Inductive Kickback

[tinman]

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

MH
If we do it your way,then the flyback current is not a current source for a load-it is a current loop back onto itself. As we wish to measure the current as a supply source from the flyback,then that source should be dissipated across a load,in this case the VR.
The CVR will show us the current flowing into the storage unit(the inductor),and it will also show us the current of that stored energy being delivered to the load. If we are to look at the current flowing into the inductor,and then the current flowing from the inductor to a load,then the diagram you posted will not show this-it will only show the current flowing through the current loop,and not to a load.  For a P/in and P/out measurement,your circuit will not show this,as there is no load,but only a loop. We want the engine to spin up the flywheel,and then for that stored energy in the flywheel to be delivered to a load-not back to the motor.

What point is there in saying the current in and out of the inductor will be the same value,when that current only loops back to where it came from?.

As i said,the current from the flyback is dependent on the load resistance. You then say it is not,and that i am wrong. But then you post a diagram showing a current loop,where the current from the flyback is sent straight back to where it came from. The current then is not a source,as it is a loop. What i was referring to is the current from the flyback being a source of current for a load,and in this case i am correct in saying that the current flowing from the inductor will be dependent on the load resistance.


Brad

[woopy]

For an inductor, you can measure the current flowing through it, and the voltage across it.

Look at the picture of the laboratory flywheel.  You can see there is a smaller disk so you can use your hands to spin the flywheel.

When you apply voltage across an inductor, the current does not flow instantly.  Instead, it slowly rises.

If you have a stationary flywheel, and then you want to make it spin by applying torque to it with your hands, it does not spin at full speed right away.  Instead, the rotational speed slowly rises.

This is the first thing you have to understand:


is like "pressure" on the coil.
Applying torque to the flywheel makes the flywheel spin, starting from zero.   The torque is a pressure on the flywheel.

Therefore voltage is like torque.

If you can understand that, it's a good start.

Hi MH

Yes very nice analogy with the flywheel.

But i have a question. If i look at the scope shot i have posted earlier, at the end of the pulse , when the voltage stops or in your analogy when the torque stops spinning up the wheel, the current fall down instantly to zero.

So to me in your analogy it would be the same as if something bloqued instantly the wheel. What happen in this case, immediately after the stop, the really high inertia of the spinning wheel will be transformed in a huge and sudden toppling a the complete system which will jump down the table. This is the flybackspike energy.

So question is,   is it possible that the "flywheel" stops instantly when the voltage stops ?

Just a supposition

Laurent

[synchro1]

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?.

@Tinman,

How do you get a 30% percent on time with a Reed switch? You are artificially starving the pulse of sufficient power to even achieve a field collapse! You need to load the coil to saturation to get a violent field collapse and current reversal, along with peak magnetic field strength for maximum rotor acceleration. Gotoluc is running a secondary coil on flyback. How is your pulse starvation approach supposed to help run his pulse motor and auxiliary Flyback power coil more efficiently? Milehigh has already exposed you as a hoaxer on the initial current fallacy.

Woopy's test was uncontrived and straight forward. You, on the other hand, have intentionally distorted the test conditions to confuse people and appear as some kind of savant. Your results are worthless. I see right through you! You're just a stinking charlatan and a retarded wanker.

[synchro1]

@Woopyjump,

"A generator in a power plant produces electromotive force by moving magnets past coils of wire; the relay coil produces electromotive force as the collapsing magnetic field moves past the wires in the coil".

[citfta]

@synchro

KNOCK IT OFF WITH PERSONAL ATTACKS OR OUR DISCUSSION IS OVER!!

@Brad

Would you please rerun your test with a longer off time to allow the current to drop to zero?