Cross-pollination thread: Ren's "On inductive discharge..."

[MileHigh]

Let's do some bridge building!

The setting:

http://www.energeticforum.com/renewable-energy/7848-inductive-discharge.html

The skinny:

So, how is it that two identical current draws can charge a capacitor to two different levels?

The answer:

Please join here and we can figure it out.

How about starting off with a simpler question:

Why does the voltage on the capacitor stop climbing when you charge it up with a Bedini circuit?

MileHigh

[MileHigh]

In a nutshell, the capacitor stops charging because the coil discharges 100% of its stored energy through the transistor instead.

As the capacitor charges up and the voltage on it gets higher and higher, that means that the potential on the transistor collector gets higher and higher.

When the coil does a discharge, it's pushing current out through the diode and then into the capacitor.  So if the capacitor is at 100 volts, during the coil discharge the diode is at 100.6 volts.

Since the transistor looks like an increasing resistance during the shut-off phase, the 100.6 volts that the transistor sees at the collector will push current through the transistor and out the emitter as it switches off.

Eventually the voltage on the capacitor gets so high that it stops charging and all of the current from the discharging inductor gets routed through the transistor while it is switching off.

The spec sheet of a transistor will tell you what average power dissipation and pulse power dissipation it can sustain without damage.

So it's not surprising that two different Bedini motors with the same average current consumption will charge the same capacitor to different voltages.

The final capacitor voltage is dependent on a few factors, the most important being the inductance of the coil, the current in the coil just before the switching, the switch-off time and the "equivalent resistance waveform" of the specific transistor(s) as it switches off.

The switch-off time is dependent on the coil geometry, the magnet geometry and strength, and the RPMs of the specific Bedini motor.

The transistor is like a dynamic resistor being whacked by the discharging coil.  And you can't forget that the discharging coil acts like a current source, which means that the voltage it generates is variable and can get very very high.  Hence the use of neons to protect the transistor.

[MileHigh]

Quoting Ren:

You say, "eventually the voltage on the capacitor gets so high that it stops charging".

Why does it stop charging? Simply because the voltage "gets so high?" Surely there is more to it than that? If the limit of the capacitor has not been reached then why does it not continue charging?

Supposing you connect a variable power supply to a 12.6 volt battery.  To be safe, between the power supply and the battery you connect a 100-ohm resistor.

If you set the power supply to 13 volts, then current will flow into the 12.6 volt battery.  If you set the power supply to 12.6 volts then the current stops flowing.

It's sort of like that when the coil discharges.  The capacitor acts like the battery, and the coil discharge cannot get "over the potential hump" associated with the charged capacitor, and therefore current will not flow into the capacitor.

The current finds another path to follow instead, and that's through the transistor while it is in the process of shutting off.

The faster the transistor shuts off, the smaller the "window of opportunity" you have for the coil to discharge through the the transistor.  Therefore, you continue to charge the capacitor.  Eventually the voltage on the capacitor gets so high that the shorter window of opportunity doesn't matter, and the high voltage cannot overcome the potential hump of the capacitor anymore.  Instead the coil discharge goes through the transistor once again with a shorter higher-voltage higher-average-power pulse.

In other words, this is a dynamically balancing system just like you see when Bedini motor stabilizes at a certain RPM.  Change one parameter and then the RPM stabilizes at a new value.  If you change a parameter associated with your Bedini motor and that affects the overall dynamics of the switching parameters and especially the timing, then the voltage on the capacitor will stabilize at a new level.

[MileHigh]

Each pulse contains the same amount of energy, so as switching time decreases, the voltage of the pulse gets higher and amount of charge (current x time) pumped through the transistor decreases.  Therefore the average power in the pulse increases as the switching time gets shorter.

The key thing to remember is that the inductor has to discharge its stored energy no matter what.  It's like an energy pump that you can't stop.   If it can't pump it's stored energy into the capacitor it will pump it into the transistor.  The faster you do the switching the pump reacts accordingly by raising the voltage high enough to discharge the energy anyways.

I assume the higher the inductance of the coil the higher it can charge the capacitor?

The higher inductance will permit you to store more energy for a given current flow.  All that means is that you have more pulse energy for the discharge.  So in principle what you say might be true, but that's all that you can say.  It's the timing and dynamics of the coil discharge cycle that really determine the final voltage on the capacitor.

Have you ever wondered what would happen if you replaced all of the components by idealized components, i.e.; perfect capacitor, zero resistance wires, ideal diode and transistor, perfect switching, etc?