Sharing ideas on how to make a more efficent motor using Flyback (MODERATED)

[poynt99]

Do as i did Itsu,and you will see if it is the negative induced voltage that is the cause for the current decrease. All you have to do is measure(scope) the voltage across the coil during the on time,with and without the rotor in play.

Seems you missed what I said in my post.

I clearly stated that you will not be able to measure any change in induced voltage across the coil because you can not isolate the coil's inductance from its resistance in your measurement. You are measuring across a relatively low impedance during the ON time, i.e. that of the battery and the ON resistance of the transistor or MOSFET, so that is why you can not see a change in the battery voltage. However, you can measure the current, and clearly it has decreased. Why? Because the voltage induced by the rotor during that portion of the cycle is in direct opposition to the battery voltage, and this reduces the current. Reduced current through the coil's DC resistance means less waste energy burned by the coil.

Starting at ground potential, and through the battery to the coil load on down to the collector, the polarities of the voltages are as follows:

- +(bat)-->+ -(coil)

[MileHigh]

Indeed, the measurable voltage across the coil is effectively clamped to the battery voltage when the MOSFET is ON.  However, the coil still can generate a "hidden" internal EMF that ultimately affects the rate of rise of current through the coil.  By observing the rate of rise of current through the coil you can make an indirect measurement of the EMF generated inside the coil that is in opposition to the battery voltage.

And like I stated a while back some of that "hidden" opposing EMF can be directly attributed to the pushing on the rotor magnets to do mechanical work, i.e.; the export of energy to the outside world with each battery voltage pulse and associated current pulse.  There are many ways to look at it and describe it - I like to use the metaphor of some of the battery voltage during the pulse being "eaten" by the export of energy to the outside word - leaving less remaining battery voltage to go towards increasing the current in the coil.  That is effectively the motor action.

[sm0ky2]

Off course, for the components to get hotter, there must flow a greater current through them, but its not easy to individually measure the current through each component.
.....
Another thing i noticed and needs further investigation is the fact that it "looks like" that the increase in current in the whole system starts immediately when the rotor is "out of sync" with the SG, so we see already an increased current while the rotor is still running.
This to me supports the  "negative-induced voltage in series with the battery voltage", or "EMF superposition" as it needs to be in sync with driving the rotor to have that effect.

Regards Itsu

yes exactly
its like when the brake pads on your car are rubbing against the rotor. it adds "friction" that heats up the system.

when the phase shift drifts in and out of synch, there are associated changes in instantaneous current, frequency, and duty cycle.
each of them can be plotted on a graph to "see" whats happening at each moment in time.
current goes up when the EM fields are working against each other, then lowers as they synch up and momentum works with it.
each "wobble" has its own forces involved, which shows up as spikes, or partial wave-forms, over each instant.
load on the rotor will create heat and increase current for the same reason.

Superposition analysis makes this process easier to deal with, and enables us to see the end result or "cumulative effects"
without all the tedious calculations in between.
kind of like a "black-box" approach

In either case we see that Both frequencies must be "synchronized" for maximum efficiency.

the "ideal" situation, would be to try to stabilize the rpm near the max. at freqs. when current (with respect to duty cycle) are at a min.

Trial & Error is not the right way to do this sort of thing, and while it might seem like the answer to just turn knobs and make adjustments until you find something that works....
This approach has, and always will only lead to more questions, and more misunderstanding.

It is better to calculate the frequencies that work together, and know that all others Don't.
and set the parameters of your system, based on That.
Only then can we establish the appropriate mathematical baseline.

[itsu]

This is true,and a very well equipped lab would be required,or maybe put the whole device in a box,and calculate all heat dissipated from each component as one device?.

I am not sure that verpies was agreeing with Poynt's theory,or the fact that verpies like to call this negative-induced voltage the EMF superposition voltage?. At any rate,if you look at Poynts own test he carried out,you can see that his theory on the negative-induced voltage being the cause of the effect is not correct. Below are his scope shot's,and you can see that while the P/in may have gone down with his simmed rotor in play,the P/out also went down(i have drawn a white line across the inductive kickback across the 200 ohm resistor). So what poynt achieved was to reduce the P/in,and at the same time reduce the P/out. So the overall efficiency remained much the same.
In my(and yours) with the rotor in play,we see a decrease in P/in,but an !increase! in P/out--as seen by the width of the inductive kickback spike in my below scope shot's,and also in the video's when using both the scope and DMM's to measure P/in and P/out.
So no-Poynts test did not show the results we are seeing.

Yes-exactly
Some of the EE guys are saying that the rotor is an energy storage device,and it is just returning it's energy back into the system.
I have asked the question many times now-->how can the rotor return more energy than it took to spin it in the first place?,not to mention the fact that there will be losses in the rotor due to windage and bearing friction. But some how we can put X amount of energy into the rotor,and get Y amount out .

Do as i did Itsu,and you will see if it is the negative induced voltage that is the cause for the current decrease. All you have to do is measure(scope) the voltage across the coil during the on time,with and without the rotor in play.

I to have seen a massive increase in current when the rotor is not in sync,and this is because the coil is now trying to suck the rotor back toward it,or pushing against the apposing magnet on the rotor. Once it becomes out of sync,it will stop very quick-or quicker than a free wheeling spin down.
At a 13% duty cycle,i cn get my rotor to !hunt!,meaning that the rotor is falling in and out of sync--you can hear it hunting. At the point where you can hear it out of sync,you also see that the input current go's up,and when it go's quiet(back in sync),the input current go's down again.


Brad

Brad,  here i have the "on time" shown when running (white trace) and when stopped (purple trace) on top of each other.
It does not show any difference, only the spike (outside the "on time") is higher.
Its point C compared to point B (ground lead) in the earlier shown diagram.


Itsu

[tinman]

Seems you missed what I said in my post.

I clearly stated that you will not be able to measure any change in induced voltage across the coil because you can not isolate the coil's inductance from its resistance in your measurement. You are measuring across a relatively low impedance during the ON time, i.e. that of the battery and the ON resistance of the transistor or MOSFET, so that is why you can not see a change in the battery voltage. However, you can measure the current, and clearly it has decreased. Why? Because the voltage induced by the rotor during that portion of the cycle is in direct opposition to the battery voltage, and this reduces the current. Reduced current through the coil's DC resistance means less waste energy burned by the coil.

Starting at ground potential, and through the battery to the coil load on down to the collector, the polarities of the voltages are as follows:

- +(bat)-->+ -(coil)

If we look at your scope shots,and the way you have tried to show the effect,then yes,youe explanation may fit. But it dose not fit in with my DUT. If we look at your scope shot's,then we can clearly see that the reduction in current flow on the P/in side has also resulted in a reduction of P/out on the output side. So if what you say is true,then i should also see a reduction on the P/out side of my DUT--which i do not. In fact,i see an increase in current flow from the output,and this can be clearly seen in the two scope shots below. You can see that the inductive kickback current flow time duration is longer with the rotor in play,but the current flow on the input side is lower.

You show less P/in and less P/out with your simulated rotor
I show less P/in,but more P/out with my real rotor.

You say to me on the other thread that-Quote:As I said Brad, the simulation could use some fine tuning to not only increase the effect, but maintain or increase the flyback power as well.

And yet my DUT is just a throw together job with no fine tuning,and i get the results you cannot--> i wounder how far i could go with some fine tuning.

Like Itsu said-the rotor seems to be putting more energy back into the system than it takes to run it. The induced reverse voltage has nothing to do wit it as far as i can see,as when the transistor switches on,the coil is shorted across the battery,and that induced reverse voltage is no longer there. Lets say that what you say is true,and as you can see in my scope shot,the voltage across the coil battery combo is at close to 10 volts when the transistor switches on. So now we dont see a reduction in voltage across the coil,as when the transistor switches on,the battery voltage is across the coil,but in stead,we see a reduction in current flow. In your sim,this results in a reduction of current flow on the output side as well--as it should,but in my DUT,we see not a reduction,but an increase in current flow from the output-all while the coil is putting energy into the rotor,so as the rotor can return this energy into the coil.

Brad