Inductive Kickback

[citfta]

This should help clear everything up:

I do like that picture!  Every time I see it it makes me feel like my brain is tilted a little.  LOL

[tinman]

Here you go Synchro-test one complete.

What do you see?
1- a 30% on time from the FG(switch closed)
2-a 70% off time(switch open)

During the 30% on time(switch closed) we can see that the voltage(yellow trace) across the inductor(supply voltage) is on the top side of the zero volt line.The current is flowing in a forward direction,as the current trace(blue trace) is also above the zero volt line.

During the 70% off time(switch open) we can now see that the voltage is inverted across the inductor/resistor series circuit-!BUT! the current is still flowing in the same direction as it was during the 30% on time,as the current trace is still above the zero volt line.
You will also notice that the current continues to flow through the load(the LED) through the complete 70% off time,and never stop's flowing right up until the next on time period. This also shows you that the magnetic field dose not collapse completely before the next on time cycle starts.

So !Bub!
Please show us where exactly the current flow reverses direction?
And please tell us how a current continues to flow(in the same direction ) during the 70% off time if the magnetic field around the inductor collapses instantaneously ?
Show us where this !new! energy is?.

[MileHigh]

Brad,

That's an awesome posting and my sincere compliments.

I will just throw in some "flywheel" colour commentary:

You can see how there is always current flowing through the inductor -> the flywheel never stops turning.

The flywheel increases in speed when you apply positive torque (positive voltage) to it.

The flywheel decreases in speed when you apply braking to it.  So you can imagine the (diode + 10 ohm CVR) acting like a caliper and brake pads putting resistance on the spinning flywheel.  When the brakes are applied that is negative torque (negative voltage.)

Bonus round:

What happens when you disconnect the 10-ohm CVR from the circuit?   Most of us know that you get a spark in the air gap because the air becomes conductive plasma due to the big negative spike of high voltage.  All of the energy in the coil is burned off in the hot plasma.  The current stops flowing through the coil.

What is the equivalent for the flywheel?

You clamp down on the brake calipers very fast and with maximum pressure.  There is a massive spike of negative torque on the flywheel when the brake pads make contact with the flywheel.  All of the energy in the flywheel is burned off in the hot brake pads.  The flywheel stops spinning.

Bonus bonus round:

The big spike of negative high voltage happens because the coil has electrical inertia in the form of flowing current.  The flowing current pushes its way through the air gap resulting in the generation of the negative spike of high voltage.

The big spike of negative torque happens because the flywheel has rotational inertia.  The rotational inertial pushes on the brake pads and that is what causes the spike of negative torque.

[tinman]

Brad,

That's an awesome posting and my sincere compliments.

I will just throw in some "flywheel" colour commentary:

You can see how there is always current flowing through the inductor -> the flywheel never stops turning.

The flywheel increases in speed when you apply positive torque (positive voltage) to it.

The flywheel decreases in speed when you apply braking to it.  So you can imagine the (diode + 10 ohm CVR) acting like a caliper and brake pads putting resistance on the spinning flywheel.  When the brakes are applied that is negative torque (negative voltage.)

Bonus round:

What happens when you disconnect the 10-ohm CVR from the circuit?   Most of us know that you get a spark in the air gap because the air becomes conductive plasma due to the big negative spike of high voltage.  All of the energy in the coil is burned off in the hot plasma.  The current stops flowing through the coil.

What is the equivalent for the flywheel?

You clamp down on the brake calipers very fast and with maximum pressure.  There is a massive spike of negative torque on the flywheel when the brake pads make contact with the flywheel.  All of the energy in the flywheel is burned off in the hot brake pads.  The flywheel stops spinning.

Bonus bonus round:

The big spike of negative high voltage happens because the coil has electrical inertia in the form of flowing current.  The flowing current pushes its way through the air gap resulting in the generation of the negative spike of high voltage.

The big spike of negative torque happens because the flywheel has rotational inertia.  The rotational inertial pushes on the brake pads and that is what causes the spike of negative torque.

I couldnt agree more MH with your pulsed flywheel analogy  . The relationship between the two are very similar.

One thing i need to say or would like to discus is this current flow from the inductive kickback in regards to the resistive load value-reference post 51. I am not seeing this on the scope. As i increase the value of the resistive load,the current flowing through the CVR decreases-along with the instantaneous current spike across the CVR. As i increase the load resistance, the current drop's,and the voltage rises. I am not seeing this instantaneous current flow remaining the same as the input current?.

Brad

[MileHigh]

Brad:

Sure, this is pretty easy to explain using the circuit that you just posted with modifications.   I will modify the schematic.

Then just run the same test, but start increasing the value of the resistor.   Say, 10 ohms, then 47 ohms, then 100 ohms, then 200 ohms.  Something like that.

You can see it's very simple, when the input pulse energizes the inductor at the end of the ON time, let's assume for example that 100 milliamps are going through the coil.

So, when the pulse goes OFF, you have 100 milliamps flowing through the coil.  We will ignore the current that also flows through the resistor.

If the resistor is 10 ohms, you should see an exponential decaying waveform that starts at -1 volt in amplitude.
If the resistor is 47 ohms, you should see an exponential decaying waveform that starts at -4.7 volts in amplitude.
If the resistor is 100 ohms, you should see an exponential decaying waveform that starts at -10 volts in amplitude.
Etc, etc.

The higher the value of the resistor, the faster the waveform will decay.

Ideally, your pulse train will have a long enough OFF time to let the current in the coil decay to zero.

So, you notice that this experiment will confirm that as long as you energize the coil for the same amount of time, and as long as there is no current flowing through the coil when you start the pulse, that the initial current flow through the coil is always the same when it discharges through different resistive loads.

MileHigh