T
Thanks, and this demonstrates how inductance is proportional to permeability. As we know, ferrites have permeability which varies enormously with temperature. Also, with some ferrites, permeability also varies with the Ampere.turns. This is a nightmare for people making resonant power supplies and who need to pretty exactly set a certain LC resonant frequency when the L is all over the place.L = N^2 Uo.Ur Ae / lg
Thanks, I am not sure its like in page 23, because our case doesn’t have the turns on the centre leg. With our case, its more like two overlapping , square-ish torroids. I don’t think the Rel1 + Rel2/2 equation applies in our case?Draw its equivalent magnetic circuit (which is like at page 23) and find all reluctances.
L=N^2/total reluctance
total reluctance=R1+Rg(if any)+R2/2 (assumed R3=R2, otherwise, R3//R2)
**broken link removed**
Thanks, in our case, we don’t have an air gap. We cant have an air gap and our ferrite surfaces are lapped and tightly clamped together. We can’t have an air gap because we only have one primary turn, and any air gap would make the inductance too low. We would then get too much magnetising current in our transformer…..(because there will also actually be a secondary, but that can have more turns than one)You simply set the inductance with the air gap, and ensure the B = 0.25T at peak I. Same for any gapped choke.
Thanks, and this demonstrates how inductance is proportional to permeability. As we know, ferrites have permeability which varies enormously with temperature. Also, with some ferrites, permeability also varies with the Ampere.turns. This is a nightmare for people making resonant power supplies and who need to pretty exactly set a certain LC resonant frequency when the L is all over the place.
Thanks, then we’ve had it, because we only have a single turn primary and we simply can’t have any air gap no matter how small, because we need to get our primary magnetising inductance up high because we only have one turn on the primary……the primary current in one of our systems for example is fixed at 1.9A RMS………..(it’s a 50khz sine). If most of this 1.9A current is magnetising current, then we aren’t going to have any “primary referred secondary current”……and then we won’t be able to get anywhere near enough secondary current for our LED luminaires…The simple answer is, no you can't. Inductance variation is one point, but first of all you don't get reasonable I²L amount without a gap. Energy is stored in the air gap, not the ferrite.
If you are using a gapped of-the-shelf core, you'll take the datasheet Al value, it's o.k. for 50 KhzThanks, the other point is that we need to know the inductance at 50kHz. -Since our inductors will have a 50kHz sine current in them.
Well, that it was what it looked like from your picture. Otherwise, put it on the other leg and recalculate total reluctance for that leg.Thanks, I am not sure its like in page 23, because our case doesn’t have the turns on the centre leg. With our case, its more like two overlapping , square-ish torroids. I don’t think the Rel1 + Rel2/2 equation applies in our case?
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