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FINAL ATTEMPT to get my head around the correct magnetising inductance

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bowman1710

Full Member level 3
Hi guys,

I'm trying to get a final idea in my head about what to do with designs and their magnetising inductance, I have put several posts on here about magnetising inductance but I feel if I put more into what I'm asking I might get more back with previously me just asking "what is the correct magnetising inductance?"

Attached is my way of thinking/working outs that I have done, with what people have said in the past it is all about flux etc first, which is how i have done it from the start. I have worked out what turns ratio I need, the minimum amount of primary turns needed with the core size and desired Sat limit and worked out the primary inductance given that amount of turns

As you can see from my design it leaves me with 4.3mH worth of primary inductance now given a switching frequency of 333KHz and a minimum input voltage of 250V with a 30W capability you would expect

Xl=2*pi*333KHz*4.3mH=9Kohm

Im=Vin/Xl=250/9Kohm=27.7mA

If you work out the input current needed as a rough ball park with 80% efficiency

Iin=(30*1.2)/250=144mA

So the Im is nearly 20% of the Iin value, so at what point do you say well.......its not going into saturation but the magnetizing current is too high. Does this need more turns? Will more turn cause too high of inductance and cause spikes on the MOSFETs? Where do you draw the line and say well it needs to be this?

You didn't tell the underlying design problem. Flyback? Forward? Push-Pull converter?

In a short, magnetizing inductance as a design parameter mainly matters for flyback design, for other design it's sufficient to know Lm > some minimal value.

Hi,

Xl=2*pi*333KHz*4.3mH=9Kohm
This formula is for sinusoidal waveform without DC offset.

Im=Vin/Xl=250/9Kohm=27.7mA
This also is for sinusoidal wavform without DC offset.
But the word "switching" tells me it is not sinusoidal. And i doubt that you actively sitch to -250V to avoid DC components.

***

For DC applications i keep on this:

1 Henry = 1 volt * 1 second / 1 Ampere. In short 1H = 1 Vs/A, or L = U * t / I (for ideal inductors)

For your application this means:
(4.3mH, 333kHz, 50% duty cycle, 250V DC)
ON time = 50% / 333kHz = 0.5 / 333000 1/s = 1.50us

The rise (delta!) in current when you switch 250V DC for 1.5us to a coil with 4.3mH is:

delta_I = t * U / L = 1.5us * 250V / 4.3mH = 87.3mA.

Delta means, if the current before switching ON is not zero, then the current at the end will not be 87.3mA.

****

I recommend you to use some simulation tool, like LTspice. There you can simulate, show how currents and voltages behave .. on a single pulse and as well how they develop druing continous switching.

Klaus

You didn't tell the underlying design problem. Flyback? Forward? Push-Pull converter?

This is for a forward converter

delta_I = t * U / L = 1.5us * 250V / 4.3mH = 87.3mA.

Delta means, if the current before switching ON is not zero, then the current at the end will not be 87.3mA

So are you saying that the magnetising current is 87.3mA

... 80% efficiency

The above current does not influence efficiency, because the energy is not lost. It is just the inductive current through the coil.

To the calculated current you have to calculate any other current, like secondary load current, core loss...

To calcualte power (efficiency) you need to calculate the average_input_current x DC_voltage.

Klaus

To calcualte power (efficiency) you need to calculate the average_input_current x DC_voltage.

Isn't the magnetising current part of the average input current?? With the inductance having some form of AC resistance surely current induced will result in losses?

Hi,

So are you saying that the magnetising current is 87.3mA
No.
When the switch is ON then the current rises form the starting value (unkonwn) with a rate of 58.2mA/us.
And as said before: this is only one component (the inductive) of the primary current.

****

Isn't the magnetising current part of the average input current??
Yes and no.
Yes, if you see it as AC. But when the switch is OFF the energy is either given to the secondary side (depends on topology) or it is given back to the primary side DC bulk capacitor.
No, if you see it as averaged DC current. Then - as long as the energy is not given to the secondary side - it is zero, becuase the current runs in both directions.

****

This is similar to a standard low frequency AC transformer.
When there is no load (and we say the loss is negligible) then there is AC current. You can measure this with an RMS meter. But there is zero average current.
But with none of both current values you can calculate the loss.

To calculate loss in an AC system you need
* the waveform of input voltage
* the waveform of input current
* and the time relation to each other (phase shift)

****

If you find it useful i can make some LTspice demos for that. But time is short.

Klaus

I recommend you to use some simulation tool, like LTspice. There you can simulate, show how currents and voltages behave .. on a single pulse and as well how they develop druing continous switching.

I have had very little luck finding magnetics models
(let alone, building them) for SPICE that can be relied
upon. If you find a good one for your core(s) of
choice, you would be lucky indeed. Or you have
something to teach the world, if you succeed at
parameterizing a "core and winding" model on your
own that matches reality at all well.

Only simple inductors operating away from saturation
have ever worked out for me, because they don't need
no schoolin'.

Hi,

sadly this is true.
The simulaton of core losses and the magnetic behaviour is very poor.

But a 4.3mH inductor on a 250V DC for 1.5us will show the current wafevorm.

Klaus

bowman1710

bowman1710

Points: 2
Bowman, I can now let you see the magnetising current.....open and trun the simulation attached of a 2 transistor forward converter ,and go "add trace", then paste I(L1) + 0.531*I(L2) in the waveform window, and you will actually see the magnetising current with your own eyes..

so to summarise, run it, right click in the waveform window, at the top, and go "add trace" then paste in "I(L1) + 0.531*I(L2) ".... then you will see the magnetising current

its just by V = Ldi/dt

Attachments

• 200W Two transistor forward converter.txt
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bowman1710

bowman1710

Points: 2
I re-read your first post Bowman, and could empathise with it......IMHO, you are making the mistake of trying to understand the theory in absolute sense, this is not the way (unless you want to be an Einsteinian scientist), you should try to just get it done so that it will work, by any naffarious means, because if it works, and it is reliable, then who cares what theory you actually understood on the way. So design something, and the doing of it, maybe several times, will drill it into you what you need to know and do. You will "come to terms" with the theory, but who cares if its understood or not.
The whole theory is just a "model" anyway.....in reality, all non-microwave engineering uses a "low frequency approximation", and current doesn't actually flow as electrons flowing along wires at all, ever, period, ..energy travels only by EM fields in all circuits, and you need 3D maths to properly analyse it....in reality...but even then, some are coming out with "string theory" instead........so we may be chasing our tails for long time, if we try and understand in absolute sense.....if you trace the theory back far enough, then eventually you bump into some maxim...and then the only reason for it is...."because the Lord made it so".

bowman1710

bowman1710

Points: 2
Magnetizing inductance and it's corresponding no load current is a pretty broad question.
It depends on current, conduction loss, frequency, core loss vs f.

If inductance is in a transformer, also the coupling factor or inverse mutual inductance ratio in order to minimize energy storage in core to efficiently transfer the energy. (e.g. 10% of max rated current (VAR) for no load excitation in a power transformer and high efficiency at max load.

If inductance is in a flyback transformer, it may be for total energy storage in the inductor rather than shared with capacitors . One chooses the energy required to transform the voltage during flyback for E=1/2LI^2. This is why flyback has limited use above 100W.

For example if you want minimum weight and maximum efficiency, one would choose the highest frequency with a core of the lowest loss at that frequency and thus low L depends on power variables for V, and I transformation as well as SRF.

If smallest size is the driving factor at lowest cost, then frequency must be compromized due to self-winding capacitance and L is a tradeoff with Rs and I demands. Thus high current must use a series resistance << load R for efficiency and this results in lower L. But too small an L increases peak/average current ratios with discontinuous modes causing more stress or cost in Caps.

For example instead of 60Hz 10kVA transformer/generators in aerospace applications, 20KHz is used and thus L can be much smaller than the typical 1~50H for 50/60Hz transformer inductance and filter chokes.

edit....

For a forward converter with low error for load regulation, this means the ratio of source ESR/step load = Load Regulation error. ( usually in %)
This source impedance is a complex value that includes all switches and d.f., winding resistance , turns ratio and ESL, effective series inductive impedance of both primary and secondary.

When multiple forward converters this is critical for regulating one output so that all outputs track well and choosing transformer coupling with the lowest inductance gets the best results that does not cause saturation or excessive ripple cap current. More sophisticated protection is needed with UVP,OVP OCP, OTP

open and trun the simulation attached of a 2 transistor forward converter ,and go "add trace", then paste I(L1) + 0.531*I(L2) in the waveform window, and you will actually see the magnetising current with your own eyes..

Is this for LTspice, what do you mean by open and run, all that happens when i open it is it comes open with WORDPAD and i get an error in LTspice, sorry :S

you should try to just get it done so that it will work, by any naffarious means, because if it works, and it is reliable, then who cares what theory you actually understood on the way. So design something, and the doing of it, maybe several times, will drill it into you what you need to know and do. You will "come to terms" with the theory, but who cares if its understood or not.

So do most people design in this way am I trying to get my head around something

When multiple forward converters this is critical for regulating one output so that all outputs track well and choosing transformer coupling with the lowest inductance gets the best results that does not cause saturation or excessive ripple cap current. More sophisticated protection is needed with UVP,OVP OCP, OTP

So basically for a forward converter a higher magnetising current is better but you don't want the think driving into saturation or causing to much ripple on the input caps?

To summarize then guys:

1) Magnetising current is dependent on the application
2) The magnetising current isnt strictly a loss aspect of a design if it is DC, it is either transferred to the secondary or given back to the primary side capacitors, if it is not given to the secondary side there are no losses
3)With a forward converter you tend to want a higher magnetising current but not to the point of saturation or excessive ripple current

Am I correct so far????

You calculations are for a sine wave, the current (83mA) does flow but its out of phase with the applied voltage, so it consumes no real power. The power loss occurs in its DC resistance and magnetic losses, which are equivalent series resistors.
Frank

to run the ltspice sims, you change them to .asc, then open them in ltspice, then run them

bowman1710

bowman1710

Points: 2
you haven't explained why magnetising current is so important in your case? when designing a transformer it is usual to design for a suitable flux swing for the operating freq, this gives the lowest turns, from there one can determine the Lmag and the Imag, if for some reason it is too large, the Imag, then the core size and/or windings need to be increased to reduce it, i.e. double the core size = 1/2 the Imag.
Obviously upping the turns and/or core size leads to a net longer wire length for the copper and more losses unless you raise the copper area, so unless minimum Imag is an absolute necessity for some reason, it pays to stick with the minimum Lmag that will give acceptable core temp rise. As the phase of the Imag is out of step with the load current, they add vectorially not scalarly, so 10A load + 1A Imag is less than 11A in the winding (sqrt(101)), so you will always get the lowest losses and smallest Tx, going with the lowest Lmag - just one of those things...

bowman1710

bowman1710

Points: 2
you haven't explained why magnetising current is so important in your case

Easy peasy,

I wouldn't say it was in my case it is important, I was just trying to get my head around whether it was deemed important or not? And how everyone went about designing transformers around the magnetising inductance. I think you have covered what I wanted to know, and agrees with someone I spoke to that knows a lot about magnetics and SMPS's. Which is basically as long as the core losses are not massive because of the flux swing and Lm doesnt result in adding over 10% to the primary current you dont need to worry about it really.

I wouldn't worry about adding 30% or more to the primary current by way of imag. Its just like a kind of slope compensation. I mean, in the ideal world, in most cases i'd have no imag but i'm quite happy to put up with it, even in large quantities, long as core losses ok, as you say.

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