I have a 15µH 1.6A inductor, this one.
The operating current is 1.3A, but the saturation current is 1.6A.
- What is the difference between these?
- Will the inductor work properly at less than 1.3A or less than 1.6A?
Electronic – How important is saturation current for an inductor
inductor
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N87 is straight-up ferrite material, not distributed air gap like powder iron types of material. Just because it's in toroidal form doesn't mean it's distributed-gap material - N87 in a toroid will saturate the same way as N87 in an E-core. There's nothing wrong with using straight-up ferrite for a boost inductor, so long as you gap it (more on this later). The fact that it's in toroidal form means you can't gap it. You might want to switch to Kool-Mu if you want to stick with a toroidal form factor.
Deriving the core inductance from the \$A_L\$ factor is reasonably good, so long as you have sufficient turns on the core and can keep the winding reasonably uniform and in as few layers as possible. Bear in mind that \$A_L\$ can vary by +/- 25% on some cores.
Boost inductors carry both the magnetizing current and energy for the load (that will be stored magnetically and delivered during the off-time.) Once the converter starts operating in continuous conduction mode (when the inductor current never goes to zero), it's even worse since you start operating on a B-H curve that doesn't reset to zero. (Bmax is still Bmax, but you now have a DC offset on which Bpeak is riding.) These are the reasons that the inductor needs air gapping - the core won't be able to handle any significant DC current without saturation otherwise.
I'm not sure I understand your test circuit. Both ends of the inductor are essentially clamped to 5V, meaning that the two capacitors (C1 and C2) contribute nothing to the simulation. If your real boost converter is arranged in this manner, it isn't a boost converter and will never work. L1 needs to release its stored energy through D1 to the load, which can never happen when D1 and the load is connected as shown. The only connection between input and output must be through L1 and D1. I would also put R1 in the source of Q1 and do a single ground-referenced measurement instead of a mathematical construction. (L1 will only saturate when Q1 is on, so measuring it when Q1 is off is irrelevant.)
Saturation current only applies to non-air cored inductors or transformers. Basically the core material (iron or ferrite) can no longer linearly increase the amount of magnetic flux with an incremental increase in current.
Inductance is defined as the amount of flux produced per amp so for twice the current you'd expect twice the flux but if this ratio starts to reduce (due to saturation) the inductance is perceived to reduce and this can cause (in some circuits) an increase in current which lowers the inductance even further leading to a rapid hard-saturation situation.
Well-known is the B-H curve and this is a good measure of where saturation occurs: -
This is a rather genralized picture - a lot of inductors are reasonably linear then have a marked flattening at the top and bottom of the curve.
wiki has a useful article about it.
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Best Answer
What they're telling you with the saturation current is that above 1.6A, the core can't take any more magnetic flux, and this translates into a change in the effective inductance when the current gets into that realm. You can put more current through the winding, and you do get more flux as a result, but the effective permeability of the core drops off.
The effect won't be a sharp break, either; you would start to see some of the effects as you get near the saturation point. If you apply a pure sine wave voltage across the coil that would result in less than 1.3A peaks, the current will also be sinusoidal, but if you run the voltage up to where the peak currents would reach saturation, the current waveform will have a significant harmonic component.