Inductors can be calculated in the same way, with the additional feature of gap length, which is calculated from desired inductance (in transformer design, the core's permeability is assumed high enough that the resulting inductance is not a concern for the user, i.e. This also applies to half wave converters (forward/flyback) in CCM, with double the turns, because the flux only goes one way.įor more detail and complete transformer formulas, try this:įor any given core geometry, you can develop a rough formula relating a proportional dimension (like overall size) to an overall property (like core product, mm^4) to calculate, in one step, what minimum size core is required for a particular voltage and current capacity. This applies most directly for forward converters at maximum duty cycle. This is handy when you need a large current and can only get a single turn foil winding to handle it.ġ/4.44 comes from: 1/4 because the magnetic flux goes from zero at the voltage peak, up to its maximum value at the voltage zero crossing, hence only a quarter of the wave needs to be considered 4/pi ~= 1.11 comes from the integral of the sine wave (this also shows up in choke-input rectifiers, because the average value of a full-wave rectified sine is pi/4 times the input peak) and sqrt(2) comes from the RMS to peak conversion for the sine wave.įor square waves, the factor is NOT 4.44, but 4.0 even, and the voltage is simply the applied squarewave voltage (Vrms = Vpk for a true square wave, so it doesn't matter). Alternately, if you want the area for a given number of turns, divide by turns instead to get A_e. Therefore, V / (4.44 * F) Phi_max, peak fluxī_max in tesla flux/area per turn (uV.s/mm^2.t are handy engineering units)įinally, divide by A_e of whatever core you're looking at to find number of turns. Easy to remember because 4.44 might be familiar from books on transformers, and if you keep track of the units, you can remember:
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