How the power transfers across the Ideal Transformer

[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.

 

[jim hardy]

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!

Are vertical scales really 10X different ? What are they -- volts ? millivolts per turn? Was this at line frequency ?

judging by shape of your curves It takes maybe 20 ma to saturate when driving center leg
more like 80 to saturate when driving an outside leg ?

When driving the Outside leg, it will saturate first because flux goes from it into the 3x greater area of the other two legs
so you have a core that's part saturated and mostly not saturated.
A picture or sketch of your setup would be informative.

How did you calculate effective permeability?

There's some error introduced by using AC meters to measure non-sinewave current after the knee. An oscilloscope would be really handy.

 

[jim hardy]

Now let us add saturation to this picture

voila, look what happens to current

observe current goes sky high when we exceed the volt-second capability of the transformer(inductor)
and observe it's late in the cycle , just before a zero crossing

in real transformers that happens when you close the switch at or real near the sinewave's zero crossing, maybe one time out of ten.

Here's a real 'scope photo from that PF thrfead i mentioned
https://www.physicsforums.com/threa...ansformer-inrush-current.811131/#post-5094110

note his 40 amp initial peak is about 160X steady state.
Note his 10 amp second peak is 40X steady state.
Remember i said a real transformer will settle itself to symmetry, but not in the first cycle as evidenced by this trace.
note also asymmetry - no spike on negative peaks
it's asymmetric meaning flux has DC content...
flux and current both get a DC component when you close near zero crossing !

that's why constant of integration and volt-second are good concepts to have firmly implanted when you start working with transformersand that's the one thought for this post.

Try a search on transformer inrush,
there are plenty of scholarly articles around with beautiful derivations.
Armed with this gut-level understanding i can accept them as works of art that i could never reproduce.That's the one thought for this post

think on it ?

old jimps correctfions are welcome.

 

[tim9000]

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.

hmm, so volt-seconds is the DC equivilant of AC's volts per Hz, but the magnetic guy couldn't determind V.s / turn only the transformer guy could work that out? The magnetic guy could only ever determine volt-sec for one turn before saturation...

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.

Ok, than so will I

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

So if you had the zero flux point at the voltage zero crossing, than it would be the same as the original picture, except the flux would be mirrored to be below, on the negative half of the graph. I this is so strange to me, to believe that the primary coil current I_p (in post #84) will only draw in one direction. If this voltage excitation was a sin wave rather than a square wave though, it would then oscillate positive and negative wouldn't it?

Are vertical scales really 10X different ? What are they -- volts ? millivolts per turn? Was this at line frequency ?

I think I calculated there were about 200 turns on the outside legs, and 840 turns on the inside leg.
The vertical scale is Volts, the horizontal scale is mA; sorry: the line frequency was 50Hz.

judging by shape of your curves It takes maybe 20 ma to saturate when driving center leg
more like 80 to saturate when driving an outside leg ?

Yes that's what I thought, so H = 0.02*840 [A/m] on the knee of the centre leg?

How did you calculate effective permeability?

I didn't, I just tried to look at some other curves on the internet to match it with mine...

There's some error introduced by using AC meters to measure non-sinewave current after the knee. An oscilloscope would be really handy

I didn't use a digital current meter (analogue), just a digital voltmeter.

When driving the Outside leg, it will saturate first because flux goes from it into the 3x greater area of the other two legs
so you have a core that's part saturated and mostly not saturated.
A picture or sketch of your setup would be informative.

It shouldn't matter, this was an experiment to understand how Mag Amp deliberate saturation behaves. Ok here is a picture of my setup:

"one (or so) thought per post"

 

[tim9000]

observe current goes sky high when we exceed the volt-second capability of the transformer(inductor)
and observe it's late in the cycle , just before a zero crossing

Other pictures we've discussed have shown the flux of a sineusoidal flux flattening off, but here it appears that the increase in current drawn due the decrease in change in flux (back emf) is exactly enough to accommodate the flux increase to maintain the relationship of it as the derivative of the flat voltage amplitude, is this because the situation above is ideal and in reality it would flatten off but due to the excitation flattening off because of the impedance drop on the primary coil, and whatever voltage excitation was left on the core, the increase in current would be exactly enough to accommodate the relationship of it being the integral? (gee this is complicated, no wonder I was confused) Hence your 'willing to draw whatever it needs to', it will draw exactly enough current to maintain the relationship, by chance, by the nature of the universe. And by another coincidence that will be by nature of circumstance the amount of voltage on the core will drop as it saturates.

note his 40 amp initial peak is about 160X steady state.
Note his 10 amp second peak is 40X steady state.
Remember i said a real transformer will settle itself to symmetry, but not in the first cycle as evidenced by this trace.
note also asymmetry - no spike on negative peaks
it's asymmetric meaning flux has DC content...
flux and current both get a DC component when you close near zero crossing !

I see the in-rush current in the CRO picture, but that's a sineusoidal supply isn't it? So why is the current only positive? Wouldn't it only be positive like that if it was a square-wave supply?

 

[jim hardy]

mm, so volt-seconds is the DC equivilant of AC's volts per Hz, but the magnetic guy couldn't determind V.s / turn only the transformer guy could work that out? The magnetic guy could only ever determine volt-sec for one turn before saturation...

Good wording , volt-second is analogous to volts/turn
V.s / turn ? Volts per turn? Remember volts per turn is at a specific frequency.
Volt-seconds per turn says what is maximum flux beyond which you're saturated.
I'm probably being sloppy again about including "per turn".

Magnetic guy doesn't know intended operating frequency . But I've seen them give that number for cores intended for line frequency.
Remember switching power supplies operate in the kilohertz's. He has to give volt-seconds( per turn,) or Bmax...
They actually give a cryptic number that's independent of core area...

So if you had the zero flux point at the voltage zero crossing, than it would be the same as the original picture, except the flux would be mirrored to be below, on the negative half of the graph. I this is so strange to me, to believe that the primary coil current Ip (in post #84) will only draw in one direction. If this voltage excitation was a sin wave rather than a square wave though, it would then oscillate positive and negative wouldn't it?

That referred to this image from post 90
and i don't understand your question see below.

So if you had the zero flux point at the voltage zero crossing,

that is the red flux trace

than it would be the same as the original picture,

that's what was the picture in post just just above, 89 , and 85 too ?
zero flux was at zero crossing

except the flux would be mirrored to be below, on the negative half of the graph.

i don't know what you mean by mirrored below; it is what it is. Mark up my sketch in Paint. Find Windows "snip" function and you can cut&paste right into paint.

I this is so strange to me, to believe that the primary coil current Ip (in post #84) will only draw in one direction.

Applause ! It is VERY counterintuitive that you can apply AC voltage and get DC current.
That's calculus for you. If calculus says it can happen, well, it probably can.
In a real transformer it is temporary

http://www.globalspec.com/reference/74817/203279/10-3-transformer-protection
but in an ideal one it would be a permanent DC offset with current swinging between zero and a large value.
i ran across it in 1974 measuring inrush to instrument circuits. Might still have a 'scope trace, but there's a better one in post 92.

If this voltage excitation was a sin wave rather than a square wave though, it would then oscillate positive and negative wouldn't it?

No ! Not in my one cycle drawing. Remember i said getting ready for a repetitive wave.

That is why solid state power relays come in both flavors - zero firing for resistive loads and peak firing for inductive loads.

Draw yourself a half cycle of AC sinewave voltage starting from zero.
Notice that it has positive value for the entire half cycle. Voltage is always positive.
Since voltage is dΦ/dt, Φ must have positive slope for the entire half cycle. It's no longer a straight line but it always points up, tapering to horizontal at the 180 degree zero crossing.
Got that ? Flux increases so long as voltage is positive.
Until that's intuitive you haven't really accepted the integral-derivative relation between sine and cosine, or between voltage and flux.

that's the one thought for this post.

Will try to do next question, but i am fading...

 

[jim hardy]

I think I calculated there were about 200 turns on the outside legs, and 840 turns on the inside leg.
The vertical scale is Volts, the horizontal scale is mA; sorry: the line frequency was 50Hz.

Yes that's what I thought, so H = 0.02*840 [A/m] on the knee of the centre leg?

good we know volts, current and number of turns. Weshould be able to get flux, and with an stimate of area , B.

Gotta run now thanks for your interest ! back in the morning.

Work on that sinewave exciting voltage plot instead of my square wave. It's important to work that derivative . integral in your head, and to believe that math says we can get DC current from an AC voltage without a diode.

 

[jim hardy]

Self saturating magamp ?

 

[tim9000]

I'm going to break the rule about thoughts/post because I feel I have no choice (almost a bit overwhelmed).

Poor wording on my part. I meant to say comparing your second sketch where the red flux triangle was offset (orange line) to be lower because it started at negative V.t, if you pushed it right to the bottom so the maximum flux was zero (at zero voltage crossing) and the minimum flux was negative magnitude of the voltage. [Was there a reason you said started from negative V.t/2 ??]
THEN it would be the same, just a mirror about the x-axis, of the original? That's what I meant by mirrored before. Hopefully by the end of this you'll figure out why this DC current from AC voltage is so confusing to me.

hat is why solid state power relays come in both flavors - zero firing for resistive loads and peak firing for inductive loads.

Could you please elaborate? So like a thyristor with different firing angle?

Magnetic guy doesn't know intended operating frequency . But I've seen them give that number for cores intended for line frequency.
Remember switching power supplies operate in the kilohertz's. He has to give volt-seconds( per turn,) or Bmax...
They actually give a cryptic number that's independent of core area...

How does he calculate volt-seconds (per turn)? Especially independant of area?

No ! Not in my one cycle drawing. Remember i said getting ready for a repetitive wave.

That is why solid state power relays come in both flavors - zero firing for resistive loads and peak firing for inductive loads.

Draw yourself a half cycle of AC sinewave voltage starting from zero.
Notice that it has positive value for the entire half cycle. Voltage is always positive.
Since voltage is dΦ/dt, Φ must have positive slope for the entire half cycle. It's no longer a straight line but it always points up, tapering to horizontal at the 180 degree zero crossing.
Got that ? Flux increases so long as voltage is positive.
Until that's intuitive you haven't really accepted the integral-derivative relation between sine and cosine, or between voltage and flux.

I get that as long as the voltage is positive, the area under it is growing, so flux is increasing (say voltage was cos, flux would be sin), but you're saying this is only the case for repetitive waves? And not for a single 360^o excitation? What's the difference, in-rush current or something? Because I don't see how if you're putting in a sine voltage, that the calculus would give a non +/- oscillating integral...in steady sate, anyway.

good we know volts, current and number of turns. Weshould be able to get flux, and with an stimate of area , B.

Gotta run now thanks for your interest ! back in the morning.

well, The dimensions of the core were roughly (mm):

Self saturating magamp ?

Not self saturating, I was putting/controlling some DC voltage on the centre leg to control the amount of current through two parallel light bulbs, the outer legs were in series that went to the bulbs (load). If you're curious, the outter coils were 0.835 Ohms each and the centre coil was 10.7 Ohm, the load was about 50.4 Ohms.

 

[jim hardy]

Poor wording on my part. I meant to say comparing your second sketch where the red flux triangle was offset (orange line) to be lower because it started at negative V.t, if you pushed it right to the bottom so the maximum flux was zero (at zero voltage crossing) and the minimum flux was negative magnitude of the voltage. [Was there a reason you said started from negative V.t/2 ??]
THEN it would be the same, just a mirror about the x-axis, of the original? That's what I meant by mirrored before. Hopefully by the end of this you'll figure out why this DC current from AC voltage is so confusing to me.

okay let me stumble along a bit further.

This might be painful, but imagine the pain to get it drawn !

This is a sine wave, just one cycle.

What do we get if we integrate it from zero to 2pi ?

we'd expect a cosine, a negative one,
because Integral of sin is - cosine + C
and here's a negative cosine, again just one cycle

In fact let us integrate sin every 90 degrees , pi/2, over a cycle
so we'll evaluate -cos from zero to (n * pi/2) for n = 0 to 4
that'd be [-cos npi/2] - [-cos0] at each n
for example at n=1 it's evaluate to [0] - [-1] = +1, because the C's cancel out
here's a table, one step at a time

n, radian...sin... cos... -cos ... ∫sin , evaluated from zero to n radians
0, 0...... 0 .... 1 .... -1..... 0

1, pi/2... ...1 ..... 0 ... ..0.... +1

2, pi ... 0 .....-1 ... +1... +2

3, 3/2 pi... -1....0 ... 0... +1

4, 2pi ..... 0 ... +1... -1... 0

Aha !
Sin swings between -1 and +1

But ∫sin when started at zero swings between zero and +2.
And that's why calculus says we can get unipolar flux by applying a bipolar AC voltage.
Over one cycle or over a zilion cycles, check it out with wolfram or a piece of graph paper .

dammit i cannot turn off bold this is really aggravating

gonna go out and throw some rocks to get over computer fristrationi hope this helps with the concept of dc from ac via calculus not silicon

remember current must be whatever is necessary to make that flux
so it'd be unipolar too in an ideal transformer

greg can you fix bold and italic so they'll turn off ?

anyhow

this sine voltage

gives this flux

old jim

 

[jim hardy]

Could you please elaborate? So like a thyristor with different firing angle?

Exactly. It fires at the peak instead of at the zero crossing, or randomly.

www.te.com/commerce/DocumentDelivery/DDEController?Action=srchrtrv&DocNm=13C3206_AppNote&DocType=CS&DocLang=EN

for some reason his current traces are reversed, read right to left . ? doggone computers.

http://www.wolfautomation.com/products/30434/peak-switching-ssr-single-phasebrcarlo-gavazzi-rm1c

 

[jim hardy]

How does he calculate volt-seconds (per turn)? Especially independant of area?

that'd be flux, wouldn't it "

There's a shorthand used by magnetics guys to avoid so many conversions ,
look at their term nh/T^2
i figured it out once long ago, don't do magnetics often enough to stay conversant in it.
just be aware there is a shorthand, you'll figure it out easily when you find "the force"..

 

[jim hardy]

I get that as long as the voltage is positive, the area under it is growing, so flux is increasing (say voltage was cos, flux would be sin), but you're saying this is only the case for repetitive waves? And not for a single 360o excitation? What's the difference, in-rush current or something? Because I don't see how if you're putting in a sine voltage, that the calculus would give a non +/- oscillating integral...in steady sate, anyway.

i think we worked this one out first post today

 

[jim hardy]

well, The dimensions of the core were roughly (mm):

so cross section of outsides is 15 X37 = 555 sq cm
and center leg twice that , ~ 1110 ?

and the fourth dot on top curve
looks like 213 volts at maybe 15 milliamps?

15 milliamps through 840 turns is 12.6 amp-turns
and 213 volts in 840 turns is 0.253 volts per turne = n dΦ/dt , 1 turn, so dΦ/dt = 0.253 webers/sec
assuming sinusoid, dsin(wt) = wcos(wt) and at 50 hz w is 100pi
so
.253 = 100pi X Φ
Φ = .253/100pi = 8.05E-4 webers

in an area of 555 1110 cm^2 , = 555 1100E-4 m^2

8.05E-4 Weber/1100E-4m^2 = .007 Tesla ?
sounds too low
Hmmm where did i go wrong ?

edit-- !

Ahhh got it - you gave millimeters and i ciphered for centimeters
make that 0.7 Tesla,

that's more like it !

 

[jim hardy]

bottom curve 35 volts in 200 turns on 555mm^2 ?
.17 volts per turn ?
what do you get for flux?

.17/100pi = 5.4E-4 weber/555E-6m^2 = .97 TDontcha just LOve it when math works out !

 

[jim hardy]

You're lucky to have a core like that with windows big enough to pass wires through.
Maybe a ten turn search coil on each one would be handy - makes enough volts to see on a meter and it's easy to figure volts per turn.
Hmm... if you made your search coils 10/pi turns your voltmeter would indicate milliwebers, wouldn't it ? 31 turns would be within 2%.
One might also factor core area into his number of turns and make the voltmeter indicateTeslas...

But that's just idle daydreaming.

This i think you'll enjoy -
Try exciting the center leg to about 1/2T
measure volts in both outside legs
Next place a single shorted turn around one outside leg, You'll be surprised how small is the spark it makes.
Then measure voltage in both outside legs again.
You'll see the shorted turn has pushed flux from one outside leg over to the other.

MMF at work !