Electric Motor Secrets

[Harvey]

Hi Robert,

As always I applaud your craftsmanship

The readings you were able to take look quite promising - but I am interested in the actual timing involved. Do you have a scope?

From what I can see in a cursory look, you have done a good job in capturing the energy and putting it to work in the third coil motor.

Keep up the good work!

[Robert49]

Hi Harvey.

Thank you for the good comment.

I seem to double the power of my motor. I intend to try driving a car alternator with it and see how it goes.
I'm retiring at the end of this year (65) so next year is full time on my projects.

Robert

[Harvey]

Thank you Robert

I know it is difficult to get the traces to light up sufficiently and still maintain a crisp defined signal especially when it is varying consistently. You have a nice shot there to show the voltage differentials. Unfortunately, I am unable to determine the precise timing and the vertical transitions are difficult to ascertain. One thing is certain however in both shots, you have 6 total firings every 5 divisions (of course we are only seeing two of the three MOSFETS) so if we had your time-base settings we could determine the RPM etc.

What I was looking for was a lengthening of the Q1(upper left) signal compared to the Q3 (lower center) as this could account for some of the increased wattage. This is because the actual on-times could differ for the two devices but the power over time could be the same even though the voltage and current is different when differentiated.

After reviewing the process involved, I believe part of the increase could be via the diodes and stray flow through Coils 1 & 2 during the Coil 3 firing. In order for this to happen, the capacitor bank drops sufficiently to allow the two diodes to conduct - so it would occur at the latter part of the Coil 3 cycle. In this situation Q1 and Q2 are both off, but the two coils will conduct in parallel through the diodes (D1 & D2 *) once the voltage is sufficiently dropped in the capacitor bank. This increases the A1 thru A3 reading but because it is a fractional part of the waveform the meters may not be differentiating the signal as well as we may like. Whatever the case, this is an undesirable result because all three coils are active simultaneously.

There are ways to keep this from happening but it would require some changes in the circuit - some means to prevent Q3 from turning on if D1 or D2 are conductive. (* I've assumed D3 is at the lamp)

I love the simplicity of the design

[Robert49]

hi Harvey

I will modify the circuit soon. I have some high speed solid state relays I can try.

There is probably something else to take into account.
Because of the size of the rotor, at all times there is a closed magnetic circuit between two of the coils which is when #1 is active there is a transformer effect with #3 and when #2 is active the transformer effect is with #1 and when #3 is active the effect is with #2...
That energy must be going through the diodes or cancelling the current from the power supply?????????????
What do you think?

Robert

[Harvey]

Hi Robert,

I have to think on this a bit - things don't come to my mind like they did when I was younger

Let's see - #3 is inversely positioned when #1 is active. However, D1 does not let that side drop much below ground so essentially we light the lamp in reverse and inversely charge the cap that is across #3 with any transformer action there.

When #2 is firing, #1 is isolated because of D2. Even though there is some transformer action there, the upper side is clamped to your B+ line through A1 and the lower side is floating even though it may drive low. I doubt that any spatial (RF or the like) transference of energy exists there.

When #3 fires we have the condition mentioned in my previous post. So any transformer action on #2 is simply additive to the conduction experienced later in the cycle. So its a bit like #1 above during the first part of the cycle and then later conducting if the low side is not too low and the capacitor bank is drained enough to allow conduction to occur.

So I don't think the large rotor is having any real adverse effect - if I'm looking at it right

It's very interesting to consider the magnet flux and timing involved

Are all the MOSFET gate signals similar or can you change the length and timing there?

[Robert49]

Hi Harvey

All gate signals are the same and all I can do is advance or retard the firing of the coils all together.

Fun to experiment

Robert

[Robert49]

Hi guys.

I have made a correction on my diagram on post #1537 . I have some numbers I'd like to share.
At 40volts input on the motor, A1 shows 550ma, A2 shows 200ma, A3 shows 348ma at 52.9volts and that is with the cap and bulb not shorted.

I added a switch on coil 3. When I switch it on, the reading on A3 changes to 552ma at 51.8volts but all other readings remain the same and the rpm goes down a bit.

Any theory ??


Robert

[Robert49]

Hello people.

I have not posted for a while, too busy building a house for a friend.
I'm doing serious testing on my best motor.
If you look at the meters, you'll see 75 volts at 1.1 amp input going into coils 1 and 2 . Coil 1 at 600ma and coil 2 at 500ma 75 volts.
Coil 3 is fed by the energy collected from the collapsing fields of coils1 and 2.
It gets 90 volts at 4 amps.
Input = 75v X 1.1a = 82.5 watts for coils 1 and 2.
Coil 3 = 90v X 4a = 360 watts.
Diode 1 shows 350ma going to cap bank from the drain of the mosfet of coil 1, and coil 2 must be doing the same.
I am shorting coil 3 with a diode and that diode is sending 250ma to the cap bank.

If I load the motor, the speed decreases and then without removing the load, it increases speed by itself.

Robert

[gyula]

Hi Robert,

Thanks for updating your project. It surely behaves unusually, not readily explainable by common knowledge. At least I am puzzled... and perhaps studying very thoroughly the flux connections of the several windings of the motor could shed some more explanation.
I guess you have already tried to extract the total energy from your setup? I mean you load the motor shaft and coil 3 and the capbank. I also guess that summing up the three output possibilities quantitatively, the total output exceeds the input. Is this so? if you have answer that is.
Or you found (if you have already done so) when loading those three outputs at the same time that the motor loses rpm and output voltages/currents drop while input current increases? So how the beast behaves in that respect?

One more thing: you wrote that If I load the motor, the speed decreases and then without removing the load, it increases speed by itself.
so I wonder how the input current changes when the event you describe happens? Will it remain the same like it was before loading the motor or it increases to a higher value?

Thanks, Gyula

[Robert49]

hi gyula.

The total power measured in the motor including input of 82.5 watts is 442.5 watts.
I slow the motor to half speed more or less by applying friction on shaft. The only thing that changes is the input amps from 1.1 amp to 2 amps.
The voltage and amps on coil 3 remain the same.

I should mention that the higher the input voltage, the higher is the ratio of power gain. At 75 volts, coil 3 gets 4.36times input. I tried it on 100 volts and coil 3 gets 5.5 times input. Almost burned the motor.
This leads me to believe that Ed Gray was right in using high voltages on his motors.
My friend Markus has a machine shop. I repair his machines and he helps me by machining parts for me. I plan to build a bigger motor on the same architecture and test it. Some time next year hopefully.
Here's a schematic of the motor and a scope shot of what is happening on the cap bank.
Thanks

Robert

[gyula]

Hi Robert,

Thanks for the additional info. Very clever setup.

Regarding the coil set 3 with its cca 4 Amper current as per measurement: I think it is reactive current with a certain phase angle to the voltage across the coil set so it is okay that Amperturns count in the performance of the motor but to estimate the power involved the phase angle ought to be considered when calculating power. I mean that perhaps further load test is needed whereby the motor shaft would drive an alternator or generator directly and then load the output with a real load.
With this, I do not mean I question the performance, I indicate that more precise measurements are needed to arrive at an answer.

Edit: on the scope shot please vertical scale/divison?

Thanks, Gyula

[Robert49]

Hi gyula

The scope is on 5volt/division.
Since I have no idea on what is happening, all input is appreciated.

Robert

[gyula]

Hi Robert,

Thanks, the step like waveform nicely shows the recharge of the capbank first from coil set 1, then from coil set 2 and then coil set 3 discharges it again to the lowest value of the step and the process repeats. (The average DC level in the capbank is surely somewhere between 170-200V as you occasionally found in tests of course.)

Well I just wanted to indicate that a more precise evaluation of the motor performance would be needed. In case of driving a generator or alternator with the motor as prime mover, then the output power is again electric like the input, and the efficiency of the generator or alternator is to be considered as a loss. For this to be meaningful, a good data on such generator characteristics should be known. Otherwise a Prony brake test could give output power estimation directly for your motor.

You wrote you had no idea what was happening (in the motor circuits I suppose) and I think you know a lot but they still may not be enough to interpret fully the measured voltages and currents. Is this so?
I think the 3-4 Amper in coil 3 can be explained partly by the small inner resistance of the capacitor bank versus that of the 75V power supply and partly by the more than twice as high DC amplitude in the capbank versus the 75V supply. Perhaps you could improve i.e. further reduce the inner resistance of the capbank if you use more parallel brances of the series capacitors, perhaps two 450V caps in series and paralleling three of such two series 'strings' would be better than say 4 or 5 such caps fully in series (unless you really need a 2000-2500V voltage rating instead of say 900V).

These are 'possibilities' on your circuit, far from complete of course.
Some questions: do you know the inner resistances of the analog ampmeters? You could check them (when they are unconnected) with a digital Ohm meter, unless they have some specifications available. Just to know what loss they may cause when Ampers flow through them (they are mainly in series with the coil sets).
What is the DC resistance of a single coil? I assume all the 12 coils are nearly identical, right? And what is the ON resistance of your MOSFET switches? These latter data would be good for the L/R time constant consideration versus the ON time of the switches.
Another question: I know you use a timing wheel to pulse the 3 optocouplers and I wonder at which clock positions the couplers are placed (thinking a clock face): say the first optocoupler for the first coil-set is placed at 12 o'clock, the second one at 3 o'clock and the third one is at 6 o'clock? (and at the 9 o'clock position there is no gap on the wheel?) Just curious.

Thanks, Gyula

[Robert49]

Hi Gyula

It was my intention to drive an alternator with the motor but the motor will not withstand that much current for long as it is wired with #23 awg wire and coil three gets hot very fast at over 4 amps.
This test will have to wait for a bigger version of the motor.

The inner resistance of the ammeters is exactly 100 milliohm.

As for the cap bank, there is already two series of caps in parallel and I found that there is a bit more energy captured with higher voltage caps.

The resistance of a set of 4 coils is 1.2 ohm

My timing wheel has four slots at 90 degrees of each other and the opto sensors are placed at 0-30-60 degrees which means each set of coils is pulsed 4 times per rotation.If you look at post #1536 you will see what I mean.

The on resistance of the mosfets is .29 ohm

I include a scope shot of the caps to show what happens when I slow the motor to about 60% speed. 5v/div
Sorry for the fuzzy scope shot..only two hands.

Thanks,

Robert

[gyula]

Hi Robert,

Thanks for the details. It is good the ammeters have only 0.1 Ohm DC resistance, meaning low loss. However the MOSFET's on resistance of 0.29 Ohm is a bit high when I compare it to the 1.2 Ohm resistance of the 4 coils, a 'rule of thumb' would be about 10% of the coils resistance i.e. around 0.12 Ohm in the present case, to reduce loss across the switch and let more input voltage drive current into the coils (coils and the switch form a voltage divider from the input 75V point of view). Especially important this switch-loss when you wish to use thicker wire for the coils to handle the bigger currents, because with thicker wire the DC resistance of the coils would be reduced, letting more voltage drop develop across the same MOSFET, this causing more dissipation loss.

I have not asked but here the inductance of the coils is to be considered too because of the L/R time constant where L is the 4 coils in series (measured by an L meter) and R is the total DC resistance of the power supply+coils+MOSFET switch(+ammeter). And why is the L/R value is important: you can read here a good explanation when the energy from the collapsing field is to be captured just after the coils current switch-off:
Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze No need to log-in, you can see the attached picture on that forum as a visitor.
And for this L/R time constant question to solve: the width of the gap on the timing wheel is to be chosen accordingly: the best would be to gradually cover or uncover the gap by say insulating tape to experiment and find the best width i.e. the best ON time for the switch, which is about 0.5 to 0.55 times the L/R time constant (half of 1.15) as described in the link.
Sorry if you are already familiar with this topic and have already tinkered with the width of the gap to get the highest captured energy from the flux collapse. If you have questions though, I try to answer.

Okay on the opto sensors placement I understand. Also okay on your capbank's two caps in series and two such in parallel and you surely know that the captured energy also depends on actual resultant capacitor value: higher residual uF value collects less energy (due to smaller DC levels) than smaller uF value if you agree. In fact the voltage level what counts first and the uF value is a secondary issue when considering stored energy in caps.

Your scope shot on the capbank voltage now shows a higher voltage swing due to the heavier discharge under a heavier load. By the way what is the normal rpm for an unloaded case of your motor at 75V input I wonder.

Thanks,
Gyula