# Does an Inductor Behave Like a Short While Saturated?

#### JackieTee

Joined May 28, 2019
11
Two different professors told me that an inductor will behave like a simple wire if it has been fed a DC current for enough time. That is, under ideal circumstances, you will see no voltage drop across it given enough time connected to a DC source.

Is this behavior what happens when the core of an inductor becomes saturated? Does your inductor behave like a simple wire when its core becomes saturated?

When I search online about inductor saturation all I get is B-H curves and discussions regarding slew rates. I have yet to see someone plainly connect an inductor's saturation to how it behaves in a simple RL circuit.

Thanks guys!

#### crutschow

Joined Mar 14, 2008
28,198
an inductor will behave like a simple wire if it has been fed a DC current for enough time
You mean DC voltage.
The current follows from the voltage.
Does your inductor behave like a simple wire when its core becomes saturated?
Yes.
When an inductor saturates, the inductance drops towards zero, so it basically starts behaving as a resistor.

#### JackieTee

Joined May 28, 2019
11
You mean DC voltage.
The current follows from the voltage.
Yes.
When an inductor saturates, the inductance drops towards zero, so it basically starts behaving as a resistor.
Thanks, Crutschow.
I don't know why this isn't explicitly stated in any of my textbooks. And, to clarify for my own knowledge sake. Under ideal circumstances, the inductor would have 0 resistance after reaching saturation. Under real circumstances, the inductor would behave like a resistor. If I were to guess, it would behave like a resistor having the same value as its DCR from its datasheet.

And yeah, I meant DC voltage. DC current is redundant - "direct current current".

#### wayneh

Joined Sep 9, 2010
17,155
One way to think of it is that, once no more energy can be put into the field, the current can pass without losing additional energy. Just like a wire with a DC resistance but no inductance. Of course the field energy is still there when the current tries to collapse.

#### JackieTee

Joined May 28, 2019
11
One way to think of it is that, once no more energy can be put into the field, the current can pass without losing additional energy. Just like a wire with a DC resistance but no inductance. Of course the field energy is still there when the current tries to collapse.
Ah, beautiful. Perfect and as I suspected! I just needed someone to come out and tell me I am not crazy. I am only 30 days into my first internship and very early into my time at university. The engineers around me didn't know the answer to this for some reason and my textbooks didn't explicitly state this relationship.

The only thing that made me suspect this was the mathematics behind the step response of an RL circuit and an application sheet saying that allowing your inductor to saturate could cause too much current to pass through it which would damage the MOSFET being used (this was an application page covering the design and theory behind a buck converter).

Thanks, guys!

#### AlbertHall

Joined Jun 4, 2014
11,599
Even if the core does not saturate applying a fixed (low) DC voltage across an inductor the current will rise - I = V / R - and after that the current will stay constant and the voltage across the inductor will be just the resistive voltage drop.

#### BobTPH

Joined Jun 5, 2013
4,038
Yes, the first two answers missed the point. An inductor develops a voltage that is proportional to the change in current across it. If you apply a DC voltage directly across the inductor, the current will start at zero and rise according to the equation:

dI/dt = V / L

For an ideal inductor (no resistance) and power supply (no resistance, able to provide infinite current) it will continue rising forever.

For a real inductor, the current will be limited by the resistance of the entire circuit (parasitic resistance of the inductor, series resistance of the power supply and resistance of the wires). Once that current is reached, there is no more change in current and no more induced voltage across the inductor. Since the current can no longer change, it is no longer acting like an inductor.

Bob

#### BR-549

Joined Sep 22, 2013
4,938
In most of today's circuits, we do not want saturation to occur. So, what you find is ways to avoid it.

But in earlier times, saturation was used both for power control and signal amplification.

Search for magnetic amplifiers. And reactance control. It's old, but it's a great technology.

Large and small amounts of current can be controlled with a coil.

And magnetic amplifiers were the first class D amps, decades before the transistor.

#### KeepItSimpleStupid

Joined Mar 4, 2014
5,090
from #7. V=Ldi/dt for an ideal inductor.

Add Rin series and make it a closer to a REAL inductor, we have Ldi/dy+I*R as the voltage across the component.

At DC, di/dt =0, thus the REAL inductor acts like a resistor.

Always watch parasitics. They are part of the real world.

Watch ideals: Power(primary)=Power(secondary) for a transformer.

A resistor is not a pure resistor when the frequency goes up. Three 1K resistors in series might not be 3K.