Confirming the Delayed Lenz Effect

[hoptoad]

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.  That's how the current flows.  The battery has no affect on what happens during the inductive energy discharge.

So the load is the switched-off transistor, the bi-filar winding B, and D1.   So that's three components in series.  The vast majority of the power dissipated in the series 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 the energy that is stored in bi-filar winding A.

MileHigh

Sorry but we'll have to agree to disagree. When the transistor turns off, current is fed by winding B into the battery. The stored energy does not create a high voltage into winding A, manifesting as a high voltage across the collector and emitter of the transistor, so long as winding B is connected as shown. You can theorize as much as you like, but actual current measurements show that winding B discharges any stored energy from the coil/s into the supply, when the transistor turns off.

Cheers

[MileHigh]

Hoptoad,

For starters, I corrected my posting and edited it to also include the battery when bi-filar winding A discharges.

When the transistor turns off, current is fed by winding B into the battery.

This is not possible for a simple reason.  Current has to go into the top of bi-filer winding A as per your schematic.  It can come from either bi-filer winding B or from the battery.  There is no opportunity for current to flow from winding B into the battery.

Note the discharging coil sets up a clockwise current flow in the left loop and a counter-clockwise current flow in the right loop.  For both of the loops, the transistor is the one thing with the highest resistance therefore it burns off most of the energy stored in bi-filar winding A.

If you have a chance you might want to investigate this again.  It looks to me like you should see a high-voltage spike across the transistor when it switches off.  That would be telling you that the transistor is being whacked by the discharging coil.

MileHigh

[hoptoad]

Hoptoad,

For starters, I corrected my posting and edited it to also include the battery when bi-filar winding A discharges.

This is not possible for a simple reason.  Current has to go into the top of bi-filer winding A as per your schematic.  It can come from either bi-filer winding B or from the battery.  There is no opportunity for current to flow from winding B into the battery.

Note the discharging coil sets up a clockwise current flow in the left loop and a counter-clockwise current flow in the right loop.  For both of the loops, the transistor is the one thing with the highest resistance therefore it burns off most of the energy stored in bi-filar winding A.

If you have a chance you might want to investigate this again.  It looks to me like you should see a high-voltage spike across the transistor when it switches off.  That would be telling you that the transistor is being whacked by the discharging coil.

MileHigh

Indeed you will see a high voltage spike across the transistor when it switches off, if coil B is not connected via a diode as shown.
But with coil B connected via the diode as shown, you will not see a great big voltage spike across the transistor, but you will see current from winding B  feeding into the supply, indicated by meters, and also because that's the only way coil B can discharge, when the transistor is off.  But don't take my word for it. Do it. Check it with meters. Theory is one thing, practical application is another. Fig 19 has been replicated by a great number of people on this forum and others, and every person who has tried it, reports the same. I_ron (who hasn't posted here for some time) tried it only very recently and reported to me via personal email, that everything I outlined would happen, did in fact happen.
A reduction in current consumption during on time, a surge of current from winding B back into the supply during off time, and an increase in motor torque. Not O/U of course, just a measurable increase in overall efficiency compared to using only a single wire drive coil.

Cheers

[MileHigh]

Hoptoad:

I'll take your word for it but it would be nice to see the data also.  It's a kind of tricky circuit because of the bifilar coil.  It's very hard to visualize this stuff in your head once you have more than just a handful of components to look at.  Perhaps one day I will try to simulate it with pSpice also.

MileHigh

[hoptoad]

Hoptoad:

I'll take your word for it but it would be nice to see the data also.  It's a kind of tricky circuit because of the bifilar coil.  It's very hard to visualize this stuff in your head once you have more than just a handful of components to look at.  Perhaps one day I will try to simulate it with pSpice also.

MileHigh

Please - as I say on my web-site - don't take my word for anything - please try it yourself.
Yes, what actually occurs with fig 19 seems counter intuitive, which is one of the nice things about it. It begs questions to be asked.

Cheers