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Effects of gaps in inductors and transformers

T

Terry Given

Tony said:
Nothing to do with a magnetic amplifier, which is
the active controlling of AC or DC power via a DC
control winding.

A ferroresonant CVT is a passive device and is a
clever design of a core, with a non-saturating
primary limb, a magnetic shunt, and a saturating
secondary limb. The winding on that secondary limb
inherently produces a constant output voltage.
The output is square, (mainly third and fifth
harmonics), and the capacitor is there to resonate-
-out those harmonics. There are even more clever
designs that use compensating windings to reduce the
secondary harmonics.

The CVT is adjusted by "shimming the shunt". I've seen
it done on some special CVT's we were buying, and that
is definitely a black art.

<http://www.research.ibm.com/journal/rd/316/ibmrd3106H.pdf>

Wow, great link, thanks!

I have the following magamp books:

Nonlinear magnetic control devices, Geyger
Magnetic amplifiers, Storm

these two have small sections on CVTs.

Magnetic amplifier engineering, Attura
Magnetic amplifier analysis, LaFuze
Magnetic amplifier circuits, Geyger
Magnetic amplifiers, Platt
Square Loop Ferrite Core Switching, Neetson
Square Loop Ferrite Core circuitry, Quartly
The Magnetic amplifier, Reyner

none of these even mention ferro-resonant CVTs.

OK its not an exhaustive list of magamp books, but its gotta be getting
close.

Cheers
Terry
 
T

Terry Given

Paul said:
OK. That is very helpful in understanding the principles involved. The gaps
I saw in some large C-core inductors I have are about 0.1", and the
laminated steel has a length of about 10". So if the magnetic path length
is increased to 1000, that is a 100 fold decrease in inductance, allowing
100 times the ampere turns, and thus 100 times the energy. This is a 100 mH
inductor rated about 10 amperes.

usually when designing a flyback style transformer, the input & output
voltages and max duty cycle set the turns ratio.

then choose a core, a switching frequency and the peak flux density (<
Bsat at max Tcore).

then select the primary turns to give the desired Bmax,

Np = Vin*Ton/(Bmax*Ae)

then choose the air gap to give the correct inductance. If the air gap
is -ve, you've picked the wrong core. And unless the gap is very small,
just assume all of the energy is stored in the gap, and use

Lmag = mu_0*Np^2*Ae/l_gap so

l_gap = mu_0*Np^2*Ae/Lmag so
 
T

Terry Given

This sounds like nonsense to me. For simple, centre-tapped windings a
high leakage inductance, corresponding to poor coupling between the
two sides of the centre-tapped winding, leads to a higher turn-off
dissipation, which can be a major source of lost efficiency, not to
mention over-heated switching transistors.

What you seem to be thinking off would seem to be shoot-through
losses, which are best dealt with by a break-before-make switch
driver.

nah, peak current mode control cures all manner of ills. And of course
some design skill ;)

Cheers
Terry
 
T

Tony Williams

[/QUOTE]
Wow, great link, thanks!

Isn't it just. Best I could find quickly.
I have the following magamp books:
Nonlinear magnetic control devices, Geyger
Magnetic amplifiers, Storm

My (one and only) mag amp bible is:-

"Self Saturating Magnetic Amplifiers"
by Lynn, Pula, Ringelman, and Timmel.
Published in 1960 by McGraw-Hill.
Lib Congress Card Number = 60-6979.

It is a serious tome, and the authors were apparently
staffers at the Westinghouse Electric Corp, Air Arm
Division, Baltimore. Inventors of "mag amp?".

Looks like I borrowed it from the Decca Radar Research
Laboratories, (in 1967), and forgot to return it. :)

Amazon has one copy, at $80.
none of these even mention ferro-resonant CVTs.

I don't see why a mag amp book should mention the
ferroresonant CVT. OK, the CVT does have a limb
that saturates, but there is no control winding
and it operates as a passive device.
 
G

Genome

[snip]
Thanks for any thoughts and discussion.

Paul

You need the following sums

L = uoueN^2Ae/Le

B = uoueNI/Le

H = NI/Le

dI = VinTon/L


You are not allowed to use one sum without thinking about how it might
affect another one. You avoid that sort of sophistry by using all the sums
to find your answer.

The B field don't depend on the core.

Choice 1) Do Sums.
Choice 2) Use sophistry to prove it does.

Subsequently

Choice 1) Problem Solved.
Choice 2) Invoke 'Ad Hominem' Attack strategy.

A gap affects ue. There is something about permeance and reluctance which is
something like conductance and resistance. The gap appears in series with
things so/but you get the answer by adding the reluctances.

Unfortunately designing magnetic components is hard because there are too
many variables to play with.

N = LI/BAe

Works

But then you might need

AwAe = LIpkIrms/BpkJK

To get a first guess at what the size of your core should be. J and K are
the guesses.... and then Bpk ignores the core loss so you might adjust
something else for another guess. J ignores the winding losses so..... you
might have to guess again.

Do you trust the software?

Leakage inductance has nothing to do with uncoupled flux. It's down to
energy stored in the field(s) between the windings, series term.

If you want to control stuff then you might control the coupling of the flux
and that is a parallel thing.

DNA
 
J

John Larkin

The plasma, and the molten steel are a resistor. The current going
though said resistor gets dissipated as heat at the resistor site,
which is the tip, and the work (weld) location.

So tha plasma is a short, and the plasma is a resistor.

Thanks for explaining that to us.

John
 
M

MassiveProng

It's mostly in the voltage drop of the arc, mainly in the cathode fall
and anode fall regions of the arc.
True, the arc will not limit current, but it has a voltage drop. The
arc somewhat simulates a hypothetical zener diode with a negative
temperature coefficient.

It also has to be in the work contact area itself as well, otherwise
there would be no heat, no puddles of steel to weld with.
 
M

MassiveProng

Correct. Missing Prong couldn't even figure that, the easy part, out.

I stated it, dumbfuck!

I also stated that despite the SMALL loss such a gap makes, the
benefits outweigh it.

You simply ignored the response, you retarded BASTARD.
What happens to flux density?
You probably wouldn't know.
 
M

MassiveProng

So tha plasma is a short, and the plasma is a resistor.

Thanks for explaining that to us.

Do us a favor, John. Go outside and get struck by lightning, and
then tell me how little difference there is between that plasma
stroke, and a short.

There are a series of resistances involved. The rod, the plasma
arc, the weld spot molten liquid puddle, and you brain, boy!
 
T

Terry Given

John said:
Correct. Missing Prong couldn't even figure that, the easy part, out.

What happens to flux density?

John

Rather than a waffly story, herewith the full technical description:

bugger all.

Cheers
Terry
 
T

Terry Given

Tony said:
Wow, great link, thanks!


Isn't it just. Best I could find quickly.

I have the following magamp books:
Nonlinear magnetic control devices, Geyger
Magnetic amplifiers, Storm


My (one and only) mag amp bible is:-

"Self Saturating Magnetic Amplifiers"
by Lynn, Pula, Ringelman, and Timmel.
Published in 1960 by McGraw-Hill.
Lib Congress Card Number = 60-6979.

It is a serious tome, and the authors were apparently
staffers at the Westinghouse Electric Corp, Air Arm
Division, Baltimore. Inventors of "mag amp?".

Looks like I borrowed it from the Decca Radar Research
Laboratories, (in 1967), and forgot to return it. :)
[/QUOTE]

:)

I sorta did that with a library book I really liked once. Told them i
lost it, and paid for it ($50 or so IIRC). miraculously turned up
shortly thereafter, but I figure its mine now, and I have the receipt to
prove it :)


and once during a pay review, when I said I wanted more and my boss said
"theres no more money" I ended up convincing him to give me a $1k per
annum book budget, I keep the books but make them available at work.
Thats why I have a copy of Snelling (and Bozorth)
Amazon has one copy, at $80.

I'll add it to my list of books to look for.
I don't see why a mag amp book should mention the
ferroresonant CVT. OK, the CVT does have a limb
that saturates, but there is no control winding
and it operates as a passive device.

Cheers
Terry
 
J

John Larkin

X drops a LITTLE BIT, you fucking retard (inferred in my post)! Now
observe what I said above about being able to drive it harder to make
up for, and even surpass your petty claim of severe losses.

Sorry, wrong guess. X, ampere-turns, will increase as air gap is
added. Magnetic path gets longer, reluctance goes down, inductance
goes down, amps increase, turns stay constant.

Did I mention severe losses? Or any losses? I don't recall doing that.

John
 
P

Paul E. Schoen

Terry Given said:
:)

I sorta did that with a library book I really liked once. Told them i
lost it, and paid for it ($50 or so IIRC). miraculously turned up shortly
thereafter, but I figure its mine now, and I have the receipt to prove it
:)


and once during a pay review, when I said I wanted more and my boss said
"theres no more money" I ended up convincing him to give me a $1k per
annum book budget, I keep the books but make them available at work.
Thats why I have a copy of Snelling (and Bozorth)


I'll add it to my list of books to look for.


Cheers
Terry

I once tried using a saturable core reactor to dynamically regulate the
current through a recloser, which changes its inductance by a factor of as
much as 3 during its operation (a steel plunger is pulled into a current
sensing coil against a hydraulic piston). It proved to be impractical
because the control current had to be changed within a few cycles,
requiring a very high power source, and also because it distorted the
waveform excessively.

The problem was solved partly by using series resistance, which greatly
stabilized the current because it was in quadrature to the changing
reactance. I obtained a patent on the improved recloser test set in 1980.
The most successful solution, however, was using digital sampling and
computing the true-RMS value over the entire decaying waveform. More
discussion of recloser testing is on my website: www.pstech-inc.com.

Here is a link to a site describing a saturable core reactor made from a
microwave oven transformer:

http://www.geocities.com/aaawelder/reactor.html

It seems the output AC current is roughly proportional to the DC current
through the saturable core. The principles are explained simply here:

http://en.wikipedia.org/wiki/Saturable_reactor
http://en.wikipedia.org/wiki/Magnetic_amplifier
http://en.wikipedia.org/wiki/Constant_voltage_transformer#AC_voltage_stabilizers

Paul
 
D

Don Klipstein

It also has to be in the work contact area itself as well, otherwise
there would be no heat, no puddles of steel to weld with.

The puddles of steel are molten by heat from the cathode fall and anode
fall mechanisms of the arc. The molten steel does not have enough
resistance to generate that kind of heat from that kind of current - see
what happens if you touch the welding rod to the work. Try that with a
voltmeter across the work and the welding rod, and see how much of the
voltage drop is in the arc and how much is not.

- Don Klipstein ([email protected])
 
P

Paul E. Schoen

Genome said:
[snip]
Thanks for any thoughts and discussion.

Paul

You need the following sums

L = uoueN^2Ae/Le

B = uoueNI/Le

H = NI/Le

dI = VinTon/L


You are not allowed to use one sum without thinking about how it might
affect another one. You avoid that sort of sophistry by using all the
sums to find your answer.

The B field don't depend on the core.

Choice 1) Do Sums.
Choice 2) Use sophistry to prove it does.

Subsequently

Choice 1) Problem Solved.
Choice 2) Invoke 'Ad Hominem' Attack strategy.

A gap affects ue. There is something about permeance and reluctance which
is something like conductance and resistance. The gap appears in series
with things so/but you get the answer by adding the reluctances.

Unfortunately designing magnetic components is hard because there are too
many variables to play with.

N = LI/BAe

Works

But then you might need

AwAe = LIpkIrms/BpkJK

To get a first guess at what the size of your core should be. J and K are
the guesses.... and then Bpk ignores the core loss so you might adjust
something else for another guess. J ignores the winding losses so.....
you might have to guess again.

Do you trust the software?

Leakage inductance has nothing to do with uncoupled flux. It's down to
energy stored in the field(s) between the windings, series term.

If you want to control stuff then you might control the coupling of the
flux and that is a parallel thing.

DNA

I don't think I need to go through all that calculation, especially when I
am using off the shelf components (but I appreciate the information). The
actual circuit performance and waveforms closely match what I see in
LTspice. I was just surprised that the toroidal inductor rated at higher
current did not produce as much output power. It may be that the smaller
inductor (Sumida CDR127/LD100) may have more effective gap and thus allow
higher current than the larger toroid (Miller 2101-H-RC). I have ordered 8
pieces of the board, and will soon be able to test them. They are designed
so I can use either the on-board SMT inductor, or an external toroid, or
both in parallel.

Basically, I can predict the maximum current in the inductor, and hence the
energy stored, vs frequency. Using LTspice with a 61 ohm load, I found that
at 200 kHz and 70% duty cycle the maximum inductor current with 12 VDC at
10 uH is 4.4A (Energy = 97 uW-sec * 0.2 = 19.4 W), and I get 40 volts (26.2
W). At 100 kHz, I can get 48 volts (37.7W) with a maximum inductor current
of 8 A (32 W). The actual inductor current in the first case, which is
running in continuous mode, includes a DC component of 650 mA from the 12
volt source. Adding that gives a power contribution from the battery of 7.8
watts in the first case and 9.4 watts in the second.

The maximum output will be generated when the inductor starts charging
again after its energy has been discharged into the output capacitor, so
there will be no "dead time". With 12 volts, the inductor charges to 8.4 A
in 7 uSec, and it takes 3 uSec to charge the output capacitor, for 70% duty
cycle. The output is about 48 VDC into 61 ohms, or 38 watts. I calculate
the average input power to be about 70% of sqrt(8.4*8.4/2) * 12 V = 35.3
watts, plus the 780mA * 12V = 9.4W from the battery, or 44.7. I'm guessing
at this, but the simulator measured input watts to be 43, so I'm close.
This is 82% efficiency.

I'm running simulations in LTspice, and I think they are pretty much
correct, but I am still a little puzzled. In the continuous mode operation
at 200 kHz, I can see the DC component through the inductor as a 388 mA
minimum current. I get input power of 28.77 W and output of 25.77 W or
89.5% efficiency. In the discontinuous mode at 100 kHz, I get 38.7 watts
out, 43.1 watts in, and 89.8% efficiency. However, I have a hard time
grasping how it can output 38.7 watts when there is almost no inductor
current (It's actually negative) 10% of the time, and peak energy of 320
uW-Sec at 100 kHz or 32 watts. Maybe I'm simplifying the calculation too
much. The true power is probably the integral of the peak energy
(0.5*I^2*L) over the entire waveform, times frequency. OK, when I do that,
I get an average of about 98 uJ, but a peak of 316 uJ = 31.6 W.

BTW, a good reference for transformers, inductors, and other AC devices is:
http://www.ibiblio.org/obp/electricCircuits/AC/AC_9.html

The entire series is very good.

Paul
 
M

MassiveProng

The puddles of steel are molten by heat from the cathode fall and anode
fall mechanisms of the arc. The molten steel does not have enough
resistance to generate that kind of heat from that kind of current - see
what happens if you touch the welding rod to the work. Try that with a
voltmeter across the work and the welding rod, and see how much of the
voltage drop is in the arc and how much is not.


The arc is as hot as the sun's surface, and where it hits the steel,
it melts the flux on the end of the rod, and the rod's steel melts,
and the arc entry/exit point at the work site heats. The fact that
the rod tip became molten is hot enough to take a small portion of the
surface, that being the work site over the edge and they meld, all in
a few fractions of a second, then on to the next bulb of hot steel off
the rod tip. Stack em along a seam and you have a weld. In this
case, an arc weld.
 
T

Tim Williams

MassiveProng said:
Do us a favor, John. Go outside and get struck by lightning, and
then tell me how little difference there is between that plasma
stroke, and a short.

Not much of one, considering the cloud started at eight hundred fucking
million volts. ;-)

Tim
 
J

Jan Panteltje

Here is a link to a site describing a saturable core reactor made from a
microwave oven transformer:

http://www.geocities.com/aaawelder/reactor.html

It seems the output AC current is roughly proportional to the DC current
through the saturable core. The principles are explained simply here:


Interesting.
Did you know that some of the old color CRT television sets used a saturable core,
'transductor' to modulate the scan amplitude to avoid pin-cushion distortion?
You could have been watching one.

There are several references about that on the net too:
http://www.bbcmicro.net/old-8bs/othrdnld/manuals/cubmanual/!cub4.htm#4
 
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