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MOT primary as hv inductor

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klugesmith

Thu Oct 15 2009, 10:20PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Crunchy Frog wrote ...
What's the purpose of the air gap, anyway?

Let's illustrate the energy storage issue with a few practical numbers.
The magnetic energy per unit volume is B * H / 2.

At a flux density B of 1.0 tesla, an air gap has an H field of B/u0 = 796000 amperes per meter, and stores energy of 398000 joules per cubic meter.
Not by coincidence, that dimensionally and numerically equals the mechanical attraction between pole faces at 1 tesla: 398000 pascals = 3.93 atmospheres = 57.7 psi. Memorable as the pressure order of magnitude that limits torque of electric motors at any scale.

At the same 1-tesla flux density, a steel core with ur of about 10000 (1e4, not 10e4!) would have H of only 80 A/m, and energy density of only 40 J/m^3.

For a down-to-earth example, let's look at the ignition coil posted by our jpsmith123 yesterday:

In very round numbers, suppose the iron core area is 1 cm^2, iron path length 100 mm, and air gap length 1 mm. To magnetize it to 1 tesla, the total MMF is HsLs+HaLa = 79.6 * 0.1 + 796000 * 0.001 = 804 ampere-turns, of which 99% is for the air gap. Energy storage is 40.2 mJ, 99% of which is in the air gap. Multiply that by frequency to get the power of a flyback converter.

The pictured product looks smaller in all dimensions than my example numbers, and would probably overheat with 800 ampere-turns regardless of the actual current to saturate the core. If the core were ferrite, the maximum flux density is on the order of 0.3 tesla, so the energy numbers would go down by about a factor of 10.

[edit] As others have said, the airgap can be used to adjust and stabilize the inductance even if energy storage is not critical. In the example, the core's permeability could be doubled or halved by a change in material, processing, or operating current and flux, and the inductance would change by no more than 1%.

Crunchy Frog

Fri Oct 16 2009, 03:18AM

Registered Member #2422 Joined: Tue Oct 06 2009, 02:41AM
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Posts: 85

2 more questions:

Does a bigger air gap then mean higher inductance?

also

What is the max input voltage (approximately) for a MOT secondary in a dc resonant charger?

Dr. Slack

Fri Oct 16 2009, 05:44AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Other things being equal, a longer airgap means lower inductance (more Ampere turns for the same flux linkage). But for your charging choke application, it's best just to design it right.

A MOT secondary is designed for about 3kV peak on it. You should be perfectly safe doubling that to 6kV in a clean environment. However, that will have used up pretty much all of your contingency, making it very vulnerable to accidental over voltages. If you have the MOTs, then running 4 in series at their design voltage will be much more robust than 2 at twice that.

Let's do a design based on some guesstimates

Your charging choke only needs to be the same voltage withstanding as your PSU, so assuming max 12kV PSU, need at least 2 MOT secondaries in series.

The choke needs to store 25% of the energy of the tank cap, I'm estimating 8J bang, so 2J in the chokes, 1J each, perhaps 8H each, so 500mA peak current.

Now analyse a typical MOT to see what inductance it delivers when gapped so that it will take 500mA without saturation

A typical domestic MOT tends to be about 1 turn per volt rms, so has about 2000T on the secondary. At 500mA, that's 1000AmpereTurns. If we want a maximum flux of 1.5T, then we need an H field given by H = B/uo = 1.5/4pi.1e-7 = 1.2M A/m. If the coil gives us 1kA, and we want 1.2MAT/m, then we need a total airgap of 1k/1.2M = 0.8mm, or a shim that's 0.4mm thick.

The MOT next to me has a core of 25mm * 70mm or 1750 u m2. That's not all iron, some is insulation, so say 1.6m m2. At a flux density of 1.5T, that's a total flux of 1.5 * 1.6m = 2.4mWeber, and as there's 2000 turns, that's a flux linkage of 2.4m * 2k = 4.8. Inductance is defined as the flux linkage per amp, so inductance = 4.8/0.5 = 9.6H.

Check the design stored energy two ways
a) electrically, 0.5LI^2 = 0.5 * 9.6 * 0.5^2 = 1.2J
b) magnetically, volume * BH/2 = gap_length * gap_area * BH/2 = 0.8m * 1.6m * 1.5 * 1.2M * 0.5 = 1.15J

They don't agree perfectly as I've been rounding off to one decimal place at all intermediate calculations, do the sums to higher precision and they will be identical. I guess this demonstrates that all* the energy is stored in the airgap.

Notice that the permeability and length of the iron path do not appear in the sums above*. That's OK when the iron has a collosal permeability, it doesn't matter much whether ur = 1k or 5k, both are "infinity"* compared to the ur = 1 of air.

I may have guesstimated your TC or MOT parameters wrongly, in which case just modify the estimates and re-run the above calculations.

* If you want to follow KlugeSmith and include the finite permeability of the iron, then like his calculations it will alter the final answer by a fraction of a percent. Getting a 400um shim thickness wrong by a few percent, or not getting the pole faces flat to a few percent of 400um after grinding the core apart is a much larger and more significant source of error.

If ever you build an inductor on a core intended specifically for inductors, a powdered iron or ferrite core with an (intentionally) very low ur like 10 or so, then you dispense with a discrete airgap as you have a "distributed" airgap between the powder grains, and do the H and B sums with the length and area and effective ur of your core.

1T === 4 bar pressure - woot - like that KS!

Crunchy Frog

Fri Oct 16 2009, 12:43PM

Registered Member #2422 Joined: Tue Oct 06 2009, 02:41AM
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Posts: 85

Wow, thanks! That's exactly what I needed.

Crunchy Frog

Wed Nov 04 2009, 03:40AM

Registered Member #2422 Joined: Tue Oct 06 2009, 02:41AM
Location:
Posts: 85

On second thought

Dr. Slack wrote ...

The MOT next to me has a core of 25mm * 70mm or 1750 u m2. That's not all iron, some is insulation, so say 1.6m m2.

Which dimensions of the MOT are these?

Dr. Slack

Wed Nov 04 2009, 08:25AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Which dimensions of the MOT are these?

Given my previous post, there's enough information to work it out from logic alone, as a puzzle; though sufficient understanding of inductors to be playing with high voltages would also make it immediately apparent. I think you will be better placed if you do some research to figure it out, than if I simply give you the answer. Sorry for the tough love.

Crunchy Frog

Wed Nov 04 2009, 04:35PM

Registered Member #2422 Joined: Tue Oct 06 2009, 02:41AM
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Posts: 85

OK I get it now.

Research FTW.

Bored Chemist

Wed Nov 04 2009, 06:21PM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

There's somethiong about the "shorted output" I have never understood.
Imagine I get a "perfect" transformer and I connect it to my 240V mains supply. This transformer has a 20:1 step down ratio so I get 12 volts out of it.
If I connect it to a 12 ohm resistor I get 1 amp through the load and so it dissipates 12W.
There's only 1 place that can have come from- it must be from the mains.
Now, since this is a perfect transformer there are no other losses.
It must draw 12W from the mains- that's 50mA.
Now I add another 12 ohms in parallel with the first.
It too dissipates 12 W and so the mains must now supply 100 mA
If I had used a 24 ohm resistor in the first place it would only have drawn 50mA from the mains.

For a perfect transformer and a resistive load the mains sees what looks like the resistor multiplied by the square of the turns ratio.
If I remember rightly that's what's called a reflected impedance.

Now if I connect a MOT (which is not a perfect transformer, but pretty good) to a short circuit and then connect the input to the mains then, from the point of view of the mains, the thing looks like a short circuit (actually a rather lower resistance than the real "short" if the transformer were perfect).

The reflected impedance of a short should be a short- anything else depends on the transformer not working perfectly. Is that the best we can do?

OK, next question.
A transformer is 2 coils of wire on a bit of iron.
One of those coils has a lot more turns than the other.
The inductance of a coil, all things being equal, varies as the square of the number of turns.

The secondary of a MOT should have something like a hundred times the inductane of the primary.
Also, because it's designed as a high voltage coil, it will be better insulated.

Surely the best way to use a MOT as a high voltage inductor is to ignore the pirmary and use the secondary- or have I missed something.
I realise that the secondary might not have the current carrying capacity you need. but if you have access to enough high voltage and high current to need to worry about that, I'm not sure you should be trying to bodge it together with an old MOT.

klugesmith

Wed Nov 04 2009, 06:56PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Bored Chemist wrote ...
For a perfect transformer and a resistive load the mains sees what looks like the resistor multiplied by the square of the turns ratio.
If I remember rightly that's what's called a reflected impedance.

Yup. At this level of model, primary and secondary inductance are infinite.

Now if I connect a MOT (which is not a perfect transformer, but pretty good) to a short circuit and then connect the input to the mains then, from the point of view of the mains, the thing looks like a short circuit (actually a rather lower resistance than the real "short" if the transformer were perfect).
The reflected impedance of a short should be a short- anything else depends on the transformer not working perfectly. Is that the best we can do?

Have you read HvWiki about transformers? MOT departs strongly from ideal transformer: smallish and nonlinear primary inductance, deliberate and substantial leakage inductance, etc. Would overheat in normal service if not for cooling fan.
The view from mains is primary inductance in parallel with (leakage inductance in series with reflected load impedance), with primary & reflected secondary winding R's worked in. I expect the circuit breaker should trip after a while, not instantly (bang) as if shorted.

OK, next question... Surely the best way to use a MOT as a high voltage inductor is to ignore the pirmary and use the secondary- or have I missed something.

Agreed, when it fits the required voltage, current, and inductance. For purposes of OP here, typical MOT secondary would have too much inductance, and would deeply saturate the core at 500 mA (when not countered by 5 amperes in primary winding). Solution was to introduce an airgap.

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