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As my late great friend Tommy used to say:
"The man who never made a mistake never made anything."
Single Phase current draw for a 3 phase output VFD: technical discussion
- Thread starter Jraef
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As my late great friend Tommy used to say:
"The man who never made a mistake never made anything."
Is it so in your case also (post#267)?
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It's the same is the vast majoity of cases.
gar
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- Ann Arbor, Michigan
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- EE
270616-0744 EDT
Smart $:
I have partially read the article you mentioned. At the beginning there are some very good statements. Later not so good. The simplified diagram shows an input inductor. There is no definition of its value, or waveforms of the current into the VFD. Some of the equations being used are ones that require a sine wave. There are undefined assumptions.
In prior posts I have provided some information from a real DC supply, including an input current waveform very peaked, meaning the RMS to average ratio is high. To some extent I provided a description of the components.
Unless your input waveforms are sine waves, and with zero phase displacement, you won't have or be close to unity power factor at the input.
For some specific VFD show us your measured input voltage and current waveforms at light load (friction and windage), and at full load.
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Smart $:
I have partially read the article you mentioned. At the beginning there are some very good statements. Later not so good. The simplified diagram shows an input inductor. There is no definition of its value, or waveforms of the current into the VFD. Some of the equations being used are ones that require a sine wave. There are undefined assumptions.
In prior posts I have provided some information from a real DC supply, including an input current waveform very peaked, meaning the RMS to average ratio is high. To some extent I provided a description of the components.
Unless your input waveforms are sine waves, and with zero phase displacement, you won't have or be close to unity power factor at the input.
For some specific VFD show us your measured input voltage and current waveforms at light load (friction and windage), and at full load.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170617-1336 EDT
Smart $:
If you have no equipment and experiments to prove what you say, then how can you back up your statements?
I doubt that you have an educational background in electrical engineering. This probably means you have not studied differential and integral calculus, differential equations, AC and DC linear circuit analysis, AC and DC machinery, magnetic circuits, semiconductor devices, and instrumentation. You also may not have run a lot of experiments in these areas.
Making the statement that a black box (in the case discussed earlier, just diodes) with no reactive components will reflect the power factor of the load back to the input of the black box makes no general sense. An ideal transformer would reflect the PF, but any linear element, resistive or reactive, or any nonlinear element, would not reflect the PF.
When you do not understand how something works (details), then it is not very feasible to predict how that something works under various conditions.
I have created some more experiments to illustrate some points. One shows the input current to a HAAS CNC mill vector drive, and it is nothing like a sine wave that it would need to be to obtain a good power factor.
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Smart $:
If you have no equipment and experiments to prove what you say, then how can you back up your statements?
I doubt that you have an educational background in electrical engineering. This probably means you have not studied differential and integral calculus, differential equations, AC and DC linear circuit analysis, AC and DC machinery, magnetic circuits, semiconductor devices, and instrumentation. You also may not have run a lot of experiments in these areas.
Making the statement that a black box (in the case discussed earlier, just diodes) with no reactive components will reflect the power factor of the load back to the input of the black box makes no general sense. An ideal transformer would reflect the PF, but any linear element, resistive or reactive, or any nonlinear element, would not reflect the PF.
When you do not understand how something works (details), then it is not very feasible to predict how that something works under various conditions.
I have created some more experiments to illustrate some points. One shows the input current to a HAAS CNC mill vector drive, and it is nothing like a sine wave that it would need to be to obtain a good power factor.
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Well, there you go. I guess you either have to take my word for it... or not. It seems you are not. So I guess we're at an impasse.
Hey! At my point in life, the only people that have educational backgrounds are people that stayed in the educational system for a lot longer than they were there as a student. The last time I saw the inside of a classroom as a student in that system was 40 years ago. Our culture was barely into the IC chip and the campus computer was a mainframe with data terminals that took up an entire building. My field was electronics.
From there I could tell you a story about how life off campus takes our knowledge and experiences down different paths... but instead, I'll shortcut it by saying all that matters is I'm here having a discussion with you about concepts that aren't seen matter of factly by what? ...say 99.99% of people out there. What does that mean to you?
I have another black box. It improves the power factor of a motor load to near 1 (assuming it was chosen wisely
) with insignificant distortion as measured at the input. All my black box contains is a capacitor bank.What device other than a capacitor can improve power factor locally? If we discuss it conceptually, a power source or storage device that can reflect the reactive current of the motor. VFD's have none other than its filter caps. Take away those filter caps and the reactive current of the motor has to be handled somewhere. That's why power companies charge some customers extra when they "use" too much.
I've never really thought of power factor as requiring a pure sine wave to achieve a 1. Considering the voltage is sinusoidal it is a common error to think the current must also be sinusoidal to get a power factor of 1. I've only ever thought of it as a number which corrects for V*A appearing as more power than is actually being used.
Anyway... I have other matters of life to attend to now, but I'm around to continue the discussion however it unfolds...
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- Retired PV System Designer
If you do the math, if the voltage is a sine wave, it is very difficult to get a power factor of 1 without having the current be a sine wave also.
Basically if (square root of the integral of current squared) times (square root of the integral of voltage squared) is equal to (the integral of current times a sine wave voltage) it is very difficult accomplish without the current also being sinusoidal.
To some extent you can get that result if the current is a chopped waveform which smooths out to a sine wave, but if the current on a large time scale is very different from a sine wave it is much less likely.
*
*More appropriately, I should have said "squared" wave given the sinusoidal voltage.
Both the voltage and current are integrated over a period of time. As long as the actual power dissipated through that period equals the multiplied values we assign the V and A for that period, we have a power factor of 1. Can a square waveform current with a sinusoidal voltage source exhibit a power factor of 1? Think about it. You already stated a chopped waveform can be smoothed out to a sine wave. With a square wave*, it is simply a matter of "pulse" timing and magnitude.
*More appropriately, I should have said "squared" wave given the sinusoidal voltage.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
Given that the voltage waveform will remain a sine wave, it is possible to have a squared off current waveform that is tuned to give you a power factor of 1. But that implies active control of the current, not just settling for the peaks-only current that comes naturally for a rectifier with only capacitive reactance in the picture.
Looking at it another way, the simple proof that the heating effect (transferred power) will be equal to the product of the RMS current and the RMS voltage relies on the fact that the current and voltage waveforms "match".
The example of displacement power factor proves that you cannot assert that equality in general. And in particular, even if the two are in phase the integral of the product will not necessarily be the product of the RMS values.
But I'm not saying it has to be a squared off wave. Just providing a [theoretical] example of it not having to be a sine wave.
Anyway, you're getting a bit off point. It's as simple as this. The motor is inductive (maybe the categorization of induction motor gave it away
). It relies [almost ] entirely on back EMF to convert electrical energy into mechanical energy. That back emf, of which the reactive current is a product, must be handled by the system somewhere. Are you saying active components can do this?A "converter" with active electronics processing can. I do it on a regular basis with the amplifiers in my vehicles... the inductive motors being acoustic drivers.
That does not answer the question
vfd with dc front end removed (rectifier, caps, etc)
only the voltage sourced inverter (and control section) remain
a bank of batteries supply the inverter
will it drive an induction motor?
I don't know for a fact. I theorize it is possible.
After giving it further thought, my reply was only considering a VFD. I know for a fact an "inverter" with a battery bank as a power supply can drive an induction motor. I installed one as part of the power backup system in a local power plant.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
The battery DC supply did the trick.
If you have an AC input and no reactive energy storage components inside the black box I do not see any way that you can source current into the motor at a time when the output voltage is at zero and still maintain a PF of 1 on the input.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170618-0724 EDT
Smart $:
I would like to know if you have studied differential and integral calculus, and trigonometry? Trig almost certainly, and calculus probably not. This information would help us talk to you.
To everyone --- I have a pet gripe about this 1.732 thing. 1732 means one of two things to me --- the birth year of George Washington (high school physics or trig was the source), or an approximation for the square-root of 3 or something relating to it such as 0.866 . Using something with the precision of about 1 part in 2000 when what is really meant is about 2, or 1 or 1.8, or 1.5 makes no sense and conveys the wrong impression that there is a connection back thru some equations that there is a square-root of 3 or 30 or 60 degree involvement.
Smart $ my last year in college was 1960, and by that point I had completed 60 semester type credit hours of graduate school classes. I have not taught any courses. I am not an academic.
On the forest and tree thing --- you can not know much about a forest unless you have extensive knowledge of the trees and other things in that forest. I try to provide details, assumptions, and other items to support comments I make in posts.
I am on a quest to try to determine if the fundamental definition of power factor is really
Power Factor = Real Power / Vrms*Irms,
the date of invention, and the inventor. I don't know when I will get or find an answer. Knowing the basic definition is critical to understanding how other equations evolved. Note: this definition applies only to a two terminal load. This equation requires no assumptions other than the two terminal requirement, and that is self evident from the equation. Any other power factor equations require assumptions, and these assumptions may not be provided.
.
Smart $:
I would like to know if you have studied differential and integral calculus, and trigonometry? Trig almost certainly, and calculus probably not. This information would help us talk to you.
To everyone --- I have a pet gripe about this 1.732 thing. 1732 means one of two things to me --- the birth year of George Washington (high school physics or trig was the source), or an approximation for the square-root of 3 or something relating to it such as 0.866 . Using something with the precision of about 1 part in 2000 when what is really meant is about 2, or 1 or 1.8, or 1.5 makes no sense and conveys the wrong impression that there is a connection back thru some equations that there is a square-root of 3 or 30 or 60 degree involvement.
Smart $ my last year in college was 1960, and by that point I had completed 60 semester type credit hours of graduate school classes. I have not taught any courses. I am not an academic.
On the forest and tree thing --- you can not know much about a forest unless you have extensive knowledge of the trees and other things in that forest. I try to provide details, assumptions, and other items to support comments I make in posts.
I am on a quest to try to determine if the fundamental definition of power factor is really
Power Factor = Real Power / Vrms*Irms,
the date of invention, and the inventor. I don't know when I will get or find an answer. Knowing the basic definition is critical to understanding how other equations evolved. Note: this definition applies only to a two terminal load. This equation requires no assumptions other than the two terminal requirement, and that is self evident from the equation. Any other power factor equations require assumptions, and these assumptions may not be provided.
.
To assist in your quest, I want to remark technology is often ahead of science and AC theory may not be an exception and so if you study the discovery of AC machinery and consequential events associated with it, you could find the origin of power factor. As for your equation for power factor, it holds good for linear loads, but not for non linear loads.
I had both trig and calculus in high school. I had to take calculus again in college because the high school course was not accredited.
Gripe noted. I doubt much else will transpire.
I'm not sure whether to provide praise or offer sympathy. Apparently, you view that as a positive note so I'll say good for you.
I'm certainly not the one to say you shouldn't look at it that way. All I'm saying for our current purpose a tree is just a tree. We do not need to know its genus. And while you may want to stroll through the forest, I want to fly over it.
Please keep us updated on your progress.
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