I have heard a lot of the time that "the core is not large enough to handle the current, and will reach saturation". What is saturation and why is it a bad thing to reach saturation?
Electronic – saturation in an inductor
inductanceinductorsaturation
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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.)
Best is to use manufacturers data.
Test in an oscillator. (See below)
Apply variable DC + AC and monitor effect on AC as DC increased.
Oscillator method - 2. Above.
Given:
A flyback converter/oscillator (eg a typical smps boost converter) operating in 'discontinuous mode'
An oscilloscope
Variable load.
Iin is a triangle wave plus an off period.
As you increase the load towards saturation the straight portion of the triangle wave will start to assume an upwards kink - ie the rate of rise of current with time will increase as you enter saturation.
Inductance will drop as the core enters saturation. More current further reduces inductance.
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Best Answer
Rawbrawb's answer doesn't explain the actual mechanism by which saturation occurs, which is a fairly easy to understand:
It helps to first understand how materials generate magnetic fields. A simple way to think of this is as each atom being a small loop of current which generates a magnetic field.
A magnetic material has huge numbers of these loops. These loops tend to align themselves into "magnetic domains", which are microscopic areas where all of the loops are in alignment. In an unmagnetized material, the directions of the domains are randomly distributed, and so there is no net magnetic field.
Applying a magnetic field to a ferromagnetic material will start to align the magnetic domains, resulting in an "induced" magnetic field from the material. Increasing the applied magnetic field will increase the amount that the magnetic domains are aligned, and so increase the induced magnetic field. This is typically very non-linear. At some point, the applied magnetic field aligns ALL of the domains, and it is no longer possible to increase the magnetic field from the material. This state is known as "saturation".