Rotating Magnetic Field's and Inductors.

[poynt99]

The scope across the coil says the induced EMF from the magnets has no effect on the peak voltage across the coil when the transistor switches on.

That may be so. However with coils it's often better to analyze current, and you can see that the current wave forms are different in each case. You can also see that the rotor would be inducing a positive voltage in the coil during the ON pulse. This manifests as a current that is in opposition to the battery's current and is why the battery current is reduced when the rotor is in place.

That is actually incorrect.
The inductance rises when there is a changing flux value in the core of the inductor/ over time. Only when the flux value is constant,is there a reduction in the inductance value. As the magnetic flux is never a constant value in the core of the inductor when the rotor is in play,then the inductance value of the inductor is indeed higher than it would be without the rotor.

Do you have data or a technical reference to back that up? It goes against the physics of how cores work. If the core's magnetic domains are anywhere other than their neutral position (i.e. non-polarized) at the instant the coil fires, then the coil's inductance will be reduced, regardless if the domains were in rotation or were static at the time. The core is partially or fully polarized in either case.

Thought experiment: Let's say that the magnets in the rotor are strong enough and the core's reluctance low enough that the core is fully saturated just at magnet/core alignment. If you were to fire the coil just before its core is saturated, and when the rate and degree of core polarization is at maximum, is the coil inductance going to be higher or lower than it is when far from saturation?

[poynt99]

I would need another inductor that has a rising and falling magnetic field to mimic the magnetic field or the rotor,and that is of the opposite field to that of the main inductor. The timing would also have to be in advance to that of the main inductor,so as the electromagnet(second inductor) induces a changing magnetic flux into the core of the main inductor before the main inductor switches on.

If this showed the same effect as the rotor,then what would the outcome be?-in that the electromagnet would consume power,where as the permanent magnets on the rotor do not.\\

Brad

I have one SS version to offer here, and you are close. Mine uses another coil yes, but it doesn't require another power source.

Your ssg rotor is an electro-mechanical energy transfer (bi-directional) and storage device. My offering does without the mechanical bit which is replaced by a resonant LC tank circuit. The bi-directional transfer bit is accomplished via transformer coupling to/from the pulse coil.

Schematic: As I don't know the specifics of your circuit and setup, I made educated guesses on values that I thought would work. With some tweaking it seems to come close. I have two of these running at the same time, one with R5=12 Ohms (with rotor), and one with R5= 12Giga-Ohms (without rotor). This way I can compare the two scenarios simultaneously as you will see.

First two scope shots: Collector voltage with and without rotor.

3rd scope shot: Pulse coil current with and without rotor. Current with rotor clearly shown to be less (Red). The purple trace is the rotor current shown in phase with pulse coil current during ON time. The rotor returns some stored energy back to the pulse coil reducing the current required of the battery.

4th scope shot: If you could scope across the coil voltage (which you can not but I can because the inductance and coil resistance are separate in my circuit), you would see the increased voltage with the rotor.

The Pin difference between the two scenarios is only about 10% less with the rotor. I'm sure with tweaking and non-linear cores (these are ideal air core), someone could make the effect more pronounced. That someone won't be myself. The effect has been verified AFAIAC, and confirms my explanation offered earlier. My Conclusion? Nothing extraordinary going on here.

So Brad, grab a MOT and have a stab at making this SS version if you are so inclined.

[tinman]

That may be so. However with coils it's often better to analyze current, and you can see that the current wave forms are different in each case. You can also see that the rotor would be inducing a positive voltage in the coil during the ON pulse. This manifests as a current that is in opposition to the battery's current and is why the battery current is reduced when the rotor is in place.

I am replying to this post,but here is a quote from your next post that needs to be looked at.

3rd scope shot: Pulse coil current with and without rotor. Current with rotor clearly shown to be less (Red). The purple trace is the rotor current shown in phase with pulse coil current during ON time. The rotor returns some stored energy back to the pulse coil reducing the current required of the battery.

So where did this stored energy come from?
If we are to assume that there should be losses due to windage and friction associated with the rotor,then the rotor should return less than it takes to drive it,and yet we see a decrease in P/in while P/out remains the same. Your own scope shot's and experiments show that the rotor(permanent magnets) are reducing the P/in because they are returning !! stored energy !! back to the system. In order for the rotor to do that,it first must have the same(probably higher due to losses mentioned above) amount of energy put into it in order for it to have energy to return--so where is this input energy to the rotor that is being returned to lower the electrical input energy of the system?. Keep in mind that the output(electrical) energy remains very constant at all times.

Do you have data or a technical reference to back that up? It goes against the physics of how cores work. If the core's magnetic domains are anywhere other than their neutral position (i.e. non-polarized) at the instant the coil fires, then the coil's inductance will be reduced, regardless if the domains were in rotation or were static at the time. The core is partially or fully polarized in either case.

In order for the magnetic domains to be aligned,then work must be done. If these domains have been aligned before the coil switches on,then less work is required from our input power to align these domains,as the PM on the rotor has already done this.

The tests i have done are as follows;
1- simply to use my inductance meter to measure inductance,and find that as i slowly bring a magnet toward the inductor,the inductance rises. Then while the magnet is stationary at the center of the core,the inductance go's back to it's original value-or very slightly lower. Then when the magnet is slowly moved away from the core,the inductance once again rises.
2- Set the coil into self oscillation (as it is bifilar,and so easy to do),then slowly bring a magnet toward the core of the coil. As the magnet is approaching the core,the frequency rises. As the magnet becomes stationary at the center of the core,then the frequency returns back to it's starting frequency,and then as i move the magnet slowly away from the core,the frequency once again rises until the magnet is far enough away from the core that no field from the magnet can induce the core-at which point the frequency has once again returned back to it's steady state.

Thought experiment: Let's say that the magnets in the rotor are strong enough and the core's reluctance low enough that the core is fully saturated just at magnet/core alignment. If you were to fire the coil just before its core is saturated, and when the rate and degree of core polarization is at maximum, is the coil inductance going to be higher or lower than it is when far from saturation?

My experiments so far show the inductance is higher.


Brad

[poynt99]

So where did this stored energy come from?

Surely you jest?

If we are to assume that there should be losses due to windage and friction associated with the rotor,then the rotor should return less than it takes to drive it,and yet we see a decrease in P/in while P/out remains the same.

Yes, of course.

Your own scope shot's and experiments show that the rotor(permanent magnets) are reducing the P/in because they are returning !! stored energy !! back to the system. In order for the rotor to do that,it first must have the same(probably higher due to losses mentioned above) amount of energy put into it in order for it to have energy to return--so where is this input energy to the rotor that is being returned to lower the electrical input energy of the system?. Keep in mind that the output(electrical) energy remains very constant at all times.

Yep.
What you need to remember is that your system is very inefficient to begin with. What is the DC resistance of your coil, 2 Ohms or so? Guess where a good portion of your input power is being burned up? In my circuit, Pin is about 3.5W, and 2W of that is being burned up in the coil's 4 Ohm resistance. Any reduction of current is going to reduced the I²R loss in that resistor. btw, your rotor is a very efficient storage device. Yeah there are friction losses, but they are relatively low.

Your rotor is making the system more efficient, or less lossy. My LC tank does the same thing. Take it as it is, there is no miracle action at hand here, just good old fashioned electrical theory. We all went down a similar road years back with Luc's Capacitor energy transfer experiments, which I also did a big paper on.

You seem so hell bent on proving that magnets do work. Well in this case the process is completely conservative, some energy in, some energy back out. There is no free lunch provided by magnets here!

In order for the magnetic domains to be aligned,then work must be done. If these domains have been aligned before the coil switches on,then less work is required from our input power to align these domains,as the PM on the rotor has already done this.

It takes a lot more work to burn that 4 Ohm resistance than it does to polarize the domains in the core. It pales in comparison. Most of your loss is in the coil resistance.

The tests i have done are as follows;
1- simply to use my inductance meter to measure inductance,and find that as i slowly bring a magnet toward the inductor,the inductance rises. Then while the magnet is stationary at the center of the core,the inductance go's back to it's original value-or very slightly lower. Then when the magnet is slowly moved away from the core,the inductance once again rises.
2- Set the coil into self oscillation (as it is bifilar,and so easy to do),then slowly bring a magnet toward the core of the coil. As the magnet is approaching the core,the frequency rises. As the magnet becomes stationary at the center of the core,then the frequency returns back to it's starting frequency,and then as i move the magnet slowly away from the core,the frequency once again rises until the magnet is far enough away from the core that no field from the magnet can induce the core-at which point the frequency has once again returned back to it's steady state.

What happens to the frequency as you are moving the magnet toward the core, then you stop before getting to it?

[tinman]

Surely you jest?
Yes, of course.
Yep.
What you need to remember is that your system is very inefficient to begin with. What is the DC resistance of your coil, 2 Ohms or so? Guess where a good portion of your input power is being burned up? In my circuit, Pin is about 3.5W, and 2W of that is being burned up in the coil's 4 Ohm resistance. Any reduction of current is going to reduced the I²R loss in that resistor. btw, your rotor is a very efficient storage device. Yeah there are friction losses, but they are relatively low.

or less lossy. My LC tank does the same thing. Take it as it is, there is no miracle action at hand here, just good old fashioned electrical theory. We all went down a similar road years back with Luc's Capacitor energy transfer experiments, which I also did a big paper on.

You seem so hell bent on proving that magnets do work. Well in this case the process is completely conservative, some energy in, some energy back out. There is no free lunch provided by magnets here!
It takes a lot more work to burn that 4 Ohm resistance than it does to polarize the domains in the core. It pales in comparison. Most of your loss is in the coil resistance.

What happens to the frequency as you are moving the magnet toward the core, then you stop before getting to it?

The frequency returns to that as if there were no PM's near it.

Your rotor is making the system more efficient,

Thank you.
Here is my very first paragraph in this thread
Quote: I posted a quick video showing how having a rotor with alternating magnetic field passing a pulsed inductor can improve the efficiency of that inductor as far as the inductive kickback output go's.


Brad