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Coupling coefficient of industrial transformers

T

Tom Bruhns

Jan 1, 1970
0
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
 
O

orvillefpike

Jan 1, 1970
0
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
 
J

John Popelish

Jan 1, 1970
0
orvillefpike wrote:
(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.
 
T

Tom Bruhns

Jan 1, 1970
0
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


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
 
T

The Phantom

Jan 1, 1970
0
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.
 
T

The Phantom

Jan 1, 1970
0
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.

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

orvillefpike

Jan 1, 1970
0
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
 
T

The Phantom

Jan 1, 1970
0
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 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.
 
O

orvillefpike

Jan 1, 1970
0
Mr. Phantom
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
 
O

orvillefpike

Jan 1, 1970
0
Also some talk about snubbers and others talk about clamps.

Thanks
 
T

Tim Williams

Jan 1, 1970
0
When you're using a half or full bridge with inductive loads, use MOSFETs or
co-pack IGBTs and forget about it. The flyback pulse harmlessly bumps into
the supply rails (peak current is lesser than or equal to the current going
through the opposing transistor when it turned off). Anything else that
happens, like ringing when neither transistor is conducting any current, is
just aesthetics.

Tim
 
O

orvillefpike

Jan 1, 1970
0
Tim,

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

Thanks
 
T

Tim Williams

Jan 1, 1970
0
orvillefpike said:
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.

Tim
 
O

orvillefpike

Jan 1, 1970
0
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.

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
 
T

Tim Williams

Jan 1, 1970
0
orvillefpike said:
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.

Tim
 
J

jasen

Jan 1, 1970
0
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
 
O

orvillefpike

Jan 1, 1970
0
Tim,

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

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

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

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

Thanks

Jasen,

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

Thanks
 
O

orvillefpike

Jan 1, 1970
0
Mr. Phantom

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

I have made some test and took some measurement on various transformer
and here is what I got.
For my Micro Wave Transformer, the resistance of the primary winding; .
1 Ohms, the current in the primary winding at 125 Vac; 5 Amps, the
impedence of the primary; 25 Ohms, the inductance of the primary; .066
H, the inductance of the secondary (3:1 winding ratio); .007 H,
coupling coefficient according to Mr. Phantom's method; .97.
For my 15 Kva transformer, resistance of the primary winding; 1.5
Ohms, the current in the primary winding at 125 Vac; .8 Amp, the
impedence of the primary; 152 Ohms, the inductance of the primary; .4
H, the inductance of the secondary (5:1 winding ratio);
..016 H, the coupling coefficient according to Mr. Phantom's method; .
9875. Does anybody see anything wrong with these numbers?

Thanks
 
O

orvillefpike

Jan 1, 1970
0
Looking at your numbers below, it seems that you probably don't have an
inductance meter, but are deriving the inductance of a winding from the ratio of
the voltage across it and the current through it.

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

Start out with not much DC current and calculate the inductance by that
method. As you increase the DC current, measure and calculate the inductance
again. You can detect the onset of saturation as the value of DC current where
the inductance begins to decrease. Saturation is a gradual thing, of course,
and you will see the calculated inductance gradually decrease as the DC current
increases.
When you were making these measurements on the primary of the two
transformers, the secondaries were unloaded, right?

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


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
 
R

Rodger Rosenbaum

Jan 1, 1970
0
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.

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.
 
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