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I have to design a transformer with defined parameters such as my primary and secondary voltage, rated power of the transformer, the flux density and the lowest frequency \$f\$, are there other formulas to estimate the core area required? As far as I am aware, designing it using the induction law below requires the number of turns \$n\$ which is at this point still unknown?

\$U = 4.44 \cdot f \cdot n \cdot B \cdot Q_e\$

where \$f\$ is the lowest frequency, \$n\$ is the number of turns, \$B\$ the magnetic flux density and \$Q_e\$ is the cross-section area of my core.

I read a pretty old article on transformer design (https://archive.org/details/funkschau-1958-heft-3/page/116/mode/2up) and came across this formula used to calculate/estimate the core area required. The article (page 117) is in german, the part below is translated using chatgpt.

If one wishes to determine the iron core cross-section by calculation, a specific peak value of the magnetic flux density in the iron must be assumed. Furthermore, the frequency and the current density in the copper also have an influence, since these determine the window area, which is related to the core width in a fixed proportion. For these relationships, the following formula applies: $$ Q_{e} = \sqrt{\frac{14 N}{\frac{f}{10}\cdot\frac{B_m}{10000}\cdot i}}$$

where \$ Q_{e}\$ is the core area in cm\$^2\$, \$ N\$ the rated power in W, \$f\$ the frequency in Hz, \$ B_{m}\$ is the magnetic flux density in Gauss and \$i\$ the current density in A/mm\$^2\$."

Unfortunately the article doesn't really explain the origin of formula. Since 1 Gauss = 0.0001 T, this would explain the division with 10000. However, I have no idea where the square root or the 14 comes from. Does anyone have any idea?

** I am aware that the usual approach is to design based on an already-available core but this is not the case here. The core material would be ferrite for high frequency usage (500 Hz to 4-5 kHz), and the power will be around 500 W. but I assume the material is irrelevant.

Thank you!

Edit: added the correct link and corrected the second formula below.

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  • \$\begingroup\$ You appear to be going round in circles without really focusing on what you are trying to achieve. You might also be looking for a simple formula that gives you an answer but, in my experience, it's an iterative process to get to the correct number of turns and core size. I can't help with the German translated formula because I don't recognize it but, I can say that you need to start focusing on the n umber of turns and try a few scenarios out. \$\endgroup\$ Commented 20 hours ago
  • \$\begingroup\$ I am unable to find the quoted (translated) section. It seems erroneous. It's also inconsistent (where is \$i\$ in the formula?). I do not see \$N\$ in the definitions block. This all seems to be hallucinated and should be struck from the post. || Given the setting (vacuum tube era electronics), and diagrams of typical laminated-iron cuts, should we assume you are after low-frequency design equations? But you put "ferrite" in the tags. Which one is it? The priorities are not the same; and unified equations are quite lengthy. \$\endgroup\$ Commented 16 hours ago
  • \$\begingroup\$ I guess the frequency is not 50 or 60 or 400 Hz but much higher. So what is the frequency used for your design? Do you want to use a ferrite or iron-core? How much power? You get the needed core weight from power, frequency and maximum flux density of the core material. Primary current is calculated from primary power and voltage. From current and core material you get the voltage per winding and the number of windings. Next is the maximum resistance of a winding and the wire diameter. The needed area for the windings is determined from diameter and the isolation thickness. \$\endgroup\$ Commented 15 hours ago
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    \$\begingroup\$ Hi @TimWilliams, thank you for the heads up, I accidentally added the wrong link, sorry! it's corrected now (page 117, eq. 19). I am planning to design one with ferrite, for frequencies above 500 Hz up to 4-5 kHz. Power will be around 500 W. \$\endgroup\$ Commented 15 hours ago
  • \$\begingroup\$ If the windings do not fit in the selected core, you have to repeat the calculations using a larger core. Extra windings area is needed for the air gaps and isolation. If the transformer gets too hot due to the loss of the transformer, you need a larger core and thicker wires. A lot of experience in transformer design would help, but a difficult design may require more than one prototype for a solution of the design problem. \$\endgroup\$ Commented 15 hours ago

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Transformer core design without number of turns available

Your title (above) is problematic because, you cannot design a transformer core without knowing the number of turns. Forget about the secondary winding; that comes good after you design the primary and, whatever you might have learned or might have been taught, the primary winding is still just a plain ordinary inductor. So, the first part to designing a transformer is to forget that it's a transformer and, instead, design an inductor where that inductor also serves as the primary winding.

Ideally, you would want the inductance to be high so that when you apply an AC voltage to it, the current is low. Remember I said we are designing an inductor first and foremost so, I didn't refer to that current as the magnetization current (although that is what it becomes when we add a secondary and think about it as a transformer). But from now, I will call it magnetization current because it is this current that magnetizes the core and creates magnetic flux in the core.

So, what determines that flux level and, how do we convert magnetization current into flux? The starting formula is what we call magneto motive force (MMF). It equals the number of turns multiplied by the current that flows in that coil.

$$MMF = \text{current x turns} = I\cdot N$$

This is only part of the story; we need to calculate the \$H\$ field (magnetic field strength) in the prospective core and, this is MMF divided by length of the magnetic field path that circulates in the core: -

enter image description here

Image from Encyclopedia Magnetica - Magnetic path length.

$$H = \dfrac{MMF}{\ell}$$

A few words about the above. H is made smaller by having a longer magnetic field path. Smaller H leads to lower flux density and this is a good thing. However, a longer path length leads to a lower primary inductance and, this is not ideal given that we want high inductance to avoid a significant level of core saturation. So, generally, when we make the magnetic path longer we also increase the cross sectional area of the core so that the inductance isn't compromised too much.

Then we have a formula that most folk will have seen and relates to what is known as the BH curve. That curve is the amount of flux density within a core that exists due to a driven level of H-field. It relates the permeability of the core to B and H: -

$$B = \mu H = \mu\cdot\dfrac{I\cdot N}{\ell}$$

Hence, higher permeabilities and/or shorter path lengths lead to a higher peak flux density (B) for a given number of turns and magnetization current. And, for ferrite materials, we tend to work with an upper limit on B of 200 mT as a guideline.

But, it still all comes back to the number of turns on the core and how much inductance that creates. Most cores have an \$A_L\$ figure that tells you the inductance for 1 turn and, if you double the turns the inductance becomes 4 times greater hence, the working inductance of a given coil is: -

$$L = N^2\cdot A_L$$

And, this formula leads to one of the important aspects about the design of the primary winding. If you double the number of turns, the inductance rises 4 times and, for a given primary voltage, the magnetization current reduces to one-quarter. Hence, the driving force behind H-field halves and, so does peak flux density. Clearly turns have doubled but, amps have quartered and, the net effect is that MMF (ampere turns) halves.

This is why you cannot choose a core without knowing the number of turns. In other words, more turns leads to less H-field and lower flux density. All of the above put into simple bullet-point context: -

  • From the inductance and voltage you can calculate Magnetization current
  • From Magnetization current and number of turns you can calculate MMF
  • From MMF and path length you can calculate H-field
  • From H and core permeability (\$\mu\$) you can calculate flux density
  • Peak flux density should be aimed to be no-more than 200 mT for ferrite cores

If B is higher than 200 mT (or thereabouts), then increase the number of turns or, lengthen the magnetic path or, increase the core cross sectional area to get more inductance (and less magnetization current). You can also raise the operating frequency if that's an option. But here, number of turns is going to be your main turn-to best-friend.

Hopefully this might help you understand why we can't design or choose a transformer core without considering the number of turns.

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  • \$\begingroup\$ I should add that 200 mT rule is for transformers only (AC). Ferrite inductors in SMPS like buck converter can very well have 1000 mT DC offset, but the AC around that is small in comparison to the DC offset. Those would however usually have an air gap. \$\endgroup\$ Commented 15 hours ago
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    \$\begingroup\$ @juhist No, they cannot; ferrite saturates at, at most, 450mT or so. Practical inductors can run up to 300, maybe 350mT, DC. \$\endgroup\$ Commented 14 hours ago
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Starting with a formula like (changing to American symbols because I'm a boorish American; N is turns, etc.),

$$V = 4.44 F N B A_e$$

(Incidentally, 4.44 is actually \$\sqrt{2} \pi\$, the peak area under a sine wave, amplitude given as RMS. The equivalent factor for a square wave is simply 4 (so using either factor for either waveform, doesn't actually make a huge difference), and note that square wave peak = RMS so we aren't measuring the square ambiguously if we leave off the suffix.)

We can substitute out N to arrive at a bulk formula for transformer design.

I'm not going to do this, actually; rather I'll do the definitional equivalent: normalize it to 1 turn, and eliminate V and I.

To do this, we need to know a few geometric factors for the core, and we need to know how much copper we're stuffing into the thing. Which means losses are a term (copper has resistance, so its amount and power dissipation matter). And then we should be mindful of core losses, but we don't have a formula handy for this. We can just guess, and assume it's alright. (The modified Steinmetz formula can in fact be used here, to give a minimum-loss design equation in closed form. For more information, see §15.4.2, Fundamentals of Power Electronics, Erickson and Maksimovic (2001).)

Geometry matters, because the core can be long and narrow, fitting many short turns, or wide and short, fitting few but longer turns. The latter are generally preferred for pulse transformers and RF baluns, where winding length limits upper bandwidth cutoff (example: P "pot" cores, binocular cores, etc.). A balanced design is preferred for power conversion, where small size and low cost are demanded, which is more or less where the cuts and stacks of conventional laminated iron come from (especially "wasteless" E-I cores, the geometry of which (and probably explanation as well) I believe I flipped past in your reference). Toroids are also popular (geometry is flexible by way of core strip width and height versus inner diameter, and assembly is relatively cheap).

So we have the winding window \$A_W\$, a fill factor \$k\$ (how much of it we can fill with copper), and some current density in that copper. We can freely assume one turn at equal current density (this of course wouldn't happen at your frequencies for a transformer of this size, were it truly solid copper, but this assumption is validated by needing many turns -- or use litz wire). Thus, $$ I = i k A_W $$ is the total current through one turn. Note that a transformer has primary and secondary so we must halve the area and dedicate it to each winding.

We have V (for N = 1, the voltage per turn) and I, which we can multiply:

$$ VI = S = 2.22 i k A_W F B A_e $$

S is not quite power, but is the apparent power capacity (in VAs) of the transformer. When PF = 1, it is power transformed.

Cores are sometimes tabulated by area product, a sort of scaling factor. We can rearrange for that:

$$ \frac{S}{2.22 i k F B} = A_W A_e $$

Cores in a given family/series have the same aspect ratio, so it's meaningful to take the square root of this.

$$ \sqrt{ \frac{S}{2.22 i k F B} } = G A_e $$

introducing G as a geometric factor, since we don't have a family to read the Aw/Ae off of.

We might further assume k = 0.7, a typical figure for round wire and tape insulation between layers. (This can be 0.3 or less for SMPS with litz wire on bobbins, up to maybe 0.9 for carefully designed thin- or no-bobbin construction using square or flat wire.) I'll also push the coefficient to the top.

$$ \sqrt{ \frac{0.64 S}{i F B} } = G A_e $$

Further pushing around coefficients, we could move a 1/10 to the denominator, and change B from tesla to gauss, and then we have the same form as the reference. Evidently their geometric factor G is around 0.0676.

You'll have to read the context, to see what range of cores, geometry and loss density they have assumed, to arrive at such a figure. Or they might not be discussed at all, but implicit in the figures (\$i\$ and \$B_m\$), in which case you need to look up material properties to see what those figures do in turn.

Applying this for typical ferrite parameters (S = 500 VA, \$i\$ = 5 A/mm^2, F = 500Hz, B = 300mT), we need \$G A_e\$ = 653 mm^2. A typical ferrite E core (one of the few shapes available in large sizes), E65/32/27, has Ae = 540 mm^2, Aw = 394 mm^2, which seems reasonable as a start. You may need the next size up.


FYI, 500Hz is well within the range of sheet steel; if you can find a source for very fine material (under 0.1mm thickness is available), quite reasonable losses can be made, while keeping build size small (B up to 1.2T may be reasonable -- though it might also be cooking pretty good by then; look up loss curves). Grain oriented material (GOSS) is preferred.

Also, cores of this size, would normally be used as several kW at high frequency. It's entirely possible your application might benefit from a more complex solution: converting to DC, to high frequency for conversion, and back to (LF) AC. This is standard fare for inverters (regular or grid-tie), motor drivers, etc. Semiconductors are so cheap these days that production unit costs would be dominated by such a component -- assuming, of course, you need enough of them that the engineering complexity of such a design can be amortized.

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For deciding the core size you have to first decide the type of core EI or EE etc. Then from that core drawing find out the available winding window area(not the core area on which the winding is done but the area of winding length x winding width) in which the copper wire will take its place. Assuming the winding window area w sq mm. Then for copper wound transformer assuming a 2.5 Ampere per sq mm current handling capacity of the copper wire, the max current that can flow in the winding area w, is 2.5w. Now assume that there is only one turn in the whole winding area. Then max power in that area will be volts per turn x 2.5w or E/N x 2.5w = P the power you want it to handle. Now you know Pand w already hence E/ N will be known. Now use E/N = 4.44fBA where B is flux density in wb/sq m A is core area in sq m, f is frequency . The core area A will be known in sq m after putting in desired values of f and B. B is 1 for iron crgo core and frequency is 50 hz for supply line mains. You can use this at sny frequency and core provided you know the core saturation flux density and frequency it can handle without much loss. For square wave E= 4 NfBA is used and for ferrite use B=0.1 Also as the winding can not take full space due to wire cross section, a multiplier fraction k1 is taken as winding constant to get the practical winding area which will be less than w. So effective window area will be k1w. Usually k1 is taken as 0.9 for the round copper wires. Similarly for Laminated core the core area A is also taken as k2 x A whete k2 is taken as about 0.9 due to reduction in actual core area due to lamination gaps.

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In my post answer I forgot to mention that it starts with the type of core selection along with approx winding window size that will fit in the required current. Then find the core area required. Reiterate to optimise. Point is that any core area can sustain the voltage provided the number of turns are enough but thats not practical as the winding area for that core may be insufficient to fit in that number of turns. Hence the core area selection is subject to the winding area and power handling both.

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  • \$\begingroup\$ I think it would be best to edit your other answer to include this additional information, and then delete this answer. \$\endgroup\$ Commented 7 mins ago

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