Inductive Kickback

[tinman]

@Tinman,

That's how to measure "Flyback" recovery; First we back the Reed switch away from the "Sweet Spot"!

LoL: Look how little "Flyback" we recover when we only charge the power coil part way!

Here is a scope shot from the pulse motor Synchro. Notice the voltage across the coil during the on time-what dose that tell you?. The current trace is now over a 1 ohm 5 watt CVR--enough power in for you now?. So now-show me where the current reverses direction?.

@MH and Verpies.
As i was saying-looking at the scope shot,you can see that the peak current during the on time is more than the instantaneous current during the off time. You can see the current drop in value at switch off.

[citfta]

Great screen shot showing exactly what we have been saying.  I have received a PM asking me to not have the thread locked so I guess if you guys want to keep it open I will ask Stefan to not lock it.  As far as synchro, guys like him always get theirs in the end.

[woopy]

Hi Guyla

I have redo the measurement so the 1 ohm sensing resistor is placed between the reed and the main coil (05 ohm and 5 mH), the probe for the blue trace is connected (not grounded ) to the output of the main coil (X1) . For the blue trace i had to invert the chanel.

-First pic is a general situation

-Second pic is the trace of the same situation but i tried to expand a max the view to see what happen when the voltage stops and the current fall straight down to zero.

Yep my trace are very different of the Tinman's. Perhaps it is due to very low resistance of my main coil and the low frequency. Tinman is running at 1000Hz ?

Hope this helps

Citfta i don't think that you have to stop this thread.

MH what about the brutal stop of the flywheel ?

Laurent

[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

I look forward to seeing the test done the way you want to do it.  It will make very little difference, because the initial current will be the same with the 10-ohm resistor or not.  In essence, that's the point of the test.  I also forgot to acknowledge your point about the average current measurement of the discharge cycle decreasing as you increase the value of the load resistance, which is absolutely true.

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.

I think that you might not be seeing the forest for the trees here.  If we ignore the diode for a second, you are just looking a the "difference" between the coil discharging through one resistor vs. the coil discharging through two resistors in series.  No matter how you look at it, the current will flow in a loop.  There can be one resistor, or two resistors in series in the loop, with the discharging inductor as the power source for the loop.

[MileHigh]

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

Yes, in your circuit the "secondary flywheel" stops right away after the capacitor finishes discharging.  The explanation for this is simple.  The secondary coil has a resistance of 210 ohms.  So that means that there is "a lot of friction in the ball bearings" for the "secondary flywheel."   If you have a coil with a very very low resistance, and you get current going through it and then you short the two terminals of the coil together, then you can assume that with a Hall sensor you will be able to detect the magnetic field around the coil for several tens of seconds or more.  Note you do not want to use a current sensing resistor because that would defeat the purpose of seeing how long current will continue to circulate in the coil.