electrofelon
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- Cherry Valley NY, Seattle, WA
Right, there it is. I knew the 1.35 value but couldnt remember the algebraic form.
Single Phase current draw for a 3 phase output VFD: technical discussion
- Thread starter Jraef
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Right, there it is. I knew the 1.35 value but couldnt remember the algebraic form.
You are refering to a six-phase rectifier DC output voltage.
The formula is Vdc = Vm x 3/pi = Vrms x 1.414 x 3/pi = 1.35Vrms
I think you meant irrelevant... but it actually is relevant... because if (which is a big IF according to the info' I gleaned from light research) the output is sinusoidal and balanced, that means the output is linear... and input power (minus losses) follows the output power.
junkhound
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- Renton, WA
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- EE, power electronics specialty
Semantics:
Do not know of any semiconductor based VFD that is 'linear' in the classic definition.
e.G. - An old Ward Leonard VFD system is linear, present day FET/ IGBT/PWM VFDs all are non-linear.
Non-linear in the same way a class D audio amp is non-linear vs. class A, B,C or A/B being linear.
What 'looks' like a sine wave in VFDs or calss D audio, is actually a bunch of non-linear pulses.
That said, an analytical model can be constructed based on linear relationship in most cases.
Do not know of any semiconductor based VFD that is 'linear' in the classic definition.
e.G. - An old Ward Leonard VFD system is linear, present day FET/ IGBT/PWM VFDs all are non-linear.
Non-linear in the same way a class D audio amp is non-linear vs. class A, B,C or A/B being linear.
What 'looks' like a sine wave in VFDs or calss D audio, is actually a bunch of non-linear pulses.
That said, an analytical model can be constructed based on linear relationship in most cases.
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mike_kilroy
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- United States
69/26.4 = 2.6
actually 69/23 = 3.0 Remember 'service factor' is always 1.0 when motor is put on a vfd.
mike_kilroy
Senior Member
- Location
- United States
So I guess your answer to my earlier question is, "Never talk about the relationship of input current to motor current on a 3 phase VFD fed by a single phase source, because you have no way of knowing exactly what it will be since it is a non-linear system; it's unknowable." Is that right? :blink:
I think I'll just try to skirt the subject to avoid sounding pedantic (or more pedantic than I already sound to most electricians), and simply stick to telling them to double the current rating of the VFD to the motor FLA and leave it as a DC bus ripple issue, since that's a better reason anyway. Then when they ask me what the input current will be, I'll tell them I have no way of knowing; could be 1.5x, could be 2.43x, could be 1.732x, who knows...
You might also consider changing your replies when asked what the 3 phase input current is to a VFD. Since it is also the exact same harmonic rich non linear incalculable waveform as 1 phase input, just adding 2 more humps to the total. If 1 phase input current is an unknown, so must be the identical 3 phase current input. And certainly stop trying to relate power in to power out in either system.
gar
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- Ann Arbor, Michigan
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- EE
170519-0900 EDT
The original thread, now only in the history books, was about using a particular 3 phase VFD, motor, and fan (the whole system together, and without any major changes) in a new application where only 1 (single) phase was available.
The new application would seldom be loaded very much, when heavily loaded it is not known what the magnitude would be, but duration is possibly less than a minute.
Various suggestions were provided.
Clearly input power is greater than output power and I doubt that system efficiency is of any concern.
Almost certainly there is a DC bus with energy storage (capacitors).
The part of the system after the DC bus will perform exactly the same for both the 1 and 3 phase applications so long as DC ripple is not a problem.
We probably don't care what the AC input current looks like to the power company.
Thus, our concern probably relates to diodes and capacitors in the VFD as to how heavily we can load the drive.
I think the end result for the original poster was to try it. Without any real analysis the system will likely work for normal conditions. If when heavy loading is required, then is there any real penality to limiting the VFD to a level that does not exceed the VFD's capability?
The two threads have gone all over the place on comments without definitions and assumptions.
Various responders throw out constants and equations without really understanding their derivation, assumptions, and proper use.
For example a textbook equation for a 3 phase full wave bridge rectifier has the average DC output voltage as 1.35*E. E is defined as the effective voltage. In other words the RMS voltage of a sine wave of peak value, Vp = 1.414*E. Note: this DC average voltage is 1.35/1.414 = 0.955 of the peak voltage, and therefore a moderately low ripple without filtering.
The equation 1.35*E is the average DC voltage to a resistive load, no filtering. Add a large capacitive output filter and the equation is ideally 1.414*E. Ideal means 0 voltage drop in the rectifier. If the rectifier circuit has a large output capacitor, then the 1.35*E equation produces a value somewhat close, but is not the correct equation to use.
A Simpson 260 in AC mode, assuming no DC component, full wave rectifies the AC input and measures the average value. Meter deflection is approximately linear with this input. For a sine wave the full wave rectified average value is 0.636 of the sine wave peak. Thus, a factor of 0.707/0.636 = 1.112 exists between RMS and average. Simpson calibrates the meter scale so that in AC on a sine wave the value displayed equals the RMS value of the sine wave. For most other waveforms the displayed reading will be in error. This is another illustration of where adequate definitions, and assumptions need to be provided.
.
The original thread, now only in the history books, was about using a particular 3 phase VFD, motor, and fan (the whole system together, and without any major changes) in a new application where only 1 (single) phase was available.
The new application would seldom be loaded very much, when heavily loaded it is not known what the magnitude would be, but duration is possibly less than a minute.
Various suggestions were provided.
Clearly input power is greater than output power and I doubt that system efficiency is of any concern.
Almost certainly there is a DC bus with energy storage (capacitors).
The part of the system after the DC bus will perform exactly the same for both the 1 and 3 phase applications so long as DC ripple is not a problem.
We probably don't care what the AC input current looks like to the power company.
Thus, our concern probably relates to diodes and capacitors in the VFD as to how heavily we can load the drive.
I think the end result for the original poster was to try it. Without any real analysis the system will likely work for normal conditions. If when heavy loading is required, then is there any real penality to limiting the VFD to a level that does not exceed the VFD's capability?
The two threads have gone all over the place on comments without definitions and assumptions.
Various responders throw out constants and equations without really understanding their derivation, assumptions, and proper use.
For example a textbook equation for a 3 phase full wave bridge rectifier has the average DC output voltage as 1.35*E. E is defined as the effective voltage. In other words the RMS voltage of a sine wave of peak value, Vp = 1.414*E. Note: this DC average voltage is 1.35/1.414 = 0.955 of the peak voltage, and therefore a moderately low ripple without filtering.
The equation 1.35*E is the average DC voltage to a resistive load, no filtering. Add a large capacitive output filter and the equation is ideally 1.414*E. Ideal means 0 voltage drop in the rectifier. If the rectifier circuit has a large output capacitor, then the 1.35*E equation produces a value somewhat close, but is not the correct equation to use.
A Simpson 260 in AC mode, assuming no DC component, full wave rectifies the AC input and measures the average value. Meter deflection is approximately linear with this input. For a sine wave the full wave rectified average value is 0.636 of the sine wave peak. Thus, a factor of 0.707/0.636 = 1.112 exists between RMS and average. Simpson calibrates the meter scale so that in AC on a sine wave the value displayed equals the RMS value of the sine wave. For most other waveforms the displayed reading will be in error. This is another illustration of where adequate definitions, and assumptions need to be provided.
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gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170519-1524 EDT
First, to add to my previous post.
If the 3 phase rectifier was a 3 pulse (half wave so to speak) instead of 6 pulse, then the calculated DC average voltage is 1.17*E with a resistive load. But change to a capacitor input filter on the rectifier output with low ripple and the equation has to become 1.414*E. The same as a 6 pulse loaded with a capacitor input filter.
Second, a rough experiment on a DC power supply relative to diode power dissipation vs load current.
This power supply consists of a Stancor RT transformer with a center tapped secondary feeding two stud mounted diodes on a heat sink connected to a capacitor input filter. Two 5 ohm load resistor were use, one alone, and then the two in parallel. The resistors are of nominal tolerancre, probably 10%. The capacitor is 53,000 ufd.
Diode temperature rise was measured via a Fluke IR meter looking at the end of the diode stud. This was the hottest available location and should have good thermal conductivity to the diode chip.
DC load current was 3.03 or 6.09 A. Ambient temperature was 69 F.
At 3.03 A the stud temperature rise was (81 - 69) = 12 F.
At 6.09 A the stud temoerature rise was (98 - 69) = 29 F.
The rise ratio is somewhat above 2 (2.4), but nothing like 4 to 1 as an I^2 would predict. We know there is a series resistive component to a diode in addition to its fixed voltage component. Thus, the result is expected. This is a simplistic analysis but useful.
AC ripple voltage on the DC at 6 A was about 0.17 V with DC being 16 V.
.
First, to add to my previous post.
If the 3 phase rectifier was a 3 pulse (half wave so to speak) instead of 6 pulse, then the calculated DC average voltage is 1.17*E with a resistive load. But change to a capacitor input filter on the rectifier output with low ripple and the equation has to become 1.414*E. The same as a 6 pulse loaded with a capacitor input filter.
Second, a rough experiment on a DC power supply relative to diode power dissipation vs load current.
This power supply consists of a Stancor RT transformer with a center tapped secondary feeding two stud mounted diodes on a heat sink connected to a capacitor input filter. Two 5 ohm load resistor were use, one alone, and then the two in parallel. The resistors are of nominal tolerancre, probably 10%. The capacitor is 53,000 ufd.
Diode temperature rise was measured via a Fluke IR meter looking at the end of the diode stud. This was the hottest available location and should have good thermal conductivity to the diode chip.
DC load current was 3.03 or 6.09 A. Ambient temperature was 69 F.
At 3.03 A the stud temperature rise was (81 - 69) = 12 F.
At 6.09 A the stud temoerature rise was (98 - 69) = 29 F.
The rise ratio is somewhat above 2 (2.4), but nothing like 4 to 1 as an I^2 would predict. We know there is a series resistive component to a diode in addition to its fixed voltage component. Thus, the result is expected. This is a simplistic analysis but useful.
AC ripple voltage on the DC at 6 A was about 0.17 V with DC being 16 V.
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Power in = power out less losses
conservation of energy
and the output fundemental better contain >95% of the power or the motor would run at an indeterminate speed
Generally the output current is a close approximation to a sinewave because of the chopping frequency (typically a few kHz) and the smoothing effect of load inductance.
But that isn't what the topic is about.
You said the same about my post that basically said the same thing as Ingenieur's. Perhaps you should read the thread title. It appears you cannot make the correlation between output and input currents.
Correct. Unless you know component values and do the appropriate calculations. There is no one size fits all magic conversion factor.
Do we not know for a fact the RMS input current cannot be less than the square root of 3 times the RMS output current?
mivey
Senior Member
Jraef's premise in this thread is that the diode current is equal sqrt(3) times output current.Do we not know for a fact the RMS input current cannot be less than the square root of 3 times the RMS output current?
mivey
Senior Member
I did not see the other thread before it got gutted but did you show an example calc where the input current does not equal output current times sqrt(3)?
Yes.Do we not know for a fact the RMS input current cannot be less than the square root of 3 times the RMS output current?
For both you and Mivey.
Simple example.
On no load, a typical cage induction motor runs at about 30% FLC. Say 30A for a 55kW, 400V motor. The input current in such circumstances is very much lower. Certainly under 10A.
Diode current? I thought it was input current.
I thought we were talking about sizing a VFD, i.e. its input current rating so as to effectively power a motor up to FLA. I think input current when the motor is not fully loaded is trivial in that pursuit.
I was simply giving an example, one on the limits, of why you can't use a fixed conversion factor.
For many, if not most three phase drives, you won't be able to get maximum rated output current if your supply is single phase.
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