How the power transfers across the Ideal Transformer

[The Electrician]

hah, ok, it's a bit of a mystery proof.
That does sound like a useful book though.So an under-used transformer will have most of the losses in teh core, and a fully loaded transformer will have most of the losses in the copper. So are you saying the designer of a distribution transformer will try and average out where the losses in the TX are, based on how loaded the TX is? For instance, say it was at rated power 100% of the time, THEN you'd want copper and iron losses to be equal, but say it was at rated power 20% of the time and over rated at 80%, then you'd have a preference to minimising copper losses? Conversely if it was at rated power 20% of the time and under rated power 80%, then you'd have a preference to design it to minimise core losses?
Is that how I should be interpretting your statement?

Not quite. It wouldn't be a good idea to run a transformer over its rating 80% of the time.

If it's running at less than full rated power most of the time, which is typical for distribution transformers, you would use less copper, so the copper losses would be higher at full rated power.

There can be other considerations that determine the allocation of copper and core losses.

For example, here's a transformer from a 4000 watt sine wave inverter:

This transformer has 12 watts of core loss at no load, and at full load. The copper loss at full load is about 200 watts. The reason for this disparity is that customers seldom run their inverters at full load, and they want very low no load and light load losses, core and copper combined. The no load losses determine how fast your battery will run down when only the nite lite is being powered.

 

[tim9000]

Not quite. It wouldn't be a good idea to run a transformer over its rating 80% of the time.

If it's running at less than full rated power most of the time, which is typical for distribution transformers, you would use less copper, so the copper losses would be higher at full rated power.

There can be other considerations that determine the allocation of copper and core losses.

For example, here's a transformer from a 4000 watt sine wave inverter:
View attachment 88865

This transformer has 12 watts of core loss at no load, and at full load. The copper loss at full load is about 200 watts. The reason for this disparity is that customers seldom run their inverters at full load, and they want very low no load and light load losses, core and copper combined. The no load losses determine how fast your battery will run down when only the nite lite is being powered.

Interesting.
Yeah I see at 80% would be a bad idea, I was just going for proportions. So the main point is that equal copper and Iron is only ideal if the thing is actually going to be used at rated, i.e. that rule is FOR rated efficiency. You never want to run it over rated, or atleast not for long, and If it is under rated a lot of the time, it makes sense ot have better efficiency at under rated, than actually AT rated.
Makes sense
Cheers!

 

[tim9000]

I think I have a better way to pose my question about whether the dΦ/dt of a rectified mag amp or one that just sees both parts of the AC supply makes any difference:

I know rectifying it is better from the stand-point of no magneto-striction, but I was wondering if there were any magnetic down-sides? So to speak.

 

[jim hardy]

Great input from Electrician !

{o me o my, c'mon alleged brain work on one thing at a time.........
got the baseplate level and nailed in for first row of siding just before dark.

I've wanted for weeks to post a "how to wire up a washing machine motor" , and lo- somebody asked.}

Okay back to inductors.

I thought you had it pretty well surrounded in this old post 107

your saturable reactor is well saturated by 2 vdc across the control winding
i don't know how many amp-turns that is though.

I've printed this from post 180 several times, but am unable to relate the numbers to the graph.

Column V mag Amp stays less than 4V but graph goes to near 70.

You asked where do you sit on the BH curve as you adjust DC.

From Post 187

With extreme DC control current, there is never enough load current to push the outer legs out of saturation. So you're way out on the "wings" sweeping current back and forth but not changing flux appreciably(hence hardly any voltage appears)
With zero DC control current you're on the steep part of the curve sweeping current back and forth and changing flux appreciably (hence voltage appears).

But in that snip from post 180 , columns Excitation, V L and I (A) are all practically constant.

so i have not folowed some of your thought process since then and have probably answered what i thought were your questions but probably weren'tnow on to more recent posts

 

[jim hardy]

now to this one , i rather muffed

. look at that picture(264 & 268)

With R = zero, we have opposing mmf's because all 3 windings are trying to push flux down. So total current will be set by Rload.
How will current divide between the center coil and the outer ones ? I'm not sure.
With your postulated turns numbers, center and series combination of outer legs would have same volts per turn wouldn't they ? Inferring same flux? Seems to me current would have to reverse in center leg... Since primary current is set by Rload it's operating as a current transformer.

Aha , yes of course
rectified AC has AC fundamental at twice frequency, and DC offset.

Flux will flow in direction of dominant amp-turns so may not be in direction of the arrows . Arrows show direction of MMF.
If we disallow current in center leg by setting R high
that thing should saturate itself because DC component of load current pushes flux down outer and up center legs.
If we allow current in center leg, its MMF opposes outer leg MMF ?
Like any good self-saturator, control winding MMF current opposes load current MMF. But usually the control source is independent of load. This one's more like a feedback winding?

I think i have that one right now. Sanity check ? Tim ? Electrician ?

 

[jim hardy]

From post 274

Hey Jim, Thanks for the reply. That's ok. But I'm hoping I can get the last few curiosities tied up soon because time is starting to run out.

That's a good answer to illustrate the importance of flux in preserving the current ratio (of a CT), my mind is still a bit like a sieve in this field sometimes. it's okay i struggle too

How important is low flux (small magnetising current) though in a VT? in a plain power transformer not very , it just wastes some heat to push the core out past the knee twice per cycle... and makes it draw harmonic current. In a metering transformer where you want accurate reproduction of Vp you'd want to operate the transformer further below its knee. That'll reduce magnetizing current but the objective was mainly to keep transformer in its linear range.

Let me see if I've got this right/straight:
I remember from our previous discussion that if the magnetising current is high (such as from saturation) than the increasee in current will be on the ?? Verb alert ? maybe "show up as" ? larger voltage drop on the resistance of the coil, and so less induced voltage on the primary and secondary.

So is it fair to say that while the impedence of the magnetising branch is linear, the amount of reduced voltage on the primary and secondary will be linear due to it will be primary resistance. While it's not what you'd call good, it is atleast predictable, so not a big deal...Actually hang on, if You want a more efficient transformer, have less resitance in the copper and less reluctance in the core, for less flux, because the lower the reluctance the lower Rc but the higher Xm, so bigger inductance means less current.
You have things moving in the right directions...
And/But with a CT the current ratio is parramount so you don't want to be having to account for magnetising current (as Ipri = Iin - IM) in your equation, that would be unprofessional for a manufacturer, take a look at phasor diagrams for a CT...
so you have to a big core (less reluctance) for bigger Xm, for less flux.
So the MMF might not be low, but the Net MMF will be low. you got it
How am I travelling?

You're doing well. A CT is just a very good PT that's operated with its secondary shorted, as near as is practical.
Primary mmf is canceled by secondary mmf.

Magnetizing current causes phase shift between primary and secondary so they want it small which means like you said, a good core.
http://www.crmagnetics.com/assets/technical-references/analysis_of_ct_error.pdf

 

[tim9000]

i've printed this from post 180 several times, but am unable to relate the numbers to the graph.
Column V mag Amp stays less than 4V but graph goes to near 70.

So the vertical axis is V3, and the horizontal is Vdc. But the table only has the saturation data points, the data points while it was amplifying were from Vdc = 0 to 0.9V but I didn't include them...probably due to space limitations.

Flux will flow in direction of dominant amp-turns so may not be in direction of the arrows . Arrows show direction of MMF.
If we disallow current in center leg by setting R high
that thing should saturate itself because DC component of load current pushes flux down outer and up center legs.
If we allow current in center leg, its MMF opposes outer leg MMF ?
Like any good self-saturator, control winding MMF current opposes load current MMF. But usually the control source is independent of load. This one's more like a feedback winding?

Yeah I agree with that, I'd like to point out that it's a ficticious design. I was trying to replicate Steiner's parallel Amp, rectification, but for a Series Amp, so you think this will operate similarly to that?

rectified AC has AC fundamental at twice frequency, and DC offset.

Does it?? I can imagine that it would have a DC, but the frequency surprises me, Are you addressing/refering to my picture in #283? So are you saying if you supply a mag amp with rectfied ac, the fundamental will make it's reactance twice as large??

 

[tim9000]

You're doing well. A CT is just a very good PT that's operated with its secondary shorted, as near as is practical.
Primary mmf is canceled by secondary mmf.

The way I rationalise "If you open circuit the secondary of a CT since there's heaps of turns on it, it's like a big inductor that wants to keep it's current flowing so it induces a big voltage and might arc." As in I personify the inductance like it has a disire. I remember ages ago you said something like 'since the other flux opposing it disapears...' so I assume the way you rationalise it is that 'if the secondary flux suddenly dissapears the flux from the primary can increase very fast causing a big dΦ/dt'
But I'm assuming the same ISN'T true if you open circuit the primary passing through it? because ampere's law around the primary doesn't quite work out the same.One thing that bothered me was that when you increase the cross sectional area of a TX and the reluctance drops, so Rc drops, I'm assuming the core power loss drops also? Ic²*Rc = core really power loss
Because even though the magnetising current I am <90^O has dropped, decreasing the reluctance by increasing the cross sectional area has increased the magnetising current Ic < 0^O through Rc
Or might the core loss actually go up if you decrease the reluctance?

 

[tim9000]

Again, why was the 840 turn curve different from the 400 and 200 ones:

??

Also I reitterate the question priorities of designing for current and dΦ/dt or inductance when rectified:

Rectified AC has a fundamental twice the frequency? Does it?? I can imagine that it would have a DC, but the frequency surprises me, Are you addressing/refering to my picture in #283? So are you saying if you supply a mag amp with rectfied ac, the fundamental will make it's reactance twice as large??

I think I have a better way to pose my question about whether the dΦ/dt of a rectified mag amp or one that just sees both parts of the AC supply makes any difference:

I know rectifying it is better from the stand-point of no magneto-striction, but I was wondering if there were any magnetic down-sides? So to speak.

At the risk of sounding like a broken record I'm going to put this to you again:

I gather when you're designing an inductor (putting an amount of turns on it) the amount of current through it is equally as important as the amount of V.s on it, however I recall you saying you saw no purpose for designing an inductor to operate at μmax, and I see that if you have a set V.s and you want a big inductance, than you may as well operate down the bottom of the BH curve. But/And I raised the point that if you were opperating at a specific current through the inductor, than wouldn't μmax be a good point to be operating at, because that way you're getting current through it, you're getting your inductance, and you're getting value out of your steel. I was wondering what your response was to that? (as well as the top posts)

Because you previously said you saw no problem in, if one wanted a bigger inductor, they could just wind more turns around it, and I can see that would be fine provided you were designing it for V.s, but if you were designing it for a Current, than doesn't that luxury evaporate? -> Say you've hit the wall as far as increasing the side of your core goes. You've got the current going through it, and it's hunky dorry, down the bottom of the BH curve. You think, "ok room to play with", and you wind more turns on it, B goes up, then isn't μmax the last port of call, it will give you the best inductance and you got more turns on it for the same current?

 

[jim hardy]

I think I have a better way to pose my question about whether the dΦ/dt of a rectified mag amp or one that just sees both parts of the AC supply makes any difference:
View attachment 88880
I know rectifying it is better from the stand-point of no magneto-striction, but I was wondering if there were any magnetic down-sides? So to speak.

Rectifying just flips the negative half cycles
so slope at every instant has same absolute magnitude in both waves
from 0 to pi radians the waves are identical and have same slope
but note from pi to 2pi, slopes have same magnitude but opposite sign

so would it be rigorous use of language to say dΦ/dt was same for both waves?
RMS of both waves would be same, presumably any function that squares would give same result for both waves

but i'd be very cautious about claiming "same" dΦ/dt for them.

 

[jim hardy]

So are you saying if you supply a mag amp with rectfied ac, the fundamental will make it's reactance twice as large??

be careful with that thought.
Look at Fourier of full wave rectified sinewave

http://www.reed.edu/physics/courses/Physics331.f08/pdf/Fourier.pdf
rectifying your sinewave it changes it from a plain single frequency to a summation of even harmonics
and the line frequency fundamental disappears !

So your AC meters measuring V and I are going to see that summation...

Impedance at each harmonic is jωL at that frequency
but remember a real core has frequency dependent losses(eddy currents) , so L itself if a f(frequency)

so Z will be larger but 2X larger would be a fluke.

Not meaning to be difficult or contrary, just be careful about taking too big leaps of thought.

 

[jim hardy]

One thing that bothered me was that when you increase the cross sectional area of a TX and the reluctance drops, so Rc drops, I'm assuming the core power loss drops also? Ic2*Rc = core really power loss
Because even though the magnetising current I am <90O has dropped, decreasing the reluctance by increasing the cross sectional area has increased the magnetising current Ic < 0O through Rc
Or might the core loss actually go up if you decrease the reluctance?

Two components to core loss, hysteresis and eddy currents.

Usually it's all lumped and ascribed to Steinmetz, fB^1.6

Steinmetz's equation, sometimes called the Power equation,[1] is a physics equation used to calculate the core loss of magnetic materials due to magnetic hysteresis. The equation is named after Charles Steinmetz, who proposed a similar equation in 1892.[2] The equation is as follows:

[PLAIN]https://upload.wikimedia.org/math/b/b/5/bb544bd4abdacdf8ce85f1cb2e3ea3de.png[/QUOTE]
https://en.wikipedia.org/wiki/Steinmetz's_equation (i was taught fB^1.4, maybe his 1892 number?)
That's loss per unit volume; volume and B are both proportional to area so i think losses will go down because of the 1.6 exponent

NASA separates them here
http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660001049.pdf

and

So does Rc get larger or smaller ?

Seems to me if the core gets less lossy, R_C will increase and I₀ decrease , ?
If the core becomes easier to magnetize, X_M will increase and I_M decrease, ?
It'd be approaching ideal...

 

[jim hardy]

Again, why was the 840 turn curve different from the 400 and 200 ones:

??

How far back was that post ?

i thought it was length of magnetic paths

but only 147.5 when driven from center ?

 

[tim9000]

How far back was that post ?

i thought it was length of magnetic paths

View attachment 89063

but only 147.5 when driven from center ?

I think you might be onto something, I can totally see that it takes less amps to get the same flux when there are more turns, but just looking at the term how does magnetic length effect 'V.s/N'? Because there's not 'length' in that term, also as far as I can see those curves are just flux Vs. Amps.

 

[tim9000]

Thanks for the reply Jim!

Rectifying just flips the negative half cycles
so slope at every instant has same absolute magnitude in both waves
from 0 to pi radians the waves are identical and have same slope
but note from pi to 2pi, slopes have same magnitude but opposite sign

so would it be rigorous use of language to say dΦ/dt was same for both waves?
RMS of both waves would be same, presumably any function that squares would give same result for both waves

but i'd be very cautious about claiming "same" dΦ/dt for them.

So then when you rectify it, the supply becomes the sum of the even harmonics, so how does that change the dΦ/dt for each hump? I imagine it would increase it?

Yeah I can see they'd have the same RMS. I'm really interested in 'dΦ/dt' because of the implications on Inductance.
So from pi dΦ/dt is negative, and from 2pi dΦ/dt is positive, so for instane in the equation E = NdΦ/dt
what implications does this positive only (due to rectification) have on E?
Thanks

 

[tim9000]

H'mm I'll have to get my head around Steinmetz, thanks.

Seems to me if the core gets less lossy, RC will increase and I0 decrease , ?
If the core becomes easier to magnetize, XM will increase and IM decrease, ?
It'd be approaching ideal...

Is that simply to say as the Reluctance of the core DECREASES then the Xm and Rc will BOTH INCREASE?

 

[jim hardy]

I think you might be onto something, I can totally see that it takes less amps to get the same flux when there are more turns, but just looking at the term how does magnetic length effect 'V.s/N'?

?? Isn't V.s flux ? V = -N dΦ/dt ; Vdt = -N dΦ ; ∫Vdt = -N ∫dΦ ; Vt -= NΦ ; for t in seconds s , V.s / N = Φ

Φ = μ μ₀ N I A_rea / L_ength
V.s / N = μ μ₀ N I A_rea / L_ength

V.s /n = 1/L_ength X (N I μ μ₀ A_rea )

check my thinking ? And my algebra ?

 

[tim9000]

?? Isn't V.s flux ? V = -N dΦ/dt ; Vdt = -N dΦ ; ∫Vdt = -N ∫dΦ ; Vt -= NΦ ; for t in seconds s , V.s / N = Φ

Φ = μ μ₀ N I A_rea / L_ength
V.s / N = μ μ₀ N I A_rea / L_ength

V.s /n = 1/L_ength X (N I μ μ₀ A_rea )

check my thinking ? And my algebra ?

Yeah I see where you're coming from:
I = Φ * Length / ( μ μ₀ .N. Area)
But...OOOH, ok, what I was confused about was, why the two bottom curves were flattening off at a lower flux, but that's because their cross sectional area is half! Does that sound right?
So they need more current to get to the same flux because the length is longer, but also since Area is half, it requires half the flux to saturate.

That finally makes sense, thanks.

what did you make of my question in #295 and my assessment in #296?

 

[jim hardy]

But...OOOH, ok, what I was confused about was, why the two bottom curves were flattening off at a lower flux, but that's because their cross sectional area is half! Does that sound right?

Sure does ! How'd i miss that ?

So then when you rectify it, the supply becomes the sum of the even harmonics, so how does that change the dΦ/dt for each hump? I imagine it would increase it?

i guess i was looking at slope of voltage...
dΦ/dt IS voltage (per turn).
so dΦ/dt also becomes the sum of even harmonics

what does that series evaluate to ? I think it'll be less than 1 so dΦ/dt decreases a little ?

Yeah I can see they'd have the same RMS. I'm really interested in 'dΦ/dt' because of the implications on Inductance.
So from pi dΦ/dt is negative, and from 2pi dΦ/dt is positive, so for instane in the equation E = NdΦ/dt
what implications does this positive only (due to rectification) have on E?
Thanks

E an dΦ/dt are one and the same, scaled by N_{Turns ?}

 

[jim hardy]

bedtime here

 

[tim9000]

i guess i was looking at slope of voltage...
dΦ/dt IS voltage (per turn).
so dΦ/dt also becomes the sum of even harmonics
View attachment 89117

what does that series evaluate to ? I think it'll be less than 1 so dΦ/dt decreases a little ?
E an dΦ/dt are one and the same, scaled by N_{Turns ?}

Ok than there will only be a possitive induced E.
Hmm, well if you differentiate that, the DC flux (2/pi) drops out, I imagine you'd be left with
dΦ/dt = 4/pi * [sin(2wt)/(w*2*(2²-1)) + sin(4wt)/(w*4*(4²-1)) + ...]
Is that what you had in mind? How's my maths?
So since inductance depends on dΦ/di what does this rectified supply dΦ/dt mean for the inductance?P.S
Was my understanding in #296 correct: "Is that simply to say as the Reluctance of the core DECREASES then the Xm and Rc will BOTH INCREASE?"Thanks again Jim

 

[tim9000]

Sure does ! How'd i miss that ?

i guess i was looking at slope of voltage...
dΦ/dt IS voltage (per turn).
so dΦ/dt also becomes the sum of even harmonics
View attachment 89117

what does that series evaluate to ? I think it'll be less than 1 so dΦ/dt decreases a little ?E an dΦ/dt are one and the same, scaled by N_{Turns ?}

Hmm, well if you differentiate that, the DC flux (2/pi) drops out, I imagine you'd be left with
dΦ/dt = 4/pi * [sin(2wt)/(w*2*(2²-1)) + sin(4wt)/(w*4*(4²-1)) + ...]
Is that what you had in mind? How's my maths?
So since inductance depends on dΦ/di what does this rectified supply dΦ/dt mean for the inductance?P.S
Was my understanding in #296 correct: "Is that simply to say as the Reluctance of the core DECREASES then the Xm and Rc will BOTH INCREASE?"
Also that hyperthetical in #289, if you were designing an inductor to run as a current, 'wouldn't it then be advantagious to run at mu max?'

Thanks again Jim

 

[jim hardy]

so for instane in the equation E = NdΦ/dt
what implications does this positive only (due to rectification) have on E?

Full wave rectification is equivalent to taking absolute value, as in that snip

rectify sin(wt) and you get abs(sin(wt) , the two vertical lines (ascii character 7C ? html &#124 ?) meaning absolute value ...

Ok than there will only be a possitive induced E.

hmm i guess that's so - voltage applied never goes negative so neither can counter-EMF.
Once again we're at that line where ideal and real world things behave differently
but yes,
applying a wave with DC component to an ideal inductor will give ever increasing current and flux, ie slope term includes a constant,
and that expression in the snip for E (and for dΦ/dt) includes a constant 2/pi .
So, flux is the integral of the snip.

Hmm, well if you differentiate that, the DC flux (2/pi) drops out, I imagine you'd be left with
dΦ/dt = 4/pi * [sin(2wt)/(w*2*(22-1)) + sin(4wt)/(w*4*(42-1)) + ...]
Is that what you had in mind? How's my maths?

that term for flux is ALREADY differentiated - Faraday says E = - N dΦ/dt

so i don't understand why you differentiated it.
But it looks like you integrated the summation terms , anyhow? D(sinwt)= wcoswt not coswt/w
had you integrated the 4/pi instead of differentiating it you'd have got the term for ever increasing flux.

So since inductance depends on dΦ/di what does this rectified supply dΦ/dt mean for the inductance?

back to basics.. Inductance is flux linkages per amp, L = NΦ/I
so how does inductance depend on dΦ/di ?
Φ = I * L/N
dΦ = dI * L/N
Neither L nor N is a function of I
so dΦ/dI = constant

An inductor will show twice the reactance at twice frequency and so forth,
so its impedance to a rectified wave will be different from its impedance to a not-rectified wave
but it should soon saturate because of the DC term.

I hope i read your questions right.

 

[jim hardy]

P.S
Was my understanding in #296 correct: "Is that simply to say as the Reluctance of the core DECREASES then the Xm and Rc will BOTH INCREASE?"
I agree with that statement. Losses get smaller. Look at the exponents in that NASA paper, and old Streinmetz...Also that hyperthetical in #289, if you were designing an inductor to run as a current, 'wouldn't it then be advantagious to run at mu max?'
Hmm advantageous how? Trade off is between core mass, amount of copper, to achieve desired inductance and losses?

 

[tim9000]

Hmm advantageous how? Trade off is between core mass, amount of copper, to achieve desired inductance and losses?

Well if there's a current running through it, there has to be some flux, so presumably it isn't right down the end of the BH curve. So wouldn't you be getting more inductance out of the copper and steel? But at higher core losses? So say you're getting the current you want through your inductor, and you're pretty far down the BH curve, close to B = 0, then couldn't you take some of the steel out, reducing the cross sectional area, increasing B, up the point where RMS B the operating point, is at maximum permeability, the maximum gradient of the BH curve. Then have you not made a saving on steel by removing some, but also the inductance has stayed the same because the drop in Area has been met with an increase in permeability? [this is an important point for me and I want to extrapolate on depending on what you say]

I hope i read your questions right.

Yeah that was a pretty good reply...actually that was a really good reply, thanks. So my supply is abs[sin(wt)] but I should have integrated it rather than differentiated (sorry I hadn't had much sleep), so the DC would still be present and would be a slope gradient. So it's not a good idea to run a mag amp rectified for a long period of time? Otherwise it will self saturate due to the DC presents, unless you're injecting a counter DC offset?

so dΦ/dI = constant

I did not realize that, so the distinction is that the inductance isn't changing due to rectification, the reactance is, because the effective frequency is the sum of the even harmonics? (which is higher than an AC exictation)

Correct?

 

[jim hardy]

so the distinction is that the inductance isn't changing due to rectification, the reactance is, because the effective frequency is the sum of the even harmonics? (which is higher than an AC exictation)

Correct? I agree.

So it's not a good idea to run a mag amp rectified for a long period of time? Otherwise it will self saturate due to the DC presents, unless you're injecting a counter DC offset?

Back to basics...
DC applied to ANY inductor(of finite value) will cause current to rise until IR drop in the wire limits current.
In a magamp that's okay because once it saturates, the load sets the current. That's how self-saturating magamps work, and recall as we mentioned pages ago the DC in control winding offsets the DC component of rectified Vsupply.

What should NOT be handed a rectified wave is a transformer or inductor that you do not intend to saturate.

Well if there's a current running through it, there has to be some flux, so presumably it isn't right down the end of the BH curve. So wouldn't you be getting more inductance out of the copper and steel? But at higher core losses? So say you're getting the current you want through your inductor, and you're pretty far down the BH curve, close to B = 0, then couldn't you take some of the steel out, reducing the cross sectional area, increasing B, up the point where RMS B the operating point, is at maximum permeability, the maximum gradient of the BH curve.
Certainly, you could do that.
Then have you not made a saving on steel by removing some, yes you have
but also the inductance has stayed the same because the drop in Area has been met with an increase in permeability?

Does % increase in μ equal % decrease in area?
L_before= μ_before μ₀ N² A_{rea before} / L_ength

L_after = μ_after μ₀ N² A_{rea after} / L_ength

Equate those two

μ_before μ₀ N² A_{rea before} / L_ength = μ_after μ₀ N² A_{rea after} / L_ength

divide out common factorsμ_before A_{rea before} = μ_after A_{rea after}
means that
μ_before / μ_after = A_{rea after} / A_{rea before}
so you'd decrease area by same proportion as permeability increased.
If you know permeability precisely enough to hit that mark, your proposition could work.
But i think your core loss will be higher.

You had some curves of inductance versus excitation way back there, which ought to be same shape as permeability versus B ?

[this is an important point for me and I want to extrapolate on depending on what you say]

test those extrapolations by good thought experiment...

 

[tim9000]

Thanks for the reply! I suppose this is a three part post:

test those extrapolations by good thought experiment...

Actually I DID try that thought experiment (changing Area only) already, and it seemed to me that when I changed the area, the permeability responeded in proportion and inversely. So Did I do it right, does permeability and area ballance out to maintain the same inductance?
Because a mag amp is only interesting in the area of BH curve from max permeability and higher B values, isn't it? Because there's no inductance/reactance you can get below max permeability that you can't get at some point from max permeability and upwards...to infinity H.

But when I Increased turns number only, I saw a decrease in the amount of flux in the core linearly due to V.s/N, so the permeability will linearly drop down from mu max to mu initial but the inductance will increase exponentially because propotional to N^2.
I think it hit initial permeability and sat there as I kept increasing turns.

Also, if I have a resistance for 200 turns, should I expect the resistance for 840 turns (assuming its the same wire) to be 'the resistance'*(840/200) ??

Thanks mate

 

[tim9000]

In addidtion to my last post. Also another couple of things,
-so when you take some steel out, the core loss propably goes up, does the copper loss change?

-I just want to Iron this out: so specifically Mr Steiner's Amp, did it saturate BECAUSE of the DC component from rectification, or was that just a helpful coincidence that there was another DC component?

-And what would I type into google to find this:

Was it from Fourier? Or something else

Cheers

 

[jim hardy]

Actually I DID try that thought experiment (changing Area only) already, and it seemed to me that when I changed the area, the permeability responeded in proportion and inversely. So Did I do it right, does permeability and area ballance out to maintain the same inductance?

Well i dont' know. You didn't show your work. Can you put numbers on it ?

Because a mag amp is only interesting in the area of BH curve from max permeability and higher B values, isn't it? Because there's no inductance/reactance you can get below max permeability that you can't get at some point from max permeability and upwards...to infinity H.

You have some thought in mind that i just don't see.
Sure, if you're on the cusp of some curve looking out, it's downhill both ways.

But when I Increased turns number only, I saw a decrease in the amount of flux in the core linearly due to V.s/N, so the permeability will linearly drop down from mu max to mu initial but the inductance will increase exponentially because propotional to N^2.
I think it hit initial permeability and sat there as I kept increasing turns.

okay. Did you plot that ?

Also, if I have a resistance for 200 turns, should I expect the resistance for 840 turns (assuming its the same wire) to be 'the resistance'*(840/200) ??

of course provided you don't look at too many significant digits. As the coil gets fatter (more layers) .each turn must be longer to encircle the larger coil diameter.

 

[jim hardy]

In addidtion to my last post. Also another couple of things,
-so when you take some steel out, the core loss propably goes up, does the copper loss change?

Core loss per pound goes up but pounds of core went down... What's exponent of B in loss term ?

-I just want to Iron this out: so specifically Mr Steiner's Amp, did it saturate BECAUSE of the DC component from rectification, or was that just a helpful coincidence that there was another DC component?

without something to prevent it, saturation will follow from application of rectified AC.

I'll look for that i think i searched on Fourier rectified

 

[jim hardy]

here it is without the derivation

http://people.clarkson.edu/~jsvoboda/Syllabi/EE221/Fourier/FourierSeriesTable.pdf

 

[tim9000]

Well i dont' know. You didn't show your work. Can you put numbers on it ?You have some thought in mind that i just don't see.
Sure, if you're on the cusp of some curve looking out, it's downhill both ways.okay. Did you plot that ?

What I specifically mean by that is modulating the 'change in flux' is the most important propety of the core of a saturatable reactor, and we can get any value of dΦ/dt from max permeability upward (obviously with a TX you actually want larger B because the TX equation voltage is a function of B)

I was hoping you wouldn't ask about the data, I'll see if I can find it, it was a bit 'wishy washy' because it was a thought experiment, if I can't find it I'll knock up a similar graph...
Ok I found the turns one, the data was taken from the 840 curve, I took area to be as A = 1, the horizontal axis is turns number and the vertical is the interpolated mu:

You can see the inductance in this one goes only up

I'll make a changing Area one, now for you.

EDIT:
Ok this is embarrassing/strange: on re-calculation it seems that as the area increases although the permeability does drop, it is well outweighed by the increase in area:

So I must have been wrong before?
So if you increase area by ANY AMOUNT the inductane will go up, regardless?

Is this what you reckon? Seems weird to me because
permeability is B/H
and B = Flux / A
and so L is proportional to A * (Flux / A.H)
you'd think the A would cancel out and L would just be a function of flux and H...
I suppose last time I didn't consider H changing too?? And I suppose it should?

 

[tim9000]

Core loss per pound goes up but pounds of core went down... What's exponent of B in loss term ?

Sorry what do you mean the 'exponent of B in loss term'? I don't understand what youre asking, sorry.

Also, did you mention the change in copper loss?

without something to prevent it, saturation will follow from application of rectified AC.
I'll look for that i think i searched on Fourier rectified

But what I'm asking is, is THAT the actual mechanism by which Steiner designed his Amp to operate, the operation principal of the amplification: it is the force that pushes it the flux down, and allows him to bring it back up again, restorting dΦ/dt to block the current off?

 

[tim9000]

here it is without the derivation

http://people.clarkson.edu/~jsvoboda/Syllabi/EE221/Fourier/FourierSeriesTable.pdf

Ah Ok, so it does come From Fourier, thanks.

 

[jim hardy]

Ok I found the turns one, the data was taken from the 840 curve, I took area to be as A = 1, the horizontal axis is turns number and the vertical is the interpolated mu:

You can see the inductance in this one goes only up

hmmm and mu went down

So if you increase area by ANY AMOUNT the inductane will go up, regardless?

Tim that's what i expect because i just don't see very much change in slope of BH curve ubtil you approach the knee.
But I've always just thought it was so and never investigated it like you have. You are in a position to assert based on your work.

Is this what you reckon? Seems weird to me because
permeability is B/H
and B = Flux / A
and so L is proportional to A * (Flux / A.H)
you'd think the A would cancel out and L would just be a function of flux and H...
I suppose last time I didn't consider H changing too?? And I suppose it should?

H is amp-turns per meter, and for some reason they call it amps per meter , i suppose because physiciists might imagine it as flowing in a sheet instead of individual discrete wires..

So sure, H is proportional to amp-turns so if you have just one solitary amp but double the turns you've doubled H.

Remember definition of Inductance, it's so beautiful and solves so many problems
flux linkages per ampere , linkages means product of turns and the flux they encircle, Turns X Webers
L = NΦ / I
multiply numerator and denominator both by N
L = N²Φ/ NI

H = NI/L_ength
so L = N²Φ L_ength/ H
it's got a length term but not an area one. Length term comes from H, amps/meter if current is a sheet, amp-turns per meter if it's divided up among discrete wires(all in series for a traditional coil).

though i never thought of doing that manipulation.