Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Core loss in Alloy powder core sounds too low?

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
Hi,

We are doing Boost PFC.
115VAC, 120khz = fsw, 270W = Pout, 390Vout
Inductor is 60 turns of ECW round a micrometals HF130125-2 core. (460uH at zero Amps)

HF130125-2 core (Page 57)
https://www.micrometals.com/design-and-applications/literature/

As you can see, there as a large amount of inductor ripple current at the peak of the mains. dI = 4.1Apkpk and dB = 0.293T (pkpk)
However, the core loss calculation from page 31 of the above Micrometals catalog gives only 137mW of core loss. (and thats an abs max, since as you know, the Bpkpk gets less as one moves away from the mains peak)

Doesn’t this sound just way , way too low?
 

Attachments

  • PFC inductor current.jpg
    PFC inductor current.jpg
    87.7 KB · Views: 47

Easy peasy

Advanced Member level 5
Joined
Aug 15, 2015
Messages
4,242
Helped
1,335
Reputation
2,670
Reaction score
1,712
Trophy points
1,393
Location
Melbourne
Activity points
23,705
1669686826887.png


100kHz, ac swing 1000G = 0.1T, losses are ~ 1200 mW/ cc x 5.48 cc = 6.576 W

so for a larger swing of 0.15T each way the losses will be ~ 10W.

[ do not confuse Gauss with Tesla ].
--- Updated ---

1T = 10^4 Gauss
 

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
Thanks very much indeed Easy Peasy, you're quite right i was out by an order of 10 with G to T conversion.
And at 200kHz, i am getting 12.31W of core loss......i believe this is too much for this core.......this isnt going to go down well with cust...who sent this in to to us be "made to work".....all the enclosure is made around this size inductor, and now its going to have to be made bigger, would you agree?
...The thing is , AYK, this value of Bpk only applies at the mains peak......so i believe it will be some 0.63x this over the mains half period
 
Last edited:

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
Thanks, i am finding by going up to 70 turns from 50, we can make the core loss less. RMS current is 2A7.
Will be a squeeze to get 70 turns on this core though.
Also, going up to 200kHz switching frequency helps the core loss.
 

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
Woops, i was using the MS130125-2 core values but its actually a HF130125-2 core...so had to re-run the numbers.
Literally the only way to get the core loss calculated is with the attached excel....which takes quite some time.....literally mapping out the u vs H graph as a number of straight line sections......then producing a dB by dH table, and getting ur at every H value, and calc'ing dB from dB = uo.ur.dH......etc etc......i am amazed the micrometals datasheets dont allow you to do this more easily....and when youre using 2 stacked cores, you have no option but to use the attached excel type method, would you agree?

With the same PFC, at 120kHz, and 115VAC in, full load out, and with 2 stacked HF130125-2 cores with 40 turns, the core loss is 9.2W...SO THATS 4.6W for each core.

...putting 50 turns on the same 2-core-stack cuts the overall losses down to 4.7W.....now sounding reasonable.....just a Nervous breakdown for the cust who did the enclosure snug for one core.
 

Attachments

  • Mag inc coeffs _HF_130125_2_pfc 115vac _2_120khz_REAL_2CORE.zip
    63 KB · Views: 28
Last edited:

Easy peasy

Advanced Member level 5
Joined
Aug 15, 2015
Messages
4,242
Helped
1,335
Reputation
2,670
Reaction score
1,712
Trophy points
1,393
Location
Melbourne
Activity points
23,705
work out the H for the max peak current, and hence Bmax = Uo . Ur Hmax

same for the min of the current ripple at the mains peak, this gives you Bmax & Bmin at the mains peak

thus Bpk for losses = ( Bmax - Bmin ) /2

pretty straight foward ... ?
 

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
Thanks, i thought because ur varies with current, it couldnt be done like that?
So it was considered we have to calculate Bmax by small-stepping H up to H max.......then going up point by point, calculate dB =uo.ur.dH....then sum up all the dB values, and get Bmax like that....kind of like a discrete integration?
I think the way that you state is the way when ur is constant as H rises?
 

Easy peasy

Advanced Member level 5
Joined
Aug 15, 2015
Messages
4,242
Helped
1,335
Reputation
2,670
Reaction score
1,712
Trophy points
1,393
Location
Melbourne
Activity points
23,705
It is perfectly straight forward, at the high current you get a value of the % Uo ( really Ur - a typo on the graph ) from knowing H,

say this is 50% for type 125, the B is then [ù.H] = Ur ( = 0.5 * 125 ) x Uo ( = 4.pi.E-7 ) .Amps. turns / Lmag

all SI units.

Do the same for the trough of the switching ripple at peak mains, the B may be similar or only a little less, as at the higher current the Ur is less so the H is less, so the B will be less than for a linear system.

all clear now ?

Use these 2 values to compute the pk-pk B ripple and then the pk only, for loss calcs, then multiply by 0.707 for the average losses over an half cycle, which are also the long term losses in the core in watts.
--- Updated ---

Just for clarity - the H shown in the graph is in Orsteds, which are not SI units - so you have to convert.

1 Oe = 1000/4π ≈ 79.5774715 A/m in terms of SI units.

or 100Ampere.Turns/Lmag(m) = 1.257 Oe. 1000 ampere.turns/Lmag(m) = 12.57 Oe

10,000 Ampere.turns/m = 125.7 Oe
--- Updated ---

Do the same for the trough of the switching ripple at peak mains, the B may be similar or only a little less, as at the higher current the Ur is less so the H is less, so the B will be less than for a linear system.
This should read ... the Ur is less, so the B will be less than for a linear system.
 
Last edited:

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
Thanks Easy Peasy, the method that you depict, and with which i find agreeable......comes out as 6W of core loss for...
HF130125-2, in a PFC of 115vac, 390vout, 260Wout, 120kHz. (50 turns)

However, the dB/dH method, depicted on this forum in 2020, gives 9.84W....

And the Micrometals website core loss calculator gives the core loss as 8W

I am not sure which one is correct out of these 3?

Micrometals core loss calculator:-
The Micrometals core loss calculator is a little bizarre.....
It asks for "inductor peak current"......but the current they are asking for there must surely be the mains peak input current.(?)
Also, the RDC changes when you change the switching frequency, which cant be right.
Also, it gives a value called "Bpk fswitch" and says this is "0.863T".......this really doesnt make any sense at all from the data that was put in (and is shown above).
The micrometals calculator also gives a value called "Ipkpk(A)" and says this is "1.8A"....which doesnt make any sense.
______________________________
I then used the Micrometals "inductor designer" tool, and again, when it says "inductor peak current", it obviosuly means "peak mains input current"...and when it asks for "inductance at peak inductor current"...it really means "Inductance at the peak mains input current level"....i am not totally sure, but this seems right.

...For my spec of 260Wout at 115vac input and 390vout, and inductor of 300uH at the peak mains input current level, it offers me a design with MP130125-2 torroid, with 57 turns of 2 strands of AWG17...and tells me i have 0.16W of copper loss.......if they think i am going to wind 57 turns of two strands of AWG17 round a MP130125-2 core they have got another thing coming!!!.......i will pick a much thinner wire, and only one strand!!!....because i can easily afford to dissipate way more than 0.16W in the winding!

...so the micrometals web inductor designer seems a little bizarre !!...and may well be giving wrong answers from the first glimpse.
 
Last edited:

Easy peasy

Advanced Member level 5
Joined
Aug 15, 2015
Messages
4,242
Helped
1,335
Reputation
2,670
Reaction score
1,712
Trophy points
1,393
Location
Melbourne
Activity points
23,705
you need to work out the ave L at peak mains, as this will be a minimum, the di will be a maximum - i.e. you cannot assume a linear inductor. This may affect your loss calculations.
 

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
Thanks, the attached is the worst case inductor current ripple....it uses a linear inductor, but its value is that of the non-linear inductor at the "real" inductor peak current....so it gives the worst case delta I, and the worst case delta B.

Taking measurements from this, the "Bpk" is 0.06T (ie calc'ing B at i(pk) and i(pedestal)......from the micrometals equation, this means a core loss of 1.15W. That just sounds way too low....and thats a maximum measurement because the absolute minimum value of inductancee was used for it.

The micrometals inductor designer says its 3W of loss. Though their calculator seems to assume that the inductor has a value of "what it would be at the mains peak input current level" throughout the switching period......which of course, it does not.....its less at the inductor current peak.

I am of the opinion that the micrometals calculator is not correct.

The taking of B at the peak and pedestal currents, and then working from there, gives a too low value of core loss.......this was what prompted the post above from 2020.

So again i believe that the dB/dH method is the way forward.....as described in the post from 2020. That is, you can only calculate the Bpkpk by doing the "discrete integration" method of small stepping the H, and finding dB/dH.......and then finding what delta B you accumulated over that dH.........then doing this so you have the accumulated B at every current level....(or therabouts as it would take to time infinite if was took very small steps).

I think this is the crux of it, quoting from 2020...
The problem is the difference between DC and AC permeability. To get the correct AC permeability from the B/H curve, you need to calculate the slope delta B/delta H.

The strange thing about the micrometals website is they recomend 57 turns of 2 strands of AWG17 round an MP130125-2 core.....this wouldnt fit round that core...and if it could, the proximity losses would then mess it up.
--- Updated ---

AYK, the key point, is that if you have a non linear inductor...then we must find Bpk, by calculating the accumulation of mag flux density as current is increased. The only realistic way to do it is to small-step the current upwards.......then calculate the dH each step......then calculate the ur at that current....then calc dB=Ur.uo.dH.....then take another small step up in current, then re-calculate the dB for that small step....then keep doing this.....and eventually you can read off your accumulated B for say 5A of current.

....AYK, this is the pennance that we serve for using non linear inductors....discrete integration effectively.
--- Updated ---

....What's notable is that even the Micrometals website dont calculate like this.....they simply make the assumption that one has chosen an inductor with high enough inductance, that the inductor current has very very small i(ripple)......this obviously makes for errors when one is using a small inductor (pushing it), and then one does have high i(ripple), and one ends up far from the "safety" of the micrometals web calculator.......
 

Attachments

  • MP torroid.jpg
    MP torroid.jpg
    347.6 KB · Views: 33
Last edited:

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
___--___--___--___---___
If we look at the attached excel, the green columns show the build up of B as current increases in the torroid (non linear) inductor.

In the “accumulated B” column, the accumulated B builds up as the current rises…this is the correct behaviour, since B is akin to energy and one cannot just suddenly get rid of energy.

The “Simple B” column shows a simple calculation of B from B=uo.ur.H and takes the new ur as that at that current level…..

…This “simple B” column is quite clearly incorrect….as the current rises above 10A, the “simple B” starts going down!! This cannot be correct…..the B must go up, since you cannot just suddenly get rid of B, because B is a form of energy…….if the B really went down, then what happened to that energy?…the energy stored in the magnetic field…it cannot simply be gotten rid of…..conservation of energy is the law of physics at play here.
Would you agree with the above?
It must be so, if it were simple, then the Micrometals web site would show it so....but Micrometals do not show it at all...they do not offer a bona fide calculator for core losses
__--___--__---_
--- Updated ---

Ive actually done the core loss calculation using the "accumulated B" method...and it comes out very close to the micrometals inductor analyzer. (2.4W vs 2.8W).
So i am thinking maybe just maybe the approximations they take in their web calculator kind of cancel each other. It seems strange that the two methods give the same result...and tends to make you think that they must both be correct?
 

Attachments

  • MP130125_2 Torroid.zip
    52.6 KB · Views: 21
Last edited:

Easy peasy

Advanced Member level 5
Joined
Aug 15, 2015
Messages
4,242
Helped
1,335
Reputation
2,670
Reaction score
1,712
Trophy points
1,393
Location
Melbourne
Activity points
23,705
B can go down easily if Ur falls due to increasing current

because B = Ur . Uo . H, and H = ampere-turns/mag-length

Energy = 0.5 B . H . Volume core, so if B falls due to falling Ur, the energy stored is less, or a negligible increase.

For best design and lowest losses - your pk-pk I ( and therefore ~ B in this case ) should be in the region of +/- 18% - 35% max above and below the mean at that point.

Your delta I ( & therefore B ) is too high.
 

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
Thanks,

Micrometals have very kindly just replied to our email....as follows.....

In a PFC application, the core losses will continuously change as the input voltage changes. To get the average core losses, one has to calculate the quasi steady-state core losses at different points along the input sine wave and then average the results. Our inductor design tool performs these calculations automatically. I have attached an Excel file which calculates the approximate core losses in a PFC application using this method. I have set up the Excel file to calculate the losses for the MP-130125-2 with 57 turns using your parameters. You can calculate different core shapes, turn numbers and materials by changing the input variables in the top section of the Excel file. I hope the calculations are useful to you.
....As such, i am wondering at what actually this technique is, it doesnt sound like simply taking Bpk and B pedestal and getting the difference between them.
The results of their web core loss calculator are very similar to the dB/dH method put forward in 2020.
 

Easy peasy

Advanced Member level 5
Joined
Aug 15, 2015
Messages
4,242
Helped
1,335
Reputation
2,670
Reaction score
1,712
Trophy points
1,393
Location
Melbourne
Activity points
23,705
what are the results ? - you omitted to re state them
--- Updated ---

and - what are the results for a design with +/- 20% ripple at peak mains .... ?
 

cupoftea

Advanced Member level 5
Joined
Jun 13, 2021
Messages
1,679
Helped
41
Reputation
82
Reaction score
87
Trophy points
48
Activity points
8,833
Thanks, my apologies, a cust is coming in thurs...expects their board that they mis-designed to be fixed. So we must do PFC 115vac in 220vdc out, 550wout, 120000hz.
We only have a couple of HF106160-2 cores.
So must now work on that one.
I am thinking 2 stacked and 30 turns of 2 strands of 0.5mm ECW.

Micrometals gives that 3.5W of core loss (presumably shared between both cores), and 2.3W of copper loss.
 

LaTeX Commands Quick-Menu:

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top