Inductive Kickback

[tinman]

Yes it is in series but it is also included in the input circuit current and the scope shot includes that too, no?

Yes it dose,but we can use the scope shot i posted above with the circuit to separate the two--the on period and off(flyback) period. The power calculations for dissipated power for the LED should be made from the 70% off period. The inductor provides the power for the LED(our current source)during the off period. We measure that power by way of calculating the current flowing through our 10 ohm resistor,and the voltage across the LED--then multiply that result by 70% to get our average power dissipated by the LED over 1 cycle. But when i do the very same with my FG,i find that i only need 25% of that calculated power to get the same voltage across the LED,and for the LED to put out the same light.

This can only mean that !although! the scope show's us the current being delivered to the load,it's actually not there,and this is what i was trying to explain to MH as to why you cannot measure current flowing out from the inductor during the flyback period,as it is a looped system.

This very thing happened not so long ago,when i was posting some measurements from the HTT test,and things didnt add up when the current was calculated across a CVR,and the power dissipated across the load resistor. When we added the two resistor values together,and the voltage across those two resistor's,we ended up with a different power value than we did when we calculated it the other way.

I will do up a video tomorrow night after work,and show you what i mean.

Brad

[gyulasun]

Okay Brad, thanks and probably using a separate CVR directly in series with the LED too and check the current pulses across it may help to estimate only the LED dissipation. This way the input coil current (during ON time) cannot be sensed by the separate CVR.

Gyula

[woopy]

Hi Gyula and all

I redo the scope measurement you have proposed and i checked all connections and changed some old crocodile pinch.

Now i got a very nice trace where we can clearly see that the current does not fall down vertically as per my first erroneous measurement.

So effectively the "negative voltage" seems to  "brake"  strongly the current  (flywheel) which goes extremely fast to zero in my particular arrangement.

Thank's to all of you  for your help.

Hope this helps

Laurent

[gyulasun]

Hi Laurent,

Very good, thanks for all your efforts. I edited your expanded scope shot to include some lines and explanations, those may be useful for interested readers  (schematic is here: http://overunity.com/16203/inductive-kickback/msg466434/#msg466434 )

Your scope shot nicely shows that from the moment the reed switch switches the current off, the coil current (yellow trace) remains above the zero line i.e. its original polarity is maintained and its amplitude reduces to zero as the flyback pulse with its negative polarity lasts.

Maybe your questions have been answered in connection with this inductive kickback topic?   

Gyula

[MileHigh]

I dont think we need the math (although the math trace works quite fine on my scope)function to work out our P/in and dissipated power when the wave forms are very clean.

I am using an LED for a load for the very reason that the I and V curve is non linear.
It al;pws me to increase or decrease the inductive kickback current while maintaining very close to a set voltage during the off cycle. I can then measure the light output from the LED via way of a light box-->a small solar panel inside a box with the LED,and then we place a resistor across the solar panel output. I can now see if the LED's output rises with the increase of current,while the voltage remains much the same across that LED. For this particular LED,it seems that the optimum voltage is 3.2v,and any additional current makes no difference to the output of the LED. The other thing is i can now see what happens when i increase the voltage on the input--this is how i increase the current on the flyback. Now we have a situation where only the current on both the input and output rises,but both the input and output voltage remain the same,even though i am increasing the voltage on the input. On the flyback side this is expected with the LED in play,but why dose only the current rise on the input,and not the voltage?.

Well, I can see from your scope shots that you always have current flowing in the coil.  Personally I find that to be a more complex measurement problem as compared to pulsing the coil with voltage, and then during the OFF time you observe the coil completely discharge its stored energy.  That should be a simpler measurement problem to tackle.   With current always flowing in the coil during the energizing cycle and during the discharge cycle, two things are happening at the same time during the energizing cycle, - 1) the energizing of the coil, and 2) the discharging of the coil though the LED.  I would have to think for a while on how to make the energy measurements in a case like that.   Although the LED appears to be clamping the coil discharge through the LED to a fixed voltage, we know it's not truly doing that.  I also see how your function generator output (?) droops s bit as the coil current increases which also introduces nonlinearities.  It all depends on how accurate you want your measurements to be.

Your observations are all good.  You are getting the feel for how you can play with the coil excitation voltage and the ON time for the excitation to literally "dial up" any final current for the coil you want.   Likewise you can "dial up" the discharge time through an LED + resistor or just a resistor by choosing whatever components you want for the load.   By playing with the inductance of the coil, the ON energizing time, and the ON drive voltage, and the nature of the load, you can dial up just about any type of pulse that you want.

For example, if someone said to you that a certain battery can be pulse charged from a coil, but the maximum initial current flow at the start of the pulse was one-half amp, you can design your pulse to do exactly that.

MileHigh