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saturable reactor from three-phase reactor ?

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Tiberius

Thu May 15 2008, 07:30PM

Registered Member #1484 Joined: Wed May 14 2008, 03:24PM
Location: Cary, NC, USA
Posts: 27

On the surface it appears to me that a three-phase reactor with a three-leg core would be a good start for a saturable reactor -- remove the center winding and rewind for control, while the two remaining phase windings become the load windings.

Would a three-phase reactor function as a saturable reactor if given this treatment?

Where can I find more information about the design and operating characteristics (input volts vs output reactance, rise/fall times, load coil turns to control coil turns) of a saturable reactor?

Steve Conner

Fri May 16 2008, 09:14AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

1. Probably, although a saturable reactor shouldn't have an air gap, and a line reactor probably will, so you'll need to get rid of the air gap. You want it to saturate, and the gap will stop it from doing so.

2. Mostly trial and error, although googling never hurt anyone.

As a starting point, the load windings should have about the same number of turns as if you were making a transformer to run off the same line voltage. If you used fewer turns, the reactor would saturate and let power through even before you started applying power to the control winding.

I don't know how to calculate the number of turns and current needed for the control winding, but I suspect it has to do with an ampere-turn balance. So if your control winding had the same number of turns as your load winding, the output current may be roughly equal to the control current. This is why saturable reactors usually have a control winding with lots of turns of fine wire: to avoid the need for a low-voltage, high-current control supply.

Bear in mind that the saturable reactor is a non-linear device, and probably (though I haven't got proof of this) works more like a phase-angle lamp dimmer than an actual variable inductor.

Tiberius

Mon May 19 2008, 04:37PM

Registered Member #1484 Joined: Wed May 14 2008, 03:24PM
Location: Cary, NC, USA
Posts: 27

So I've thought about this some more and discovered the obvious contradiction. There will be no reactor with reasonably high current windings *and* a high impedance as it would have no need for high current windings.

It looks like I'm going to have to rewind the load coils as well, and right now I'm looking for a target of 1H inductance across the two windings. This is to limit load current to around 600mA at an input of 240V, limiting a 60:1 step-up transformer to a theoretically non-lethal 10mA on the output.

I've been playing with some numbers to get an approximate idea of how many turns I'm going to need using this little calculator here:

This calculator is suggesting that 11 turns of #2 AWG (diameter 0.003272) on a 4" by 4" square core with relative permeability 40000 (the figure I've found for silicon steel/"electrical" steel) yields an approximately 1H inductance. If I must maintain a volts-turns ratio to prevent saturation it seems that the inductance will be *significantly* greater than the minimum required to achieve limiting of 600mA in the scenario above. Approximating the volts-turn ratio as 1 results in a calculated inductance of nearly 500H! Can these figures possibly be accurate?

Dr. Dark Current

Mon May 19 2008, 04:51PM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

You must take care not to saturate the core by the load winding itself, 11 turns will saturate it at 240V for sure.
If your core is 4" by 4" (this is the 'leg' size right?), then I would guess something like 2-3V/turn to be the maximum volt per turn voltage (so you'll need some 80-120 turns maybe).

Dago

Mon May 19 2008, 05:00PM

Registered Member #538 Joined: Sun Feb 18 2007, 08:33PM
Location: Finland
Posts: 181

Tiberius wrote ...

This calculator is suggesting that 11 turns of #2 AWG (diameter 0.003272) on a 4" by 4" square core with relative permeability 40000 (the figure I've found for silicon steel/"electrical" steel) yields an approximately 1H inductance. If I must maintain a volts-turns ratio to prevent saturation it seems that the inductance will be *significantly* greater than the minimum required to achieve limiting of 600mA in the scenario above. Approximating the volts-turn ratio as 1 results in a calculated inductance of nearly 500H! Can these figures possibly be accurate?

I'm fairly sure something is wrong with your calculation, 11 turns cant in any way yield 1H of inductance with your core. More like hundreds if not even thousands of turns are required for 1H.

Steve Conner

Mon May 19 2008, 10:54PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Tiberius wrote ...

So I've thought about this some more and discovered the obvious contradiction. There will be no reactor with reasonably high current windings *and* a high impedance as it would have no need for high current windings.

That's why you'll need to take the core of a line reactor apart and reassemble it without the air gap, like I said.

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