What Happens When a Transformer Core Reaches Saturation?

In summary, the conversation discusses the topic of transformer saturation and its effects on the voltage and current in a transformer. It is mentioned that when a transformer saturates, the permeability drops and the flux does not increase much, causing the voltage on the secondary to become a square wave. There is also discussion about using transformers in saturation to produce voltage pulses near the peaks of the source voltage waveform. The concept of inductance and its relevance to transformers is also addressed, with the conclusion that it is not possible to have a transformer with no induct
  • #1
tim9000
867
17
Over the years I've learned a lot about TXs etc. But I'm a bit rusty. Any thoughts on general reminders about core saturation.

I know there are huge issues, but the specifics I can't remember that well. I vaguely remember hearing something like when there is a DC component on a transformer the hysteresis curve would get shifted up. But what's the difference between say if you were using an air core or a saturated core in practice? (as far as V and I are concerened)
So when the core saturates and the permeability drops away to near air permeability and the flux isn't increasing that much or hardly at all, what's the big deal? Still better than using an air core isn't it? Or are there spikes in current or something I'm forgetting that happen at saturation that are dangerous, that wouldn't happen if it was just a linear air core?

CheersEdit: actually one thought occurs to me, if the primary voltage is going up, but the dΦ/dt isn't going up much because it's saturated, then V2 = N2*dΦ/dt
the voltage on the secondary will be kind of a square wave won't it?
And I suppose there is little back EMF on the primary so it will draw heaps more current?
But how does this contrast to if the TX was DESIGNED to be air core, instead?
 
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  • #2
I'm not sure, but isn't there some non-linearity near saturation? While this might be useful in a few applications, in most applications it will lead to unwanted signal (or even power:nb)) mixing.

Plus there's the loss of the typically carefully designed efficiency with the magnetic flux spreading all cattywhumpas.

I like your idea of using a transformer in saturation to square up a wave. Clever.
 
  • #3
Jeff Rosenbury said:
I'm not sure, but isn't there some non-linearity near saturation? While this might be useful in a few applications, in most applications it will lead to unwanted signal (or even power:nb)) mixing.

Plus there's the loss of the typically carefully designed efficiency with the magnetic flux spreading all cattywhumpas.

I like your idea of using a transformer in saturation to square up a wave. Clever.
Yeah I looked into it and there are transformers that do actually do that, although it's a bit complicated.

But I'm still confused about what is happening on the secondary, so the current drawn from the primary spikes and the excitation voltage on the primary drops because the resistive voltage on the coil gets bigger, but rather than flattening the top off the secondary excition more, wouldn't since it's proportional to the derivative of flux the secondary excitation actually drop right away to just be really low...
 
  • #4
I'm not sure the phasing is correct on this one,
saturation doesn't have to occur at sinewave peak
but how nicely his picture conveys the derivative relation between flux and Vs !
http://www.allaboutcircuits.com/textbook/alternating-current/chpt-9/practical-considerations-transformers/
Special transformers known as peaking transformers exploit this principle to produce brief voltage pulses near the peaks of the source voltage waveform. The core is designed to saturate quickly and sharply, at voltage levels well below peak. This results in a severely cropped sine-wave flux waveform, and secondary voltage pulses only when the flux is changing (below saturation levels): (Figure http://www.allaboutcircuits.com/textbook/alternating-current/chpt-9/practical-considerations-transformers/#02314.png)

02314.png
 
  • #5
jim hardy said:
I'm not sure the phasing is correct on this one,
saturation doesn't have to occur at sinewave peak
but how nicely his picture conveys the derivative relation between flux and Vs !
http://www.allaboutcircuits.com/textbook/alternating-current/chpt-9/practical-considerations-transformers/
02314.png
Hi Jim!
I did read a bit about peaking TXs but I didn't see that page, that's great. Peaking TX has an extra secondary winding that would Resonate with a capacitor, to capture the harmonics, from memory.
So that picture is pretty indicative of how any es will react in any transformer? Not just a peaking transformer?
I have some issues to nit-pick about that picture, see where the flux goes up linearly and the derivative of that is a steady es, wouldn't in reality that section of flux increase still look like a sineusoid going from -90o to 90o, and the flux wouldn't be flat, it'd just be a very shallow increase. See my interpretion of a normal saturating transformer:
my interpretation.png

I still need to read that page you linked to better, but what do you think?
 
  • #6
i agree with you

search on saturated transformer and you'll get photos of actual oscilloscope traces.

glad to see you're working derivatives in your head now ! ""Progress through patience , perfection is an asymptote..."
 
  • #7
jim hardy said:
i agree with you

search on saturated transformer and you'll get photos of actual oscilloscope traces.

glad to see you're working derivatives in your head now !""Progress through patience , perfection is an asymptote..."
Cool!
I'm happy you agree, thanks.

I have another question, which is, since (neglecting leakage flux) the Epri = N dΦ/dt
and Esec = N dΦ/dt

it is the change in Φ (which is net Φ from the Pri and Sec MMFs), which causes the back EMF on the Pri to limit the current drawn from the primary coil.
What does inductance actually have to do with a transformer? it seems like it's just faraday's law (as stated above) that explain transformer operation, I can't imagine how stored energy in inductance is relevant? I know that some magnetising current is needed to create the flux, but is this relevant to inductance?

Is inductance or stored inductive energy relevant to a TX, or is it just a side affect? Could you have a TX with no inductance and still set up the flux?
 
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  • #8
What is inductance ? The physics definition ?

Inductance is flux linkages per ampere,
which is NΦ/I

Doesn't that determine what is magnetizing current?
Try and work that out with algebra.

What's inductance of an ideal transformer primary ? Number, please...
 
  • #9
jim hardy said:
What's inductance of an ideal transformer primary ? Number, please...
Infinite
jim hardy said:
What is inductance ? The physics definition ?

Inductance is flux linkages per ampere,
which is NΦ/I

Doesn't that determine what is magnetizing current?
Try and work that out with algebra.
Edited: Yeah so that's what I'm now thinking, it's not possible to have a TX with no Inductance because the higher the inductance, the higher the flux (Li/N). So if there is no leakage flux to dissipate energy in eddy currents in objects near by. What happens because of, or to, the energy stored in the inductance of the core?
Does it go back to the primary, get transferred to the secondary, or get burnt in magnetising losses?

EDIT: (Relevant extract from a message)
[
As far as I've been thinking about it, you need a little bit of magnetising current to set up the flux (which means that the PF will be less than 1), But I'm not sure how much? I always thought the amount of flux is dependant on the reluctance of the core (but that's only if you know the MMF), then the amount of flux in the core would determine how much the back EMF was, thus determining how much real in-phase current the primary side would draw? But maybe I'm thinking about it wrong if you only know the applied voltage: is the amount of magnetising current that the TX needs determined by the inductance reactance: Imag = Vpri/jω(L)
thus Φ = I*L/N,
thus Back EMF = N*dΦ/dt

What do you think?
Also, since there is in reality a leakage flux, does this mean that the bigger the leakage reactance the lower the power factor?
]

Thanks!
 
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  • #10
keep going
Φ = I*L/N
so EMF = N*dΦ/dt as you said
EMF = N * d(I*L/N)/dt
L and N are both functions of geometry and iron properties not of time so they're constants

EMF = (N*L/N)* dI/dt = the familiar E=LdI/dt, so we have agreement with a well known formula. That sort of confirms our logic.

tim9000 said:
Also, since there is in reality a leakage flux, does this mean that the bigger the leakage reactance the lower the power factor?

Seems so to me, it's in series with the winding so adds voltage drop IjXleakage, shifting phase a little.
 
  • #11
jim hardy said:
keep going
Φ = I*L/N
so EMF = N*dΦ/dt as you said
EMF = N * d(I*L/N)/dt
L and N are both functions of geometry and iron properties not of time so they're constants

EMF = (N*L/N)* dI/dt = the familiar E=LdI/dt, so we have agreement with a well known formula. That sort of confirms our logic.
Seems so to me, it's in series with the winding so adds voltage drop IjXleakage, shifting phase a little.
Ok, let's just map this out to refine the process>
There are two ways of modeling the TX:
Through MMF if you know the current through the primary coil.
Or
Through knowing the applied voltage to the TX primary coil, by which is determined in this order:
Know the inductance so work out the magnetising current, know the magnetising current to find the flux, find the flux to find the primary Back EMF and the secondary excitation voltage.

How's that? But that still doesn't address if the 'Energy' stored in the inductance does anything, or where it goes?

Cheers
 
  • #12
tim9000 said:
How's that? But that still doesn't address if the 'Energy' stored in the inductance does anything, or where it goes?

?? You're not thinking again...

That's basic to ohm's law for AC .
When Z is reactive, voltage and current are not in phase .

Draw yourself two sine waves one above each other 90 degrees out of phase.
Look at instantaneous power flow
Observe that
whenever volts and amps have same sign their product is positive, ie power flows into the reactance
when their signs differ power flows back out of the reactance

that's what reactance does, cycles power back and forth between itself and source,

Have you never made this sketch?

inductive-reactive-power.png

hence the old adage "Inductors and capacitors draw only imaginary power:.. "
Hogwash !
They draw very real power but it's only borrowed for a quarter cycle then returned.
So it averages zero.
 
  • #13
jim hardy said:
?? You're not thinking again...

That's basic to ohm's law for AC .
When Z is reactive, voltage and current are not in phase .

Draw yourself two sine waves one above each other 90 degrees out of phase.
Look at instantaneous power flow
Observe that
whenever volts and amps have same sign their product is positive, ie power flows into the reactance
when their signs differ power flows back out of the reactance

that's what reactance does, cycles power back and forth between itself and source,

Have you never made this sketch?hence the old adage "Inductors and capacitors draw only imaginary power:.. "
Hogwash !
They draw very real power but it's only borrowed for a quarter cycle then returned.
So it averages zero.
Ha Ha! Brilliant.
NO! I've never actually made that sketch, but I like it a lot! Ok, humour me though, say there are two inductors, one inductor is the inductance of the leakage flux and one inductor is the inductance of the core.
So there are two stored inductive energies. Both these energies come from the primary excitation, so are both these stored energies returned to the primary excitation supply?
Cheers!
 
  • #14
ps that sketch is at http://www.electrical4u.com/electric-power-single-and-three-phase/
 
  • #15
jim hardy said:
ps that sketch is at http://www.electrical4u.com/electric-power-single-and-three-phase/
That's actually going to be handy for me to revise, cheers.
tim9000 said:
Ha Ha! Brilliant.
NO! I've never actually made that sketch, but I like it a lot! Ok, humour me though, say there are two inductors, one inductor is the inductance of the leakage flux and one inductor is the inductance of the core.
So there are two stored inductive energies. Both these energies come from the primary excitation, so are both these stored energies returned to the primary excitation supply?
Although in saying that, I guess one thing that also bugs me is, there are formulas for magnetic (hysteresis and eddy current) loss are caused from the flux that is in the inductor, so not all the energy can be returned to the primary (if that's where to goes to), unless the inductor energy and core losses are different...?
 
  • #16
tim9000 said:
Both these energies come from the primary excitation, so are both these stored energies returned to the primary excitation supply?
Cheers!
Of course, inductance is inductance...
tim9000 said:
there are formulas for magnetic (hysteresis and eddy current) loss are caused from the flux that is in the inductor, so not all the energy can be returned to the primary SOURCE !
tim9000 said:
unless the inductor energy and core losses are different...?

Of course, that's why we use the concept of ideal circuit elements so we can think one step at a time.
A real inductor is not quite lossless.
So we add to our ideal inductor an ideal resistor in series to represent resistance of the wire,
and another in parallel to represent losses from yanking the iron molecules around,
and we arrive at a model that's closer to reality.

You need to practice step-wise thinking. :smile:
 
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  • #17
tim9000 said:
That's actually going to be handy for me to revise, cheers.

Although in saying that, I guess one thing that also bugs me is, there are formulas for magnetic (hysteresis and eddy current) loss are caused from the flux that is in the inductor, so not all the energy can be returned to the primary (if that's where to goes to), unless the inductor energy and core losses are different...?

Inductor energy always comes back (at least ideally). Look at Jim's graph. Inductors borrow energy for a time then return it. (Some tiny amount of the flux links to other circuits as a parasitic transformer or RF noise, but that is rarely modeled.)

Core losses are resistive. Core losses include eddy currents and the like which create I2R losses as heat. They include I2R losses in the windings. (Technically those aren't core losses, but they are real resistive losses.) So an inductor continually borrows and returns the same energy, but each time this happens parasitic core losses take their toll. Generally it's not good to have purely reactive power flowing around causing I2R losses.
 
  • #18
Jeff Rosenbury said:
Inductor energy always comes back (at least ideally). Look at Jim's graph. Inductors borrow energy for a time then return it. (Some tiny amount of the flux links to other circuits as a parasitic transformer or RF noise, but that is rarely modeled.)
Yeah, the problem I was conceptually having was with where the energy goes. Jim confirmed that it goes back to the primary source.
as for your next paragraph I'll address that with my next reply to Jim.
 
  • #19
jim hardy said:
Of course, inductance is inductance...Of course, that's why we use the concept of ideal circuit elements so we can think one step at a time.
A real inductor is not quite lossless.
So we add to our ideal inductor an ideal resistor in series to represent resistance of the wire,
and another in parallel to represent losses from yanking the iron molecules around,
and we arrive at a model that's closer to reality.

You need to practice step-wise thinking. :smile:
Ah of course *>hits head<*, I totally forgot about the parallel resistance to the magnetising Xm, oh dear, my mind is like a sieve when I don't practice this in a few months. (yeah so the Rc takes care of the magnetising losses. I suppose if you wanted to improve the Ideal TX model to make it more realistic, you could put a resistor in parallel to the Leakage Reactances to model any small losses caused by the leakage flux inducing magnetic losses in the real world outside of the TX.

Thanks for confirming that the inductance is only relevant ot the primary side of the ideal TX.
Hmmm...
 
  • #20
Seems thinking has converged ... rewarding to see "the light come on", isn't it ?
 
  • #21
It's worthwhile spending the time to become conversant in the model of a real transformer

this one at Wiki is just one of many similar
https://en.wikipedia.org/wiki/Transformer

500px-TREQCCT.jpg

when you can explain what physical effect each component represents ,
and how you might measure it
you are well equipped in the transformer department.
 
  • #22
jim hardy said:
Seems thinking has converged ... rewarding to see "the light come on", isn't it ?
Rewarding coupled with a feeling of foolishness :p
 
  • #23
dont be so hard on yourself. Enjoy your success !

help somebody else. " Pay it Forward" as the kids say (ahh, Helen Hunt - sigh)

old jim
 
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What causes core saturation?

Core saturation occurs when the magnetic flux density reaches its maximum limit in the core of an electrical device. This can be caused by an excessive input voltage, too many windings in the coil, or the use of a magnetic material with a low saturation point.

What are the consequences of core saturation?

When a core becomes saturated, it can cause distortion in the output signal and decrease the efficiency of the device. It can also lead to overheating and potential damage to the core and other components in the circuit.

How can core saturation be prevented?

To prevent core saturation, it is important to select a magnetic material with a high saturation point and to design the device with the appropriate number of windings in the coil. Additionally, using a feedback control system can help regulate the input voltage and prevent saturation.

What are some common signs of core saturation?

Some common signs of core saturation include distortion in the output signal, increased heat in the core, and a decrease in the efficiency of the device. In extreme cases, the device may experience a complete failure.

Can core saturation be reversed?

Core saturation cannot be reversed once it has occurred. However, steps can be taken to prevent it from happening in the future, such as using a different magnetic material or adjusting the design of the device. In some cases, the device may need to be replaced if the saturation has caused irreversible damage.

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