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I am building an RL series circuit with a toroidal inductor that I am winding on my own. The toroid is made of steel (low carbon, I believe, but not 100% sure) and has about 280 turns of 22 gauge wire. The inner diameter of the toroid is 4.25 in. and the outer diameter is about 5.25 in. The resistor is 0.6 ohms. The circuit is diagrammed below.

schematic

simulate this circuit – Schematic created using CircuitLab

I know that according to Maxwell's equations, $$\mathbf{B}=\mu(\mathbf{H} + \mathbf{M})$$ where B is the magnetic flux density in Teslas, \$\mu\$ is the permeability of the material, H is the magnetic field strength in Henries/meter, and M is the magnetization of the material in Henries/meter. Note that $$\mu = \mu_0\mu_r$$ where \$\mu_0\$ is the permeability of free space and \$\mu_r\$ is the relative permeability of steel, which I take to be about 1000 at DC. I also know that, in ferromagnetic materials, permeability is a function of the frequency of the input signal as well as of the magnitude of H (link 1, link 2).

A few questions:

  1. How can I calculate/experimentally determine the minimum voltage/current with which I need to drive the circuit in order to saturate the magnetic flux density inside the toroid? I want to be able to run an AC signal (< 100 Hz) through my circuit for several minutes such that the magnetic field reaches saturation but without causing the circuit to overheat.
  2. Is it possible to apply such a high voltage/current that the relative permeability decays to 1? If so, does this mean that at very high voltages, the field wouldn't saturate since the permeability is too low?
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  • \$\begingroup\$ Are you talking about silicon steel, used in transformers, or just plain steel? \$\endgroup\$ – Sparky256 Mar 11 '18 at 23:31
  • \$\begingroup\$ Just plain steel! I can have the exact specs for you tomorrow morning (in about 13 hours). \$\endgroup\$ – Vivek Subramanian Mar 11 '18 at 23:36
  • \$\begingroup\$ Would be something like this: mcmaster.com/#wire-rope-links/… \$\endgroup\$ – Vivek Subramanian Mar 11 '18 at 23:39
  • \$\begingroup\$ Silicon steel, normally laminated with lacquer in transformers, is very had to machine or even drill, yet it makes a very strong wire for suspension bridges. If it is easy to drill, it is low-carbon steel 1018, or something like it. \$\endgroup\$ – Sparky256 Mar 11 '18 at 23:48
  • \$\begingroup\$ magnetic field reaches saturation but without causing the circuit to overheat That is slicing it thin, with contradictory results. They are not exclusive events. \$\endgroup\$ – Sparky256 Mar 11 '18 at 23:50
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1.

I made an inductor tester when I was designing large coils for an arc welder PFC stage. It consisted of a large capacitor=C1 that charges slowly through a resistor R1.

A large IGBT (shown as a switch, SW1) was connected in such a way to switch ON when a button was pressed. This IGBT would connect the capacitor to the unknown inductor.

A current transformer (AM2) was used to measure the current through the inductor and this current was plotted on an oscilloscope.

As we know, the equation for an inductor is V = L*dI/dt Where L is the value of the inductor, V is the voltage across the inductor, and dI/dt is the rate of change of the current through the inductor. (Ignoring resistance.)

schematic

simulate this circuit – Schematic created using CircuitLab

In this tester the voltage is essentially constant -- the capacitor is large enough so that it does not appreciably discharge during the test.

This means that the product of L*dI/dt should be a constant. But as you know, the inductance will decrease when the current rises in an inductor.

This is observed on the oscilloscope as a linear ramp L*(dI/dt) when the test begins. As the inductance goes DOWN, the slope of the current (dI/dt) increases.

The point where the slope of the line goes non-linear is the point where the inductor begins to saturate.

You can measure the inductor value by picking two linear values and inserting it into the equation, L = V*dt/di.

2.

I believe you are defining core saturation. The core will definitely be saturated. It's not the voltage that causes the core to saturate. It's the level of current in the inductor, analogously the intensity of magnetic flux.

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  • \$\begingroup\$ If you have a bench supply with enough current to saturate the core, then this works fine by just plugging the banana plug in quickly (for SW1) and having your scope across the clip-leads resistance to sense the series current. Do have a flyback diode across the coil for when you unplug it - a 5" toroid packs a few joules and recent power supplies from CN are not so robust. \$\endgroup\$ – Henry Crun Mar 12 '18 at 1:15
  • \$\begingroup\$ @HenryCrun I like your inductance meter solution. I tend to forget about saturable reactor circuits. They were used a lot before transistors were around. (And still useful today) \$\endgroup\$ – Martin Klingensmith Mar 12 '18 at 12:56
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Here's one approach, based on the old DC controlled variable reactance technique. The point of this technique is to let you use your DC bench supply to saturate the core, while only needing some low power AC method to measure the inductance/reactance.

Put two separate, non-overlapping windings (with the same number of turns) on your core. e.g each covering 1/4 or the circumference. Connect them anti-phase in series, to your DC variable power supply. Because these two windings are anti-phase, AC will be cancelled out. Because the two windings are physically separate parts of the core, the DC will saturate the core region under the winding (making a virtual airgap).

enter image description here You only need enough turns, so that your DC current can saturate the core. If you use 5 turns on each winding, and 10A saturates the core, then a 280 turn coil will saturate at 5*10/280Amps

Now you can put another winding/s on that you use to measure the AC inductance/reactance/transformer coupling with. You can measure the inductance of this winding with any other approach e.g inductance meter. This AC winding doesn't need any particular number of turns, it just needs to be measurable for you.

You will increase the DC current from your power supply until the inductance drops by what you define saturation as (eg to 1/2). You are of course measuring DC saturating current.

If it was a pure AC current it will be peaking at 1.4x the RMS value. So you might say that the AC.sat.RMS = DC.sat/sqrt(2). i.e if DCsat=10A, AC=7.1Arms. However saturation is not an absolute hard edge: it remains up to you to define how much saturation is the limit. When you are using AC most of the waveform is below saturation, so you might say that 9Arms is OK for you.

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  • \$\begingroup\$ Thanks, @Henry, but could you please clarify your last paragraph? Specifically, how can I determine for a given AC frequency what current would saturate the core if my input signal is DC? \$\endgroup\$ – Vivek Subramanian Mar 12 '18 at 18:43

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