Electronic – Properties of inductors with a magnetic core

Yes, a inductor sortof resists current changes, just like a capacitor resists voltage changes. In fact, inductors and capacitors are current/voltage mirrors of each other. The way I like to think of inductors in circuits is that they give inertia to current. They don't of course, but it seems a useful conceptualization technique.

In the schematic without the diode, if everything starts out at 0 and the switch is closed, the current will be a exponential decay toward Vs/R. Initially all the voltage is accross the inductor, and in the steady state there is 0 voltage accross it.

The interesting stuff happens when the switch is opened. At any one instance, the inductor will maintain its current constant. This includes the instance the switch is opened. Without the diode, there is no obvious path for the current. The inductor voltage will increase to whatever maintains the current thru it.

A mechanical switch works by touching together two conductors. When the switch opens, the conductors move away from each other. This can't happen instantly, so when the switch first tries to stop the current thru it, the contacts will be very close together. It won't take much voltage to cause arc over. Once the arc is started, the gas between the contacts becomes a plasma, which has high conductivity. The arc can therefore continue for a while as the contacts move farther apart. During this time, the voltage accross the switch isn't zero, so the inductor current decreases. As the contacts move further apart, the arc voltage increases, decreasing the inductor current more rapidly.

Eventually the current is low enough that it can't sustain the arc and the switch finally opens for real. At that point there is little energy left in the inductor. The only place for that current to go is onto the inevitable parasitic capacitance accross the inductor and other parts of the circuit. Every two conductors in the universe have some non-zero capacitance between them. This capacitance is small, and therefore the voltage will rise quickly. This also decreases the current in the inductor rapidly. Eventually a peak is reached where the voltage on the capacitance actually starts to push the inductor current the other way. In a perfect system, all the energy on the capacitance would be transferred to the inductor as current, but this time in the opposite direction. Then it would charge up the capacitance again in the opposite direction, and the whole cycle would repeat indefinitely. In the real world there is some loss, so each swing back and forth will be a little lower in amplitude as energy is lost as it is being sloshed back and forth between the inductor and the capacitance. Voltage plotted as a function of time (as a oscilloscope does) will show a sine wave with amplitude decaying exponentially towards Vs.

You can't tell by visual inspection, that's for sure because some of them are lacquered/painted and even those that aren't all tend to look dark-grey. What you are asking is really tricky to fathom because there are so many characteristics that look the same between two ferrites at one frequency but are vastly different at another. If you are still interested I'll try and say what I'd do (what I'd really do is throw all my unboxed/unmarked ferrites in the trash and buy some more).

I'd consider winding (say) 5 equally spaced turns and putting the coil in a circuit to see what its inductance was - maybe a colpitts oscillator with a few caps that can be switched in and out. Maybe even make a band-pass filter from it and see where it resonates if you have a signal generator.

First type of result this will tell you is the inductance of the wound core. Then using the squared relationship between turns and inductance you can deduce its "effective permeability". This should enable you to narrow down the type of core to a range of possibilities.

You need to be be avoiding "test frequencies" significantly above 100kHz and preferably more like 10kHz - this is to reduce parasitic capacitance giving you errors.

OK so far, you might have determined the approximate "effective permeability" of the core BUT there are plenty of suppliers toting vastly different materials that you'd have to read through to try and identify the part so I'd next consider seeing how the indctance varied with temperature.

You don't need to test over a vast range, maybe just 25ºC to 50ºC would give you a decent shot at trying to uncover the ferrite. Use the oscillator/filter idea mentioned earlier and a controlled temperature - almost certainly the inductance will rise with temperature although there are a small percentage that will stay stable or fall but this will give you another tell-tale characteristic of the ferrite.

So now you have effective permeability and some idea what its temperature characteristic looks like. Scanning through various supplier's websites might narrow down the ferrite to maybe five or ten types.

It's going to be a long process this way and you may never uncover what it is that is sitting in your junk box. I suppose if your effective permeability is low it's likely to be either very temperature stable (i.e. good for filters up to (say) 1MHz) or it could have very low losses up to over 50MHz. The temperature test that indicated hardly any change in inductance across 25ºC might tell you its a material like Ferroxcube's 3D3: -

Also shown is 3C90 for comparison. 3D3 has a flat curve of inductance/permeability against temperature; probably changing something like 5% in a 25ºC change around ambient. 3C90 probably changes about 20%. It also has a much higher permeabilty. I'd recognize these two ferrites from their characteristics!

I think I've definitely convinced myself to throw all unrecognizable ferrites in the bin.

Bottom line - if you have a target circuit try it.

EDIT Also, here's is a question/answer on EE stack exchange that might also be useful or provoke some other ideas.