Confirming the Delayed Lenz Effect

[Farmhand]

I recommend this MIT lecture video even for the layman, it helped me to understand a lot, I had to watch it several times and I have a way of being able to understand in a general way what he is talking about even though I don't understand the calculus very much at all because I've never learned it. I still found it was enlightening.

What do you guys think of the Prof. and this lecture in particular? At 36:00 minutes on he explains what I am seeing in the coils that will work to produce the speed up effect. If we go from say Omega L of 300 Ohms at 1000 rpm to Omega L of 900 Ohms at 3000 rpm then it might be enough to see the effect. Just arbitrary figures. And I might have the terminology wrong I'm just writing what I heard him say. But I did watch the entire video and watched him do all the calculations, he just lost me and I couldn't fathom it all at once. But I got what he was saying anyway I think.

Lecture
http://www.youtube.com/watch?v=UpO6t00bPb8

Cheers

[Farmhand]

MileHigh, I see now what you mean about the flywheel yes it would do that, I envision when we try to stop the wheel quickly there is a very sharp rise in something for sure so I see your point now. The flywheel would seem to be the mechanical analogy like how a mechanical hammer works, I've used one of those there neat. And the elastic bag the gaseous one. There is no doubt a flywheel is not easy to pull up quickly, but it would seem the force needs to be applied to the flywheel to get the effect but with a coil the force is removed to see the effect. When we switch off the power that is driving the flywheel it doesn't let go of it's energy catastrophically it slows down gradually.

Take a vertical water pipe of say 20 mm Inside diameter and 100 meters "high" and install a fast " valve" at the bottom end then take out 20 meters of pipe in the middle and install a coil shaped (or not even) appropriately elastic rubber tube (bag) with an inside diameter of say the same 20 mm, the top of the pipe is fitted to the bottom of a tank with X head of water available and the flow from the valve is unimpeded, the bag is so that it would burst with 1.5 times the latent pressure, at the bottom of the bag (coil) is a safety valve set to fire when the pressure becomes slightly higher than the latent pressure and the main bottom valve is rated to burst at 1.5 times latent pressure. What happens when we turn the valve on and off quickly. Would the bag not expand under the momentum of the falling water build pressure then squirt water from the safety valve to prevent rupture ?

The force would be like the hammer effect in water pipes. I imagine a water system for low voltage low frequency stuff and a gaseous system for HV HF stuff. Just what I see when I visualize.

I see your point with the flywheel though if we try to stop it. But with water or gas we just turn a handle to stop the flow and the build up happens.

Cheers

[MileHigh]

Farmhand:

I dug up a posting that I made a while ago, food for thought about the flywheel business:


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MECHANICAL EQUIVALENT CIRCUIT OF A JOULE THIEF

Imagine you go to the gym and you find an old-style exercise bicycle.  The type with a seat and pedals and a chain link to a big flywheel, like a regular bicycle.  There is a friction belt that goes around the circumference of the flywheel.  You set the tension on the friction belt to adjust the difficulty level.

Imagine the belt is completely loose.  You pedal for a few seconds and get the flywheel spinning and then you stop pedaling.  Then you add tension to the belt and the flywheel spins down and stops.  Then you loosen the belt and repeat the whole process all over again.

Even when you are completely exhausted, it's still possible for you to pedal and get the flywheel spinning if you pedal slowly and take your time to build up the speed.  Don't forget that the friction strap is loose when you pedal.

That's a Joule Thief.  You are the battery.  The flywheel is the coil.  The friction belt is the LED.

The torque that you put on the flywheel from pedaling is the battery voltage.  The torque that the flywheel puts on the belt during the braking is the coil voltage when it's de-energizing.  The rotational speed of the flywheel is the current through the coil.

I have never heard of any claims of over unity exercise bicycles and by the same token, a Joule Thief - an energizing and de-energizing inductor - is not over unity.

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In this context, is it possible to make a connection with a Joule Thief driving 100 LEDs?

Well, 100 LEDs means a larger voltage for the same current.  Since R = V/I we can say that a Joule Thief driving a 100 LEDs has a much larger "LED equivalent resistance" to drive.

Quoting myself above:

Imagine the belt is completely loose.  You pedal for a few seconds and get the flywheel spinning and then you stop pedaling. Then you add tension to the belt and the flywheel spins down and stops.  Then you loosen the belt and repeat the whole process all over again.

Adding tension to the belt means that you are putting a resistance on the spinning flywheel.  So 100 LEDs is equivalent to adding more tension to the friction belt, and the spinning flywheel will come to a stop more quickly.  You can easily see this on a scope with a real Joule Thief.

Also, you know that even if you add tension to the belt, the moment the belt makes contact with the flywheel the flywheel is still spinning at it's current angular velocity.  That explains why the 100 LEDs will light up.  The angular velocity of the flywheel is akin to the current flowing through the coil.  So even with 100 LEDs, when the inductor starts to discharge, there will still be sufficient current to light up each LED.

The more LEDs you add, the faster the current flow decreases and the shorter the time the LEDs are illuminated.  That's equivalent to adding more tension to the belt and slowing down the flywheel faster.  So, more LEDs in series = adding more tension (and resistance) to the belt.

Just for clarity:  The friction between the belt and the exercise bicycle flywheel while the friction is being applied represents the time the LEDs are illuminated and discharging the coil's stored energy.  In one case you get heat generated between the belt and the flywheel, and in the other case you get light and heat generated inside the LEDs.  In both cases you have an "energy burn."

The energy stored in the flywheel = 1/2 Moment_of_Inertia x angular_velocity-squared.
The energy stored in the coil = 1/2 Inductance_of_Coil x current_flow_around_the_coil-squared.

If you notice they are essentially the same formula.  The amount of moment of inertia of a flywheel is equivalent to the amount of inductance in a coil.  The angular velocity of the flywheel is equivalent to the current flowing through the coil.  The torque either applied to the flywheel to make it spin up faster (energizing), or produced by the flywheel when it is driving a friction load and slowing down (de-energizing) is equivalent to the voltage across the coil.

If you can visualize that it may help you visualize how a circuit with a coil operates.  My example of the exercise bicycle is an excellent analogy for the energy dynamics taking part in a Joule Thief circuit.  When you strip a Joule Thief circuit down to it's bare essentials, it's nothing more than a battery energizing an inductor followed by the inductor discharging though one or more LEDs. It's a very trivial circuit when you look at it from that perspective.

MileHigh

[Farmhand]

MileHigh, Here in these shots I think what I can see is the rotor of my motor is returning energy to the capacitors by generation through the coils. Because of the spacing of the coils when the rotor magnets pass they can at times put energy back into the electrical circuit briefly before it goes back to the rotor. This part of the generation is partly counteracted by the anti cogging effect at that part of the rotors travel I think.

In between the white lines is what I see as the mosfet "on" time just after the current drops off, but see before the current is rising in the coil before the mosfet allows energy to be input from the supply, and yet there it is, I think that is the rotor generating electricity in the coil and the current flows around through the return circuit as HopToad so kindly pointed out (thanks Hoptoad). Going that way means it must also go through the charging coil MC2 which helps turn the rotor, then into the charging capacitor/s to be switched through teh motor coil MC1 again along with what was also added from the supply. The only way the current can rise in MC1 that way when the mosfet is off is if a south magnet is passing it. As the south magnet is passing the MC1 coil and reinforcing the currents around the circuit MC2 is attracting a north magnet to it, as it just pushed the south that is passing the MC1 coil and MC1 pushed the north just before that.  If you look at my rotor timing chart you'll see what I mean, it is very confusing but I think I eventually got it fairly correct.

Still I might be a tad off on my analysis but what else would explain the current rise in a switched off coil if not from the magnet through the return circuit. The coils are kind of loosely locked together and the magnets help shape and reinforce the currents.

Also notice the RMS current values ? What does that mean ? THe current is scoped across 0.1 Ohm resistors. So that around 900 mA RMS in MC1 and 700 mA RMS in MC2, and the circuit drawing only 400 mA from the supply.

Anyway is that resonance. Or not sloshy enough. 

Cheers

P.S. and harder running I think lowers the RMS current in the coils and raises the input current. So the energy stops being returned when under proper load.

Actually I think MC2 maybe forcing the current I see in MC1, dunno that's why I asked your opinion.  It is a bit confusing.

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[hoptoad]

snip...
In between the white lines is what I see as the mosfet "on" time just after the current drops off, but see before the current is rising in the coil before the mosfet allows energy to be input from the supply, and yet there it is, I think that is the rotor generating electricity in the coil and the current flows around through the return circuit
snip...
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Yep.... KneeDeep