How the power transfers across the Ideal Transformer

[tim9000]

your math looks good
it's an approach i never thought of, figuring L from Energy
*Can you measure energy in the core ?

No I'm not measuring energy in the core, but I am getting a value for the energy from that integration formla via the simulation.

hmm we're speaking of this arrangement ?
View attachment 89841
again, reason it by taking to extremes

1. Zero DC flux, adjust AC source voltage - impedance is fairly constant at flux below saturation where slope is constant, see above second BH curve shortest red tangent (which i should have numbered 1.) .
Upon reaching saturation impedance drops as on tangents 2 and 3. Remain aware it's a non-linear device then because current departs from sine shape.
So yes, V.s per turn affects impedance, and applying enough V.s per turn effects a precipitous drop in impedance !
2. So much DC that core is utterly saturated -
now you're not sweeping flux across zero anymore, you're operating out on a wing like upper BH curve ?

Sorry, could you please remind me what the difference was between small and large signal values?

Sorry, when I said "And I can't remember, did you think that any V.s/N on the two 200 turn coils would make any difference to the impedance of the reactor?"
Judging from the rest of what you said I think you still understood what I was asking. Yeah taking it to the extremes is a good idea.

So that small signal impedance, that was the wobble on the BH curve? (The Wobble around the large signal method point on the BH curve?)

So when there is no DC flux, the wobble is the entire movement of the BH curve, sweeping up the linear region and back down negative again. But when there is DC it will wobble a bit around the DC B point. But when it is 'utterly' saturated it won't wobble at all?

So the more saturated we get, the less small signal impedance there is?

If all that's true, all that was said 'eons ago' might actually make a lot more sense to me.

So if you were to calculate the impedance of the reactor using KVL on your measurements this would give the large signal method impedance, or something else?)

Thanks!

 

[tim9000]

The reason I'm not certain the measurement using KVL of the impedance of the reactor would be just the large signal method, is because the small signal wobble would be present on each individual leg, and although both cois are in series, I'm not convinced their respective wobbles will cancel out, I have a feeling they might be additive?

 

[jim hardy]

Referring to this picture
snipped from www.mdpi.com/1996-1944/7/3/1850/pdfhttps://www.physicsforums.com/attachments/upload_2015-10-7_9-32-16-png.89846/

So the main important point is that a form factor will only distort when the permeability changes dramatically: so if the inductor is used in the linear region where the permeability changes very little, then the form factor will stay nice and sineusoidal.
Similarly if the inductor goes from extremely saturated to totally/utterly saturated then the form factor will still stay sineusoidal. Because the permeability was already near vacuume, meaning the reactance didn't change, meaning the impedance didn't change phase during the cycle, meaning the form factor didn't distort.

I think you're right.

That book i mentioned by Geyger cites some original patents. Check out USPTO.GOV. get a TIFF reader(free) so you can view their images

this is from US 743444 and shows the same thing we were doing a couple hundred posts ago

fig 2 being where the saturable reactor operates with DC. If you push it on up and right to beyond curvature, you're back to linear...

This patent , 720884 http://pdfpiw.uspto.gov/.piw?PageNu...1=0720884.PN.%26OS=PN/0720884%26RS=PN/0720884
goes through the same learning curve we have.
Interestingly he says the magnetization need not be aligned, can be perpendicular ! I never thought about that.
Anyhow i mention it because they'll make interesting references for your paper - directly applicable.

find on USPTO site this

then this

and enter the seven digit number. These are so old you have to preface with 0 to get seven digits.

 

[tim9000]

I think you're right.

At this point I'm so short on time, you thinking I'm right is good enough for me.

As for the rest, interesting, I'll go though it next time I'm waiting for the magnetic FEA simulator to take 7.5 hours to converge to a solution.

Sorry, could you please remind me what the difference was between small and large signal values?

Sorry, when I said "And I can't remember, did you think that any V.s/N on the two 200 turn coils would make any difference to the impedance of the reactor?"
Judging from the rest of what you said I think you still understood what I was asking. Yeah taking it to the extremes is a good idea.

So that small signal impedance, that was the wobble on the BH curve? (The Wobble around the large signal method point on the BH curve?)

So when there is no DC flux, the wobble is the entire movement of the BH curve, sweeping up the linear region and back down negative again. But when there is DC it will wobble a bit around the DC B point. But when it is 'utterly' saturated it won't wobble at all?

So the more saturated we get, the less small signal impedance there is?

If all that's true, all that was said 'eons ago' might actually make a lot more sense to me.

So if you were to calculate the impedance of the reactor using KVL on your measurements this would give the large signal method impedance, or something else?)

Thanks!

The reason I'm not certain the measurement using KVL of the impedance of the reactor would be just the large signal method, is because the small signal wobble would be present on each individual leg, and although both cois are in series, I'm not convinced their respective wobbles will cancel out, I have a feeling they might be additive?

 

[jim hardy]

my local salvage yard got in a real treasure

a three phase autotransformer from a Square D 8606 reduced voltage motor starter, with 240 volt 200 amp windings on two legs only - center leg is bare... !

just the autotransformer and some busbar...

That's like your three legged core except all 3 legs are same size. It's going to weigh a hundred pounds maybe two... but where else can you buy such a thing for thirty cents a pound ?

that'd be real fun to tinker with so i hope the guys didnt put it in the baler , i asked them to set it aside while i found out what it was.

Anyhow back to business

Sorry, could you please remind me what the difference was between small and large signal values?

that patent 743444 above - i'd call fig 2 "small signal" because his AC is small compared to the DC so never drives the core out of saturation.

So that small signal impedance, that was the wobble on the BH curve? (The Wobble around the large signal method point on the BH curve?)

So when there is no DC flux, the wobble is the entire movement of the BH curve, sweeping up the linear region and back down negative again. But when there is DC it will wobble a bit around the DC B point. But when it is 'utterly' saturated it won't wobble at all?

So the more saturated we get, the less small signal impedance there is?

If all that's true, all that was said 'eons ago' might actually make a lot more sense to me.

Back to the future, eh ?

So if you were to calculate the impedance of the reactor using KVL on your measurements this would give the large signal method impedance, or something else?)

not sure what conditions you have assumed...

 

[jim hardy]

i got scrambled with so much going on around home - bear with me, and repeat whatever questions are most pressing...

 

[tim9000]

i got scrambled with so much going on around home - bear with me, and repeat whatever questions are most pressing...

Ok no worries,

Back to the future, eh ?

Yeah hopefully this will be the last time.

not sure what conditions you have assumed...

Say you've got a bit of DC flux, sort of just out passed the linear region. Assume you're measuring the voltage over a nice sineusoidal supply, and the voltage over a purely resistive load (light bulbs) and the only unknown is the voltage drop over the reactor. You're also measuring the current through the lot using a true RMS meter. So that gives you you KVL impedance for the reactor. Now is that impedance going to just be the large signal impdeance?
We're measuring the reactor across the series of both outer coils.
Ages ago when I said "RMS impedance" and you said you'd never heard of such a thing, or couldn't imagine such a thing. That's what I was talking about. The impedance is going to wobble due to the small signal wobble around the large signal point out on the BH curve.

What I can't figure out is, both of the coils on the out side are in series, so do their small signal fluxes cancel out or not? I assume not because

when there is no DC flux, the wobble is the entire movement of the BH curve, sweeping up the linear region and back down negative again. But when there is DC it will wobble a bit around the DC B point. But when it is 'utterly' saturated it won't wobble at all

Is what we calculate to be the impedance of the reactor using KVL measured, including this small signal wobble? or just the point it wobbles around.

 

[jim hardy]

wow i haven't been thinking straight since those stints - too much anesthesia i think

decompressed over Thanksgiving at kids' house in Virginia(1050 miles East of here)

is your thesis submitted ?

 

[tim9000]

wow i haven't been thinking straight since those stints - too much anesthesia i think

decompressed over Thanksgiving at kids' house in Virginia(1050 miles East of here)

is your thesis submitted ?

Hey Jim, Glad you've 'decompressed' and hopefully your health is exponentially recovering every day.
Yeah my thesis was assessed, they hated my project, but at least it's all over now. I can send you a copy if you like?

 

[jim hardy]

Sorry i was out of it for a while

please send a copy

old jim

btw i acquired a 3 leg core similar to yours
weighs 300 pounds
for future tinkering - your project made me nostalgic for the one i had in 1970's

old jim