How the power transfers across the Ideal Transformer

[tim9000]

aha we crossed in the mail again

did i clarify ?

Too early to say, I think maybe so.

Okay now i have to re-orient , what's next ?

This came to me as I was laying in bed, maybe I was thinking about this point all wrong. So would you say that when you change the area of the cross section of the core, it doesn't change where we're operating on the BH curve. That infact it changes the entire BH curve, if you increase the cross sectional area (decreasing the reluctance of the magnetic path) it will scale the BH curve and make the whole thing bigger, but the operating point will actually stay the same, as far as volt*seconds is concerned. (?)

Regarding my post #124

 

[tim9000]

Sticking to the top graph on post #147 for the moment, let's talk about this 'load line' thing, so the pk-pk V is 2V, the pk-pk current is 110mA, Say we're putting on the DC control current, so average Z gets shifted, then Z is varying along the curve as it wobbles between Zmin and Zmax (in each outer leg), excitation voltage is constant, but this makes current and voltage on mag amp (by 2Vpk-pk) vary.

so magamp's impedance way out there past the knee is V/I = 2volts /0.11amp = 18 ohms ?

Doesn't seem very accurate for Z at time, that number is it pk to pk Z?
Although Assessing Z_min and Z_max I'm not really sure what you'd be able to do either, I suppose you'd have to say:
Z_rms = 18 Ohms /(2*√2)
Z_max = 18 Ohms /(2*√2) + 18/2
and Z_min = 18 Ohms /(2*√2) - 18/2
?
So current through mag amp is: I_min = V_supply / [18 /(2*√2) + 18/2] ? And
I_max = V_supply / [18 /(2*√2) - 18/2]
And Rms current through amp to load is I_rms = V_supply / [18 Ohms /(2*√2)]
??

 

[jim hardy]

Regarding my post #124

1. So what did you think permiability μ of my core might be? (I assume the value they usually talk about is μ at the knee)

hmmmm.

Center leg has 840 turns
and knee is about 300V at 27.2 ma

so X_L = V/I = 11,029 ohms
L = X_L / ω = 11,029 / 100pi = 35.1 Henry ? Sounds incredible..

Air core should be
L = μ₀N²A/L_ength
L = 4piX10^-7 X 840² X 555 X 10^-6 X2_{parallel magnetic paths} / 295 X 10^-3
L = 3.34 millihenry

35.1 measured for iron core / 0.00334 calculated for air = 10,520 relative permeability ?
what the heck is that stuff ?

Please check my arithmetic ?

I looked up AL (from your chart, the one with the 'nh' mystry terms) which is apparently the 'inductance factor'. But I'm still none the wiser as to how or what those terms are in relation to the inductance factor.

 

[jim hardy]

Crossed in the mail again
you must have mumetal or cobalt of some sort ?

gotta crash

back soon

old jim

 

[tim9000]

so XL = V/I = 11,029 ohms
L = XL / ω = 11,029 / 100pi = 35.1 Henry ? Sounds incredible..

Huh, ok, tell me what you think of this,
I was thinking take the outside leg to calculate permiability: 35V/200turns = 0.175
B_peak = 0.175/(100*pi*555^-6) = 1.004,
say the length is about 0.095*3+0.03*4+0.016*4 = 0.469m:
It should approximately be μ_r = [1.004/(840*10*10^-3/(0.095*3+0.03*4+0.016*4))] /4piE-7
=44,608

Crossed in the mail again
...gotta crash

back soon

No worries, after you get to my crossed in the mail post #152, I want to put a hypothetical to you to make sure that I can re-word my understanding of the progress and that you'll concur:
Take the graph:

But forget about the triangle on it (I can't be bothered removing it), now there's no control winding as far as we're concerned. Now if I take a point on that, say 15V, the impedance of the amp must be Z = 15/0.02 Ohms
The difference between when there is a DC offset is that you can no longer use that direct relation, you then must look at the amount of change, ala what we've been talking about. That's what I want to put to you.

The other thing I want to ask is, take the actual BH curve, as opposed to VI curve

Say I want to calculate Z of my mag amp at the midpoint and say B peaks at the knee point on the curve (possibly irrelevant), also say there's no DC control current. Let the excitation voltage be 100Vrms at 50Hz and there's 200 turns around the steel.
So how do I calculate Z? I suppose I know I because I know H? or can I do this without current?.
Z_middle = N* dΦ/dt / I = N*(Φ_mid-0) / (I*100π) ?

 

[jim hardy]

wow I'm still on 124...

I looked up AL (from your chart, the one with the 'nh' mystry terms) which is apparently the 'inductance factor'. But I'm still none the wiser as to how or what those terms are in relation to the inductance factor.

well it has units inductance per turn²

so if L = μμ₀N²A/L_ength
L/N² = μμ₀ A/L_ength

which for a specific size core with particular inductance is a shortcut to figuring inductance,
so they give that number for cores . Apparently it's not used for power transformer cores, just applications where they want a precise inductance, see snip below. I'm learning too.

Magnetics Inc is an old line manufacturer with a good reference library
i think you'll enjoy perusing their FAQ at this link
http://www.mag-inc.com/

see also http://www.mag-inc.com/design/technical-documents/general-information

 

[jim hardy]

I was thinking take the outside leg to calculate permiability: 35V/200turns = 0.175
Bpeak = 0.175/(100*pi*555-6) = 1.004,
say the length is about 0.095*3+0.03*4+0.016*4 = 0.469m:
It should approximately be μr = [1.004/(840*10*10-3/(0.095*3+0.03*4+0.016*4))] /4piE-7
=44,608

μr = [1.004/(840*10*10-3/(0.095*3+0.03*4+0.016*4))] /4piE-7 =44,608 ?
Is (0.095*3+0.03*4+0.016*4) L_ength ? in meters ?

Φ = μ_rμ₀ N I A/L_ength.
and B = Φ/A so area disappears
μ_r should be B * L_ength / (μ₀ * N * I )
What happened to I ?

i thought you energized both outside legs, each having 200 turns
so turns would be 400
and your graph shows volts on one of two coils
and since the outside legs aid not much flux goes through center leg
so i get path length excluding center leg of 110 X 2 + 37.5 X 4 = 370mm

and we need a flux with air core in order to calculate μ_r . That might be calculated , or estimated by wayyyy over saturating the core from control winding and measuring Z.

what'd i miss ?

 

[jim hardy]

trial air core calc

so from previous post

μr should be B * Length / (μ0 * N * I )

μ_r = 1.003 * 0.37 / (4piE-7 * 400 * .0875 ) = 8438

how far is 8438 from 10,520 ? about 25 % ?

check my arithmetic ?

 

[tim9000]

μr = [1.004/(840*10*10-3/(0.095*3+0.03*4+0.016*4))] /4piE-7 =44,608 ?
Is (0.095*3+0.03*4+0.016*4) Length ? in meters ?

Φ = μrμ0 N I A/Length.
and B = Φ/A so area disappears
μr should be B * Length / (μ0 * N * I )
What happened to I ?

i thought you energized both outside legs, each having 200 turns
so turns would be 400
and your graph shows volts on one of two coils
and since the outside legs aid not much flux goes through center leg
so i get path length excluding center leg of 110 X 2 + 37.5 X 4 = 370mm

Yes it was in meters, I'll use your dimensions on your drawing of a total mag path length in the core of 0.480, rather than my 0.469.
Sorry I am really out of practice and was perhaps too nonchalant.
I only energiesed ONE side coil, and left the other side and the control open circuit.
I calculated the path length at the total length, did it matter that two of the legs were in parallel?
What I was saying before was
μ_r = B/(Hμ_o) = [1.004/(840*10*10^-3/(0.095*3+0.03*4+0.016*4))] /4piE^-7 =44,608
The current above in bold, was 10mA. BUT I think that I was calculating N*I from the control winding, and instead I should have been using N = 200 instead of 840, and current from the graph, instead of 10mA, how embarassing. So forget that.

Recalculating as before, using the total length: μ_r = B/(Hμ_o) = [1.004/(200*87.5*10^-3/(0.48))] /4piE^-7 = 21,914...probably wrong beause I probably don't use the total length
But when I recalculate using only the length portion for that leg at that B:
μ_r = B/(Hμ_o) = [1.004/(200*87.5*10^-3/(0.110+0.0375+0.0375))] /4piE^-7 = 8,446

Which is the same as trying it your way, I get:
μ_r = B* Length / (μ0 * N * I ) = 1.004*(0.0375+0.0375+0.11)/(4pi*10^-7 *200*0.0875) = 8,446
Which is close to what you got, What did you think of my method?
You got a really big relative permeability (quoted below)
B_{centre leg peak} = (297/840)/(100pi*(0.037*0.03)) = 1.014
So you got a really big value for the permeability using the control widing:

so XL = V/I = 11,029 ohms
L = XL / ω = 11,029 / 100pi = 35.1 Henry ? Sounds incredible..

Air core should be
L = μ0N2A/Length
L = 4piX10-7 X 8402 X 555 X 10-6 X2parallel magnetic paths / 295 X 10-3
L = 3.34 millihenry

35.1 measured for iron core / 0.00334 calculated for air = 10,520 relative permeability ?
what the heck is that stuff ?

Please check my arithmetic ?

But when I apply the other method I get something really small for the relative permeability using the control winding:
μ_r = B* Length / (μ0 * N * I ) = 1.014*(0.110)/(4pi*10^-7 *840*0.0272) = 3,885 rel permeability
So what's going on? Bizzare

 

[jim hardy]

first, a detail from #152

Doesn't seem very accurate for Z at time, that number is it pk to pk Z?
Although Assessing Zmin and Zmax I'm not really sure what you'd be able to do either, I suppose you'd have to say:
Zrms = 18 Ohms /(2*√2)
Zmax = 18 Ohms /(2*√2) + 18/2
and Zmin = 18 Ohms /(2*√2) - 18/2
?

since Z = Volts/Amps
if we use pk=pk in both numerator and denominator
or we use rms in both numerator and denominator
we get same result
because the 2√ 2 is in both numerator and denominator so cancels
i cannot conceive of a peak to peak impedance

I only energiesed ONE side coil, and left the other side and the control open circuit.
I calculated the path length at the total length, did it matter that two of the legs were in parallel?

yes it did

flux was free to circulate through both legs in parallel
magnetic circuits are analogs to electric ones, flux divides among parallel paths
and one calculates reluctance by reciprocal of sum of reciprocals of parallel reluctances , just like resistances in parallel.
MMF is analogous to voltage
Flux is analogous to current
Reluctance is analogous to resistance
Sum of fluxes at any junction is zero just like KCL
Sum of MMF's around any path is zero, just like KVL
so you can't calculate permeability using just length of one leg

if you happened to jot down voltage on other coils at that time you can see just how much flux went through each leg,
if you short the turns on either center or right leg it'll push all the flux into the other one

so by energizing both outer coils and shorting center one you could take center leg out of the magnetic circuit,
..............................

Recalculating as before, using the total length: μr = B/(Hμo) = [1.004/(200*87.5*10-3/(0.48))] /4piE-7 = 21,914...probably wrong beause I probably don't use the total length
But when I recalculate using only the length portion for that leg at that B:
μr = B/(Hμo) = [1.004/(200*87.5*10-3/(0.110+0.0375+0.0375))] /4piE-7 = 8,446

Which is the same as trying it your way, I get:
μr = B* Length / (μ0 * N * I ) = 1.004*(0.0375+0.0375+0.11)/(4pi*10-7 *200*0.0875) = 8,446
Which is close to what you got, What did you think of my method?

Clever , but
I didnt see a number for H(mmf) or current before, maybe i missed it at 2am,
but now i think do , from your post just above

μr = B/(Hμo) = [1.004/(200*87.5*10-3/(0.110+0.0375+0.0375))] /4piE-7 = 8,446

But when I apply the other method I get something really small for the relative permeability using the control winding:
μr = B* Length / (μ0 * N * I ) = 1.014*(0.110)/(4pi*10-7 *840*0.0272) = 3,885 rel permeability
So what's going on? Bizzare

you're using way too short length, only one leg instead of the whole magnetic closed loop path ?

 

[tim9000]

if we use pk=pk in both numerator and denominator
or we use rms in both numerator and denominator
we get same result
because the 2√ 2 is in both numerator and denominator so cancels
i cannot conceive of a peak to peak impedance

H'mm ok.
Well what I mean by Zmax and Zmin...So the current through the mag amp will oscillate back and forth at excitation frequency. But impedance of each outer leg will also swing a little bit as the excitation voltage goes from one peak to the other peak, each cycle, point on the BH curve going from the left minimum offset (no red triangle) to the furthest point on the right of the curve (where there is the full triangle). Correct?
So what is Zmin and Zmax, moreover what exactly is that 18 Ohms, if there is no average impedance of each outer leg?

 

[tim9000]

Excellent picture by the way.

you're using way too short length, only one leg instead of the whole magnetic closed loop path ?

When you say I'm using too short a length, did you mean just for:

"But when I apply the other method I get something really small for the relative permeability using the control winding:
μr = B* Length / (μ0 * N * I ) = 1.014*(0.110)/(4pi*10-7 *840*0.0272) = 3,885 rel permeability"

Or for
"μr = B/(Hμo) = [1.004/(200*87.5*10-3/(0.110+0.0375+0.0375))] /4piE-7 = 8,446"

too? I'm assuming that's no good either because I didn't use the complete path length, just as the 3,885 answer I only used the length of the path where the cross section of the winding was the same.

I'm a bit stuck, because ordinarily I'd think that 'well B is the same through the core, because as the flux from the center splits, so does the cross sectional area half, so B is the same'
But when I put in:
μ_r = B* Length / (μ0 * N * I ) = 1.014*0.480 / (840*4*10^-7*pi*0.0272 = 16,952
Which...is too high? It's even higher than your:

Air core should be
L = μ0N2A/Length
L = 4piX10-7 X 8402 X 555 X 10-6 X2parallel magnetic paths / 295 X 10-3
L = 3.34 millihenry

35.1 measured for iron core / 0.00334 calculated for air = 10,520 relative permeability ?
what the heck is that stuff ?

Say I tried to use the excitation of the 200 turn coil, as in you're new picture, and tried to add the additional parallel paths, knowing that the flux will split at a 2:1 ratio through the centre leg, say y = 1.004 and we know X goes down the other outer leg, so 2X goes down the center leg, so 3X = Y, therefore X = 0.338 T, thus the centre leg has 0.676 T in it:
μ_r = B/(Hμ_o) =
[1.004*(0.110+0.0375+0.0375) + 0.676*(0.110) + 0.338*(0.110+0.0375+0.0375)] / (4piE^-7200*87.5*10^-3) = 11,627
which is close to your permeability using the centre leg impedance, but I'm still non any less baffled. Unless you think it's a reasonable replacement for the other attempts to attain the permeability using the 200 turn coil curve.

 

[jim hardy]

Say I tried to use the excitation of the 200 turn coil, as in you're new picture, and tried to add the additional parallel paths, knowing that the flux will split at a 2:1 ratio through the centre leg, say y = 1.004 and we know X goes down the other outer leg, so 2X goes down the center leg, so 3X = Y, therefore X = 0.338 T, thus the centre leg has 0.676 T in it:
μr = B/(Hμo) =
[1.004*(0.110+0.0375+0.0375) + 0.676*(0.110) + 0.338*(0.110+0.0375+0.0375)] / (4piE-7200*87.5*10-3) = 11,627
which is close to your permeability using the centre leg impedance, but I'm still non any less baffled. Unless you think it's a reasonable replacement for the other attempts to attain the permeability using the 200 turn coil curve.

Good ! That's the approach for how flux divides. I don't think the ratio center to outer will be quite 2::1, but you mage a major improvement in approach.

center leg reluctance R is in parallel with 2R X (30/125 + 1 + 30/125), the C shape to its right

 

[jim hardy]

I'm a bit stuck, because ordinarily I'd think that 'well B is the same through the core, because as the flux from the center splits, so does the cross sectional area half, so B is the same'
But when I put in:
μr = B* Length / (μ0 * N * I ) = 1.014*0.480 / (840*4*10-7*pi*0.0272 = 16,952
Which...is too high? It's even higher than your:

i remember making the same assumption about uniform flux density
what i did was cut core right in half and use path length around one half
which is more like, traversing approximate middle of the iron,

110 + 45 + 110 + 45 = 310 mm

so your result of 16,952 adjusted to my length , multiply by 310/480, gives 12,691
still high but closer

would take me a while to find my calc and see what's reason for difference

but i want to get this up , w'e many timezones apart
1pm here and i have to get some yard work done before heat of the day

EDIT i retyped 31/48 * 16952 into windows calculator and this time it reports 10, 948 which i think is close.
i don't know if i have fat fingers or a recalcitrant computre, but have developed habit that with Windows calculator i have to take the 3 out of 5 answers that agree because always get a 3/2 split.
Hence my frequent "Check my arithmetic"

windows is sentient and knows who hates it.

 

[jim hardy]

what i did was cut core right in half and use path length around one half

rason that works is we could fold right half over against left half , making two rectangles into one with uniform cross section.

Making the back to back D's into a single D twice as deep

 

[tim9000]

rason that works is we could fold right half over against left half , making two rectangles into one with uniform cross section.

Making the back to back D's into a single D twice as deep

Yeah I thought that was the reason, It did occur to me, so using that method
μr = B* Length / (μ0 * N * I ) = 1.014*0.310 / (840*4*10-7*pi*0.0272) = 10,829
Which is close enough to your value using the impedance for me to take it as gospel. I'll just put the value of 11,000 (obtained using the 200Turn) as being different because of a not quite 2:1 flux division, and say it reasonably cooberates the value. So call it somewhere between 10,520 to 10,829.

Ok, in the interests of time I think we can call permeability? And move on...or back, as it may be. So how about:

Doesn't seem very accurate for Z at time, that number is it pk to pk Z?
Although Assessing Zmin and Zmax I'm not really sure what you'd be able to do either, I suppose you'd have to say:
Zrms = 18 Ohms /(2*√2)
Zmax = 18 Ohms /(2*√2) + 18/2
and Zmin = 18 Ohms /(2*√2) - 18/2
?
So current through mag amp is: Imin = Vsupply / [18 /(2*√2) + 18/2] ? And
Imax = Vsupply / [18 /(2*√2) - 18/2]
And Rms current through amp to load is Irms = Vsupply / [18 Ohms /(2*√2)]
??

if we use pk=pk in both numerator and denominator
or we use rms in both numerator and denominator
we get same result
because the 2√ 2 is in both numerator and denominator so cancels
i cannot conceive of a peak to peak impedance

H'mm ok.
Well what I mean by Zmax and Zmin...So the current through the mag amp will oscillate back and forth at excitation frequency. But impedance of each outer leg will also swing a little bit as the excitation voltage goes from one peak to the other peak, each cycle, point on the BH curve going from the left minimum offset (no red triangle) to the furthest point on the right of the curve (where there is the full triangle). Correct?
So what is Zmin and Zmax, moreover what exactly is that 18 Ohms, if there is no average impedance of each outer leg?

Then after we've got impedance sorted I want to get back to inductance (if you please).

Thanks again

 

[jim hardy]

Sounds good.

Let me ask one more question before we leave permeability

the nature of your DC source that powers center leg...

Is it just an adjustable DC supply? Does it have a large capacitor at its output ? Was it perhaps connected but turned off or set for zero when you did your AC volts-current plot on outer legs ?

 

[tim9000]

Sounds good.

Let me ask one more question before we leave permeability

the nature of your DC source that powers center leg...

Is it just an adjustable DC supply? Does it have a large capacitor at its output ? Was it perhaps connected but turned off or set for zero when you did your AC volts-current plot on outer legs ?

The DC supply was adjustable, I suppose it would have had a large capacitor or two inside it, however I believe (from memory) the DC leg was open circuit when I took the AC current and voltage plot. And out of interest I did notice that when it was connected but turned right down the magnetic circuit was effected compared to when it was turned off. But as I said, it was actually Open circuit when I took the readings, so no impact what so ever.

CheersP.S

So how about:

 

[jim hardy]

H'mm ok.
Well what I mean by Zmax and Zmin...So the current through the mag amp will oscillate back and forth at excitation frequency. But impedance of each outer leg will also swing a little bit as the excitation voltage goes from one peak to the other peak, each cycle, point on the BH curve going from the left minimum offset (no red triangle) to the furthest point on the right of the curve (where there is the full triangle). Correct?

back to post 147...

so over a range, like the big triangle, slope* of orange hypotenuse would be an 'average' impedance, One might adjust his drawing to make hypotenuse a better least squares fit to the blue curve?
*(slope = rise/run)

So what is Zmin and Zmax, moreover what exactly is that 18 Ohms, if there is no average impedance of each outer leg?

There IS an average impedance, Δvolts / Δamps
and 18 = estimated 2 volt red line / estimated 110ma from this picture in 147

Impedance changes drastically at the knee
my magamp book uses a two segment slope volts/current curve to simplify, just two straight lines intersecting at the knee.

 

[tim9000]

back to post 147...
View attachment 87240

so over a range, like the big triangle, slope* of orange hypotenuse would be an 'average' impedance, One might adjust his drawing to make hypotenuse a better least squares fit to the blue curve?
*(slope = rise/run)There IS an average impedance, Δvolts / Δamps
and 18 = estimated 2 volt red line / estimated 110ma from this picture in 147
View attachment 87241

Impedance changes drastically at the knee
my magamp book uses a two segment slope volts/current curve to simplify, just two straight lines intersecting at the knee.

Excellent, I think I'm close to getting this now.
I need a bit of a refresher on the curve:
right so in post 147 we were using some hyperthetical example of exciting with 15VAC and having 20mA load current for zero DC control current. We then applied some DC current to offset the point to where the red mark is on the curve, where by the load current becomes dependant on the excitation V / the amount of impedance that isn't collapsed on the mag amp + the load impedance. This will wobble a bit due to the excitation fluctuating between zero and a peak.
Can you remind me how we got the magnitude of that wobble, that is, how it was about 110mA?

 

[tim9000]

I wanted to post this after I get your reply about the magnitude of 110mA in the above post, but I also need to shoot it off before I go to bed:

There IS an average impedance, Δvolts / Δamps
and 18 = estimated 2 volt red line / estimated 110ma from this picture in 147

If I take a point on that, say 15V, the impedance of the amp must be Z = 15/0.02 (Ohms) from the definition of V/I
IS THE DIFFERENCE between when there IS a DC offset is that you can no longer use that direct relation/definition, you then must look at the amount of change, ala what we've been talking about. Then we have to look at Δvolts / Δamps.
OR althernatively are you saying that Δvolts / Δamps = V / I and that taking Z = 15/0.020 is actually just the same thing? (So I was seeing a difference that wasn't there perhaps?)

In your post #147 Say I want to calculate Z of my mag amp from the actual BH, say at the midpoint of the top quadrant, and say B peaks at the knee point on the curve (possibly irrelevant), also say there's no DC control current. Let the excitation voltage be 100Vrms at 50Hz and there's 200 turns around the steel.
So how do I calculate Z? I suppose I know I because I know H? or can I do this without current?. Was dt = 100pi?
Z_middle = N* dΦ/dt / I = N*(Φ_mid-0) / (I*100π) ?
(as you can see, My basic calculus has been very neglected over the last year)

 

[jim hardy]

Okay, I'm a bit OCD so had to revisit permeability when driven from one end

from post 160

flux was free to circulate through both legs in parallel
magnetic circuits are analogs to electric ones, flux divides among parallel paths
and one calculates reluctance by reciprocal of sum of reciprocals of parallel reluctances , just like resistances in parallel.

so, since Reluctance is μμ₀L_ength / area
and the core can be represented as an assemblage of pieces with constant area
and since area, μ and μ₀ are all constant
i can easily calculate an effective path length when driven by a coil on one outer leg

it's back to back D's ,,,
this time i'll use lengths to center of iron

assume driving coil on left, work from right to left
1/(1/(110 + 37.5 + 37.5) + 1/110 + 1/110) + 37.5 + 37.5 + 110 = 227.4 ,

so your formula

μr = B* Length / (μ0 * N * I )

μ_relative = 1.003 X 0.2274 / (12.56E-7 X 200 X 0.0875 ) = 10,377

hmmm to 3 places we've got μ_relative numbers of 10,400 10,500 and 10,900

~5% spread ?
not too bad for reading off graphs , i'd say.Okay I'm satisfied - you have a great core there !
and we're getting consistent results.

Check my arithmetic ?

 

[tim9000]

Okay, I'm a bit OCD so had to revisit permeability when driven from one end

I understand, Post #170 isn't going anywhere.

assume driving coil on left, work from right to left
1/(1/(110 + 37.5 + 37.5) + 1/110 + 1/110) + 37.5 + 37.5 + 110 = 227.4 ,

I see what you're doing, It took me a couple minutes but, nice work! funny to divide the center like that.

 

[jim hardy]

I understand, Post #170 isn't going anywhere.

thanks for the patience. That one next. Post 170:::

I need a bit of a refresher on the curve:
right so in post 147 we were using some hyperthetical example of exciting with 15VAC and having 20mA load current for zero DC control current. We then applied some DC current to offset the point to where the red mark is on the curve, where by the load current becomes dependant on the excitation V / the amount of impedance that isn't collapsed on the mag amp + the load impedance. This will wobble a bit due to the excitation fluctuating between zero and a peak.
Can you remind me how we got the magnitude of that wobble, that is, how it was about 110mA?

gotta get my head around 147, too.

I hope i didn't just assume it... do we know what is your load? Known resistor or a light bulb ?

 

[tim9000]

I hope i didn't just assume it... do we know what is your load? Known resistor or a light bulb ?

The light bulbs in parallel were something over 50 Ohms

 

[jim hardy]

dang it whole post disappeared **!@@@{(*TYEW$^#%#! I HATE (PRESENT PARTICIPLE ) WINDOWS

 

[jim hardy]

from 147

back to the picture
were we using this one ?

from your 148

I was thinking this one: (it's from 140)
jim hardy said: ↑

That one was just intended to show that impedance collapses when DC pushes us out past the knee
were the load line establishes current by Vsupply/Zload.
So between the knees, magamp controls current
it gradually relinquishes control as we push it into saturation.

looks like my poor drafting skill made the two yellow lines not look equal width..

Between the knees Zmagamp is 70 volts p-p / load current p-p
beyond the knees it's 2 volts p-p / same load current, a 35:: 1 turndown.

That was purpose, to show collapse of impedance.
Load line tells us what would be load current

but for that picture i'd assumed constant load current, i think 110ma p-p
Let us assume both of thise yellow load current lines were 110ma wide,
Zsaturated = 2V/0.11amp = 18 ohms
Zbetween knees = 70 / 0.11 = 636 ohms.

 

[jim hardy]

Now make a load line for the voltage divider

I = Vsupply / (Zmagamp + Zload)

assuming your Vsupply were 70V p-p
and your load were 50 ohms
and we can swing your magamp between 18 and 636 ohms

at 18 ohms we'd have
I = 70∠0 / (j18 +50)) = 70 /( 53.1∠19.8) = 1318ma ∠-19.8°
Vmagamp will be 1.318amps X 18 ohms = 23.72
Vload will be 1.318 X 50 = 65.9
it's a phasor voltage addition , so Kirchoff will be happy provided we include the angles

and at 636 ohms we'd have
I = 70∠0 / (j636 + 50) = 70 / (638∠85.5 ) = 109.7ma∠-85.5°
Vmagamp will be 0.1097 X 636 = 69.8 volts
and Vload will be 5.48 volts
observe that your magamp can only hold off 70 volts p-p, so 110 ma was a good guess for shutoff current.

where we are on the curve depends on DC through control winding.

 

[jim hardy]

i guess i should add volts across load to the curve ?

 

[tim9000]

This is good, despite your unfortunately uncooperative computer, I feel we're making progress...you'd better read this whole post through before putting anything to paper, I've been editing it all bloody day.

i guess i should add volts across load to the curve ?

That's alright, you only left off "23.72V" under the 70V on your load line picture didn't you? I think I get that, if so.

Now would probably be a good time for me to pester you about explaining what you meant by:
" volt-seconds per turn defines a core's flux capability
volt seconds defines the ability of a winding on a core to not saturate when asked to block current"
I'm not sure I completely get that...

I should have put up more precise data before, I would have if I had of known I'd struggle the concept so much, since we need it I'll include some better numbers than eye-balling the graph. (sorry I realized I was doing this for 70V pk, not pk-pk)

I do however think I understand the load-line idea and I really like it...with a 'but':
SO what you said was for those two series 200 Turn coils when we excite with 70 V pk-pk at 50 Hz, that will get us with a peak flux density B at the knee point, with a pk-pk current of around 110mA, when no DC control, telling us the impedance of the Amp + bulbs = Z_{between knees} = 70 / 0.11 = 636 ohms. I completely agree with that method. But say I want to go past the knee point t 50V*√2 = 70V pk = 141 V pk-pk.
At 70V rms excitation with zero DC I got 67.3V rms across the mag Amp and only 66mA rms through the load. So I think it can hold off more current than you estimated:
Say we were using 70V rms. At a VDC = 424mV (not shown below) then you would get the 23.72 V across the mag amp. So I think the impedance collapses more than you reckon, see:
(I'm pretty sure the DMM gave it in rms, so) RMS Excitation:

Looking at my data for 50Vrms*√2 = 70V pk, (which is less extensive data than at 70V rms data set):

The supply would fluctuate a bit, you can see as VL + V3 are sometimes greater than V1 (VL + V3 = V1 ideally). You can see that the voltage across the amp went as low as 775mV rms and the hold off current was as low as 38mA rms! (not bad eh?)
I take it that's 0.038_rms*2√2 = 107.5mA pk-pk.

Rather than 2V for ΔV_pk-pk, should ΔV = 50 - 49.8 = 0.2Vpk-pk? [edit: not pk-pk] Because we really didn't stipulate how far the DC offset control current was being pushed out, why not go as far as the data goes?
So re-assessing the impedance:
Z_saturated = 0.2V/(2*√2*0.038)amp = 1.86 ohms [Edit: is actually 5.26 Ohms probably, because 0.2 not pk-pk]
Z_{between peak voltages} = (50*2*√2) / (2*√2*0.038) = 1315.8 ohms

I = V_{supply pk-pk} / (Z_magamp + Z_load) = 141.4∠0 / (j1.86 +50)) = 2.8 <-0.037 A pk-pk = 1A rms
...what the hell, why was my rms current apparently according to the data only 150mA rms??

EDIT:
ok, well looking at my notes they were 40W bulbs (in parallel), to be used on RMS 240V. So that's 166mA per bulb or 333mA rms total rated current through load? So 150mA may be reasonable?

I thought I had a decent glow on them by 70V rms (see data above), say I was taking the power at 15VDC at 70V:
70*0.178 = 12.46W being used by both the bulbs, do you think that would be pretty bright? I wish I took a video.
at 50V rms:
50*0.15 = 7.5 W is 3.75 W per bulb, that seems small.

What I didn't like before was that you used the shut-off current to approximate the 636 Ohms, then I think you might have used that to confirm the shut-off current's accuracy as a guess.
I'm not sure I'm happy with our method of using the shut-off current as the guess for the magnitude of the yellow when we're saturated, I'm not happy using (2*√2*0.038) = 107mA pk-pk:
Z_saturated = 0.2V/(0.107)amp = 1.86 ohms [Edit: is actually 5.26 Ohms probably, because 0.2 not pk-pk]

I am happy with using the data:
Z_{between peak rms voltages} = 2*49.01_Vrms / (2*0.038) = 1,289.7 ohms
I think the method of calculating the shut-off current as:
I_rms = 50∠0 / (j1289.7 + 50) = 38.7mA < -1.53°

V_magamp will be 0.0387 X 1289.7 = 49.9 volts rms
and so wouldn't V_load will be: 50 - 69.9 = 100mV?
The data said load was 2.32V, indicating that (50 - 2.32)Vrms / 1289.7 Ohm = 36.97mA rms
Which is not much more than 1mA different from predicted, not bad.

Given the difference in
Z_saturated = 0.2Vpk-pk / (2*√2*0.038)amp = 1.86 ohms [Edit: is actually 5.26 Ohms probably, because 0.2 not pk-pk]
I = V_supply / (Z_magamp + Z_load) = 141.4∠0 / (j1.86 +50)) = 2.8 <-0.037 A pk-pk = 1A rms to my 150mA rmsPerhaps if ΔV / ΔI was actually in Rms so
0.2/0.038 = 5.26 Ohms, so
I = V_supply / (Z_magamp + Z_load) = 50 / (j14.88+50) = 0.96 A rms...ah still too high...
From the 50V data, 0.775V / 0.15 = 5.1666, so Z = ΔV / ΔI = 5.17 Ohms, makes sense right?

KVL still not adding up because say 150mA load current WAS right:
0.15*(50_load + 5.26_{collapsed impedance}) = 8.29V, not 50V

Ordinarily I'd think I read the scale on the meter wrong...Eureka!
ok, so 50.4 Ohms was the COLD resistance, but the hot resistance using 240²/40W = 1.44K Ohms...great, all I need is a semi-linear load to make things even more complicated!

so to satisfy KVL above the bulbs would have been around 330 ohms, geez what a day.

I'm still not sold on if we need a better way to get the saturated impedance than the same ΔI as at zero control DC. Because it was so vague about how far the DC shifted out, I just took the difference between the last two data points for ΔV on the load. And I don't think it should be the load, I'd have thought the volts on the Amp itself.
I suppose that's like saying we applied 12.5V DC to the control coil (looking at between 10 & 15VDC) and got a 0.2 rmsV on the load there.
I Try, If Z_amp = ΔV / ΔI = ΔVamp/0.038 = (0.915-0.775) / 0.038 = 3.68 Ohm
Which doesn't give the right V3 mag amp Voltage, so you'd have to conclude that ΔI is wrong and find it from the data as ΔI = ΔV/Z = (0.915-0.775) / (0.775V / 0.15) = 0.14/5.166 = 27.09 mA rms
which then gives Z_amp = ΔV / ΔI = 0.14/ 27.09mA = 5.166 Ohms.
Which doesn't use the BH curve, because that 0.15 was measured current, and not from expected impedence and the BH curve, as I want it to be. Is there another way of getting ΔI using the BH curve, maybe permeability?Your thoughts?

lol, thanks

 

[jim hardy]

I feel we're making progress...you'd better read this whole post through before putting anything to paper, I've been editing it all bloody day.

Thanks.

And it looks like a head scratcher.

i'll be baaaacckkkk...

 

[jim hardy]

That's alright, you only left off "23.72V" under the 70V on your load line picture didn't you? I think I get that, if so.

Yes.
I probably should have even extrapolated the line all the way to 0 volts and zero amps..
Unlike a tube amplifier load line where load and amp are both resistive, this load line is a resistance load and an inductance amp
you saw my polar-rectangular in voltage divider arithmetic.
so i wasn't quite sure whether to draw a straight load line or attempt a curve; figured the straight line was better to demonstrate the concept.
....................................

Incandescent lamps change resistance about tenfold cold to hot so i'd have expected more like 100 ohms cold - maybe your 40 was both in parallel ? No matter, you're aware...

Now would probably be a good time for me to pester you about explaining what you meant by:
" volt-seconds per turn defines a core's flux capability
volt seconds defines the ability of a winding on a core to not saturate when asked to block current"
I'm not sure I completely get that...

Back when we were developing inductance concepts, around posts 85 to 95

remember this curve? annotated, from 85

If it had only one turn,
volt-seconds applied to that one turn would be the flux at which it saturates
e = -n*dΦ/dt
Φ = -1/n * ∫edt , n=1

when it saturates it can no longer block current

if n>1 it can hold off more volt-seconds before it saturates and impedance collapses

so volt seconds per turn describes the core's saturation flux, basically its area
and
volt-seconds describes the the ability of the inductor (core + winding) to hold off current against applied voltage, in terms of volt-second productso we could speak of a core and its windings as capable of so many webers, B X A_rea
or of so many volt-seconds .
Latter will be handy sometimes.
It's the physical concept behind volts per hertz rating of transformers, generators and motors.next we made our voltage a repetitive wave

and saw that starting the sinewave at zero gives flux a DC offset

then we let it saturate

and saw the current spike when we exceed the inductor's volt-second capability
complete with real 'scope trace of real transformer inrush in post 91

so what's going on in the magamp is
we pick an inductor with enough volt-second capability to hold off Vsupply
then we degrade that capability by driving it partway toward a knee with DC
and let Vsupply saturate it
so current peaks appear and current goes up.

Sometimes we arrange the windings so DC and load current fluxes always buck or boost, and we can push the magamp either away from or toward saturation by polarity of DC.
see gold line here

but i think yours is not wound that way. Just wanted to plant the thought in preparation for self saturating magamp with rectifiers later on.Observe this will give a non-sine current, that's why my old timey book says to use average not RMS meters
but we have what we have and will be okay if not exact. I don't think magamps is an exact science anyway.

I printed out your last post, took 3 pages, but that way i can study it without flipping screens which wears me out...

on to next paragraph of it.

Sorry for the delay - i had two things to tend to, one sixty miles North and the other 70 miles South of here. Have been only intermittent on PF last two days.

old jim

 

[tim9000]

pretty helpful explanation I think

If it had only one turn,
volt-seconds applied to that one turn would be the flux at which it saturates
e = -n*dΦ/dt
Φ = -1/n * ∫edt , n=1

when it saturates it can no longer block current

if n>1 it can hold off more volt-seconds before it saturates and impedance collapses

H'mm, I'm just thinking 'the more turns you have the less current it takes to saturate the core' but I also understand inductance proportional to n squared. Well looking at that equation I'm assuming dt is just the amount of time so far, because e is a constant so it's just like ∫dt, the flux will ramp up at the gradient -e/n

Those V I curves I took to generate the BH curve, how would I go about converting those into flux density Vs time curves?

Interesting, I never concidered that you could shift the amp away from saturation, yes I have a couple questions about rectifiers when we get to it.

I printed out your last post, took 3 pages, but that way i can study it without flipping screens which wears me out.

Ah, that is actually a really good idea I hate reading off a screen,

thanks Jim I appreciate it

 

[tim9000]

Incandescent lamps change resistance about tenfold cold to hot so i'd have expected more like 100 ohms cold - maybe your 40 was both in parallel ?

Yep, they were, and yep they did change, you'll get to that part.

I can't even really remember what I did the other day that well, but from memory I ended up using measured current, to calculate the impedance, but instead I wanted to come up with was, was very much like you're curved load-line: some sort of impedance curve VS. DC control, for a set excitation voltage, determined by the BH curve, to from that determine the load current.

I suppose that sort of mag amp load line would look something like this

made from the BH curve...
The data of one of the V I curves used in the plot was:
Energising one half small coil V:
0.8
2
5
10
13.4
15.2
20
23.77
28.2
32.2
33.99
35.15
37.4
39.17
40.22
Corresponding current:
I (mA)
2
4.8
10
17
21
23.2
30
36
44.4
57
71.5
85.5
150
260
390

 

[jim hardy]

I should have put up more precise data before, I would have if I had of known I'd struggle the concept so much, since we need it I'll include some better numbers than eye-balling the graph. (sorry I realized I was doing this for 70V pk, not pk-pk)

it's okay , I'm so fat-fingered i make mistakes with windows calculator and that's why i am so slow - takes about ten edits to get 'em all fixed. maybe i can find an old pocket calculator...

I do however think I understand the load-line idea and I really like it...with a 'but':
SO what you said was for those two series 200 Turn coils when we excite with 70 V pk-pk at 50 Hz, that will get us with a peak flux density B at the knee point, with a pk-pk current of around 110mA, when no DC control, telling us the impedance of the Amp + bulbs = Zbetween knees = 70 / 0.11 = 636 ohms. I completely agree with that method.

cool !

But say I want to go past the knee point t 50V*√2 = 70V pk = 141 V pk-pk.
At 70V rms excitation with zero DC I got 67.3V rms across the mag Amp and only 66mA rms through the load. So I think it can hold off more current than you estimated:

been wanting to look at that myself.

Would that be something like this ? 70V = 35 across each of two coils?

I interpolated 77.5 ma rms off that chart

oops, you said " I got 67.3V rms across the mag Amp "
that'd be 33.65 per coil
and that interpolates for me to 65 ma , close enough to your 66 (hooray something went right !)

So I think it can hold off more current voltage? than you estimated:

Yes , it seems to hold off 70 volts RMS pretty well

i understand now, those are RMS voltages on your graphs
i've probably mixed messages on you on that regard, trying to progress from DC to AC one step at a time.

Good for you !

Say we were using 70V rms. At a VDC = 424mV (not shown below) then you would get the 23.72 V across the mag amp. So I think the impedance collapses more than you reckon, see:

good show...

so, much of your data was taken with core pretty well into saturation

the real region of interest is the steep part of that curve, where you have lots of gain , below 2Vdc.
You said that center coil was what, ten ohms and 840 turns ? Correct my numbers...
2v would be 200 ma, which X 840 turns is 168 amp-turns ?
How does that compare to your load amp-turns,
do i remember right ~178 ma X 200 turns on each of two outer legs = 31.6 X 2 = 71.2 amp-turns?
You swamped the core, but it doesn't hurt anything... Do you have more readings in steep part of curve ?

Let me plant another graphical idea now...

In our 70 volt excitation curve we'll do two things
we'll rotate that red volts line to where it intersects our blue curve
and we'll lengthen it to reflect different excitation voltages.

Then i'll try to figure out how to add DC to it...

 

[jim hardy]

Looking at my data for 50Vrms*√2 = 70V pk, (which is less extensive data than at 70V rms data set):

The supply would fluctuate a bit, you can see as VL + V3 are sometimes greater than V1 (VL + V3 = V1 ideally). You can see that the voltage across the amp went as low as 775mV rms and the hold off current was as low as 38mA rms! (not bad eh?)
I take it that's 0.038rms*2√2 = 107.5mA pk-pk.

Yes, VLoad + Vmagamp = Vsupply but it's phasor addition, and with saturation it may be distorted from sinewave. But instant by instant KVL held ! And with enough math tools we could prove that, but even the pros don't try to analyze magamp waveforms to very great detail..

As you know , a half cycle of 50 volt sinewave has a smaller volt-second product than a 70 volt one,
so your magamp is better able to hold back current. Love it when observation and mental model agree !
To use it at higher voltages you'd add turns, or core area.

 

[jim hardy]

Now back to flux in the core

Observe that on alternate half cycles,
in outer legs, load current opposes DC flux in one and aids in the other.
The effect swaps legs very half cycle.

So one leg is pushed farther into saturation by load current
and the other is pushed away from saturation
giving the outer legs unequal reluctances..
But since their windings are in series, they have equal MMF's applied
and their different reluctances should result in different fluxes

If there's nothing to prevent it, that difference in AC flux will do its best to flow through the center leg.
That's why i asked about AC on your center coil.
And about the nature of your DC source.
If your DC source has a large capacitor on its output,
that capacitor would allow AC current to flow in center winding.
That current would block AC flux by allowing AC current and Lenz's Law to take action, forcing the AC flux on around outside path.

If instead your DC source is a true current source that disallows AC current,
then
AC flux can flow through center leg .
That let's magamp's counter EMF voltage swap from one leg to the other on alternate half cycles.

So a magamp with true DC current on center leg is a slightly different animal than one with just DC voltage on center leg.
That's why some of the old timers added a choke in series with their center leg, to block AC current in it forcing flux out where they wanted it..

Not a major issue, just don't want it to come as a surprise . I don't know what is te nature of your DC source for center leg. I think you said about 3/4 volt AC on it? So you don't have a whole lot of AC flux there at whatever condition you measured it.

whew too many ideas churning around

i still want to add DC to that last VI graph in 185

 

[tim9000]

Yes, it was 70Vrms on the whole amp, so that would have been 35rms per outer leg. I didn't think of that.
Unfortunately I only took many data points running up the linear part for 70Vrms when I was modulating the DC control, which you've seen.

Then i'll try to figure out how to add DC to it...

Great, thanks, because I really want to be able to generate a load line and/or inductance line, for the impedance or inductance of the mag amp, Vs. control voltage.
And on that note of you saying Z is the slope of the redline intersecting Zero to the curve (post #185), I thought you said that it was the tangent gradient to a point on the curve?
Because this is a point of some confusion to me, when I plotted the Z from V/I and when I plot ΔV/ΔI I (far right) get very different things:

Where the current in mA, is the X-axis.
EDIT: Woops, forgot to correct the scale for the far right, here's proper SI division:

I'm not entirely sure what the difference between it and the one in the middle (this one is the tangent gradient and the one in the middle is the gradient of the red line from zero to the curve), (i.e why this one is so spiky. Probably because it was a crude form of diffenentiating with limited data.) Maybe in essence they're the same result? Just in different forms? However I do notice that they resemble what I would expect the permeability to look like, which I would expect to be proportional to the impedance, so that's good.I'm assuming that either the corrected right graph units or the middle graph is the correct representation of V.s per outer leg, or the centre leg graph:

Where X-axis is load current mA and Y-axis is the impedance of the amp.
It tells me the maximum impedance of the core without DC control, at an amount of V.s? (or how much volts it can hold off without DC control at a V.s)
I guess I have to work out the V.s/turn then times it by 400 turns to work out the amount for the two outer legs in series, because I only took a curve of one leg excited, not both excited in series?

I was thinking I should be able look at a graph for V/Hz impedance expectation, look at the point of 70V for 50 Hz, and see it matched something like: 657.9 ohms per leg (from Z_{between peak voltages} = (70Vrms / 2legs) / (0.038 A rms) data) Then as I'm adding DC it becomes V/Hz = Excitation RMS + VDC, and I can see what the impedance will be.
Maybe make the from the centre leg data V.s/840, then make a curve to fit the outer legs V.s/840 * 400 = V.s of outer legs?

Also, have you any ideas how I would convert my V Vs I curve (far left) into a B(t) curve?

Cheerio

 

[jim hardy]

And on that note of you saying Z is the slope of the redline intersecting Zero to the curve (post #185), I thought you said that it was the tangent gradient to a point on the curve?

i was taught graphical approach to tube amplifier design so it's what i still use.

Z at any point is the slope of the curve at that curve, which the tangent line shows us.

In this image that we've used before

out at that operating point, load current is set by your lamps because the magamp's Z has collapsed ,
so Z is slope of blue line over the span of yellow current line
about the ratio of red line to yellow line lengths.

What flux is in the core at that operating point ?

Would it be fair to say it sees a DC flux maybe equivalent to what 275 ma of DC current in the load winding would produce? Extrapolate straight down from red spot on graph..
And it sees AC flux of (yellow line) / (2√2) rms.

That DC flux really came from DC on the center winding though.
With 840 turns there instead of 200, it took only 275 X 200/840 = 65.5 ma on center leg to make the DC flux. Check my arithmetic.
If center leg was ten ohms that's only 0.65 VDC across it
i don't see any DC voltages that low - if I'm reading your posts right you were WAY out on the curve for most of your readings.
But that's beside the point.

All i wanted to do was cultivate fluency with the V-I (or B-H) graph.

So far we've figured Z at some constrained point on the curve where it's comparatively linear,

In order to operate there something has to constrain our volts or our amps
and the load line demonstrates how that happens.

What if , to understand the magamp's magnetics a little better
i took the magamp with Zload of zero and just subjected it to voltage?

Now only the magamp's inductance would constrain current
and when voltage reached the knee we'd start saturating late in every cycle so current would increase

Calculating Z as the ratio of my voltmeter reading divided by my ammeter reading,
Z should be fairly constant until i reach the knee
where amps begin to skyrocket so my calculated Z goes down.
Maybe it's more intuitive to me because I've done that with analog meters and a Vatiac, and it makes a real vivid picture when the ammeter pegs for just a teeny tweak of the variac..

Anyhow that's the mental model i was trying to convey with that picture.
you have an AC voltmeter across the magamp and an AC ammeter in series with it
so by dividing their readings you calculate Z.
Now the current will be peaky and a long way from a sinewave, but you'll get a series of readings that plot a curve nonetheless.

And i see you did that.
Is middle graph a plot of your raw volts/amps from left graph?
were lamps connected or disconnected when you took that data ?

Your rightmost post looks almost like a plot of the slope of leftmost plot, which is what it should look like


As of this minute I'm puzzled by rising behavior between 0 and ~35 ma
do you think your core might have a slow liftoff like this mumetal (post 52) ?

trouble is i don't see it in your graph, even at 400%.

we're moving along

 

[tim9000]

Ok, so it is actually the tangent to the curve, and the right-most graph is more accurate, but you can take the gradient of the red line passing through the zero point as a rough graphical method.

As of this minute I'm puzzled by rising behavior between 0 and ~35 ma
do you think your core might have a slow liftoff like this mumetal (post 52) ?

(you can see in the data of the curve cut-&-pasted in post #184) Certainly looks like it has a positive second derivative at first, if I had to describe it I'd say it has a 'less slow liftoff than the mumetal, and that the liftoff was more prolonged' so that the transition looks smoother than with the mumetal.
The first derivative looks exactly like what I'd expect the permeability of a ferris metal to look like.Is there a profile for the resistance of an incandescent for Ohms V.s heat or voltage? ( I could probably deduce and make one from the data I took I suppose)
I'm not sure if you've been pondering, but I'm eager to know what you thought of my idea; So did you think it was feasible to make some sort of load curve/line based off the BH curve, and one for the light bulb load. Then I could say: "right'o, I've got me excitation voltage a constant set known value, I've got my load line-curves (knowing my point on the load-curves caused by the DC), now I can work out using KVR the load current for the value of how much control flux I'm injecting"?

How I would convert my V Vs I curves (for 200 turns or 849 turns) into a B(t) curve?
The data I have is V Vs I, so for a frequency, didn't you say that they were the same as DC curves? How do I make these curves into V(t) DC curves?The right-most load-curve from the 200 turn:

for the 200 Turn and the (corrected) 840 turn impedance:

are probably no good because the number of turns for the impedance curve I'd need would be 400 turns?
But I haven't figured out how any of these match up with the data we've been discussing.
Possibly because converting V.s and V.s/N from my data still confuses me. And I need a V.s for a 400 turn core to make an apropreate load-curve to match my data? (hence my theory that I need to convert my existing curves into an equivilant)EDIT: WOOPS, THOSE Ohms in the graphs SHOULD BE MUCH BIGGER THAN THAT, I HAD IT RIGHT THE FIRST TIME, I SHOULD HAVE /1000 NOT x1000

[
I remembered that you said V.s/Turn gave the property of the core's flux capability, I had a crack at seeing if they'd be the same:
At Peak impedance seen on graph correlated Vrms/Hz :

From straight V/I 200 turn curve:
20V/(50*2*pi*200) = 0.000318

From tangent V/I 200 turns curve:
(10+(13.4-10)/2)V / (50*2*pi*200) = 0.000186
For peak impedance of 850 Ohms

From straight V/I 840 turn curve :
213.8V / (50*2*pi*840) = 0.00081

From tangent V/I 840 turns Curve:
(35.88+(103.8-35.88)/2)V / (50*2*pi*840) = 0.00026465
for peak 22640 Ohms

But as you can see the V/Hz are not the same values, should I just disregard the 200 turn set and say that the tangent V / I for the 840 turn curve is probably the most accurate?

Anyway according to that at 0.00026465 (V.s/Turn I assume) the impedance of the control coil circuit is 22640 Ohms, but as we saw, in the data, the peak impedance of the two series 200 turn coils was 1289.737Ohms at 50Vrms at 50Hz. Yet when I put in the V/Hz = V.s/Turn = 50/(2*pi*50*(200+200 turns)) = 0.00039788

So it seems the control coil has a different peak impedance than the two series outer coils, but I'd of thought the V.s/Turn would have been the same because I thought that was a property of the core and not the core + coil.

So I'm not sure what's going on?]
Also could you explain to me again why it's the tangent to the curve, and not just the value of V/I ?

Thanks!

 

[jim hardy]

wow I'm going to have to print your data

it's almost 2am here , i think I'm about done for the day

back tomorrow

jim

 

[tim9000]

wow I'm going to have to print your data

it's almost 2am here , i think I'm about done for the day

back tomorrow

jim

Also please see the Edit to my last post.

I thought this may be of help, I took extracts from my large data set where I was exciting the two series 200 turn coils, for values where the DC was off, remember V3 is the voltage across the Amp:

And the gradient of which (using the two methods) Is:

Also could you explain to me again why it's the tangent to the curve, and not just the value of V/I ?

I can kinda see why it's the tangent on an actual BH curve because
V/N = dΦ/dt
but even then I'm not sure about the ΔI and not just straight I.
I see the connection of ΔV/ΔI even less so to the V vs. I curve.

 

[tim9000]

As for generating a B(t) method, I thought I could use V.s/AN = V/wAN curve, but instead of the applied V, I could just choose an arbitrary voltage for all the points of B then

I was thinking I could generate the B(t) curve, for example 50V DC by Time at point = B_{at point} / [840*A*2*PI*50]
or a 1V DC curve by Time at point = B_{at point} / [840*2*PI*50]

Just to double check, V.s = V/(2*pi*freq) right??
Why not just V/Hz?
Because time period = 1/f
so V/f would be [V.s]

 

[jim hardy]

After a dozen false starts I'm back. Sorry, just couldn't settle on a direction and got frustrated at myself.
A new touchpad driver saved this computer from Maxwell's Silver Hammer...

(you can see in the data of the curve cut-&-pasted in post #184)

from 184...

I can't even really remember what I did the other day that well, but from memory I ended up using measured current, to calculate the impedance, but instead I wanted to come up with was, was very much like you're curved load-line: some sort of impedance curve VS. DC control, for a set excitation voltage, determined by the BH curve, to from that determine the load current.

I suppose that sort of mag amp load line would look something like this

made from the BH curve...
The data of one of the V I curves used in the plot was:
Energising one half small coil V:
0.8
2
5
10
13.4
15.2
20
23.77
28.2
32.2
33.99
35.15
37.4
39.17
40.22
Corresponding current:
I (mA)
2
4.8
10
17
21
23.2
30
36
44.4
57
71.5
85.5
150
260
390

wasn't sure what the columns were, ac volts across and ac amps through the little coil ? For various DC through center leg ?
That curve looks like what i'd expect it to.

don't know what direction to go from here.

if i understand we have this

and we started out with 0.8vac and 1ma ac = 1250 ohms for coil
and at 20 volts across coil we had 21 ma = 952 ohms for coil
and at 40.2 volts we had 390 ma = 103 ohms

but your curve goes clear to zero ohms and has DC on horizontal axis

i know that excel spreadsheets allow one to plot stuff with one click
but I'm not sure what I'm looking at.

 

[tim9000]

After a dozen false starts I'm back. Sorry, just couldn't settle on a direction and got frustrated at myself.
A new touchpad driver saved this computer from Maxwell's Silver Hammer...

Glad to have you back safe and sound...and with a new driver, though I'm not sure I get the 'silver hammer' remark.

If you're talking about the "Resistance Vs. DC V for 70Vrms Excitation" curve, than it does go to zero, that curve wasn't made from the same data that I cut and pasted that was used for the BH curve. That curve in post #184 was made from setup in the picture you drew in your last post # 194. The curve in post # 184 was V3/I for a variety of DC voltages with a fixed 70VAC. Where V3 is the voltage on the mag Amp (the two 200 turn coils).
And yes that's correct about the columns, that the data I cut-and-pasted was measureing the AC current as the AC Voltage was increased for only one excied small coil, like in the pictures in post #163 and post #172. There was no DC and the other 200 Turn was OC.

 

[jim hardy]

Maxwell's Silver Hammer is a weird 60's Beatles song

Bang! Bang! Maxwell's silver hammer
Came down upon his head.
Bang! Bang! Maxwell's silver hammer
Made sure that he was dead.

i think it's about how life seems to beat on us with its unexpected events

 

[tim9000]

Maxwell's Silver Hammer is a weird 60's Beatles song
i think it's about how life seems to beat on us with its unexpected events

Glad I asked, I just youtubed it. Yeah it was alright. I really only like 'a day in the life' and 'dear prudence'.

 

[tim9000]

See BH in post #193.
After calculating V.s/N as (V_rms*√2*2) / (π*50)
the V.s/N is smaller than B calculated via EMF equ. By a factor of 159.258 (top picture)
And when I include the area of the core, B is smaller than V.s/N by a factor of 0.17678 (bottom picture):

 

[jim hardy]

[just moving it to this thread...

QUOTE="tim9000, post: 5208207, member: 480143"]No the current doesn't, I know the voltage will start to move from being over the ideal TX to be over the resistance of the coil (as the current spikes), so V.s on the core won't be sineusoidal as the core saturates. But you can see from the graphs in post # 198 that the shape of the N.s/N curve is identical to the BH curve, so what does that mean?
Also, I don't see why B = Vrms/(2*π*f*N*Area) From the EMF of a TX equation (which is essentially Faraday's equ.)
But if: Φ = V.s/N
than why V.s/(N*A) = Vrms*√2*2*/(N*π*50*Area) = VPK*2*/(N*π*50*A) isn't equal to B?[/QUOTE]
okay will look after breakfast..
this curve from 198 ?

i have to catch up with what you did in 193
TTFN

 

[tim9000]

[just moving it to this thread...

QUOTE="tim9000, post: 5208207, member: 480143"]No the current doesn't, I know the voltage will start to move from being over the ideal TX to be over the resistance of the coil (as the current spikes), so V.s on the core won't be sineusoidal as the core saturates. But you can see from the graphs in post # 198 that the shape of the N.s/N curve is identical to the BH curve, so what does that mean?
Also, I don't see why B = Vrms/(2*π*f*N*Area) From the EMF of a TX equation (which is essentially Faraday's equ.)
But if: Φ = V.s/N
than why V.s/(N*A) = Vrms*√2*2*/(N*π*50*Area) = VPK*2*/(N*π*50*A) isn't equal to B?

okay will look after breakfast..
this curve from 198 ?

i have to catch up with what you did in 193
TTFN[/QUOTE]
No worries, take you're time, take 8 hrs, I'm not tired but I think I should 'hit the hay' it's 2:20am on my end.

Cheerio