Author Topic: Inductive Kickback (Read 159764 times)

MileHigh · « **Reply #60 on:** November 20, 2015, 10:03:54 PM »

Quote from: verpies on November 20, 2015, 09:52:28 PM

No, the instantaneous power dissipated into the load (the 100Ω resistor) is I²R in both cases ...or 100W.
That would be the case in any constant current source.

The formula P=V*I or P=E²/R could be applied if V represented the magnitude of a voltage source.
In Milehigh's example, there are no such sources.

Ha! Ha! Verpies you tripped up!

The initial instantaneous power for the 100-ohm load is 100 watts. The initial instantaneous power for the 5000-ohm load is 5000 watts.

But of course the total energy in the current pulses for both cases will be the same - no OU.

verpies · « **Reply #61 on:** November 20, 2015, 10:06:55 PM »

Quote from: MileHigh on November 20, 2015, 10:03:54 PM

Ha! Ha! Verpies you tripped up!

Yes, I did but I will make an example of myself how quickly such mistakes should be admitted.

verpies · « **Reply #62 on:** November 20, 2015, 10:11:58 PM »

Quote from: MileHigh on November 20, 2015, 09:55:56 PM

I just posted that Tinman was wrong about the current flow in an inductor. Chances are if I didn't post the correction, then nobody would have corrected him.

You'll get no argument from me.
I did not even read that post of his, but if I did I would've objected to his statement about the load resistance affecting the current of a constant current source (an inductor in this case).

citfta · « **Reply #63 on:** November 20, 2015, 10:20:18 PM »

I'll second that idea of admitting mistakes quickly. Milehigh I wasn't meaning to say you were wrong with what you said it was the way you said it that could lead some (or many) to believe that you believed in OU from the discharge of a coil. I am pretty sure you didn't mean that.

woopy · « **Reply #64 on:** November 20, 2015, 10:37:30 PM »

Thank's a lot Citfta

Finally somebody answering directly my questions.

So let's go further if you don't mind.

in my question 3
you seems to agree that the current builds up effectively the magnetic field which propels and gives the kinetic energy to the rotor.
But is the current or his energy still stored in the magnetic field after the magnetic field has propelled the rotor ?? Any entropy ?

So in my question 4
you seems to say that the input energy could be used 2 time ?

Thank's very much for your input and for this thread , because i have so much to understand

Laurent

MileHigh · « **Reply #65 on:** November 20, 2015, 11:46:35 PM »

Laurent,

I my opinion, if you can understand how an inductor and a mechanical flywheel are directly related, that will make all the difference. So the challenge is to just understand that first, and then once you understand the concept, then you can apply it to understanding circuits.

You saw the "Woopy's Linear Track Experiment" drawing that I made?

Look at the attached image that I found on Google. This is just a teaser, does the image look familiar?

MileHigh

citfta · « **Reply #66 on:** November 20, 2015, 11:49:59 PM »

Quote from: woopy on November 20, 2015, 10:37:30 PM

Thank's a lot Citfta

Finally somebody answering directly my questions.

So let's go further if you don't mind.

in my question 3
you seems to agree that the current builds up effectively the magnetic field which propels and gives the kinetic energy to the rotor.
But is the current or his energy still stored in the magnetic field after the magnetic field has propelled the rotor ?? Any entropy ?

The magnetic field is maintained as long as there is power applied to the coil. When power is no longer applied the magnetic field begins to weaken as it tries to maintain the current through the coil. If you continue to apply power after the current has reached the maximum value which is also when the field has gotten the strongest then any power after that is wasted as heat because of the resistance of the wire in the coil. So you want the pulse to be just long enough to cause the current to reach its peak and that will also mean the field has reached its peak. Then when you turn off power to the coil the kickback will be the strongest and can thus give the most power to the cap.

So in my question 4
you seems to say that the input energy could be used 2 time ?
Yes by capturing that kickback spike and reusing it you can make your system more efficient. I think Luc just posted some more videos about that today.

If you have time try the test I proposed for you earlier with a 2 channel scope. You can see the polarity switch with scope channel that is across the coil and you can see the current continue flowing through the resistor with the scope channel across the resistor. If you can set this up with a 555 timer or some other means to fire the coil regularly it will be easier to get the scope picture you want.

Thank's very much for your input and for this thread , because i have so much to understand

Laurent

MileHigh · « **Reply #67 on:** November 21, 2015, 12:06:42 AM »

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:

Applying voltage to a coil makes the current start to flow, starting from zero. Voltage 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.

shylo · « **Reply #68 on:** November 21, 2015, 01:02:50 AM »

Hi all,
If you take a single coil , and pass a magnet over it , as the magnet approaches it induces a potential (voltage) correct?
There is no current if the ends of coil aren't connected (open) correct?
If not open but connected to a load current will flow. correct?
But when the magnet is approaching the coil it builds , hits it's peak then builds in the opposite direction, short the coil right after the peak (disconnect load and short coil) then reconnect load.
That gives twice the work out of the coil.
Now have recovery diodes and caps to collect the short, for both pulses.
Passing a magnet over a coil produces two pulses.
When the magnet is at the center of the coil that is where you short the coil leads together removed from the load , reconnect and do it again, before the passing magnet leaves(or stops influencing the coil)
Then short again.
artv

synchro1 · « **Reply #69 on:** November 21, 2015, 02:23:47 AM »

What's wrong with this picture? The arrow in (c) looks llike it's moving in the same direction as (b) through the inductor. This is an illusion caused by confusing path with current direction.

citfta · « **Reply #70 on:** November 21, 2015, 02:36:14 AM »

YES!! Now you have it! Of course with no where to go except through the switch you will get arcing at the switch that way. That is why we are looking at using the energy from the collapsing field to charge another cap. We don't waste the energy and the reed switch will last longer too.

Great!! I am so glad you got it.

Respectfully,
Carroll

synchro1 · « **Reply #71 on:** November 21, 2015, 02:39:31 AM »

Quote from: citfta on November 21, 2015, 02:36:14 AM

YES!! Now you have it! Of course with no where to go except through the switch you will get arcing at the switch that way. That is why we are looking at using the energy from the collapsing field to charge another cap. We don't waste the energy and the reed switch will last longer too.

Great!! I am so glad you got it.

Respectfully,
Carroll

@Citfta,

You just turned that smart!

tinman · « **Reply #72 on:** November 21, 2015, 02:42:12 AM »

Quote from: MileHigh on November 20, 2015, 09:31:50 PM

When you disconnect the power source from the coil, whatever the magnitude of the current flowing through the coil is, that's the initial amount of current that will flow through the load.

If you have one amp flowing through the coil and you disconnect your power source and the load is a 100-ohm resistor, then initially the current flow through the load will be one amp and the voltage across the load will be 100 volts. If the load is 5000 ohms, then initially the current flow through the load will still be one amp and the voltage across the load will be 5000 volts.

Quote

You are right about the current not changing direction but you are not right about the magnitude of the current.

You do understand that i am referring to the average current over one cycle ?.

tinman · « **Reply #73 on:** November 21, 2015, 02:45:33 AM »

Quote from: verpies on November 20, 2015, 10:11:58 PM

You'll get no argument from me.
I did not even read that post of his, but if I did I would've objected to his statement about the load resistance affecting the current of a constant current source (an inductor in this case).

Once again,i am referring to average current over one cycle.

synchro1 · « **Reply #74 on:** November 21, 2015, 02:47:28 AM »

This should help clear everything up:

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Author Topic: Inductive Kickback (Read 159764 times)

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