How do rotary converters work?

Well, the title about says it all. I have some electrical background but have not yet found a concise explanation of how a 3 phase motor can create 3 phase power from a single phase input. Lots of stuff out there on how to make a converter though. Can anyone provide an understandable explanation of how this principle works without resorting to calculus or worse? Thanks.

Randy in Everett

This entire forum of Practical Machinist basically discusses how to build or buy and use rotary phase converters and VFDs (variable frequency drives).
You might want to view a schematic I put together for a 30 hp rotary converter on that forum. It is under the thread called "Integrated panel rotary phase converter design", and the document can be viewed by clicking here:

http://www.practicalmachinist.com/phaseconverter.pdf

I'll not go through the whole process of designing a converter, but here's brief synopsis of how they work:

Basically, a rotary phase converter consists of taking a three phase induction motor and running it on single phase power using two legs, and then taking the third leg for three phase off of the running motor stator windings. You end up with the two 'live' legs from single phase 240 volt supply, and the third or 'manufactured' leg comes from the motor windings, where the third leg is generated in the motor windings by magnetic action of the rotating rotor. A rotary phase converter is quite literally a rotating transformer.

You then take these three wires and connect them to a three phase machine motor, which will start and run just as on regular three phase from the power company.

The biggest job in building a rotary converter is getting the converter motor (called the 'idler') started. A three phase motor will not start when supplied with single phase; you must either get the shaft up to speed mechanically, or use a starting capacitor for a second or two during starting. Once running, a three phase idler motor will run on single phase, and generate three phase in the process. Since the idler is not loaded mechanically, the energy comes out of it in the form of amps on the generated third leg at 240 volts.

The final stages of building a rotary converter involves wiring run capacitors between one or both of the single phase legs and the manufactured leg. This supplies reactive power to that leg, which increases its voltage and also improves power factor.

Hi There,

Does your system provide true, balanced 3 phase output?

Just Wondering,
Webb

The diagram referred to above does not produce the most balanced output. To get that, run capacitors can be installed from the other single phase leg to the generated leg. If that is done, the load machines will sense true balanced output.

Unfortunately, it's still not as good as power company three phase power. It's similar to being supplied three phase with an 'open delta' connection, in which the final voltage regulation between phases is dependent on the load. That's about as good as a rotary converter can do.

However, by oversizing the idler motor with respect to the load, the above objection is markedly reduced, so that if you are running a 5 hp lathe with a 10 hp rotary idler, the lathe probably cannot tell the difference between converter three phase and utility three phase.

Hi B. Nelson,

Thanks for your reply. What do you know about power factors? If the rota-phase is tuned to 2 HP mill and then you switch to a 1 HP lathe, won't the power factor go down and a loss of efficiency?

I don't see how one can run a motor at full load on a rota-phase without it eventually overheating (unless all three legs are balanced and in proper 120 degees phasing). And if one has a shop where the number of machines running vary and the HP of those machines vary, how can one tune it to be efficient?

Regards,
Webb

Usually a lightly loaded rotary converter will run at a lower power factor (and hence less efficiency) than a fully loaded converter.

Fortunately it's both easy and cheap to correct power factor with additional capacitors. Although there are methods to calculate the angle between the lagging current and the voltage and compute the capacitance in either mfd or (usually with motors), in kvars, I don't go to the trouble to do this. Instead I add run caps between both the A and B legs, and the B and C legs to shoot for good current balance among all three phase legs. If the single phase supply current on the A and C phases is still too high compared to what the motor should be drawing, then I'll add power factor correction capacitors between legs A and C.

I almost never have to do this when building a balanced rotary converter, where there are run caps from both the A and C phases to the B phase.

Your last comment is a very good one as it raises the issue of the tradeoff between running your converter at full load (with good power factor and poorer voltage/current regulation), versus running it at partial load (with good voltage/current regulation but poorer power factor). You can't have it both ways.

My solution to this tradeoff in some shops which have widely varying loads is to a) use dual rotary idlers, and switch one on or off to more properly match the load being run; or b) build an UNbalanced rotary converter, and add the balancing run caps at the LOAD motors. I will admit this is a highly unorthodox practice, but it works well to provide for better regulation with varying machine loads.

Hi B. Nelson,

Thanks for the reply. That more or less jives with what I understand about rota-phase converters. But I am still unclear as to what kind of phase angles do you get out of rota-phase. I don't own a oscilloscope to check this myself but I have checked phase-to-phase voltages and phase to neutral voltages. I find I can get pretty good phase-to-phase voltages but the phase-to-neutral voltage will be high on the generated leg (I like to call L3 and I believe you refer to as "B").

I just seems that unless the three phases are properly spaced, there is going to be degradation of torque and the magnetic flux in the motor will be unbalanced and some of the power will be wasted; creating heat with reduced or little benefit to the output of the motor. so, even though the motor is using its full load amperes, will you get full HP output? I have trouble understanding that one. Any insights?

Regards,
Webb

Balancing the phase-to-phase voltages to within 10 percent (better yet, to within 5 percent for CNC applications) is possible.

Using a "clamp meter" on the idler motor to ensure that the phase currents are all within the nameplate rating is a very good idea.

With an unbalanced converter, the highest idler motor phase current is most likely going to be that which is associated with the manufactured phase, and that is really what limits the usefulness of an unbalanced converter.

With a balanced converter you can achieve ... by experiment ... voltage and current balance. And that should be the goal, if at all possible.

I completely agree with Peter's statement. That's exactly how I 'tweak' a balanced design converter when finished building it, adding or subtracting run caps to get current balance.

It can be shown using phasor algebra that with an induction motor if the voltage imbalance goes much over 10%, the current imbalance can approach 100%. This means the current in one line is zero, so you could actually disconnect that leg from the running motor and not affect its mechanical output at all!

For something like a big submersible pump running continuously heavily loaded, it's a great way to burn up the motor. Fortunately for lathes and mills, the situation is not quite so critical and they can tolerate voltage imbalance of 10% without burning out.

The line to neutral voltages are really not useful in balancing a converter. Of course, the high leg ('B' phase or L3 manufactured leg) will always be at a higher voltage than the other two legs, by a factor of 1.732.

Hi Guys,

Let me see if I understand. We can tune the rota-phase with cap.s until we get a good phase-to-phase voltage balance and a good phase amperes balance at load. If we accept that the phase-to-neutral voltage of the generated leg (B phase or L3) is going to be higher than the other two by a factor of the square root of 3; then the phase angles of the three phases cannot be equally spaced (120 degrees apart).

Now the poles inside a 3 phase motor are spaced for a 3 phase supply that has its phase angles equally spaced at 120 degrees. And we are supplying that motor with a three phase power that does not have its phase angles equally spaced.

I understand that the motor will run and may draw its nameplate full load amperes. But because the phase angles are not equally spaced, won't there be some loss of output HP and/or torque? Won't there be some excess heat generated by the flux imbalace (or flux timing imbalance if you prefer). If not, why not?

Thanks for your information!
Regards,
Webb

Your reasoning would be correct if you were dealing with a grounded wye configuration, where the voltage to neutral (or ground) was equally 120 volts.

When you build an ordinary rotary phase converter which operates on ordinary household power in the USA, that converter is being supplied single phase with 240 volts, which comes from a center-tapped single phase transformer where the neutral (center tap) is not only grounded, but supplies 120 volts on either side for your ordinary household receptacles.

When you hook up your idler, your 'manufactured' leg or 'B' phase is connected delta with respect to the two single phase 'hot' legs ('A' and 'C' phases) which supply the idler.

This asymmetrical geometry of the connection, which is called a 'high leg delta' connection, has absolutely NOTHING to do with any imbalance between either the voltages or the currents between the A, B, and C phases of the converter output. It ONLY has to do with the voltages of the phase legs to the NEUTRAL. Your three phase machines do not use the neutral. There is no problem. Phases A, B, and C are spaced equally at 120 degrees from each other in the high leg delta connection.

The phase-to-phase angles *could* vary except for the fact that the magnetic poles impressed upon the rotor of the induction machine are in a 120-120-120 degree electrical relationship because the A and C (and B) windings are always in a 120-120-120 degree mechanical relationship.

Consequently, the B phase, the manufactured phase, will also be in a 120 degree relationship to the other, non-manufactured phases.


[This message has been edited by peterh5322 (edited 05-21-2003).]

Hi Guys,

Thanks again for the reply. I appreciate the mentoring. I've done some research on high leg delta power systems and I think I understand better what is going on. But I am still confused about one thing. If the input supply from L1 and L2 is single phase with a phase angle of 180 degrees, and it is connected to the idler motor and all the other motors in the 3 phase system when on. How come there isn't a conflict between the 180 degree (L1 to L2)supply and the 120 degree (L1 to L2) 3 phase system? I would think some sort of isolation would be a good idea. Any comments?

Thanks Again!
Webb

That's the magic of what the idler motor does. By virtue of the 120 degree winding separation described above by Peter, the rotor induces current in each of the three windings, EACH PHASE of which is 120 electrical degrees separated from the other. There are THREE CIRCUITS.

When you say the A-neutral-C phases are 180 degrees apart, you are referring to TWO CIRCUITS, A-neutral, and neutral-C. As confusing as it my seem, the A-neutral-C phases are STILL 180 degrees apart even when the A-C phases are part of an A-B-C high leg delta connection!!!

Not meaning to be difficult or 'smart', but three phase electricity IS a learning challenge without doubt.

All,

One would have to draw a complete "phasor" diagram, showing both the phasors representing the *actual* inputs, L1 and L2 and also the *effective* outputs, A, C and B.

L1-N-L2 is the 120/240 volt single-phase input system.

(Remember that A is really L1, while C is really L2 while N is "ground").

Now, where the magic of vector mathematics, and the representation and the behavior of phasors comes in is what happens with B.

There is a secondary three-phase system, which is superimposed onto the primary single-phase system.

This secondary system consists of A (identical to L1, remember), C (identical to L2, remember), N' (to be described later), and B.

This secondary system is a 138/240 Wye three-phase system.

N' is related to N as follows:

1) the phasor N'-N is orthogonal (90 degrees) to the single-phase system, and

2) the magnitude of N' - N is 0.289 "per unit", while

3) the magnitudes of A - N', C - N', and B - N' are each 0.578 "per unit".

Since the L1-L2 voltage is 240 volts (N is not actually used in rotary converters), A - N' = C - N' = B - N' = 138 volts, while N' - N is 69 volts.

So, the manufactured three-phase system is really a 138/240 Wye system with a "virtual neutral" (which is not utilized nor is it usually accessible) which is 69 volts above "ground".

The load motor can't tell the difference between a 138/240 Wye system and it's equivalent 240 Delta system, so the load motor believes and behaves as if it was being supplied by a balanced 240 volt three-phase Delta system consisting solely of A, C, and B.

(There are some advanced techniques which can make use of N' for debugging purposes only. N' is at the T7-T8-T9 point within the idler motor).

Peter.



[This message has been edited by peterh5322 (edited 05-24-2003).]