Coupling coefficient of industrial transformers

Also some talk about snubbers and others talk about clamps.

Thanks

When you're using a half or full bridge with inductive loads, use MOSFETs or co-pack IGBTs and forget about it. The flyback pulse harmlessly bumps into the supply rails (peak current is lesser than or equal to the current going through the opposing transistor when it turned off). Anything else that happens, like ringing when neither transistor is conducting any current, is just aesthetics.

Tim

-- "Librarians are hiding something." - Steven Colbert Website @

Tim,

In my circuit's simulation, the voltage goes above the rail, with flyback diodes in the IGBT. Some of the articles I have read have snubbers with the flywheel diode. They talk about in rush current (di/ dt) at turn-on, voltage spikes (dv/dt) at turn-off that can cause false triggering and/or destroy the flyback diode and the IGBT. I have been reading the article that T.J. Byers wrote in Nuts and Volts about snubbers but it seems to missing some information. I sent him an email asking him about my particular design and he said that he would use a clamp circuit, to add to my confusion. What did you use, as an isolator, between the turns of the inductor you made with strips of cooper pipe as the conductor?

Thanks

Wait, and NO CAPACITOR after the inductor!?

Are you doing this in shorting- or open-circuit commutation?

If you have the bridge supplied by an inductor, obviously the rails will be VERY squishy, and yes, you will have *considerable* overshoot. This energy is absorbed with a diode to dump the energy into a capacitor, with the average power dissipated by a resistor.

No. Not big. Small, like a few uF. Fast capacitors. As close to the bridge as possible, if you are going constant voltage.

I hope you know (or realize now) what a difference it makes in the bridge being constant current (i.e. inductor or resistor or CCS supplied) versus constant voltage.

What rail? Your IGBT simulator models OBVIOUSLY aren't co-pack models if the output voltage is ever going outside of the supply or ground rails the IGBTs are connected to. An inductor-supplied bridge will of course exceed the *supply* voltage, but not the rail voltage.

Three things kill IGBTs:

1. Excessive voltage (causes avalanche or something)
2. Reverse voltage (no idea what it causes, but probably something nasty)
3. Excessive current * voltage (standard power dissipation problems)

Number 2. is fixed with reverse diodes sufficiently close to the transistors, 3. by limiting current (a desat detector, for instance), but 1. has to be fixed by clamping the voltage to some absolute standard (a constant voltage bridge with flyback diodes does this by default) or otherwise absorbing enough under all conditions that it cannot cause breakdown.

Tim

-- "Librarians are hiding something." - Steven Colbert Website @

there's likely less than 1% leakage, but how clean is your 125V?

maybe get a variac and bring the voltage up slowly.

Bye. Jasen

Tim,

In my circuit, I have an inductor and a capacitor after the inductor. I don't have a capacitor before the inductor like in your circuit for your induction oven.

What do mean, if I run my test in open circuit or short circuit commutation? Are you talking about the load on the secondary of the transformer? I have a .1 Ohm resistor as the load to simulate welding at around 60 Amps.

I don't know the difference it makes in the bridge being constant current versus constant voltage.

When you are talking about the rail voltage, are you talking about the voltage after the AC has been rectified and filtered. What is your definition of supply voltage as opposed to rail voltage? The AC, rectified and filtered, makes a "DC" voltage that isn't very smooth but for welding purpose is seems that it should be OK.

Thanks

Jasen,

I am going to have to make some tests to try and figure out the leakage inductance. It seems, from the simulations, that it doesn't take much leakage inductance to greatly reduce the efficiency of the transformer, so anything less than 1% would make big difference. I don't know how clean is the 125 V. but it is in a residential area so I "assume" it is reasonably clean.

Thanks

Mr. Phantom

Will the method of looking at the inductor's waveform, with an oscilloscope, still work if the inductor is used as a filter with rectified DC? Tell me if I got this right, if the inductor saturates, it will no longer filter the current. The way I see it, once it saturates, if the current keeps increasing, the current's ripple will keep increasing.

I have made some test and took some measurement on various transformer and here is what I got. For my Micro Wave Transformer, the resistance of the primary winding; .

1 Ohms, the current in the primary winding at 125 Vac; 5 Amps, the impedence of the primary; 25 Ohms, the inductance of the primary; .066 H, the inductance of the secondary (3:1 winding ratio); .007 H, coupling coefficient according to Mr. Phantom's method; .97. For my 15 Kva transformer, resistance of the primary winding; 1.5 Ohms, the current in the primary winding at 125 Vac; .8 Amp, the impedence of the primary; 152 Ohms, the inductance of the primary; .4 H, the inductance of the secondary (5:1 winding ratio); .016 H, the coupling coefficient according to Mr. Phantom's method; . 9875. Does anybody see anything wrong with these numbers?

Thanks

Looking at your numbers below, it seems that you probably don't have an inductance meter, but are deriving the inductance of a winding from the ratio of the voltage across it and the current through it.

The scope method doesn't work as well in an inductor with a large DC component. What you can do, however, is to measure the AC voltage across the inductor and the AC current through it, take the ratio and get the AC impedance. Make sure that your meter doesn't measure true RMS AC+DC; you don't want it to see the DC component. If it does, you will need to put a capacitor in series with it to measure only the AC component.

Start out with not much DC current and calculate the inductance by that method. As you increase the DC current, measure and calculate the inductance again. You can detect the onset of saturation as the value of DC current where the inductance begins to decrease. Saturation is a gradual thing, of course, and you will see the calculated inductance gradually decrease as the DC current increases.

When you were making these measurements on the primary of the two transformers, the secondaries were unloaded, right?

The MWT would seem to be well into saturation, but otherwise the numbers seem reasonable.

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M. Rosenbaum

I don't have an inductance meter, all I have is an oscilloscope, a 2 Amp adjustable DC power supply, a Wavetek digital multimeter and a clamp on Amp meter. Do these meter measure AC+DC? Since I don't have a large DC power supply, I guess I'll have to make these measurement in the circuit.

Yes these measurement are make with the secondary unloaded and the impedence of the secondary is calculated according to the turn ratio.

Are you saying that the MWT is in saturation because it draws 5 Amps. I thought so too, the value of H is around 2,600. For laminated steel, it starts to saturate at around 1,500 I think. Before I removed the magnetic shunt in the core of the MWT, the current was around 3 Amps which put H at around 1,500. Is there a problem if the transformer is saturated? Does it reduce the coupling coefficient? If I put the appropriate air gap in the core of the transformer, to bring the value of H at around 1,500, will it cause a problem in the performance of the transformer?

Thanks

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The clamp on by its very nature can't measure DC, so you're all right there. Set your Wavetek to measure AC volts and connect it to your DC power supply with the supply set to put out maybe 5 volts (with the range of the Wavetek set appropriately to measure 5 volts). If the Wavetek reads essentially zero, then it's not a true RMS AC+DC meter, and you're good to go.

If the transformer core is substantially into saturation, the unloaded primary current will be high, as you have noted in the MWT transformer. The coupling coefficient will be reduced somewhat. The additional IR drop in the wire of the primary will reduce the efficiency of the transformer (increased temperature rise as a result) and will degrade its regulation. But, it will still perform as a transformer.

Gapping the core of a transformer will cause performance degradations similar to those caused by saturation. The no load excitation current will go up, causing extra losses.

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M. Resenbaum

have to take a few turns out of the primary (which I don't want to do) or lengthen the core (which would be complicated) or reinstall the magnetic shunt, am I right? Is there other way to achieve that?

I often see air gaps in large inductor, my guess was that they did that so they didn't have to make a large steel core. I thought I could do the same in a transformer.

Thanks again

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To reduce the flux density in the core you must *add* turns to the primary.

Or you must increase the *cross-sectional area* of the core. Lengthening the core (making the magnetic path length longer) will not change the flux density in the core much (if the primary turns and cross-sectional area remain unchanged) but it will cause the no-load exciting current to increase.

The flux density in the core leg upon which the primary is wound is determined by the cross-sectional area of that leg and the volts per turn of the primary winding. Reinstalling the shunt will reduce the effective path length in the core as seen by the primary. The core leg where the primary is wound will still see the same flux density, but the shunt will reduce the flux density in that part of the core beyond the shunt. Thus there will be less volume of core with the same flux density as the primary leg itself. That part of the core beyond the shunt will not be as far into saturation, and this fact plus the reduced effective path length for the primary will cause the exciting current to decrease. But, the part of the core upon which the primary is wound will be just as far into saturation as when there is no shunt if the primary turns remain unchanged.

Inductors have gaps because the inductor is intended to store energy and an ungapped core doesn't store much energy. Most of the energy is stored in the gap of an inductor.

Transformers are designed so that as little energy as possible is stored in the core. When energy is stored in the core, the exciting current goes up. This is what is wanted in an inductor, but not in a transformer.

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M. Phantom

I took my MWT that drew about 5 amps and I removed the magnetic shunt, the transformer still drew about 5 amps.

There was 115 turns on the primary, I added 33 turns so now it has 148 turns, now the primary of the transformer draws only .75 amp.

Since I want to make an inductor out of it, by adding the appropriate air gap, and since I want to pass 15 to 20 amps in that inductor, is there a way to predict it's characteristics as the current increases or is it done by the "trial and error method".

Thanks

Obviously, it was very much into saturation. By adding turns, you have brought it substantially out of saturation.

Are you planning to use this inductor you're making as a welding inductor?

You should read up on "Hanna curves". Here are some links to get you started:

and this one may point you to more info than you wanted:

and finally, for more detail, go to the library and get the following papers:

Design Relationships for Iron-Core Filter Chokes, Theron Usher, p. 454, Transactions of the AIEE, Sept. 1957.

Design of Air-Gapped Magnetic-Core Inductors for Superimposed Direct and Alternating Currents, Anil K. Ohri et al., p. 564, IEEE Transactions on Magnetics, Sept. 1976.

Simplify Air-gap Calculating, Warren A. Martin, p. 94, Electronic Design magazine, April 12, 1978.

I am planning on using a similar inductor in the filter that I want to use to make DC before chopping it and feeding it, through an H bridge, to the primary of the power transformer. I have tested about 10 MWT and, from my calculation, at about 3 amps it's just before it starts saturating. Some MWT draw as much as 6.5 amps. with no load at the secondary. I guess they are made cheap and whoever makes them don't care much about performance. I'll take a look at reference you gave me when I have a minute.

Thanks again

M. Phantom

I have looked at the links you sent me. Once I have found the characteristic of the inductor in AC, how can I transpose them in DC when I want to use that inductor as a filter? If, for example, I have an inductor with no air gap, how can I test to see if it is saturating, what frequency do I use if the AC has been rectified? I haven't found any simple explanation in the links you sent me.

Thanks again

I think my questions are not very clear. Once I have the AC characteristic of an inductor, like value of its inductance, how does relate to its DC characteristics? When filtering DC, does the inductance value decrease when it saturates like in AC? What frequency do I use to make my calculations in DC since I am rectifying AC? Would it be 120 Hz? Thanks again

Yes, it does. As you pass DC through an inductor, its inductance varies in a way that depends on the particular core material. Using gapped silicon steel, you will probably get an inductor whose inductance increases a little at first as you gradually increase the DC in the winding. But, eventually the DC will push the core into saturation and the AC inductance (that's a little redundant) will decrease drastically.

For making your calculations in DC, you use a frequency of DC, zero Hz. For your ripple calculations 120 Hz is probably the place to start, although there will be harmonics that a very careful analysis should consider.

At 0 Hz, the inductance is 0 Henry, so it's like a straight wire?

How do I calculate the inductance value if it's filtering DC plus a AC ripple at 120 Hz?

Thanks