How the power transfers across the Ideal Transformer

[tim9000]

Sorry, I've had a lot going on, I've been getting heaps of data for my thesis. I have a long train trip to make on tuesday the 28th, so expect a reply before wednesday the 29th.
I'll formulate my reply on the train then.
Cheers!

 

[jim hardy]

Not to worry I'm behind too... two more graphs to go.

old jim

 

[jim hardy]

Last thing we did was look at this graph where current is independent(controlled) variable
and flux is dependent(observed)

That's an easy experiment to perform because we have adjustable current sources.If we wanted to swap our thinking around and plot current(observed) versus flux(controlled)
we could simply pick numbers off the graph and replot it

or we could just swap the ordinate and abscissa (rotate the graph)
Paint has a "Flip" button...

to get things going the right way though , incrasing to right and up, also flipped the text.
At my modest skill level, fixing that distorted the text and shrunk the graph
but here's my rendition.

okay now we have a picture.
If i could sweep flux(density) from zero to ~1.1 T , the range i picked two charts earlier, i should get this curve for current.
But how would one run that experiment? I do not know of an adjustable flux source.
I could measure flux by integrating voltage from a search coil and adjust current to get desired flux, plotting the results,
but that is a work-around if not outright cheating.
How do we design a lab experiment to sweep flux?

While we ponder that, consider this digression:
This B-H curve , or H-B curve, is obviously some sort of polynomial.
If the curve has n points, a quadratic of order n+1 can hit them all but will overshoot drastically in between them. That's why one does a least squares curve fit and settles for an approximation.
That curve has two inflection points so would take at least a third order equation to describe it.
But it's do-able with the tools you young fellows have today. In my working days i did them in Basic on a TI99 i'd brought from home .Back on track now...
Hmmmm... Flux is integral of voltage?
Φ = ∫vdt ?
Would Mother Nature help me out here ?
What if i picked a v that has a simple integral...one that results in a sweep ?
Integral of a constant k is a ramp kt , and that's a sweep.
Φ = ∫vdt , make v constant and Φ becomes linear with time, Φ = v∫dt = vt
i think that means i could express flux in units of volt-seconds.

So my lab experiment would be:
1. set up a recorder to graph current on vertical axis
and time on horizontal axis
2. Start the recorder.
3. Apply a 1 volt step to my coil
Current should increase at a modest rate until i reach ~ 1 T, the knee of my core, and increase rapidly thereafter.
If that worked, i'd connect my search coil-integrator to a second channel on the recorder , just to show flux is linear with time. Winding resistance will come into play at high current, limiting how far i can go .

That's my one thought for this post - how to sweep flux.
I just knew that integral-derivative relation would be good for something !

Volt-seconds, eh ? That's a number that is significant for transformers.

 

[jim hardy]

ps Faraday included a minus sign, and i have been woefully un-rigorous in honoring that. Kindly forgive...

old jim

 

[tim9000]

Last thing we did was look at this graph where current is independent(controlled) variable
and flux is dependent(observed)That's an easy experiment to perform because we have adjustable current sources.If we wanted to swap our thinking around and plot current(observed) versus flux(controlled)
we could simply pick numbers off the graph and replot it

View attachment 86447

or we could just swap the ordinate and abscissa (rotate the graph)
Paint has a "Flip" button...
View attachment 86444to get things going the right way though , incrasing to right and up, also flipped the text.
At my modest skill level, fixing that distorted the text and shrunk the graph
but here's my rendition.

View attachment 86448
okay now we have a picture.
If i could sweep flux(density) from zero to ~1.1 T , the range i picked two charts earlier, i should get this curve for current.
But how would one run that experiment? I do not know of an adjustable flux source.
I could measure flux by integrating voltage from a search coil and adjust current to get desired flux, plotting the results,
but that is a work-around if not outright cheating.
How do we design a lab experiment to sweep flux?

While we ponder that, consider this digression:
This B-H curve , or H-B curve, is obviously some sort of polynomial.
If the curve has n points, a quadratic of order n+1 can hit them all but will overshoot drastically in between them. That's why one does a least squares curve fit and settles for an approximation.
That curve has two inflection points so would take at least a third order equation to describe it.
But it's do-able with the tools you young fellows have today. In my working days i did them in Basic on a TI99 i'd brought from home .Back on track now...
Hmmmm... Flux is integral of voltage?
Φ = ∫vdt ?
Would Mother Nature help me out here ?
What if i picked a v that has a simple integral...one that results in a sweep ?
Integral of a constant k is a ramp kt , and that's a sweep.
Φ = ∫vdt , make v constant and Φ becomes linear with time, Φ = v∫dt = vt
i think that means i could express flux in units of volt-seconds.

So my lab experiment would be:
1. set up a recorder to graph current on vertical axis
and time on horizontal axis
2. Start the recorder.
3. Apply a 1 volt step to my coil
Current should increase at a modest rate until i reach ~ 1 T, the knee of my core, and increase rapidly thereafter.
If that worked, i'd connect my search coil-integrator to a second channel on the recorder , just to show flux is linear with time. Winding resistance will come into play at high current, limiting how far i can go .

That's my one thought for this post - how to sweep flux.
I just knew that integral-derivative relation would be good for something !

Volt-seconds, eh ? That's a number that is significant for transformers.

Very pretty, I verymuch look forward to reading it on the train. Also, I may be able to post some nice magnetic curves soon myself. I think I'll also have a confirmation to ask about how I've been taking some of my data.
TTY soon!

P.S
"please forgive unrigourousness"
-Jim you can get away with anything

 

[tim9000]

I'll be honest, I started doing CAD on the train and I fell asleep, I'm only up to post #61. But as Jim's philosophy goes, the fewer thoughts at a time the better anyway, so maybe it's for the best.

If a sinusoid of voltage is applied to a winding on a core, the flux will be sinusoidal even when the core is well into saturation, provided the resistance of the wire making up the winding is zero. If the winding resistance is not zero, then the voltage seen by the core will be reduced by the I*R voltage drop in the winding due to the current drawn through that winding's resistance. If the core is saturating, the peaky current will cause the voltage seen by the core to not be sinusoidal, and, of course, the integral of that non-sinusoidal voltage will not be a sinusoid either; this is where reality differs from theoretical perfection.

This sentence is really where the money is at for me. Puts many pieces of the puzzle together. So when it saturates you draw more current (spikes), meaning the voltage on the primary resistance of the winding detracts from the excitation on the core itself, meaning that the voltage on the core isn't sineusoidal, meaning that the (flux) integral of that non-sineusoid will not be a sineusoid either?

So a DC curve really is just that, 'DC excitation'?
I know we're talking DC, but just for a second I want some thoughts on my method of calculating BH curve using measurements of AC excitation V and I with an open circuit secondary. I figure that B is proportional to
Voltage/(Area*NumberOfTurns*radial frequency),
and H is NumberOfTurns*Current (I_pri), so I can plot the BH curve this way, instead of with a fluxmeter. But is this really just the top half/quater of the hysteresis curve?

Now
if Inductance is flux linkages per ampere,
which is flux per amp-turn,
is it apparent that for our mu-metal core
inductance would calculate out to a function of current?

amps ...flux ... inductance
8 ... 1 ... 1/8 = 0.125
10 ... 1.04 ... .. 1.04/10 = 0.104
30 ... ~1.1 ... 1.1/30 = 0.037
60 .. 1.18 .. ... 1.18/60 = 0.02
a six to one turndown.
And that's why we operate most inductors below the knee, where our approximation of constant μrelative is close enough

Another brilliant piece of the puzzel.
So the maximum point of inductance would be something like at the top of the linear part, something like L = 0.8/3?
Ok, that's why we operate inductors below the knee, but can you remind me why we operate transformers just below the knee and not lower down, why is this getting the best 'value for money out of our steel'?
This thread surpassed my expectation of how much I was looking forward to getting back to it! Thanks

 

[jim hardy]

Glad to see "the light coming on" for you.

Will be back after run some errands and think a bit more.

It's almost time to move from DC flux to AC . Don't forget about "Constant of Integration" from calc class...

yes one thought at a time.

 

[jim hardy]

So when it saturates you draw more current (spikes), meaning the voltage on the primary resistance of the winding detracts from the excitation on the core itself, meaning that the voltage on the core isn't sineusoidal, meaning that the (flux) integral of that non-sineusoid will not be a sineusoid either?

Yep ! Well worded !

 

[jim hardy]

I know we're talking DC, but just for a second I want some thoughts on my method of calculating BH curve using measurements of AC excitation V and I with an open circuit secondary. I figure that B is proportional to
Voltage/(Area*NumberOfTurns*radial frequency),
and H is NumberOfTurns*Current (Ipri), so I can plot the BH curve this way, instead of with a fluxmeter. But is this really just the top half/quater of the hysteresis curve?

Yes. Assuming you're using AC meters there's no sign
so you'll get a curve relating open circuit volts to magnetizing current
of course with no signs it's a one quadrant curve , as you said half/quarter.

that's how one tests a current transformer to make sure it's not got shorted turns.
Or an unknown transformer to see what voltage it's good for.

It's really heartening to see your interest , not to mention progress .

 

[tim9000]

It's really heartening to see your interest , not to mention progress .

It would be much harder without your guidance.

Hmmmm... Flux is integral of voltage?
Φ = ∫vdt ?
Would Mother Nature help me out here ?
What if i picked a v that has a simple integral...one that results in a sweep ?
Integral of a constant k is a ramp kt , and that's a sweep.
Φ = ∫vdt , make v constant and Φ becomes linear with time, Φ = v∫dt = vt
i think that means i could express flux in units of volt-seconds.

Ok, so you need a high-speed graph to record a DC flux sweep (current over time) of the core.

What about that graph of yours:

So the maximum point of inductance would be something like at the top of the linear part, something like L = 0.8/3?
Ok, that's why we operate inductors below the knee, but can you remind me why we operate transformers just below the knee and not lower down, why is this getting the best 'value for money out of our steel'?

Also (reminder needed) purtaining to my 'value for money' question above: so the primary current will spike either when the secondary load is drawing too much current OR when the core is saturating, because the dΦ/dt has reduced the back-emf on the ideal core?

 

[jim hardy]

So the maximum point of inductance would be something like at the top of the linear part, something like L = 0.8/3?
Ok, that's why we operate inductors below the knee, but can you remind me why we operate transformers just below the knee and not lower down, why is this getting the best 'value for money out of our steel'?

well 0,5 T looks like ~2.2 amps, 0.23 h
maximum inductance is down in the linear region where curve is steepest, remember it's flux per amp-turn...

So a transformer(or motor) with way more steel than it needs will operate cooler and draw less magnetizing current.
From a buyer's perspective that's what i'd want
but in the shoes of somebody who has to beat his competition's price, i couldn't afford to buy and ship that extra steel and the extra copper required to encircle that larger than necessary core. Unless i was selling to audiophiles...

 

[tim9000]

well 0,5 T looks like ~2.2 amps, 0.23 h
maximum inductance is down in the linear region where curve is steepest, remember it's flux per amp-turn...

So a transformer(or motor) with way more steel than it needs will operate cooler and draw less magnetizing current.
From a buyer's perspective that's what i'd want
but in the shoes of somebody who has to beat his competition's price, i couldn't afford to buy and ship that extra steel and the extra copper required to encircle that larger than necessary core. Unless i was selling to audiophiles...

We "crossed in the mail"

well 0,5 T looks like ~2.2 amps, 0.23 h
maximum inductance is down in the linear region where curve is steepest, remember it's flux per amp-turn...

So a transformer(or motor) with way more steel than it needs will operate cooler and draw less magnetizing current.
From a buyer's perspective that's what i'd want
but in the shoes of somebody who has to beat his competition's price, i couldn't afford to buy and ship that extra steel and the extra copper required to encircle that larger than necessary core. Unless i was selling to audiophiles...

Ah, yeah, so it's about material minimisation.

YES! The point where the linear part is steepest is point of maximum inductance! That's what I theorized a few weeks ago!

EDIT: I'll save my next question for after your reply to me before we crossed in the mail to keep the 'thought stream' clear.

 

[jim hardy]

aha you posted while i was typing. Did i answer your question?

Ok, so you need a high-speed graph to record a DC flux sweep (current over time) of the core.

What about that graph of yours:

i'm between oscilloscopes right now

drew this in paint

of course that assumes ideal wire with no resistance so all the applied voltage causes induction dΦ/dt
and current becomes whatever is necessary to push that flux around the core.

We okay so far ?

 

[tim9000]

aha you posted while i was typing. Did i answer your question?i'm between oscilloscopes right now

drew this in paintof course that assumes ideal wire with no resistance so all the applied voltage causes induction dΦ/dt
and current becomes whatever is necessary to push that flux around the core.

We okay so far ?

I think so (on both counts), so there would be no practical use using a magnetic core lower down the curve than the steepest point of the linear section, because it would have a low B and a lower inductance? You could say that an operating point at the steepest point of the linear section would be ideal.You did say it was ok to say that H = NumOfTurns* open circuit current, for the method I was doing my BH graph didn't you?

 

[jim hardy]

Also (reminder needed) purtaining to my 'value for money' question above: so the primary current will spike either when the secondary load is drawing too much current OR when the core is saturating, because the dΦ/dt has reduced the back-emf on the ideal core?

Yes for a load spike primary and secondary current will look alike
saturation spike has a distinct shape and timing.
There was a thread a while back with actual 'scope traces

 

[tim9000]

Yes for a load spike primary and secondary current will look alike
saturation spike has a distinct shape and timing.
There was a thread a while back with actual 'scope traces

So they look differently, but have the same root cause?

 

[jim hardy]

I think so (on both counts), so there would be no practical use using a magnetic core lower down the curve than the steepest point of the linear section, because it would have a low B and a lower inductance? You could say that an operating point at the steepest point of the linear section would be ideal.

I think we're okay there, except ideal meaning best behaved transformer to you - bean counters will see ideal as one pushed nearer the knee.
Remember our model pages back ::: An "ideal" transformer has a BH curve that's infinitely step - zero magnetizing current.

 

[jim hardy]

So they look differently, but have the same root cause?

their commonality is reduction of flux ,
from load current mmf cancelling out primary mmf
or from increased magnetizing current reducing voltage applied to Xpri as you said

 

[jim hardy]

wow you were ready, weren't you ?

 

[jim hardy]

Wanting to lead into Ac behavior but not quite sure of best approach

If we have to back up and try another so be it
but here's one start

no saturation, flux and voltage both start from zero.

Integral of a square wave is a triangle wave.
Flux increases whenever voltage is positive, decreases whenever voltage is negative.
Peculiar - voltage is bipolar but flux never makes it to negative.

one thought per post.

 

[jim hardy]

See you in the morning !

old jim

 

[jim hardy]

You did say it was ok to say that H = NumOfTurns* open circuit current, for the method I was doing my BH graph didn't you?

H is strictly speaking per unit length of core, but we haven't dimensioned the core so amp-turns works for me.

 

[tim9000]

Yeah 'ideal' from an engineering point of view, not a bean-counter point of view.

wow you were ready, weren't you ?

I tried holding off replying for a bit so the conversation could reach steady-state, ha ha.

their commonality is reduction of flux ,
from load current mmf cancelling out primary mmf
or from increased magnetizing current reducing voltage applied to Xpri as you said

H'mm, do you mean recution of change in flux?
Ah I'm having a brain failure: So when the core is saturating the magnetising current is going up because it needs more current to...get the same change in flux? What wording would you use to say why the magnetising current goes up? Also when it's saturating, I suppose the amount of current going into the ideal core has dropped because the voltage on the ideal core has dropped because of the voltage drop on the primary coil impedance has increased?

I've been looking forward to our foray into AC because I have an interesting comparison question to ask about inductance soon.

See you in the morning !

old jim

Hopefully I'll speak to you in your morning, which will probably be late evening for me.

 

[jim hardy]

H'mm, do you mean recution of change in flux?

yes, reduced dΦ/dt better describes why counter-emf fell which allowed more primary current..

Ah I'm having a brain failure: So when the core is saturating the magnetising current is going up because it needs more current to...get the same change in flux? What wording would you use to say why the magnetising current goes up?

dosn't this picture answer that ?

how about
As the saturating core becomes increasingly unwilling to accept flux, more current is necessary to push flux through it. ?

Also when it's saturating, I suppose the amount of current going into the ideal core has dropped because the voltage on the ideal core has dropped because of the voltage drop on the primary coil impedance has increased?

You're back to our transformer model ?

Yes, X_M's increased current I_M increases the drop across R_P as you said earlier. So E_Pdrops.
I'm still thinking freeze - frame, instantaneous, like DC ... that's valid because at any instant AC has only one direction.
If you apply DC to a transformer it will soon saturate .

 

[jim hardy]

Observe that flux indeed is integral of voltsdt
so we could use volt-seconds as easily as Teslas or Webers

transformer datasheets often give the number of volt-seconds that will drive the core to saturation. It eases some calculations.
If it's just a core you're looking at , that'd volt-seconds per turn.

Is it becoming clearer now why volts-per-hertz is such a useful derived term?

no saturation, flux and voltage both start from zero.

Integral of a square wave is a triangle wave.
Flux increases whenever voltage is positive, decreases whenever voltage is negative.
Peculiar - voltage is bipolar but flux never makes it to negative.

But, voltage swapped polarity before we reached saturation and flux started integrating back down.
At end of cycle, flux had the value zero
and the integral of a symmetric wave over a whole cycle is zero
so for me the math and my mental image agree. (That's sort of unusual - i usually have to struggle to get there)

Had our voltage not changed polarity before we got to saturation, current would have gone sky high. That's why transformers have a volts per hertz rating.

one thought per post

old jim

 

[tim9000]

A very good morning to you Jim.

As the saturating core becomes increasingly unwilling to accept flux, more current is necessary to push flux through it. ?

I don't really like saying 'unwilling' because it's like it has a desire, I know you were probably speaking more to the finite amount of magnetic regions becoming less avaliable. I suppose in the back of my mind I am also thinking like why would there be more current through a saturated inductor, and all I can think is the deminished change in flux increase permitts current to increase (which in turn increases the flux evermore slightly). Which would be analogous in a way to increasing the secondary load on a TX which further de-saturates the core yet has the same affect. [Although according to you a different shape]

[Btw: The fact that there is no percievable advantage for a core below the maximum inductance (change in flux/current: steepest linear part of the curve) will be very useful in my project]

Ok, to AC:
So volts per Hertz is a good indication of flux density.

transformer datasheets often give the number of volt-seconds that will drive the core to saturation. It eases some calculations.

That's going to be useful to remember, what did you mean by "if it's a core you're looking at that's volt-seconds per turn'?

I take it that the diagram represents excitation voltage. This is what one of the things I was going to get at before: So the area of the integral is zero, so the flux goes to zero, I get that mathematically, but in reality, say I could see the current flowing AC into the coils of the transformer, and see the arrows of the flux moving and getting stronger over the half of the cycle, surely the flux would change direction in the core? Otherwise it just goes in the direction of which way the initial AC magnitude was going?

P.S, These are the curves I plotted the other week, they are from the same E core, the one at the top is the centre leg, which was twice as thick as the outter legs, the bottom is one of the side legs. I think the steel might have been M4 grain oriented silicon steel, I measured it as 0.27mm thick. From my brief looking on the net I think the max relative permiability or affective relative permiability or whatever is 14, but I'm not sure, what do you reckon(?):

Thanks!

 

[jim hardy]

I don't really like saying 'unwilling' because it's like it has a desire, I know you were probably speaking more to the finite amount of magnetic regions becoming less avaliable.

okay, i do tend to anthropomorphize which is academically, well , sloppy.
For whatever reason the incremental flux per ampere gets smaller. I envision the magnetic regions like springs, they can be aligned by applying mmf but will snap back if mmf disappears.
When they're nearly all aligned you've achieved saturation. That's the reason for the knee and the nonlinearity of BH curve.

one thought per post, but another post will follow this one in just a minute.

 

[jim hardy]

I suppose in the back of my mind I am also thinking like why would there be more current through a saturated inductor, and all I can think is the deminished change in flux increase permitts current to increase (which in turn increases the flux evermore slightly).

Let us be careful to keep AC and DC analysis separate.
Your description is quite right for AC.

Which would be analogous in a way to increasing the secondary load on a TX which further de-saturates the core yet has the same affect. [Although according to you a different shape]

Indeed , remember from many pages back that an increase in load current, that is secondary current, is immediately reflected in primary.
In contrast saturation doesn't happen until the volt-second limit for that core has been reached.
Load can change anywhere in the AC cycle
Saturation can only occur after volts have been too high for too long.

what did you mean by "if it's a core you're looking at that's volt-seconds per turn'?

Some transformer core datasheets give a number that's the volt-seconds to reach saturation assuming one turn. That's because the magnetic guy designing the core has no idea how many turns the transformer guy who buys the core from him will put around it.

 

[jim hardy]

I take it that the diagram represents excitation voltage. This is what one of the things I was going to get at before: So the area of the integral is zero, so the flux goes to zero, I get that mathematically, but in reality, say I could see the current flowing AC into the coils of the transformer, and see the arrows of the flux moving and getting stronger over the half of the cycle, surely the flux would change direction in the core? Otherwise it just goes in the direction of which way the initial AC magnitude was going?

You're speaking of this diagram ?

I get that mathematically, but in reality, say I could see the current flowing AC into the coils of the transformer, and see the arrows of the flux moving and getting stronger over the half of the cycle, surely the flux would change direction in the core? Otherwise it just goes in the direction of which way the initial AC magnitude was going?

Yes, and thank you.

I sneaked that in on you. It's a subtlety both of calculus and of inductors.
Look at the starting point. Flux and voltage are both zero.
Recall from calculus that integration yields a "constant of integration" which must be included in any initial value solution.
So, Φ = ∫vdt with voltage constant at V will yield Φ = Vt + C, C being the constant of integration.
I think of C as the starting point.
We started at zero flux and zero voltage. Indeed for that condition flux goes only one direction, as shown.
That's because the integral Vt evaluates only from zero to +Vt then back down to 0.

What if we'd started instead with some negative flux ?
Excuse me for a few minutes while i draw that in Paint.

 

[jim hardy]

okay the orange is flux starting from negative.
Observe green current trace also needs to be moved down to be symmetric about zero.
Were i more fluent with paint id've removed the red flux trace.

looking more like what we're accustomed to seeing ? Symmetric about zero ...
a real inductor will make itself symmetric because of losses in copper and iron
but we're not there yet, we're doing ideal . It directly applies to startup transient for real transformers

.

i'll look at your next question now.