Coupling coefficient of industrial transformers
orvillefpike
commercial type, transformers like the ones used in shops, like a
three phase 15 Kva or a 30 Kva?
Thanks
Tim Williams
Tim
--
"Librarians are hiding something." - Steven Colbert
Website @ http://webpages.charter.net/dawill/tmoranwms
"orvillefpike" <orvill...@yahoo.com> wrote in message
news:1175963054....@w1g2000hsg.googlegroups.com...
Tom Bruhns
So, measure one.
I'd bet it's higher than Tim's guess. I've measured RF transformers
with better coefficients than .99, and with the much higher
permeability of mains-frequency cores, it should be easy to get over .
99. Admittedly, the RF transformers are wound specifically for tight
coupling.
Cheers,
Tom
orvillefpike
How do I measure it?
In the simulation, if the transformer has .990, it's a lot different
than .995, for example.
Thanks
BFoelsch
wrote:
Exactly what are you trying to calculate/simulate? Coefficient of
coupling is not a useful concept when discussing power transformers;
it is assumed to approach 1.
The Phantom
wrote:
I can't speak to the size transformer you have cited, but a 3 kW single
phase transformer I have on hand has a measured k of .999883
orvillefpike
> On 7 Apr 2007 09:24:14 -0700, "orvillefpike" <orvillefp...@yahoo.com>
> wrote:
>
> >Would anybody know what is the coupling coefficient of step-down,
> >commercial type, transformers like the ones used in shops, like a
> >three phase 15 Kva or a 30 Kva?
> >Thanks
>
> I can't speak to the size transformer you have cited, but a 3 kW single
> phase transformer I have on hand has a measured k of .999883
Because I am feeding it with a square wave, the shape and the
amplitude of the output is very different whether the coupling
coefficient is .990or .995.
Thanks
john jardine
"The Phantom" <pha...@aol.com> wrote in message
news:k0gi13l861fuolmhh...@4ax.com...
Thanks for that figure!.
It's the first time I've -ever- seen a real "k" quoted for power equipment.
When forced to, I use an arbitrary value of 0.999 as it's physical symptoms
seem to correspond well with reality but have [was!] always been leery of
working with a number so near to 'perfection'.
--
Posted via a free Usenet account from http://www.teranews.com
orvillefpike
> wrote:
>
> >Would anybody know what is the coupling coefficient of step-down,
> >commercial type, transformers like the ones used in shops, like a
> >three phase 15 Kva or a 30 Kva?
> >Thanks
>
> I can't speak to the size transformer you have cited, but a 3 kW single
> phase transformer I have on hand has a measured k of .999883
Also, the smaller the coupling coefficient, the bigger the leakage
inductance, which causes voltage spikes. Once I know the magnitude of
the voltage, I can design a snubber circuit if there is a need for it.
Thanks
The Phantom
wrote:
There have been threads on this topic fairly recently.
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')
I just did this measurement on a 25 VA filament transformer and got k =
.99767
Tom Bruhns
OK, if your simulation shows different results if it's .99 versus if
it's .995, that's exactly a clue how to measure it. Simulate a
circuit you can measure, and trim the simulation until it matches the
measurement.
One way to do this: remember that the coefficient of coupling is the
fraction of magnetic field shared by two coils. If the two coils have
exactly the same number of turns, and you apply 1V to L1, assuming
that you load L2 very lightly, and that the resistance of L1 is low
enough that there is insignificant drop in that resistance due to the
current through L1 when it's excited, the voltage you'll see at L2
will be just the coefficient of coupling. But of course there's no
guarantee the number of turns will be EXACTLY the same. However, if
you measure L2 with L1 excited, and then measure L1 with L2 excited,
you can resolve both the coupling coefficient and the turns ratio.
You can add measurements to resolve other things: you can change
frequency to see effects due to resistance of the windings and
capacitance across the windings.
Another way: you can measure the leakage inductance and figure out
the coupling from that. It likely will be important to also know the
AC resistance of the windings at the frequency of measurement, though.
I don't claim to have given you a recipe here...only hints. Keep your
wits about you and account for parasitic effects like winding
resistance and capacitance, and possibly even nonlinearities in the
core.
Cheers,
Tom
orvillefpike
> wrote:
>
> >Would anybody know what is the coupling coefficient of step-down,
> >commercial type, transformers like the ones used in shops, like a
> >three phase 15 Kva or a 30 Kva?
> >Thanks
>
> I can't speak to the size transformer you have cited, but a 3 kW single
> phase transformer I have on hand has a measured k of .999883
How did you come up which such a precise number?
Thanks
orvillefpike
> On 8 Apr 2007 12:46:42 -0700, "orvillefpike" <orvillefp...@yahoo.com>
>
> - Show quoted text -
Would this method work if I don't feed the transformer at its nominal
voltage, because this transformer was ment to be connected at 600
Volts on its primary side.
Thanks
The Phantom
Because the permeability of the silicon steel core varies somewhat with flux
density, k will vary a little with excitation level, but I think you will get
usable results with a reduced excitation level.
The Phantom
Don't be fooled by the 3 leading 9's. That number only has 3 significant
digits.
And with an iron core transformer, such a measurement is probably not
repeatable to 3 digits, but that's the result of the measurement at the time.
Temperature and magnetic history of the core can affect the measurement.
orvillefpike
> On Apr 8, 7:58 am, "orvillefpike" <orvillefp...@yahoo.com> wrote:
>
>
>
>
>
> > On Apr 7, 10:15 pm, "Tom Bruhns" <k...@msn.com> wrote:
>
> > > On Apr 7, 9:24 am, "orvillefpike" <orvillefp...@yahoo.com> wrote:
>
> > > > commercial type, transformers like the ones used in shops, like a
> > > > three phase 15 Kva or a 30 Kva?
> > > > Thanks
>
> > > So, measure one.
>
> > > I'd bet it's higher than Tim's guess. I've measured RF transformers
> > > with better coefficients than .99, and with the much higher
> > > permeability of mains-frequency cores, it should be easy to get over .
> > > 99. Admittedly, the RF transformers are wound specifically for tight
> > >coupling.
>
> > > Cheers,
> > > Tom
>
> > How do I measure it?
> > than .995, for example.
> > Thanks
>
> OK, if your simulation shows different results if it's .99 versus if
> it's .995, that's exactly a clue how to measure it. Simulate a
> circuit you can measure, and trim the simulation until it matches the
> measurement.
>
> fraction of magnetic field shared by two coils. If the two coils have
> exactly the same number of turns, and you apply 1V to L1, assuming
> that you load L2 very lightly, and that the resistance of L1 is low
> enough that there is insignificant drop in that resistance due to the
> current through L1 when it's excited, the voltage you'll see at L2
> guarantee the number of turns will be EXACTLY the same. However, if
> you measure L2 with L1 excited, and then measure L1 with L2 excited,
> You can add measurements to resolve other things: you can change
> frequency to see effects due to resistance of the windings and
> capacitance across the windings.
>
> Another way: you can measure the leakage inductance and figure out
> AC resistance of the windings at the frequency of measurement, though.
>
> I don't claim to have given you a recipe here...only hints. Keep your
> wits about you and account for parasitic effects like winding
> resistance and capacitance, and possibly even nonlinearities in the
> core.
>
> Cheers,
>
> - Show quoted text -
Correct me if I am wrong but I would think that it's even harder to
"measure" the leakage inductance. I don't think that you could
"measure" the leakage inductance without figuring it out from other
measurements in some kind of test under certain condition.
Thanks
Tim Williams
Tim
--
"Librarians are hiding something." - Steven Colbert
Website @ http://webpages.charter.net/dawill/tmoranwms
"orvillefpike" <orvill...@yahoo.com> wrote in message
news:1176332328....@d57g2000hsg.googlegroups.com...
Tom Bruhns
As Tim wrote, short the secondary, measure the primary. That's not
quite all there is to it, since you can't really short the secondary
inductance; you're putting a resistance equal to the winding
resistance across it. But yes, you can do it if you think about it
carefully.
I think the measurement of the secondary voltage with the primary
excited, and vice-versa, is a better way, though there you technically
need to compensate for the drop in the resistance of the excited
winding because of the current through the winding. That is, the
voltage across the pure inductance is less than what's applied to the
winding. I'm not sure you got a proper answer to the question about
how to measure the coupling so precisely. Consider if the transformer
is 1:1; you could connect the windings so that you only have to
measure the difference between them to know how much lower the
secondary is. You do need to account for the case where the
transformer is, say, 1.001:1 turns ratio. Then when you reverse the
windings, be sure that you know which winding has the higher voltage.
You loose the polarity of the difference when you're only measuring an
AC amplitude. If the transformer isn't 1:1, you can still do it if
you use an accurate voltage divider... -- I haven't actually done
this with mains-frequency transformers, so I may be missing some
practical aspects...I normally work with things at 100kHz up into many
MHz, where there are ways to deduce the coupling, also, and I do have
some experience with those.
Cheers,
Tom
The Phantom
wrote:
>
>Correct me if I am wrong but I would think that it's even harder to
>"measure" the leakage inductance. I don't think that you could
>"measure" the leakage inductance without figuring it out from other
>measurements in some kind of test under certain condition.
>Thanks
Go look up the thread with the subject line:
"About Leakage Inductance in Transformers"
that I started on March 1, 2007. Read my first post and Tony Williams'
response to my questions.
Each winding in the transformer has a leakage inductance associated with
it, and determining the individual leakage inductances from measurements is
difficult with an iron cored mains frequency transformer. Shorting one
winding and making measurements at another winding gives a result that
combines the effect of the separate leakage inductances, and very often
this is all that is needed.
The Phantom
Are you referring here to the post where, after I gave the OP a k of
.999883, he said:
"How did you come up which such a precise number?"
I suspected that he saw the number .999883 and thought, six significant
digits. I wanted him to know that that value only has three significant
digits.
If the question is, how did I get even a three significant digit
measurement, that's easy to do with good DVM's using the method I described
to him (and which you are recommending).
On the other hand, trying to get more than about 1 significant digit by
some method involving leakage inductance will be difficult with a mains
frequency iron core transformer.
But if you ask me if I believe the k value of .999883 is *accurate* to 3
significant digits, that's another question. The OP didn't ask that; he
just asked how I got "such a precise number", and I wanted to be sure that
he understood that .999883 is not *precise* to 6 digits. :-)
Trying to get repeatable measurements from a mains frequency iron core
transformer is not easy. I find that if I just try to measure the
self-inductance of a winding at 60 Hz and some excitation level, the
reading will drift for many minutes, sometimes taking 5 minutes or more
before the measurement is stable to 3 digits. Apparently the initial
transient of connecting the meter tweaks the core and it takes a while to
relax, and if the transformer has just been connected to line power, it can
take even longer!
I mentioned that flux density at the excitation level of the measurement,
temperature and magnetic history of the core could affect measured k. How
much would depend on the particular transformer, of course.
The 1943 book I refer to in the earlier thread says it well (about a method
for measuring leakage inductance):
"...this method is inherently inaccurate when used with *measured* values
of the self- and mutual inductances of iron-core transformers. The leakage
inductance of one winding of such a transformer often may be as small as
0.2 per cent of its self-inductance. For example, if the self-inductance
of winding 1 is 10 henries, its leakage inductance may be about 0.02 henry.
If the value of the leakage inductance is to be determined from Eq. 91, to
the nearest millihenry--or within about 5 per cent of its true value--the
value of the self-inductance must measured to the nearest millihenry, or
within 0.01 per cent of its true value, and the mutual inductance must be
measured with the same per cent accuracy. Such precise measurements are
impossible with iron-core transformers."
Tom Bruhns
...
> On the other hand, trying to get more than about 1 significant digit by
> some method involving leakage inductance will be difficult with a mains
> frequency iron core transformer.
>
...
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
orvillefpike
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
John Popelish
(snip)
> 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.
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.
> How would I
> know if it's saturating or close to saturating.
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.
Tom Bruhns
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
John Popelish
> 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.
I already knew that. How is it working 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.
Is there an air gap on one or both sides of the shunt
blocks, or are they part of the core laminations?
> Once the magnetization field
> is established, I believe that no matter what the demand is on the
> secondary side, it cannot saturate the transformer.
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.
> 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.
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.
> 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.
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.
> Thanks
>
orvillefpike
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
John Popelish
> 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.
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
> 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....
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.
> 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.
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.
> 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.
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.
> I have about 20 MOT and
> they all have this magnetic shunt in them.
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.
orvillefpike
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
John Popelish
> 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.
Did you choose Courier font to make the drawings? Notepad
works just as well for this task.
> In one of the two posts, the drawing looks decent, at my
> end anyway.
Did the one I corrected look ok at your end?
> 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?
One does not analyze saintly behavior. One just admires it.
> 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.
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).
orvillefpike
> works just as well for this task.
I used Times News Roman font.
> Did the one I corrected look ok at your end?
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.
> 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).
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
jpop...@rica.net
> > Did you choose Courier font to make the drawings? Notepad
> > works just as well for this task.
>
> I used Times News Roman font.
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.
http://www.tech-chat.de/aacircuit.html
The web [age is in German, but the program displays English.
> > Did the one I corrected look ok at your end?
>
> 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.
That is what I drew with a fixed width font.
> > 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).
>
> 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?
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
1/8th inch gap between the halves. I also drilled the center leg of
both stacks with a pair of small holes, to make a narrow spot that
would saturate before the rest of the core, effectively increasing the
gap, to lower the inductance in a step, rather than having it collapse
as the whole core saturated, all at once. I normally use it with the
welder set to around 100 amps.
When I break the arc, it pulls a pretty long spark before quenching.
JeffM
>Actually my drawing, on the forum looks OK,
>
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
>yours looks all screwed up,
>
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.
*THIS* is what MOST people see:
http://groups.google.com/group/sci.electronics.design/msg/ce7b4d35448181b9?dmode=source
http://groups.google.com/group/sci.electronics.design/msg/ce7b4d35448181b9&fwc=1
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
orvillefpike
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
JeffM
>M. JeffM
>What's Usenet and what is a newsreader?
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
orvillefpike
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 Phantom
wrote:
>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
The idea is to short one winding and *measure* the inductance and
resistance at the other winding with a meter, such as this:
http://www.bellnw.com/products/0650/
>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.
You will get better results if you use this method rather than trying to
measure the leakage inductances.
>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.
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 (http://www.variac.com/) so that you can apply a low
variable voltage to any winding of a transformer under test.
>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.
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.
>
>Thanks
The Phantom
wrote:
What kind of semiconductor devices are you using in your bridge? Are
they MOSFET's? What is the manufacturer and part number?
orvillefpike
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 Phantom
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.
>The semiconductor I am using for my tests are IGBTs from International
>Rectifier part# IRG4PC50KD.
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.
orvillefpike
> The method of measuring the voltage on each winding while exciting the
> other is for finding thecouplingcoefficient.
>
> Using the variac is for detecting impending saturation.
Sorry, I wasn't paying attention.
> 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 thetransformer, etc., can cause spikes
> that may need to be snubbed.
In the simulation I made with the transformer's coupling coefficient
of .99, which should be realstic, the voltage spike goes above the
voltage DC supply, not by much, still under the IGBT's rated voltage,
but I would rather be safe than sorry. In real life that voltage could
be above the rated voltage of the IGBT. I have been reading a
technical paper on "Switching Voltage Transient Protection Schemes for
High Current IGBT Modules" (tpap-6.pdf) from International Rectifier,
but it is not clear to me and it looks like a copy that was made in
the '50s. I don't realy know if I need a snubber circuit and the few
articles I have been reading are contradicting. Some say you need a
snubber to protect the IGBT from voltage spike some say you need a
snubber to protect from di/dt.
Thanks again
orvillefpike
Thanks
Tim Williams
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 @ http://webpages.charter.net/dawill/tmoranwms
"orvillefpike" <orvill...@yahoo.com> wrote in message
news:1177702171.8...@s33g2000prh.googlegroups.com...
orvillefpike
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
Tim Williams
> 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.
1. The voltage should be going no more than, say, 1-1.5V above the rail.
This is the forward voltage of the co-pack diode; see data sheet. On the
oscilloscope it may appear more due to inductance (in which case, you are
only looking at the wrong ground point, relative to the transistor). I
would say no more than 5V overshoot, or you have something really fucked up
with your bridge, which must be low inductance and well bypassed with low
inductance capacitors.
2. The co-pack diode, or an externally added one, certainly ought to be the
same rating as, or better than, the transistor it's protecting.
3. dV/dt shouldn't be causing any troubles whatsoever because your gate
driver should have a low impedance output.
> What did you use, as an isolator, between the turns of the inductor
> you made with strips of cooper pipe as the conductor?
Insulator, not isolator. Paper. Heavy mylar tape would be better, but I
don't have any.
orvillefpike
> This is the forward voltage of the co-pack diode; see data sheet. On the
> oscilloscope it may appear more due to inductance (in which case, you are
> only looking at the wrong ground point, relative to the transistor). I
> would say no more than 5V overshoot, or you have something really fucked up
> with your bridge, which must be low inductance and well bypassed with low
> inductance capacitors.
>
> 2. The co-pack diode, or an externally added one, certainly ought to be the
> same rating as, or better than, the transistor it's protecting.
>
> 3. dV/dt shouldn't be causing any troubles whatsoever because your gate
> driver should have a low impedance output.
Tim
My simulation circuit is very simple. It has a 125Vac 60Hz supply, a
diode bridge, a capacitor and inductor filter, an IGBT H-bridge
feeding a transformer and the secondary of the transformer is
connected to a resistor. The IGBTs have a freewheel diode. I can only
go so big for the capacitor in the filter.
When there is no leakage inductance, in my transformer, there is no
voltage kick back and the voltage doesn't go above the rail. If there
is a 1% leakage inductance, the voltage goes above the rail by about
40 Volts. Since I don't exactly now the specs of my transformer, I
don't know exactly what to expect in terms of the size of the voltage
spike. I would like to avoid learning about it by blowing IGBT.
I don't mind adding a snubber, if I need one, but I don't know much
about how to design them and I have tried adding them to my simulation
circuit but they don't seem to make a difference. If I could find a
simple article on the use and design of snubbers, it would help. I
know there is probably no simple explanation about snubbers but the
few articles I have read about them have different approach .
Thanks
Tim Williams
> My simulation circuit is very simple. It has a 125Vac 60Hz supply, a
> diode bridge, a capacitor and inductor filter
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.
> an IGBT H-bridge
> feeding a transformer and the secondary of the transformer is
> connected to a resistor. The IGBTs have a freewheel diode. I can only
> go so big for the capacitor in the filter.
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.
> When there is no leakage inductance, in my transformer, there is no
> voltage kick back and the voltage doesn't go above the rail. If there
> is a 1% leakage inductance, the voltage goes above the rail by about
> 40 Volts.
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.
jasen
>
> My simulation circuit is very simple. It has a 125Vac 60Hz supply, a
> diode bridge, a capacitor and inductor filter, an IGBT H-bridge
> feeding a transformer and the secondary of the transformer is
> connected to a resistor. The IGBTs have a freewheel diode. I can only
> go so big for the capacitor in the filter.
> When there is no leakage inductance, in my transformer, there is no
> voltage kick back and the voltage doesn't go above the rail. If there
> is a 1% leakage inductance, the voltage goes above the rail by about
> 40 Volts.
there's likely less than 1% leakage, but how clean is your 125V?
> Since I don't exactly now the specs of my transformer, I
> don't know exactly what to expect in terms of the size of the voltage
> spike. I would like to avoid learning about it by blowing IGBT.
> I don't mind adding a snubber, if I need one, but I don't know much
> about how to design them and I have tried adding them to my simulation
> circuit but they don't seem to make a difference. If I could find a
> simple article on the use and design of snubbers, it would help. I
> know there is probably no simple explanation about snubbers but the
> few articles I have read about them have different approach .
maybe get a variac and bring the voltage up slowly.
Bye.
Jasen
orvillefpike
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
orvillefpike
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
Rodger Rosenbaum
>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.
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.
>
>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?
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.
>
>Thanks
>
orvillefpike
> 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.
> transformers, the secondaries were unloaded, right?
>
> The MWT would seem to be well into saturation, but otherwise the numbers seem
> reasonable.
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
Rodger Rosenbaum
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.
>
>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?
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.
>
>Thanks
orvillefpike
>
> - 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
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.
>
>Thanks again
>
orvillefpike
> >> >the value of H at around 1,500, will it cause a problem in the
> >> >performance of thetransformer?
>
> >> current will be high, as you have noted in the MWTtransformer. Thecoupling
> >> primary will reduce the efficiency of thetransformer(increased temperature
> >> rise as a result) and will degrade its regulation. But, it will still perform
> >> as atransformer.
>
> >> those caused by saturation. The no load excitation current will go up, causing
> >> extra losses.
>
> >> >Thanks- Hide quoted text -
>
> >> - Show quoted text -- Hide quoted text -
>
> >> - Show quoted text -
>
> >M. Resenbaum
>
>
> - 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
<SNIP>
>
>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%20wound%20components.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
> 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
> 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%20wound%20...
orvillefpike
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
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
>On May 27, 1:52 pm, orvillefpike <orvillefp...@yahoo.com> wrote:
>> 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?
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.
>Thanks again
orvillefpike
>
> - 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
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.
>
>Thanks
orvillefpike
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