Single phase induction motor speed control

[John HaineParticipant @johnhaine32865]

As I mentioned in another thread I thought I would do some investigation of the electrical characteristics of the single phase motor used on my Quorn to assess the possibility of controlling its speed. This is a Parvalux motor of the type that was recommended by the casting supplier (Model Engineering Services?) – it's an SD 13 rated at 150 W (about 1/5 HP) mechanical power, capacitor start and run, 3 wire. The reversing circuit recommended and used is this:

So depending on the switch position the capacitor (8.4 mfd 440V working) is in series with one or other of the coils. Since there is no preferred direction the coils must be the same, the motor is symmetrical. I measured the DC resistance across the plug pins, it is 35 ohms for each winding.

Next I connected it to the mains and measured the voltages across the mains, the winding straight across the mains, the other winding in series with the cap, and the capacitor itself. These were:

VW1 242 V (mains)

VW2 288 V

VC 396 V.

The supply current according to my plug-in power meter was near enough 0.8A (but I'm not very confident of its calibration).

Interesting that the capacitor voltage is significantly higher than the supply – that's because there is a partial resonance with the inductance of the winding, and shows that the capacitor working voltage needs to be significantly higher than mains.

The phasor (sort of vector) sum of VW2 and VC has to add up to VW1 since they share terminals, so by applying a bit of trig we can calculate the phase of the winding voltage VW2 compared to VW1.

Edited By John Haine on 27/01/2018 14:22:48

[John HaineParticipant @johnhaine32865]

[John HaineParticipant @johnhaine32865]

The formula is the cosine rule,

angle = arccos[(VW1*2 + VW2^2 – VC^2 )/(2*VW1*VW2)]

This gives 96 degrees for the phase between the voltages on the two windings when running. So in effect this is a 2-phase motor and the capacitor is chosen to give 90 degrees phase shift between the voltages. It is interesting that the capacitor voltage is much higher than mains, and the voltage on the second winding is higher too but by not by much. The capacitor is nearly resonant with the winding inductance, which is why there is a voltage multiplication.

I have a little clip-on current transformer rescued from a failed "Efergy" electricity power monitor, which isn't that well calibrated – I have more confidence in my voltmeter's calibration. But using that I could measure the current in the two windings, estimating them as 1.09 amps in the cap-fed and 0.83 in the other. (As a check, knowing the volts on the cap, the frequency and its value, I calculated the current in it as 1.04 amps which is very close.) So it looks like the winding fed by the capacitor is contributing significantly more power than the other.

Of course the currents in the capacitor fed winding and the other are not in phase so one can't just add them to find the supply current.

 

Edited By John Haine on 27/01/2018 14:38:55

[John BaronParticipant @johnbaron31275]

One way of controlling the motor speed is to vary the frequency of the supply voltage. This can be done by rectifying the mains to turn it into DC, and then chopping up the DC to provide pulses at a high frequency. Say 20 KHz. By modulating the pulses at 50 Hz plus or minus a few cycles will allow the motor speed to be changed up or down.

[John HaineParticipant @johnhaine32865]

Quite. Also known as a VFD. What I'm trying to do is evaluate how well a motor of this type will perform with a variable frequency drive. See this post.

I'm just going to wrap a cold towel round my fevered brain for a bit then return to the calculation.

[Howard LewisParticipant @howardlewis46836]

Interesting; being a simple mechanical engineer, expected the capacitor to charge to peak voltage, so that

VC = Supply x 1.414 (240 x root 2 = 339) rather than 396. Don't believe, (SINCERELY HOPE NOT) that the mains delivers 280 volts RMS!

Possibly the Inductance is causing the circuit to resonate? What say you, electronics gurus?

Howard

[Martin ConnellyParticipant @martinconnelly55370]

Howard, the capacitor and winding circuit can be considered to be a resistance, an inductance and a capacitor in series. The thing they have in common is that the current passing through them. However the voltages across them will be different for each component. In the resistor the voltage and current will be in phase, in the inductor the current lags 90° behind the voltage and in the capacitor the current leads the voltage by 90°. Plotting these phased voltages should give the vector sum of the applied voltage. It may sound complicated but I remember it from A level physics. Plotting the voltages can be quite revealing and it does not surprise me that there are voltages higher than the applied voltage in the circuit.

Martin C

[Martin ConnellyParticipant @martinconnelly55370]

John, your two measured currents are close to 90°out of phase, you can use Pythagoras to calculate the approximate vector sum as 1.37 amps. This is over 250 Watts so I think the clip on ammeter may be too inaccurate.

Martin C

[EmgeeParticipant @emgee]

John, see the link below for manufacturer's data on the SD13.motor.

**LINK**

Emgee

[John HaineParticipant @johnhaine32865]

Thanks Emgee, got that already.

Some experimentation with the clip on sensor showed it was picking up stray fields from the motor giving errors as big as the apparent measurement! So have to ignore the current readings.

What I'm really interested in is what happens when the frequency is changed – how much will the phase angle in the winding fed through the capacitor change?

I can calculate the relative phase angles of the other voltages and with a bit of assumption I can calculate the effective "Q" of the resonance of the C with the winding inductance – it comes out at 1.3 which is quite low. This is a bit suppositious because the apparent inductance probably depends on the speed. But anyway that probably is enough to say that the motor would tolerate a +/- 20%-ish change in frequency (say 40 to 60 Hz) (and so speed) without a huge change in available torque and power. But the interesting thing is that the capacitor-fed winding must be contributing more power than the other.

Since this is effectively a 2-phase motor a better way to change its speed would be to use a 2-phase VFD of course, since its windings are clearly identical, and eliminate the capacitor. I suspect that only rather small motors will be capacitor run like this, because at higher powers the cap will have to handle a large continuous current and would get hot. That's probably why larger motors switch the cap out of circuit once started. I have one of those, I'll see if I can make similar measurements.

[John OlsenParticipant @johnolsen79199]

If I needed to try to speed control this motor, I would just connect it to a VFD without the capacitor, eg Connect the common point of the windings to one phase, and each end of a winding to each of the other two phases. This will give the correct Voltage across each winding, and a phase difference between the windings of 120 degrees instead of 90 degrees, close enough that it should start and run fine. (The capacitor will never actually give 90 degrees anyway.) The usual point about watching the motor temperature at low speeds would still apply.

If you vary the frequency with the capacitor in place, the reactance of the capacitor will be higher at low frequencies, which is not what you need. The ideal size of capacitor would need to change with the frequency

I have a vague idea that Newton Tesla were at one time supplying a two phase motor and VFD for their Myford conversion set, I don't know if that is still the case, but you could ask them.

If you bring a series circuit into resonance, the voltage across each component can be much higher than the supply voltage.

John

[John HaineParticipant @johnhaine32865]

A capacitor can give 90 degrees with the right value – after all I measured 96 degrees which is closer than 120 degrees using 2 phases of a 3 phase supply. Obviously the ideal capacitor varies with frequency but the question is with a given capacitor how much variation should be possible? It seems like the Q in my case is quite low so one could expect a reasonable range.

As John says it should be possible to connect the motor to 2 phases of a 3 phase VFD (provided that the VFD can accept the rather large load imbalance that results). A bit of figuring shows that the starting torque of the motor will be reduced by a bit over 30% doing this, which may or may not be significant depending on the application. It would perhaps be nice if there was an option on VFDs to obtain 2-phase quadrature drive rather than 3-phase, provided the output circuit would stand it.

[Author]

Posts