# Sychronous Pernament Magnet Alternator under Load

1. May 22, 2014

### R.G

Hi
I have a question about a Synchronous Permanent Magnet Alternator (Small Wind Turbine Alternator)
I want to understand the ratio between RPM and the Voltage from the Alternator.
I can't find any equation for that ratio , and I read somewhere on the web , that if I have a load ,than the Voltage for specific RPM will change, it doesn't sound right to me. if I have a load , the RPM will change due to EMF , and with it the voltage(as consequence of lower RPM) until the Power from the generator equals the load ,am I wrong?

thank you
R.G

2. May 22, 2014

### jim hardy

There's a phenomenon called "Armature Reaction" that causes the effect you describe.

Does your unit's instruction sheet give a number for perhaps "load regulation" or "synchronous impedance" ?

Does it have a voltage regulator ?

search on the terms above should take you to some theory pages.

Basically here's what goes on:
Look at this neat animation Sparky came up with (wish i could embed it)
http://www.ece.umn.edu/users/riaz/animations/alternator.html

For your unit the rotor is a permanent magnet, not the elctromagnet shown.
When current flows in the armature winding it creates a MMF that opposes the field's MMF. So the strength of the magnetic field is reduced. So voltage goes down.
An active regulator would compensate for that effect, either by applying out-of-phase load current to strengthen the field, or in a wound rotor machine by increasing field current.

If one wants to be picky, the rotor in that demo is a quarter turn off which is probably why there's no units given on his graph. One assumes it's volts, more likely it's flux $\Phi$. Recall volts is d$\Phi$/dt which for a sinewave is just a 90 degree (1/4 turn) phase shift.
But at this stage, that's a detail for the future. Not labelling the graph leaves him mathematically okay, while he still paints a good mental picture for the beginner. .

old jim

3. May 22, 2014

### cabraham

Under load the voltage will change even if rpm stays constant. The stator winding has lots of inductance. When unloaded, the flux rate of change wrt time multiplied by turns equals voltage. But when load current is drawn, the inductive reactance of the stator winding drops some voltage and the terminal voltage is less than open circuit value. In an alternator or generator, permanent magnet or field wound, the stator inductive reactance can be 50%, or more of the load resistance at full load. Reactance of 100% is not uncommon.

So if your open circuit voltage is 10 volts, then a 1.0 ohm load is added, with a 1.0 ohm inductive reactance in the stator winding (100% of load R), the terminal voltage falls to 7.071 volts. The 1.0 ohm resistance phasorially adds to the 1.0 ohm reactance in quadrature results in an impedance of 1.414 ohms. The current is 7.071 amps.

For a field wound unit, the voltage regulator increases field current until the terminal voltage attains its regulated value of 10.0 volts, or whatever it is desired to be. A permanent magnet alternator does not use this method, when load current increases, terminal voltage drops.

Claude

4. May 22, 2014

### jim hardy

Indeed, our central station generator was 171%.

@ R.G. : Another measure of it is "Short Circuit Ratio". Search on the terms, and with cabraham's fine introduction above to the calculations, you'll be all set. You could even test your unit to measure its characteristics if you have voltmeter and ammeter and a way to spin it at known rpm's..

5. May 22, 2014

### cabraham

Thanks for sharing that Jim. You also make very good contributions to this forum. So if we have a reactance of 171%, then for a 1.0 ohm load value, the reactance is 1.71 ohm. The impedance is 1.981 ohm. So for full load, the current is 10.0V/1.981 ohm = 5.048 amps. The terminal voltage is 5.048 volts. So with 171% reactance in the stator winding, under full load nearly half the open circuit voltage is dropped across the stator inductive reactance.

A field wound alternator equipped with a regulator would boost the field current to a higher value so as to produce an open circuit voltage of 19.81 volts. When loaded with 1.0 ohm, the terminal voltage is 10.0V with current of 10.0A. However, if the field current is at this high value loaded, then the load is suddenly removed, the terminal voltage jumps from 10.0V up to 19.81V. This is known as a "load dump".

Hopefully this gives you a good start. I'd suggest reviewing some peer-reviewed uni level texts on motors and generators for more details.

Claude

Last edited: May 22, 2014