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There's a lot of documents that talks about dimensioning of the transformer for a SMPS power supply (in this case one quadrant only is not of my interest).

this site cites the following:

The main constraint in all cases (except for saturable inductors) is that peak magnetic flux density Bmax should not approach the core material's saturation flux value Bsat.

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Contrary to popular misconception, Bmax does not depend on the magnetic material properties or air gaps. It does not depend on the transferred power neither. However, for thermal reasons, we have to limit ohmic losses in the wires. Most textbooks provide formulas for estimation of the core size based on the product of magnetic cross-section area by the window area available for the winding. Unfortunately, this method is not very helpful because these formulas are based on pretty much arbitrary selection of current density and on an assumption of a certain window utilization (fill) factor.

So, I calculated the Bmax for my project, and it gives-me similar results to that of poweresim.com. But with the AwAe method it requires more than 4 times the core size!

Anyway, the Bmax formula being close to the results, it does not have the current included in the formula. So that any power (respecting the Vpk and f in the formula) could be used with the same Ae transformer.

Is the core size not dependent on H, just B? That is, having enough space for windings for the projected currents/potentials and getting Bmax below saturation Bsat, and dis-considering hysteresis, eddy-current losses, what should I consider?

If not what should I consider then? If I try to calculate H, it will give-me values really above saturation (from B-H curve), for near most materials, so I think thats wrong.

Diego C Nascimento
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I don't know if you have miscalculated but it works like this.

H is ampere turns per metre and the metre part is the mean length around the core that the magnetic lines of flux follow: -

enter image description here

So, amps (peak) x turns (primary) divided by core length gives you H (peak). Note that the "amps" number chosen is the magnetization current i.e. NOT THE SECONDARY LOAD CURRENT (referred to the primary).

Next you have: -

B = \$\mu H\$ where \$\mu\$ is the effective magnetic permeability of the transformer core in absolute units. Effective permeability depends on several factors such as length of core (mentioned above), the cross section, the material type and if any gaps are present.

The manufacturer of the core will tell you the effective permeability of the core you choose. You may also choose to gap the core and this dramatically changes effective permeability.

Next, look at the data sheet for the core and see what flux density will be produced for the given H field. If it looks like it's beginning to heavily saturate then increase turns a tad.

If turns are (say) doubled, primary inductance quadruples and current quarters. This is useful because, for a given primary drive frequency, if you double the turns you half the H field. If really pushing the core you will have to consider gapping.

Andy aka
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  • Read, studied, calculated, applied, but still the results don't match. With lets say $ \frac{32 \times 60}{0,0184} = ~104347 \space H(A/m) $ ! With this I will need a core bigger than a line frequency transformer, even trough I know of projects that uses this core with 6Kw easily. What I need to apply seems to be the relative permeability to get a new H(A/m) value? The manufacturer only provides a fixed $ \micro e $ or one related to temperature. Even trough it is an integral and so gives even more bigger values. – Diego C Nascimento Sep 07 '15 at 06:42
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    Where does the 32 and the 60 come from - one of them is current but i suspect you have used the full load current referred to the primary - you should use the no-load current for calculating flux density because load ampere-turns in secondary exactly cancel load ampere-turns in primary. – Andy aka Sep 07 '15 at 09:59
  • Yes 32 A is the full load current in the primary, in fact the peak is 50 A. Secondary current is 200 A. I have read your other response and read some papers about magnetization current. Will take a read again. – Diego C Nascimento Sep 08 '15 at 01:07
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    It's the no-load current flowing in the primary that determines saturation. – Andy aka Sep 08 '15 at 09:30
  • So I can assume a square wave of Vin amplitude and the primary inductance (or the magnetizing inductance is different?) to calculate the no-load current? – Diego C Nascimento Sep 08 '15 at 12:25
  • With no load connected to the secondary, the inductance measured (or calculated) on the primary IS the magnetization inductance. The current that flows is the magnetization current. If your SMPSU is a non resonant type then you can assume a square wave as your input voltage. – Andy aka Sep 08 '15 at 16:31
  • Yes, its hard switched (but magnetizing inductance current will be a sawtooth, summed with the load current, results in a trapezoidal waveform). Alright. I have doing a lot of work, and seems that the magnetizing current is in the $ Im < 1 A $ range (what saturates the core anyway). Other tools suggests that the EE80 could take 4kW, probably I need a small gap then (with the inefficiency, EMI, and higher switching losses of the gap). But for now I need to calculate the new B/H curve. You say I need to look at the catalog for B/H but as Bsat does not change I can use $ B = μH $ no? – Diego C Nascimento Sep 09 '15 at 10:18
  • But this seems to use SI units. So lets take as example μ = 1745. H = 1 Oe = ~79A/m. This gives B = ~136000 T?! Something wrong. – Diego C Nascimento Sep 09 '15 at 10:25
  • How many turns? How many amps and what is the mean length of the core. I'm reading stuff I can't comment on. The BH curve remains the same ONLY the effective permeability changes. – Andy aka Sep 09 '15 at 11:08
  • Sure, but you will need more field force (H) to reach the same density (B), at least not integrating with time. Is this I know, the permeability changes the "slope", reducing induction as effect 1. The core Le = 184mm, Ae = ~400mm², and I probably use 2 so to double the Ae and Al, getting the same inductance with half the turns and field force. With 4 cores I get near (>7000 W), but I don't know what I'm making wrong in the formula above, so to calculate the total magnetic path permeability including the air gap. – Diego C Nascimento Sep 09 '15 at 11:25
  • I have given you an answer that I believe is correct - I don't know about the other formulas and I can't check your work using my formulas because I know nothing about the inductance of the primary, the turns, the frequency or the voltage applied. I can't give you any more help and I'm not going to comment on Aw and Ae formulas because I don't use them. If you tell me mag current, number of turns and effective core length I can give you H. Start there. – Andy aka Sep 09 '15 at 11:53
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    Or to formulate better: Primary inductance $ Lp = Al \times N² = 9000 \times 3600 = 32mH $ Magnetization current $ Imag = \frac{V \times t}{L} = \frac{285 \times 0,000025}{0,032} = 0,22A $ Magnetization force $ H = \frac{I \times N}{Le} = \frac{0,220 \times 60}{0,0184} = ~717 A/m $ – Diego C Nascimento Sep 09 '15 at 12:54
  • NO - Le is 0.184 m = 184 mm = 18.4 cm. This reduces B by 10. – Andy aka Sep 09 '15 at 13:10
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    Sorry, so much unit conversion. $ H = \frac{I \times N}{Le} = \frac{0,220 \times 60}{0,184} = ~71 A/m (~0,89 Oe) $ well below saturation point. Thanks for your help. I will take a look at the permeability, seems that in $ B=μH -> μ = μm = Kmμ0 $. – Diego C Nascimento Sep 09 '15 at 13:26