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This is probably a stupid question, but please bear with me. I'm making a boost inductor for PFC and I just wanted to make sure I'm calculating for the voltage correctly. Is the equation:

\$N = \frac{V\times10^8}{4fA_eB} \$

Or does that only apply to transformer windings?

If it does apply to inductors, should \$ V =V_{boost}\$ or rectified \$ V_{in}\$?

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  • \$\begingroup\$ Neither. V should be the voltage "across" the inductor. Anyway, I wouldnt calculate the number of turns with Faraday equation. Instead, if you have calculated the required inductance (L) then go for \$L=A_L \cdot N^2\$. Wind a few (at least 10) turns to the core, measure its inductance and calculate \$A_L\$ from the equation I gave. Then calculate required N for required L. \$\endgroup\$ – Rohat Kılıç Jun 13 '17 at 4:21
  • \$\begingroup\$ While inductance is incidentally relevant, it's needed for ripple calculations, the core volt.second_max is is vital, to avoid the inductor saturating. This cannot be measured with an \$A_L\$ estimate, only with knowledge of permitted Bpeak and \$A_e\$ as above, or by measuring inductance with varying current and seeing where it collapses. \$\endgroup\$ – Neil_UK Jun 13 '17 at 4:56
  • \$\begingroup\$ @iuppiter Please show said core with such a ridiculously low saturation current. Most cores for PFC are irin powder and work up to and beyond 1 T. \$\endgroup\$ – winny Jun 13 '17 at 5:09
  • \$\begingroup\$ @winny I was actually planning on adding an air gap to increase the reluctance in order to use a higher \$B\$ which was suggested to me by another user. I'ts that I already have a bunch of E cores on hand and I really didn't want to wait another week or spend the money to buy cool μ cores. \$\endgroup\$ – iuppiter Jun 13 '17 at 5:12
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    \$\begingroup\$ No, air-gap gives you higher H. \$\endgroup\$ – Neil_UK Jun 13 '17 at 7:13
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If your core Bfield is swinging linearly between + and - \$B\$ Tesla peak, at a frequency of \$f\$ Hz, in a core of area \$A_e\$ square metres, then the peak square wave voltage per turn is \$4 f A_e B\$ in volts. I'm not going to try to guess in what unit system your \$10^8\$ factor might be appropriate. As you see, keep it all in SI and it tends to be a bit easier.

Obviously you can invert that to get a Bfield slew rate, when a certain volts/turn is applied to the inductor. This is the way round to use the relation when you are figuring out how much the field slews when you connect Vin or Vboost to your inductor for a certain time.

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  • \$\begingroup\$ I found that equation in a manufacturer datasheer where \$B\$ is in gauss (which I found weird) and \$A_e\$ is measured in \$ cm^2\$(which I also found weird). I'm guessing that's where the \$10^8\$ factor comes from. \$\endgroup\$ – iuppiter Jun 13 '17 at 4:45
  • \$\begingroup\$ You'll notice that in the answer, all I did was nail the units and measures of all the terms you used. I hate conversion factors. I hate equations without defined terms. I hate it when equations only hold up for certain waveforms, or certain measures like peak, rms, or peak to peak without spelling out which. \$\endgroup\$ – Neil_UK Jun 13 '17 at 4:54
  • \$\begingroup\$ it seems to work for transformer windings where \$N = \frac{400V\times10^8}{4\times 100kHz\times 1.78cm^2 \times 1800 gauss}\$ returned 31.21098627 turns. Does that sound right? \$\endgroup\$ – iuppiter Jun 13 '17 at 4:55
  • \$\begingroup\$ In the datasheet the equation was actually for \$B_{max}\$, I just turned it around. Can I get rid of that stupid \$10^8\$ and just use teslas and \$m^2\$ ? \$\endgroup\$ – iuppiter Jun 13 '17 at 4:58
  • \$\begingroup\$ @iuppiter who cares whether the equation in stupid units is right? Do it in furlongs, fortnights and hundredweights for all I, or anybody else in their right mind, would care. Use SI. If you care to repeat the calculation using square metres for Ae, and Tesla for peak field, then I might be bothered to comment. \$\endgroup\$ – Neil_UK Jun 13 '17 at 5:00

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