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