Inductive Kickback

[MileHigh]

And what about the flywheel?  Well, if you are getting the flywheel analogy, then figuring out what will happen with your electrical circuit becomes a no-brainer when you do the same experiment in your head with a flywheel.

Recall, that current flow is equivalent to the rotational speed of the flywheel.

So, here is the equivalent experiment done on a flywheel:

You always start with the flywheel spinning at 100 RPM.  Then you apply the brakes with different brake pressures.  First you apply the brakes gently (10 ohms.)  Then you apply the brakes with increased pressure.  Then you apply the brakes one more time with even more pressure.

You observe what happens to the flywheel when you try different brake pressures.  That's it.

P.S.:  Obviously, the higher the brake pressure, the higher the initial negative torque (negative voltage) on the spinning flywheel.

[MileHigh]

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

The "secret" is that for all practical intents and purposes they are literally identical.   They have identical sets of equations that describe their respective behaviours.

What it means is that Mother Nature decided to keep it simple.  It turns out that electrical components act in the same way as physical things that we see and use in our daily lives.

[tinman]

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

Should the circuit not be like the one below,where we can measure the current flowing through the inductor during the on time,and then measure the current flowing through the VR during the off time? This would show us what current flows through the inductor during the on time,and we can then see what current is flowing through the load during the off time,and how the different resistive loads change the flyback current delivered to those loads. This is what i was referring to,as we are looking into the flyback power delivered to various loads.

What you have depicted in your diagram is a current loop.

Brad

[MileHigh]

Brad:

You can experiment with that if you want but do you really need it to be that complicated?   I think the use of the diode has merit if your signal source is your unamplified signal generator and you will be using low resistance values.

We know that the current going through the coil is going to be an increasing ramp that will eventually level out.   Then, when the applied voltage to the coil goes OFF, the voltage across the variable resistor will tell you the current going through the coil provided that you measured the value of the variable resistor beforehand.  Do you see what I mean?  You can make an argument that the 10-ohm CVR is redundant.  The variable resistor is the CVR with the caveat that you cannot see the current increasing through the coil during the energizing cycle.  However, do you really need to see the increasing current through the coil while you are energizing it?  If you always energize the coil with the same pulse timing, then the final current through the coil at the instant the energizing pulse goes OFF will be the same.

MileHigh

[MileHigh]

The attached graphic should make everything much clearer.