Coupling coefficient of industrial transformers

[orvillefpike]

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.






- Show quoted text -- Hide quoted text -

- Show quoted text -

M. Resenbaum

From what I know, to get the transformer away from saturation, I would

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

 

[The Phantom]

M. Resenbaum

have to take a few turns out of the primary

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

(which I don't want to do)
or lengthen the core

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.

(which would be complicated) or reinstall the
magnetic shunt, am I right? Is there other way to achieve that?

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.

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.

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.

 

[orvillefpike]

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 atransformer.






- Show quoted text -- Hide quoted text -

- Show quoted text -

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

 

[The Phantom]

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.

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?

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

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

http://www.fnrf.science.cmu.ac.th/theory/magnets/Producing wound components.html

http://www.edn.com/article/CA426069.html

http://www.eastern-components.com/pdf/ferrite-int/tech5.pdf

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

http://www.epanorama.net/links/componentinfo.html#coils

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.

 

[orvillefpike]

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?

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

 

[orvillefpike]

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

 

[orvillefpike]

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

 

[The Phantom]

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?

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.

What frequency do I use to make my calculations in DC since I am
rectifying AC? Would it be 120 Hz?

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.

 

[orvillefpike]

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.




- Show quoted text -- Hide quoted text -

- Show quoted text -

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

 

[The Phantom]

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

Like a straight wire with resistance.

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

You will need to use the DC current value to determine where on the
hysteresis loop you are operating, and then from the slope of the B-H curve
at that point you can calculate the "inductance". I put the word
inductance in quotes because when you are operating an iron core inductor
near (or into) saturation, the inductor operation becomes non-linear and
the concept of inductance is no longer appropriate. If the ripple current
is nearly as large as the DC component, then one could say that the
inductance is not constant. It's as though it varies from a small value
when the operating point is well into saturation as the ripple current
reaches its maximum, to a larger value when the operating point comes out
of saturation and the ripple current reaches its minimum. You could use
the average value of the "inductance" in calculations for a first try at
analyzing the inductor. The usual methods of analyzing a circuit where
linearity applies can no longer be used to give accurate results. You
would have to simulate in Spice. Then you don't even specify "inductance";
you use a non-linear inductor model and specify its physical parameters.

A certain amount of trial and error might work best for you. Use the
dimensions of the core and the known B-H curve for silicon steel to
calculate the flux density in the core for your desired DC current. Pick a
number of turns so that the DC current plus the expected maximum ripple
current won't push the core too far into saturation. If the size wire you
have to use to get that many turns on the core is too small, then the
operating current may cause it to overheat, and you will need a bigger
core. If the wire isn't going to overheat, then see if you get enough
ripple reduction in actual operation. If not, then, again, you will need a
bigger core. And, of course, the air gap will greatly affect all of this.
It's not easy, especially when you're trying to use a core you already have
on hand, instead of selecting one that calculations show will meet your
requirements.

 

[orvillefpike]

M. Phantom

I have an air gap in the inductor, I had to put one because the core
can't handle much more than 3 or 4 amps before saturating and I want
to flow 15 to 20 amps. All I want to do is filter the rectified AC. If
the inductor is not big enough, it's not a big deal, I can modify it
or put another one, I just want to try and avoid the problems that
could happen if it goes into saturation.

Thanks