How the power transfers across the Ideal Transformer

[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

 

[tim9000]

In that last post, there is an implicit understanding of mine that flux = V.s/N and I'll throw some numbers in for colour: Take the knee point flux density;
V_rms*√2*2 / (π*f*A*N) = 296*√2*2 / (π*50*0.00111*840) = V_pk*2 / (π*50*0.00111*840) = 5.7163 T
But we previously determined using the EMF equ that the knee point was about 1T

To point out, I was also able to cobble together a VI curve in post #192 of the series (200T + 200T circuit) mag amp voltage, for when there was no DC. I then took the gradient using the two methods (using zero to curve gradient [straight V / I] or tangent to curve) to show the impedance of the mag amp as the excitation voltage was increased.

To add to the question you moved over threads and the graphs in post #198,
is it fair to say that 'for each material core, there will be a specific V.s/N at which the is the V.s/N where maximum inductance can be achieved, regardless/irrespective of reluctance or turns number'?

And I'm still confused as to why Z = ΔV/ΔI at a point, and not just V/I (which is the gradient from zero to point on curve)
i.e. what the difference in compared methods of plotting Z curve in posts #190 and #192 between tangent Z and direct Z = V / I

Thanks

 

[jim hardy]

from 198

After calculating V.s/N as (Vrms*√2*2) / (π*50)
the V.s/N is smaller than B calculated via EMF equ. By a factor of 159.258 (top picture)

159.256 is soooo close to 50Π
it's got to be simple.

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]

Volts/hz and volt-seconds are so closely related i never thought about calculating volt-seconds from frequency..
Volts/hz is handy for inductance based equipment like motors and transformers that have iron cores.
The reason for inventing the term "volts/hz" is that, for a sinewave of given amplitude, there's some frequency below which you'll exceed the volt-second capability of the inductor and saturate it.

Just to double check, V.s = V/(2*pi*freq) right??

I have to check. I remember so little I'm condemned to a life of figuring out basics.

were applied voltage a 50 hz sine wave instead

volt seconds = ∫₀^0.01 sin(100Πt) dt = - 1/100Π * [cos(100∏t)^0.01 - cos(100∏t)]₀
= -1/100∏ * [ cos(∏) -cos(0) ] = -1/100∏ * [ (-1) - (1) ] = 2/100pi = 1/50pi
hmm that's 1/Πf not 1/2∏f
check my arithmetic, though
again volt-seconds depends on where in the cycle we start and how far into the cycle we proceed.

You'd design an inductor for whatever volt-seconds you want it to withstand without saturating.
Observe you could make a transformer that won't saturate on being energized at a zero crossing. That's a handy trick to know about when working on power systems supplied by an overload intolerant source like an inverter --- use a transformer at half its voltage rating (maybe 70%) and you'll ever get that awful one-time-out-of -ten inrush.one thought for this post

 

[tim9000]

159.256 is soooo close to 50Π
it's got to be simple.

When I do V.s/N = 2*√2*V_rms/N
it equals B!???
I was doing V.s/N = 2*√2*V_rms / (N*π*50)
which was 159 times bigger.
What the hell is going on?

 

[tim9000]

When I do V.s/N = 2*√2*V_rms/N
it equals B!???
I was doing V.s/N = 2*√2*V_rms / (N*π*50)
which was 159 times bigger.
What the hell is going on?

In addition to not knowing teh causality.
By dropping the π50 the 400 turn series coils and the single 200 turn coil are still 0.49316 factor, out by:

are you sure dropping the 50π isn't just a coincidence/ red hering?

 

[tim9000]

You may recall I've made mention that I'm still confused regarding ΔV/ΔI
I can kinda see why it's the tangent on an actual BH curve because
V/N = dΦ/dt
However the VI curves I took my excitation V goes up only, it didn't waver, as did my current, now if I take V/N = Area*(B2-B1) points on the BH curve that V/N number will decrease as the curve flattens. So that V is what? the voltage on the core excluding coil? Regardless if I take Area*(B2-B1) / (current2 - current1) from the BH curve it will give me numbers which don't mean anything to me, don't look like impedances I recognise.
As I've said if we were taking the direct V / I of the V vs. I curve I could see how that would give us the impedance of the mag amp, but I don't see why the tangent to the curve is relevant...[Anyway I posted some Z comparison curves before.]

Thanks

 

[jim hardy]

And I'm still confused as to why Z = ΔV/ΔI at a point, and not just V/I (which is the gradient from zero to point on curve)
i.e. what the difference in compared methods of plotting Z curve in posts #190 and #192 between tangent Z and direct Z = V / I

Hmmm maybe there's not such a fundamental difference after all, provided we are mindful of what our meters are telling us.
Here the meters will report volts and amps , and below the knee volts and amps will both be decent sinewaves.
Impedance should be pretty linear below the knee because the red and blue lines don't diverge very much hence volts/amps should give fairly constant Z.
Beyond the knee current increases rapidly., the red and blue lines diverge , so what are our meters reporting?
The ammeter is now handling a very nonsinusoidal wave
voltmeter still sees a sinusoid provided there's no load impedance , which is how i think you (dammit italics stuck on again) took your readings
so volts/amps gives a number that's related to imedance . Impedance is some weighted average of the slopes along the blue line
but we can't see that relation with just meters
so we work with what we have, bearing in mind we're limited by using meters rather than oscilloscopes.

Our meters can only see the red lines.Awareness of that let's us characterize our inductor with the measuring devices at hand .

Red line is large signal response
small signal response will fall closer to your blue line.
That should i think be the difference between your V/I and delta oops tangent curves.

 

[tim9000]

Hmmm maybe there's not such a fundamental difference after all, provided we are mindful of what our meters are telling us.
Here the meters will report volts and amps , and below the knee volts and amps will both be decent sinewaves.
Impedance should be pretty linear below the knee because the red and blue lines don't diverge very much hence volts/amps should give fairly constant Z.
Beyond the knee current increases rapidly., the red and blue lines diverge , so what are our meters reporting?
The ammeter is now handling a very nonsinusoidal wave
voltmeter still sees a sinusoid provided there's no load impedance , which is how i think you (dammit italics stuck on again) took your readings
so volts/amps gives a number that's related to imedance . Impedance is some weighted average of the slopes along the blue line
but we can't see that relation with just meters
so we work with what we have, bearing in mind we're limited by using meters rather than oscilloscopes.

But even fundementally, even if we had perfect equipment, I still don't understand the tangent slope, or the red line slope, what the difference between the two means, or rather, specifically why the tangent to the curve is V / I, any more than the red line is Z.
Case-in-point see post # 205.

 

[tim9000]

last point before I go to bed, I was thinking I should be able to get a load line of the mag amp using the inductance relation:
Z = 2*pi*freq* N²*A*µ / length = 2*pi*freq* N²*A*(B/H) / length
which would have the same shape as the µ curve, which is proportional to the gradient of the B curve.
But what has been irking me is that I can't seem to figure out (probably because I dont' understand Volt-seconds well enough) how to get some sort of impedance per turn value curve for the core. Like a load line derived from the BH curve, to give me the impedance of the core at a specific flux density, then I can just work out what the impedance of the core will be for a given control DC current.

 

[jim hardy]

hmm 203 -205 apeared while i was typing 206.

When I do V.s/N = 2*√2*Vrms/N
it equals B!???
I was doing V.s/N = 2*√2*Vrms / (N*π*50)
which was 159 times bigger.
What the hell is going on?

I always have to go back to basics , that's why i;m so slow

volts per turn is indeed a measure of flux
e= sin ωt
and e = -n dΦ/dt
edt = dΦ
∫dΦ =∫edt
Φ = ∫sin wt = 1/ω cos wt
at 50hz
Φ = 1/100pi * cos wt
since cos = just sin shifted by 90 deg
i'll write jsin in its place
Φ= 1/100pi * jsin wt
Φ = 1/100pi * j * e where e is volts in one turn

now from 201

In that last post, there is an implicit understanding of mine that flux = V.s/N and I'll throw some numbers in for colour: Take the knee point flux density;
Vrms*√2*2 / (π*f*A*N) = 296*√2*2 / (π*50*0.00111*840) = Vpk*2 / (π*50*0.00111*840) = 5.7163 T
But we previously determined using the EMF equ that the knee point was about 1T

flux = V.s/N
flux at any instant is volt-seconds per turn, and volt-seconds is 1/Πf only when sinewave started at zero and you're at 180 deg...see 202
but you're looking for a continuous sine function not just one point in time ? i think that's the stumbling block.
isn't flux sinewave 1/100pi * volts per turn ? (100pi = 2 pi X freq) ? check my arithmetic above.

296 volts / (840 turns) X 1/100 pi ) = 0.00112 weber , in 0.00111 square meters = 1.01 T
since that's sinewave RMS volts it'd also be RMS flux, cosine. Peak √2 greateri still haven't resolved the 159X disparity - did it self announce yet ?

 

[jim hardy]

last point before I go to bed, I was thinking I should be able to get a load line of the mag amp using the inductance relation:
Z = 2*pi*freq* N2*A*µ / length = 2*pi*freq* N2*A*(B/H) / length

that's Zmagamp , 2pi X freq X inductance

which would have the same shape as the µ curve, which is proportional to the gradient of the B curve. ?? It's just a number not a curve unless you make something a variableBut what has been irking me is that I can't seem to figure out (probably because I dont' understand Volt-seconds well enough)
volt-seconds is simply what the integral vdt evaluates to given some starting and ending point,
it has the subtlety (fortuitous quirk if you will) that it happens to numerically equate to amount of flux induced in an inductor that's been subjected to that voltage. I think they're not so important for AC operation excpt to understand why saturation occurs.

how to get some sort of impedance per turn value curve for the core. Like a load line derived from the BH curve, to give me the impedance of the core at a specific flux density, then I can just work out what the impedance of the core will be for a given control DC current.

Okay i think we're about ready to attack that.
Given the vagaries broached in 206 regarding red vs blue lines

what if we had an AC current in the outside coils
at the same time as a DC current in the center ones so both outer legs have DC offset

such that our AC applied voltage-current curve looked something like this pink line ?
Of course the actual AC current is zero-centered, but it's working the core out past knees , on its flat part.

What would the meters see ?

? two or three volts AC, a couple hundred milliamps AC ?
AC meters don't know anything about the curve. They have tunnel vision.

Now let us change the DC bias so the core is pushed over the knee

aha now we've raised impedance by shifting DC. Saturable reactor operation.

Current will be no longer sinusoidal, voltage probably won't either .
So our meters will be off by some unknown form factor, but with careful logging you can get repeatable results.

One might estimate the impedance at any point as the slope of the missing leg of that pink triangle
but what does he assume for volts and amps?

My old textbook makes a table of results from actually drawing those pink triangles .on an idealized BH curve

Then it plots a family of volts versus load amps curves for each value of DC.
but they haven't made that step real clear
and I'm working on figuring it out.
Their approach is graphical not calculation.