I need to design some \$3.7 mH\$ coils for two low-pass filters and a \$5\mu H\$ coil for a boost converter. But am having some trouble finding suitable ferrite cores, since there will be a significant amount of current running thru all of them. So I thought I'd ask here for some guidelines about calculating and choosing the right ferrite cores for my applications.

First, I am aware of the equation: $$B=\mu_0\times\mu\times\frac{N\times I}{l_e}$$ where \$B\$ is magnetic field strength, \$N\$ is number of turnes, \$I\$ is current thru the coil and \$l_e\$ is effective length of the core.

However I was suggested by an employee at one of the company that produces ferrite cores, that whilst this is a valid equation to use, for high flux densities I am better of using one of these equations: $$U_{rms-sine}:=B_{max}\times A_e\times N\times f\times\pi\times\frac{2}{\sqrt{2}}$$ $$U_{rms-square}:=B_{max}\times A_e\times N\times f\times\pi\times 2$$

Where \$B_{max}\$ is typically used around 300mT, \$U\$ is the applied voltage and \$f\$ is the frequncy of the applied voltage. And I should calculate for \$A_e\$, for effective core size.

From this on I've lost contact with him, so am kinda lost. While I trust him, about these equations, I am also unsure (shouldn't they depend on current?) on how to approach this problem, and solve it.

Here are some numbers with what I am working.

Low-pass filter:

  • current thru the coil maximum at around 7.5A
  • voltages across the coil will be from 12V to maximum around 27V (square wave will be applied to the filter)

Boost converter:

  • input voltage of maximum 15V (square wave)
  • input current of maximum 15A

These numbers already include some margin.

Also if I will get some large values for the core size can I connect more of them together?

How, parallel or serial? By first equation I should connect them in series (lesser turns=lesser B), but by second/third I should connect them in parallel (more turns = smaller Ae), so which is it?


Low-pass filters are in a H-bridge, for class D amplifier. And boost converter is what will be powering it.


Frequency of the square wave applied to the low-pass filters will be around 20kHz. And frequency of the square wave produced by boost converter will be somewhere between 100kHz and 500kHz.

  • \$\begingroup\$ The ferrite manufacturer should give you B/H curves, you shouldn't need to resort to the full equations above. From the curves you can find Imax for your core, divide by the number of turns if you're winding it yourself, and you're done. The real question is what is Imax in your circuit, and for that you probably want to simulate it. \$\endgroup\$
    – tomnexus
    Jun 23, 2015 at 18:43
  • \$\begingroup\$ @tomnexus The only thing I could read from the B/H curves are B/H values when the core enters saturation and its no longer usefull. Plugging Imax,ui,N and le into the first equation I get B and I should choose such values that the calculated B is lower than Bsaturation. And since such a core would be big/expensive I though I could use more cores. But here is the problem, by first equation I should put the cores in series(lower N means lower B, and should get it below Bsaturation), but by second/third, I should put them in parallel(bigger N means lower Ae) and this seems contradictory to me... \$\endgroup\$
    – Golaž
    Jun 24, 2015 at 11:24

2 Answers 2


Lets start with a few basics here

$$V = N \dfrac{d}{dt} \Phi = N \cdot A_e \dfrac{d}{dt}B$$

Here \$ B \$ is flux density and for a typical ferrite running at 20kHz I would personally be designing for a flux density swing of around \$200 \text{ mT}\$

Assuming the input voltage is is 27 volts peak to peak maximum, 50% duty cycle @ 20kHz \$ dt = 25 \mu s\$

We now know everything except \$A_e\$ and \$N\$ so we can rearrange the above equation to find \$ N \cdot A_e\$

$$N \cdot A_e = \dfrac{V \cdot dt}{dB} = \dfrac{27 \cdot 25 \times 10^{-6}}{0.2} = 3.38 \times 10^{-3}$$

Note that this product is using Area in SI units (\$m^2\$) and the data sheet probably gives \$A_e\$ in \$\text{mm}^2\$ so we want \$3380000\text{mm}^2\$

In theory any combination of \$N \cdot A_e\$ will work but we note the more turns we have the more area is needed for the windings and the size of the wire we need is set by the current it needs to carry. A typical figure for wire size is to assume a maximum current density of \$ J= 4.5 \cdot \dfrac{\text{amp}}{\text{mm}^2}\$

Given our current of 7.5 amps this requires a wire diameter of \$ \sqrt{\dfrac{4 \cdot I}{J \cdot \pi}}\approx 1.5\text{mm}\$

So we can have a core with a lot of turns and small \$A_e\$ or few turns and big \$A_e\$ the most economic design will be the smallest core into which the turns will fit and if we assume that the wire packs poorly the area required for a single turn is \$A_t = (1.5 \text{mm})^2 = 2.25 \text{mm}^2\$

We can thus work out the area required for the winding \$ A_w = N \cdot A_t \Rightarrow N = \dfrac{A_w}{A_t}\$

Substituting this into the equation above

$$N \cdot A_e = \dfrac{A_w \cdot A_e}{A_t}=\dfrac{V \cdot dt}{dB} \Rightarrow A_w \cdot A_e = \dfrac{A_t\cdot V \cdot dt}{dB} = 7605000 \text{mm}^4$$

Where \$A_e\$ is the core effective area and \$A_w\$ is the available winding area taking into account the bobbin.

Now the coil you have designed will have much more inductance than you want so you will need to introduce an air gap. I usually don't try to calculate this but ask my winding house to gap the core, in the center leg to give me the inductance I want.

The boost converter can be designed in a similar way but I'd ask further advice as at such high frequencies skin and proximity effects are significant.

  • \$\begingroup\$ Very detailed, thank you very much. Just out of curiosity where do you buy the cores from? Do they design the cores even for small quantaties, like 2...? What do you think will be the price for a core like this? \$\endgroup\$
    – Golaž
    Jun 25, 2015 at 12:27
  • \$\begingroup\$ @Golaž Most people use standard sized cores companies like ferroxcube make a wide range. \$\endgroup\$ Jun 26, 2015 at 12:17

To select a suitable core you should refer to the manufacturer data. Some of them publish very extensive datasheets and application notes.

For example, some time ago I came across this very good FERROXCUBE soft ferrites and accessories databook. Inside it you'll find extensive data on both ferrite materials and ferrite cores, with application hints and formulas. Probably not a step-by-step guide to spec an inductor, but most of the data is there.

For example, this is an excerpt of the section on 3C90 material:

enter image description here

and this is an excerpt of the section on a specific toroidal core type:

enter image description here

  • \$\begingroup\$ I am aware of these graphs and information. But this still doesnt help me in calculating the needed core size. I am lost in the equations I posted. \$\endgroup\$
    – Golaž
    Jun 24, 2015 at 11:16

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