Thread: Single Phase current draw for a 3 phase output VFD: technical discussion

1. Single Phase current draw for a 3 phase output VFD: technical discussion

In the interest of fairness, I am opening a new thread here based on side band discussions that I had removed from a previous thread as not being germane to the OP's question, and I have closed that thread because the OP's question was already answered before this tangent ensued.

So as a service to valued members who think I am wrong, I am allowing them to show me how that is, because I have been teaching this very issue for 10+ years and if I am indeed wrong, I need to change my thinking.

So here is my stated premise on a subject that is basically unique to phase conversion equipment.
Given: Voltage is equal (230V for argument sake, because that is primarily where this takes place). Motor is 3 phase, lets say the load is 10kW to try to make things easier with nice round numbers. Input to the VFD is 230V single phase, output of the VFD is 230V 3 phase. VFD input power factor is .95

My premise: Input current TO THE VFD is equal to the Output current to the motor x 1.732, because of the following well established and published formulae:

Three phase; I = W/V x PF x 1.732 ergo I = 10,000/230 x .95 x 1.732 = 26.42A

Single phase; I = W/V x PF ergo I = 10,000/230 x .95 = 45.76A

45.76/26.42 = 1.732

So tell me where and why this does not apply.
Last edited by Jraef; 05-16-17 at 02:42 PM.

2. Originally Posted by Jraef
In the interest of fairness, I am opening a new thread here based on side band discussions that I had removed from a previous thread as not being germane to the OP's question, and I have closed that thread because the OP's question was already answered before this tangent ensued.

So as a service to valued members who think I am wrong, I am allowing them to show me how that is, because I have been teaching this very issue for 10+ years and if I am indeed wrong, I need to change my thinking.

So here is my stated premise on a subject that is basically unique to phase conversion equipment.
Given: Voltage is equal (230V for argument sake, because that is primarily where this takes place). Motor is 3 phase, lets say the load is 10kW to try to make things easier with nice round numbers. Input to the VFD is 230V single phase, output of the VFD is 230V 3 phase. VFD input power factor is .95

My premise: Input current TO THE VFD is equal to the Output current to the motor x 1.732, because of the following well established and published formulae:

Three phase; I = W/V x PF x 1.732 ergo I = 10,000/230 x .95 x 1.732 = 26.42A

Single phase; I = W/V x PF ergo I = 10,000/230 x .95 = 45.76A

45.76/26.42 = 1.732

So tell me where and why this does not apply.
Very simple. You are using linear equations.
They are not applicable in non-linear systems.

And thank you for resurrecting a topic that some mod decided to decimate.
Life moves on.

Or, to quote the Rubaiyat of Omar Khayyám,
The moving finger writes
Nor all thy Piety nor Wit.
Shall lure it back to cancel half a Line,
Nor all thy Tears wash out a Word of it.
Last edited by Besoeker; 05-16-17 at 03:36 PM.

3. gar
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Jraef:

The original thread's primary question was about using a three phase VFD on a single phase source. You correctly pointed out that derating was required because of possible problems with diode capacity, and separately output ripple on the DC bus because of capacitor size.

Fundamentally the question is how heavily can the drive be loaded when fed from a single phase source. The number 1.732 has nothing to do with this question.

A lot has to do with diode current in the VFD as designed. Also tolerable ripple voltage on the DC bus.

Most likely without any study of the VFD it can support an output load 50% of its three phase rating when supplied from single phase of the same nominal voltage. I believe you essentially said that. How much more can it be loaded without problems?

Certainly one needs to consider actual diode current and its relationship to diode characteristics, such as http://www.vishay.com/docs/93516/vs-50pfrwseries.pdf , and heat sinking, and the actual ambient temperature the drive is subjected to.

The diode current is not sinusoidal as someone suggested. Following is an example. Its total on time is about 2.5 mS from a period of 8.33 mS. The DC load was 5 ohms across a 53,000 ufd capacitor. That time constant is about 0.25 seconds, and pulsed every 8.3 mS.

4. I understand the non-linearity of how diodes conduct. How does that change the input current equation? Is it because the diodes conduct for longer on a single phase source rather than pulling current at a higher amplitude? If so, how is that possible? The diodes are the same, the forward conduction voltage is the same. ALL of the energy going to that bus must come through the diodes, there will only be 4 diodes feeding the DC bus from a single phase source. How does the current flow through those individual devices not change?

5. Originally Posted by Jraef
I understand the non-linearity of how diodes conduct. How does that change the input current equation? Is it because the diodes conduct for longer on a single phase source rather than pulling current at a higher amplitude? If so, how is that possible? The diodes are the same, the forward conduction voltage is the same. ALL of the energy going to that bus must come through the diodes, there will only be 4 diodes feeding the DC bus from a single phase source. How does the current flow through those individual devices not change?
As I posted before there is no simple answer. The current waveform depends on circuit component values. Bottom range VSDs may have no DC choke and little line inductance. Currents can be very peaky - I did present a waveform to show that. Some have a DC choke which makes the current longer period and less peaky.

The upshot is that the input Irms is different even if the output current is the same.
That's why you can't apply a fixed muliplier.

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Originally Posted by Besoeker
Very simple. You are using linear equations.
They are not applicable in non-linear systems.
But if we assume the VFD is 95% efficient (which most respectable VFD's would be), power out is still going to be pretty close to the power in. Even in a non-linear circuit.

And if we have an input inductor on the VFD, won't the input current be pretty close to a sine wave, instead of the capacitor input waveform Gar posted?

I would expect Jraef's equations to be pretty good approximations for practical purposes. And even better approximations if we account for those VFD losses.

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I'm not smart ... or stupid ... enough to disagree with your results, but input PF probably is near 0.95 with single phase, I really don't know. But the motor side won't be that high I don't believe. But input I(rms) won't have a good correlation to PEAK diode heating with the periodic non-sinusoidal current flow.

That said, rules of thumb are often useful, and whether the real number is 1.414, 1.5, 1.732, or 2 ... or something somewhere between, doesn't negate its usefulness.

Originally Posted by Jraef
VFD input power factor is .95

Three phase; I = W/V x PF x 1.732 ergo I = 10,000/230 x .95 x 1.732 = 26.42A

Single phase; I = W/V x PF ergo I = 10,000/230 x .95 = 45.76A

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Won't Diode heating be function if Irms? Is not rms the heating amount as opposed to Iaverage?

since Irms is the heating equiv to dc, then the real power in should be same as real power out, assuming PF=1 and eff=10%, which is fine assumptions for this discussion I think...

Thus Jraef equations should be 100% accurate.

I think some are confusing avg and rms?

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9. gar
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The issue from the earlier and now closed thread was could a particular three phase input VFD, motor, and fan be used in a different application with single phase power input.

The real question boils down to --- how much does the system have to be derated and does that provide sufficient power to satisfy whatever the new requirements may be? I don't believe this was ever really answered other than to try it.

The amount of derating is not determined from any of the constants 1.732, 3.1416, 0.866, 0.707, 1.414, 1.0. 0.636, 0.318, 0.5, 3.0, or many others. Nor do we care what the single phase power input is for the power range indicated.

Will the proposed VFD do the new job is the question?

.

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That could have been the question though. But for purposes of simplifying, people live using rule of thumbs or plain derating factors/multipliers. 1.732, 2X. . .

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