Question about single phase to three phase conversion

[SunnyBoyNY]

But each of the two capacitors will have differing values, is that not enough to produce three phase angles?

First, if the two capacitors are connected to a DC link (which is a voltage source) then they are slaved to it. Voltage on them would not have any phase shift. There would need to be an inductive element between the voltage source and these capacitors to produce phase shift.

Also, the phase shift would appear only at a certain frequency - so you would need to pull current at a certain frequency to have the two cap voltages phase shifted 120d to each other.

At zero frequency the caps would resonate with their phase inductors. Provided non-zero resistance, it's a second order damped system. When the initial ringing dies out, both caps will have the same DC voltage as the source.

 

[Bararontok]

First, if the two capacitors are connected to a DC link (which is a voltage source) then they are slaved to it. Voltage on them would not have any phase shift. There would need to be an inductive element between the voltage source and these capacitors to produce phase shift.

Also, the phase shift would appear only at a certain frequency - so you would need to pull current at a certain frequency to have the two cap voltages phase shifted 120d to each other.

At zero frequency the caps would resonate with their phase inductors. Provided non-zero resistance, it's a second order damped system. When the initial ringing dies out, both caps will have the same DC voltage as the source.

The source is a 1-phase AC source, not a DC source.

 

[NascentOxygen]

No, the intention was to use the device to convert 1-phase to 3-phase current. The supply would be split into 3 sets of terminals, 1 with no capacitor and the other 2 with capacitors of different ratings to give each one a different phase angle.

A capacitor in series with the mains is a dodgy arrangement that allows a motor to see a phase with a leading angle, but that won't be anywhere near 120° as far as I can see, though I'm no expert. It relies on a characteristic of the motor winding. If you use a different motor, you'll need a different capacitor or performance will suffer even more.

You won't generate 120° lead just by adding a series capacitor to your single phase supply. It may be possible to generate 120° lead and lag using a more complex passive network, carefully designed, but it will fall apart as soon as you try to draw anything but miniscule current from it. You may be able to power a thimble-sized miniature model 3ɸ motor as a demonstration, but nothing of any use, is my thinking.

If it was as simple as you picture it, then there would be no need for the big, heavy duty complex circuits that are employed to do the task.

 

[Bararontok]

The capacitors of the design are in parallel to the load. Though it may be correct that the capacitor would have to have a different value for different motors but the device can have a wattage rating label placed on it to ensure that the supply is not overloaded since other types of power supplies already have these labels anyway.

 

[NascentOxygen]

The capacitors of the design are in parallel to the load.

If the mains see the capacitors in parallel with the load then those capacitors can do nothing more than PF correction. The 1ɸ mains has to see them in series with something to achieve a phase lead.

 

[SunnyBoyNY]

The source is a 1-phase AC source, not a DC source.

I assume you will rectify the single phase AC source to DC and put an energy buffer to the link as 1 phase connection is not capable of delivering continuous power as the three phases are.

Simply, if you motor takes 3 kW, then it is 3 kW continuous. Torque is dependent on the rotor/stator flux linkage and stator current. With sinusoidal windings the equations will look like this:

<br /> Va = X*sin(ωt)\\<br /> Vb = X*sin(ωt+\frac{2 Pi}{3})\\<br /> Vc = X*sin(ωt-\frac{2 Pi}{3})\\<br />

Current are:

<br /> Ia = Y*sin(ωt)\\<br /> Ib = Y*sin(ωt+\frac{2 Pi}{3})\\<br /> Ic = Y*sin(ωt-\frac{2 Pi}{3})\\<br />

Thus the power to the rotor is:

<br /> Pr = Va*Ia+Vb*Ib+Vc*Ic = \frac{3}{2}XY<br />

As you can see the rotor power is not dependent on phase.

On the other hand, provided that you want to pull power from a 1-phase connection with unity power factor, the available power is:

Pin = V*sin(wt)*I*sin(wt)=VI*sin^2(wt)=VI/2*(1-cos(2wt)).

Therefore you need an energy buffer.

I would strongly suggest you do some reading on the topic.

Excellent books are:

Principles of Power Electronics by Kassakian, Schlecht, Vergese from MIT
Fundamentals of Power Electronics by Erickson and Maksimovic from Colorado
Ac Electric Machines and Their Control by Torrey from Union