Coupling coefficient of industrial transformers

On Apr 12, 1:34 am, The Phantom wrote: ...

...

Thanks for your thoughtful postings on this. They've given me some good food for thought, some things to ponder. I especially appreciate the comments about the practicalities of measuring iron-core transformers.

Cheers, Tom

M. Bruhns

I thought of doing like you suggested, using the simulation and comparing with the actual circuit, but I didn't think it could work because I want to know "k" to figure out what kind of voltage kick- back the leakage inductance would send back to the "H" bridge. Since that kick-back might be large enough to destroy the bridge, I was using the simulation to avoid destroying expensive IGBTs. In the simulation, if "k" is equal to 1 there is no kick-back, if it is .99, the kick back is quite large.

M. Phantom

I read a few of the post you recommended and I read other post on "Leakage Inductance" and a lot of them revolve around shorting the secondary. By doing this, how will it not destroy the transformer, unless you use lower voltage than its nominal voltage. I could connect the 10 Kva transformer I have on 125 Vac on the primary and measure the OCV of the secondary and vice versa if this method is good enough. I have a microwave transformer, that I rewound the secondary with 40 turns of # 10 wire, to make some test. I would like to know "k" for that transformer but I couldn't feed the secondary with 125 Vac, the current would be too high and if I use a lower voltage coming from a small transformer, like lets say a 16 Vac, 40 Va transformer, it will need to be able to supply more current than it was designed for if it's connected to the secondary of the MOT. I have a question about MOT, when I connected the transformer to 125 Vac, with the secondary disconnected it drew about 3 Amps., if I removed the magnetic shunt, I believe it would be called, between the primary and secondary winding, the current goes to 5 Amps. How would I know if it's saturating or close to saturating.

Thanks

orvillefpike wrote: (snip)

I would think that the current would increase very much less than that. The shunts come into play when the secondary is overloaded or at least, heavily loaded. With no secondary load, the shunts (and their air gaps) are paralleled with solid core material with no air gap, so the shunts are almost invisible as far as the flux they carry.

Use a Variac to raise the voltage slowly to normal, while measuring the primary current . As saturation begins, the current will rise much more than in proportion to the voltage, as it does with a linear inductance.

You can reduce the saturation by adding some turns to the primary.

A couple of comments:

First, I think you will find it easiest to measure your transformer's coupling coefficient in the way that I believe "The Phantom" posted originally. Quoting what he wrote: "To measure the coupling coefficient (of an iron core transformer) without making an inductance measurement, do this:

Apply rated voltage (sine wave) at one winding, and measure the open circuit voltage at the other, getting the ratio V2/V1'. V1' means that winding 1 was excited.

Now excite winding two and measure the open circuit voltage at the other winding, getting the ratio V1/V2'.

The coupling coefficient is very nearly SQRT(V2/V1' * V1/V2')

The turns ratio is very nearly SQRT(V2/V1' / V1/V2') "

Second, your H-bridge design should be such that it can withstand a moderate amount of leakage inductance in the transformer it drives. Why would you not want it to be able to do that? And is it really so difficult? I assume the kickback is at turn-off of an H-bridge device, and is largely from an inductance that has considerable capacitance associated with it, being part of the windings of a line- frequency transformer, so dV/dt should not be 100V/nanosecond as it can be when power mosfets are driving a high frequency inductive load with very little capacitance. If that's the case, don't either the substrate diodes in your H-bridge devices, or diodes you've purpously put into the circuit, handle it nicely? You do want to have a large enough capacitance on the DC supply to the H-bridge so that the kickback can be absorbed into it. It's certainly not as big a problem as trying to use the driver bridge to dump the stored inertial energy in a large DC motor and its rotating load, where you might have to absorb hundreds of joules.

Perhaps you could provide more information about the problems you see in your simulation. Perhaps the solution is not to demand an unreasonably high coupling coefficient in the transformer, but rather in the proper design of the driving circuit.

Cheers, Tom

M. Popelish

You're the one that told me that I should bring my topic, on "Power Inverter Design". to this forum after M. Sennewald kicked my topic out.

The MOT with, what I believe is called the magnetic shunt, looks like this. ____________ I I I I_____ I_____ I I I I I_____ I_____ I

The MOT, with the shunt removed, looks like this. ____________ I I I I I I I I I I_____ I_____ I

There is no air gap in the transformer. Once the magnetization field is established, I believe that no matter what the demand is on the secondary side, it cannot saturate the transformer. I believe that the only way to saturate a transformer is by having the magnetization current to large for a particular core. Correct me if I am wrong.

M. Bruhns

For me, the method described by "The Phantom" is much easier for me. Why would people promote the shorted secondary method. I made simulation, driving the primary of the transformer, at 40 Amps and at 240 Vac. In real life, I probably wouldn't drive the transformer at much more than 20 Amps or 25 Amps, I wanted to have a safety factor. When the "k" factor is less than 1, the amount of energy, going through the freewheel diode, when one side of the bridge is turned off and before the other side is turned on, is quite large and I am afraid it would destroy the diode. Some of the methods, described in International Rectifier's application notes, to reduce the kick-backs, include having the least amount of leakage inductance possible in the transformer, slowing down the turn on and turn off time of the "H" bridge, at the cost of efficiency of the bridge and if all else fails, then design a snubber or a clamp circuit.

Thanks

The idea is to short one winding and *measure* the inductance and resistance at the other winding with a meter, such as this:

You will get better results if you use this method rather than trying to measure the leakage inductances.

Are you doing all this for a hobby purpose, or is it for your job? If it's for your job, they your employers should buy (or maybe they already have) a variable transformer

You can use a variac to apply a voltage to one winding of the transformer (with *none* of the other windings shorted, of course). Slowly turn up the voltage while measuring the current, as John Popelish described. When you begin to saturate the core, the current will rise more rapidly.

Another way to detect saturation is this: if you have an oscilloscope, put a 5 or 10 watt, 1 ohm resistor in series with the winding that you are driving. Look at the voltage across the resistor with the scope. When the voltage applied to the winding is low, the current (as measured by the voltage across the resistor) will have a sinusoidal wave shape. Slowly turn up the voltage, and when you reach saturation, the wave shape will begin to have a distorted, "peaky" wave shape.

What kind of semiconductor devices are you using in your bridge? Are they MOSFET's? What is the manufacturer and part number?

I already knew that. How is it working out?

Is there an air gap on one or both sides of the shunt blocks, or are they part of the core laminations?

Current in the secondary produces a magnetic field that reduces the flux in the core, reducing its tendency to saturate. However, the reduced flux swing through the primary reduces the back EMF produced by the primary so the primary current increases. The shunts limit how low the magnetic flux in the primaries part of the core can be pushed down by the secondary current, by providing an alternate flux path for the primary flux that does not pass through the secondary turns.

That is the way I understand it. The flux has to be changing to produce a voltage per turn. The integral of voltage per turn and time per half cycle defines how much total flux swing that instantaneous rate of change will accumulate each half cycle. If you want to lower the peak flux each half cycle, you either make the cycle shorter time, or lower volts per turn, or both.

If you knock the shunts out, you will increase the coupling coefficient a bit. They are there to effectively add an inductance between primary and secondary, as a crude current limit in the microwave oven application.

M. Popelish

Wow, my drawing really came out crappy. I'll try it again This is the core of the MOT with the shunt ____________ I I I I_____ I______I I I I I_____ I______I

This is the core of the MOT with the shunt removed ____________ I I I I I I I I I I______I_____I

I hope this looks better.

Although the people that replied to my questions are very knowledgeable, the subject didn't seem to interest as many people as on the LT Spice forum, but there is, still, always somebody that can answer my questions.

Originally, I thought of asking my questions on LT Spice forum because it is the forum, that I found, that had the biggest member list and a lot of brilliant people too of all the electronic forum that I found, but I didn't know that it was illegal to ask these kind of questions on that forum....

There is no air gap between the shunt and the main core, at least, not that I know of. When I took them out, so I could put more turns or the #10 wire, I had to use a punch and a hammer and really whack it because it was pressed in pretty tight. I figured that the shunt where used to sort of "increase" the length of the core which increases its resistivity and therefore reduces the current needed to magnetize it, without having to build a large transformer. I have about 20 MOT and they all have this magnetic shunt in them.

Thanks

Are you composing this picture with courier or other fixed width (per character) font? I think I can figure out what you are trying to show:

" This is the core of the MOT with the shunt _____________ I I I I_____ I _____I I I I I______I______I

This is the core of the MOT with the shunt removed _____________ I I I I I I I I I I______I______I

The moderator fights a valiant fight to keep the posts on the topic of the use of LTspice. When you get to the point of working out a spice model for your transformer and driver, it will be on topic, there.

If they were made of separate pieces of metal, then there is at least a small effective gap for any flux that has to connect them to other pieces of core material. The gaps may have been filled with epoxy or other hard material. But even a tight fit has less magnetic continuity than a single piece of metal has.

Increasing the effective length of the flux path raises the total current (ampere turns) it takes to saturate that path. But the shunts provide a parallel path that lowers the reluctance (magnetic resistance), slightly lowering the ampere turns it takes to saturate the core (though that has nothing to do with their purpose). Think of the path that passes through the secondary and wraps back around the outside to the bottom of the primary as one path, and the shunts are a second path that bypasses the part through the secondary. Parallel resistors have lower total resistance than either resistance, alone.

That is because the magnetron load has a negative resistance characteristic that would run away if the transformer didn't have the positive impedance of the current limit in series with it.

M. Popelish

My first "drawing" was made, along with the text, in Word, the second post was made the same way but, once pasted it the forum, was corrected. In one of the two posts, the drawing looks decent, at my end anyway. I'm only kidding about M. Sennwald, he has been very helpful to me and I see that he is very helpful to everybody else on the forum. It is hard to understand that somebody so knowledgeable would spend so much time helping others. Who is this guy, what does he do for a living, what are his motivations? I used the core of a couple of MOT that I rewound with # 6 wire, added an air gap and used them as inductors for the output of my diode bridge that I used in some test that I made with my welding machine. I first used a regular Lincoln 180 Amps that I outfitted with a huge diode bridge and these inductor to try DC welding but I have no way to know if the inductors I made are doing their job. I don't know if they are really smoothing the current or if it's saturating. When I weld with or without the inductor, I don't see a difference. The only thing I noticed is that when I weld, if there is a piece of metal near the inductor, it will stick to it. Thanks

Did you choose Courier font to make the drawings? Notepad works just as well for this task.

Did the one I corrected look ok at your end?

helpful to me and

One does not analyze saintly behavior. One just admires it.

I made a large inductor (and rectifier) to add to my AC stick welder, but it works completely differently with the choke. It definitely puts out less current at a given setting, and the arc is quieter and the slag splatters a lot less. And, of course, it makes a difference in the weld penetration if the work or the rod is positive.

Either your inductor is not large enough, is saturating, or something else is wrong. I hope you do have the choke downstream of the rectifier (between the rectifier and the arc).

I used Times News Roman font.

Actually my drawing, on the forum looks OK, yours looks all screwed up, but that's OK, I guess you get the picture. The drawing with the magnetic shunt looks like a window with 4 square panes and the one without the magnetic should like 2 long panes.

Yes the inductor is between the diodes and the arc. It could be that the choke isn't big enough, since I used ordinary #6 wire with insulator, I could only fit around 15 turns of wire. How is your choke constructed?

Thanks

That font uses a variable space per character, so they do not line up in a nice XY grid. If you construct and view with courier (or fixed width) the drawings always look the same to everyone. I'll bet you can set your newsreader to fixed width font and all the drawings everyone posts here will look fine. There is also a neat schematic drawing program that does all the work for you that you can download for free.

That is what I drew with a fixed width font.

Mine is made with two stacks of E laminations about 4 by 6 inches. Each stack is about 2 inches thick. Both stacks are wound with 80 turns of 1/4 by /18th rectangular enameled wire. I intended to run them in series for low current and in parallel for high current, but I have always run them in series, so that makes 160 tuns. There is a

When I break the arc, it pulls a pretty long spark before quenching.

This is NOT a Web-based **forum**. This is NOT "Google Groups". This is USENET: http://66.102.9.104/search?q=cache:Gnnj10AfkBUJ:bibleocean.com/OmniDefinition/Usenet+follow-Usenet-customs-and-*-rules+*-a-service-for-*-*-*-*+hide-the-fact-*-*-*-they-are-*-on-Usenet+concerns-*-*-*-*-about-the-Google-interface+what.is.wrong+*-*-*-*-*-*-*-*-*-*-*-*-*-is-now-legendary+*-*-savvy+Google-cannot-muster-*-*-*-*-*-*-*-*-*-*-*-*-*+*-*-policed&strip=1

As Popelish has been trying to tell you, ONLY IF YOU USE A PROPORTIONAL FONT.

MOST people don't access Usenet thru Google Groups. Most people don't see what you see. Most people use a **newsreader**. Most people set their newsreaders to display in monospaced type.

DON'T use M$Word for ASCII art. Use NotePad (a Courier font). Cut & paste that to the posting page and don't worry how it looks on Google's proportionately-spaced page. Google's latest "improvement" removed the fixed-width/proportional font option.

If you are going to continue to use Usenet, you might think about getting a real newsreader

--like using Thunderbird/SeaMonkey to access Usenet. http://66.102.9.104/search?q=cache:aHUJnR199ugJ:en.wikipedia.org/wiki/List_of_news_clients+*-Gravity+Xnews+Pan+trademark+Disclaimers+last.modified+*-Agent+nonprofit+Thunderbird

M. JeffM

I guess I'm green at this. What's Usenet and what is a newsreader?

M. Popelish

This is a big inductor. Where do you get that kind of enameled wire? Is there an easy way to measure or figure out the value of the primary and secondary inductance of a transformer? Also, is there a way to "see" if a transformer or a choke is saturating when in a circuit. Because I want to make big chokes I want to know if the air gap that I add is working according to calculation as opposed to "real life" situation.

Thanks

Click the links I provided and read the pages.

Here's ANOTHER page you need to read: http://66.102.9.104/search?q=cache:ohQq8nqqThwJ:groups-beta.google.com/support/bin/answer.py?answer=46492%26topic=9253+remove-*-*-*-irrelevant+Usenet+*.*.*.*.*-*-*-*-relevant.*+2006+be.interested+your.reply+Summarize.what.you're.following.up+before.the.original.*.*+*-for-your-readers

Mr Phantom

This is a hobby project. A couple of years ago, a friend of mine, developed a kind of magnetic amplifier designed to couple electrical networks on a very large scale. It is meant to be used by the utility companies to connect the high voltage main power line to the street grid's lower voltage or to connect one utility company to another and synchronize them with almost no loss. The idea is a connection that is almost 100% efficient. He is working with a local university. He needed some way of controlling one of the winding at different frequencies and with different pulse width. I made a little circuit for him with a PIC Microcontroller. The deal was that I would assist him in the logic part and he would help me in the power part. Once I delivered my part, it was almost impossible to get him interested in my project so that's why I'm seeking higher intelligence on this forum. I don't have a Variac but I could probably borrow one from my friend. I mostly want to know if an inductance saturates because I made a few inductor and since I used MOT's core I had to add an air gap. According to my calculation with the current I intend to feed through the inductor, the air gap should be .035", but in real life, it might be too big or not big enough. From want I understand, if the air gap is too small, the inductor will saturate and, if it is too big, it's effect will be reduced. I guess the method with the Variac is better than the method of feeding one winding and measuring the voltage on the other side and vice versa? The semiconductor I am using for my tests are IGBTs from International Rectifier part# IRG4PC50KD. I understand that these might not be the best for what I am doing but I got them real cheap on E-Bay and I got them to try my prototype to see if it had any chance to work.

Thanks

The method of measuring the voltage on each winding while exciting the other is for finding the coupling coefficient.

Using the variac is for detecting impending saturation.

In another post you said, "When the "k" factor is less than 1, the amount of energy, going through the freewheel diode, when one side of the bridge is turned off and before the other side is turned on, is quite large and I am afraid it would destroy the diode."

These parts have a built-in anti-parallel diode. The diode is rated to carry the same current as the IGBT in the forward direction. If a high voltage is applied to the IGBT-Diode combination when the IGBT is turned off, that can damage the part. But, in an H-bridge configuration, the decaying current from the leakage inductance is clamped to the power supply voltage, and the current won't be any larger than the forward current in the IGBT was. Thus, no damage will occur from that cause, because the diodes are rated to carry that current. This all assumes that you don't short-circuit the output while the bridge is running.

However, stray inductances in your wiring arrangement, such as excessively long wires among the IGBT's and the transformer, etc., can cause spikes that may need to be snubbed.