# Boost Inductor Voltage calculations

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 $V_{in}$?

• 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. – Rohat Kılıç Jun 13 '17 at 4:21
• 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. – Neil_UK Jun 13 '17 at 4:56
• @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. – winny Jun 13 '17 at 5:09
• @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. – iuppiter Jun 13 '17 at 5:12
• No, air-gap gives you higher H. – Neil_UK Jun 13 '17 at 7:13

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.
• 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. – iuppiter Jun 13 '17 at 4:45
• 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? – iuppiter Jun 13 '17 at 4:55
• 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 $m^2$ ? – iuppiter Jun 13 '17 at 4:58