Confirming the Delayed Lenz Effect

[Mk1]

@all

I have found this ... http://www.youtube.com/watch?v=v8TbqZa8li4   atracting a stone with a coil.

I also made a coil a while back that did not show load on the circuit , https://www.youtube.com/watch?v=w54lxNS3Hus&list=UULKKCauubuWLRdng24R-cKQ&index=4

Its a simple joule thief with a pickup coil , a led on the output of the coil and on the joule thief , usually doing so dims the light output of both led , but not in this case because the led is connected backwards.

Mark

[conradelektro]

Sorry to distract from the magnetisation and coil topic.

I could design a commutating circuit with one hall sensor, two P-Channel and two N-Channel MOSFETS (some sort of transistor H-Bridge).

See the attached circuit, scope shots and photo. The scope shots are over the drive coil. Looks like some very clean switching (commutation) of the drive coil.

The results of a first test:

12 Volt, 17 mA, 70 Hz (0.2 Watt, 4200 rpm)

20 Volt, 25 mA, 140 Hz (0.5 Watt, 8400 rpm)

I still have a mechanical problem, the plastic axle is not straight (causes vibrations). I will make a brass axle which should be better.

This seems to be a low power drive circuit to produce considerable spin. One can now try to find some "Lenz free coils" which produce more than 0.2 Watt or 0.5 Watt from the 4200 rpm or 8400 rpm.

May be people who build similar set ups can publish their power requirements for their spinners?

Greetings, Conrad

[conradelektro]

Other MOSFETS:

I switched to the AUIRF9Z34N (P-Channel) and the AUIRFZ34N (N-Channel) because they have less resistance when switched on. This produces a little bit higher rpm value (about 10%) for the same power input. Theire 55 V Drain to Source braekdown Voltage is high enough because the drive coil is switched (commutated) cleanly, only small spikes.

Greetings, Conrad

[hoptoad]

hi Hoptoad
Are those "totallydamped" pages yours ?
If yes, Ill like your opinion about the possible damage of transistor (MileHigh states)
in the circuit I posted earlier (No 19)
thanks
cheers

Yep, those are my pages. If following figure number 19, with one coil winding connected back to the supply via a diode (in feedback mode), the cemf spike will be damped by the battery and should alleviate any potential harm to the transistor. If the circuit is not using one winding as feedback, or there is not even a diode straight across the driving coil winding, there is a possibility of transistor damage if you are using high impedance coils or supply voltages greater than 12 volts. However, if you have connected the windings as shown in fig 19 then there is no real threat of potential transistor damage.

Cheers

[MileHigh]

Hoptoad:

Sorry but I am going to correct you.

In figure 19 when the transistor is on the current from the battery flows through bi-filar winding A only.

When the transistor is switched off the bi-filar winding A becomes the power source and it has to discharge its stored energy.  The "load" in this case is the switched-off transistor, bi-filar winding B, and D1, and this is in parallel with the battery.

[I am editing here to add the battery to the discussion when the bi-filar coil A discharges.]

Note the current can also flow through the switched-off transistor and through the battery.  So when the bi-filar winding A is discharging the current can take two paths.

So the load is the switched-off transistor, the bi-filar winding B, and D1 and the battery.   So that's three components in series in parallel with the battery, which is a fourth component.  The vast majority of the power dissipated in the load will go into the component that has the highest resistance.

The component with the highest resistance is the switched-off transistor.  So when the transistor switches off, it instantly gets whacked with almost all of the energy that is stored in bi-filar winding A.

MileHigh