracy of these CT's on your system. Would you please let us know the mfg of the CT and part number? Also, have you done any comparisons with a TRMS clamp on meter (model number would be great here to please so I can look up stats) and the current displayed by the CT(whatever CT is hooked too)? Thanks Buzzp
Hi Guys, this has made for some interesting reading, and triggered something which I noticed recently. I have about 230 VFD's (525V, 50Hz)in my plant, and we are doing our own repairs and fault finding (not relying on supplier contracts etc..)
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!
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!
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