Wednesday, 28 September 2016

switch mode power supply - What is the current limit/saturation of series inductors?


I am currently trying to increase a 12Vdc 30A (from a computer power suppy) to 24Vdc 12A. The circuit I have used is based on a application note for the NCP3063 boost circuit with an external mosfet, but I am having problems trying to find a large inductor which is capable of handling the average inductor current of 25.7A.


So, to increase the current through my inductor, could I simply put two inductors in series? If they both had a saturation point of about 10A, would that then give me twice the inductance and twice the current limit?



Answer



Two inductors in series would still saturate at 10A and also you'd have twice the inductance. Two inductors in parallel would share the current reasonably well but now the net inductance has halved because they are in parallel.



This would mean you choose inductors that are twice the inductance and this has a net benefit because, for a given ferrite size/mass, getting twice the inductance only needs 1.414 (sq root of 2) times the number of turns and flux saturation is driven by amperes x turns.


So you started with 1 x 10uH inductor saturating at 10A and you end up with two parallel inductors of 14.14uH each saturating at 7.07A but because they are in parallel the net saturation current is 14.14 amps.


Small print: Assumes that the two inductors are well-matched i.e. reasonably share the load current and that they don't leak enough flux to each other to make calculations more complicated.


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