Inductor charging - stupid question

[fxeconomist]

Sorry for this guys, but I really need to know...

As I understand, an inductor acts as an open circuit while it charges.
Now, as the current doesn`t actualy flow thru it while charging, can it actually charge while the circuit is open ?

I am looking to pulse current for less than it is the full cycle, but pulse it with a single wire from a PWM pulser. So on one side of the inductor the wires connect from both inductor ends to the load, on the other side of the inductor one wire would go to the pulser. We pulse for less time than it takes the inductor to charge - and since it acts as if the circuit is open, it should not care if it really is. On the end of the pulse inductor should deliver full inductive kickback to load while not endangering the pulser. Can this work ?

[bistander]

...
an inductor acts as an open circuit while it charges.
...

Incorrect premise.
bi

[fxeconomist]

Mm, so disappointed. I was hoping it works this way, cause I don't know how to protect the PWM pulser from the spike while delivering the spike to the load.

[tinman]

Mm, so disappointed. I was hoping it works this way, cause I don't know how to protect the PWM pulser from the spike while delivering the spike to the load.

When your PWM switches off (becomes open circuit), the current flowing through the inductor will continue to flow in the same direction, but the voltage across the inductor will invert. If you run a wire from the positive output of your PWM to the negative side of your load, and then series connect a diode between the negative side of your pwm output, to the positive side of your load, the load will clamp the inductive kickback spike, and will keep your pwm safe, provided your load is sufficient to do so, and you place the diode in the correct flow position.

As I understand, an inductor acts as an open circuit while it charges.

Yes and no.
This is a tricky one, but the overall result is that it does take energy to !charge! an inductor.
By charge, i am guessing you mean-to build the magnetic field around the inductor.
So here is where it gets tricky. It does not take energy to build the magnetic field, but energy is transformed into heat due to ohmic losses in the inductors windings resistance, and a small portion to eddy currents, due to the changing magnetic field. It is actually the building magnetic field that slows the buildup of current flow through the inductor (CEMF). If there was no changing magnetic field, the current flow through the inductor would reach it's maximum value instantly. It also takes no energy to maintain a magnetic field. All constant energy flowing through the inductor is transformed into heat, through the same ohmic losses.

Hope that helps.

[fxeconomist]

Hope that helps.

Thanks! But given the difficulty of all of this, I decided for a change. After I made the unipolar generator and managed to read the amps in mid june, I've had a lot of troubles and misfortunes. Installing a coil in the middle of the generator is really something difficult. So I decided to alter the experiment, and instead of a coil, I will use a capacitor. It won't be as good, it will not give the voltage I wanted but much less, and filtering with diodes will not be possible, but perhaps I'll get to see something on the oscilloscope - if the dielectric induction synergizes with the unipolar induction.