What happens to inductor stored energy as saturation occurs?

[Circlotron]

Say we apply a DC voltage to an inductor and watch the current increase in a more or less linear manner. The stored energy also rises as time goes by. Then all of a sudden the inductor saturates and the inductance decreases so we take measures to cause the current to level off to a constant value. What happens to the energy that has been accumulating in the inductor? Seeing the inductance has dropped, it would appear the energy has disappeared somehow.

[Siwastaja]

Think about saturation as inability to take in any further energy, and dropping inductance (it's not like it suddenly drops to zero anyway) a consequence of this feature.

[mawyatt]

As the effective inductance levels off due to saturation the stored energy also levels off, additional inductor current then produces heating. As a simple model of an ideal inductor Leff with a series resistance Reff, these both depend on the inductor current with Leff decreasing with applied current as saturation is approached, and Reff increasing as saturation is approached.

Another way of thinking is the Inductor "Q" begins to decrease as saturation is approached, with "Q" as the inductors "ability" to store energy vs. energy dissipation, which is also a valid use of "Q" over the usual omega*L/R.

Best,

[coppice]

Say we apply a DC voltage to an inductor and watch the current increase in a more or less linear manner. The stored energy also rises as time goes by. Then all of a sudden the inductor saturates and the inductance decreases so we take measures to cause the current to level off to a constant value. What happens to the energy that has been accumulating in the inductor? Seeing the inductance has dropped, it would appear the energy has disappeared somehow.

Just think of it like any other store that fills to overflowing. The content are still there. You just can't pack any more in. Stop trying to push more energy in, and what was stored starts flowing out.

[Kleinstein]

The formula of 1/2 L * I² is only valid for a constant inductance L. When going to saturation the old energy is still there in the magnetic field it is just the increasing the current even further will only add little more energy. Usually L does not go all the way to zero, but the air part still stay there.
The power related to the current times voltage at the inductive part still gets stored.

[T3sl4co1l]

*Incremental* inductance drops. The average drops much slower; energy continues being stored, but at a slower rate for every increment of current (or in this case, faster by increment of flux..!).

Tim