# What would be the consequence of the inductor saturating?

The following is a section from a book where it is talking about decoupling by LRC at supply terminals. I highlighted the sentence where I didnt understand:

I read that if an inductor has an iron core, increasing the current after a point saturates the magnetic flux. Thats what I know about sauration of an inductor with an iron core.

But what is this highlight talking about? Is that assuming that the inductor has an iron core? And most importantly what would happen if the inductor saturates in that context?

• Any core except for air/vacuum will saturate at some point. A bit over simplified you could say your inductance drops to zero at this point. If you have both a DC and AC component, you must compute the peak of the sum of the two. – winny Jan 9 '19 at 18:21
• Oh so inductance becomes zero after the saturation point(after some DC CURRENT value)? Thats the point of the text I think then. Thanks – pnatk Jan 9 '19 at 18:27
• @panicattack Not quite zero. Instead, it's probably better to imagine that the core transitions (saturation usually isn't a sudden event) so as to appear to have a vacuum or "air" core. (Not so unlike the idea, and I'm not suggesting a reality here -- just a thought experiment, of a capacitor's dielectric gradually appearing more like a vacuum or air dielectric as the capacitor charges up.) – jonk Jan 9 '19 at 18:38
• Jonk is correct. Hence oversimplification. But not due to DC current value, peak current! DC plus peak of the AC. – winny Jan 9 '19 at 19:01
• What Jonk said. I haven't seen it in this context, but it's pretty apparent when you make a boo-boo in a switching regulator (or amplifier) design. Inductor current that should be climbing steadily over time due to applied voltage will instead climb steadily for a while, and then shoot up. You usually know you've hit this operating point by the loud "pop" sound of transistors shattering, or smoke, or one of the other many non-standard physical effects that make power electronics fun. – TimWescott Jan 10 '19 at 1:11

You are showing a picture with a frequency response, but this sort of problem cannot be analyzed like this, anymore. An .AC analysis linearizes the circuit, but a saturating inductor is anything but. So, the only way to see it is to simulate it in time-domain. Or on the breadboard, be sure to bring protective gear...
Here's an attempt in LTspice using the nonlinear Chan core, with some bogus values (for exemplification, only). The first picture shows what an .AC analysis would look like:
Which shows a nice, smooth response, two docile poles. If you switch over to a .TRAN analysis, and make it a frequency sweep (from fmin to fmax) with a high enough amplitude (see vhigh and vlow):
Below is the whole simulation, and the above three show details in three sections around the beginning, middle, and end. You can clearly see that the beginning (3rd row) and the end (1st row) are low in distortions, but the middle is not only distorted, but also high valued. As @winny and @jonk say, in simplistic terms, the inductance drops, hence the current rises, but relative to the waveform shape. And the transition is related to the hysteresis of the core. You can think of it in terms of a tanh() curve: the slope around zero is smooth and quasi-linear, but when you get to saturation, the value along the x-axis (magnetic field strength) rises very abruptly with the slightest y-axis (magnetic flux density).