Electric Motor Secrets

[Peter Lindemann]

Great Work!!

Hi gyula

I did not mean "input power" . I meant effective power inside the motor.

I think we agree on everything without knowing. To make an analogy, we are in the same car going in the same direction but one of us is looking through the windshield and the other through the rear window.

In any case I think I should do the torque test as Peter Lindemann showed in his video .

Thanks

Robert

Robert,

Sorry I've been gone a few days. A lot has happened in this thread. 69% recovery in your tests is very good and on the high side of what I was saying is to be expected. As you move toward the dynamometer test, just realize that this motor has a completely different speed/power curve than a standard DC traction motor. Your best power measurement will NOT BE at 1/2 of idle speed, but will probably appear at a higher speed. Start by loading it to about 80% speed and see what you get. Then load to 70% speed, then 60% speed, etc. Mechanical energy production is inversely proportional to air gap, so cutting the air gap in half will double the torque (as a general rule).

Currently, your air gap is quite large for a motor operated on magnetic attraction, as .5mm is about 0.019 inches. Torque would be 4 times more if the total gap was about 0.005 inches. In the Flux Motor model we built in Santa Barbara in 1983, we were testing air gaps in the 0.002 range with extremely excellent results

I love your design and implementation, though.

Best regards,
Peter

[Robert49]

Oh, indeed, you used effective power inside the motor and somehow my mind read it as "input power" ... my bad.

The power inside the motor indeed increases whenever you utilize the collapsing field of the coils because the current from that field flows also through the coils, not only via the bulb load.

Earlier you wrote that "this motor has no dead time: there's always current flowing into it because the pulses are overlapping."
Was this so also in your latest tests too? I mean by manipulating this and the on time you could also increase efficiency.

Gyula

Gyula

After I posted "no dead time" I modified the timing wheel and now there's a 5 degree dead time. I had to do this because the overlapping pulses were creating drag and the amperage was too high.
I intend to make a variable timing wheel so I can adjust it.

Thanks

Robert

[gyula]

Oh, indeed, you used effective power inside the motor and somehow my mind read it as "input power" ... my bad.

The power inside the motor indeed increases whenever you utilize the collapsing field of the coils because the current from that field flows also through the coils, not only via the bulb load.

Earlier you wrote that "this motor has no dead time: there's always current flowing into it because the pulses are overlapping."
Was this so also in your latest tests too? I mean by manipulating this and the on time you could also increase efficiency.

Gyula

[Robert49]

Hi gyula

I did not mean "input power" . I meant effective power inside the motor.

I think we agree on everything without knowing. To make an analogy, we are in the same car going in the same direction but one of us is looking through the windshield and the other through the rear window.

In any case I think I should do the torque test as Peter Lindemann showed in his video .

Thanks

Robert

[gyula]

...
I just tried it and they both agree on 2.7amps

Thanks

Robert

Hi Robert,

Because the bulb was connected to the right hand side, A1 measured the total input (1.8A) and did not measure the sum of A2 and A1 (2.7A), right? So the input power can be calculated as 50V*1.8A=90W in this test. So I agree with your test.

HOWEVER, in your post #1584 you wrote this:

"If I put the bulb wire before A1 , I get 3.55a on A1 which means the effective power going in the motor is 130v x 3.55a =461 watts ??"

And this is the input power calculation what I do not agree with because in that setup you considered the SUM of A1 and A2 when figuring out input power.
First the bulb was connected to the right hand side of A1 and input current was A1=2.2A from V1=130V and A2=1.35A from V2=230V. So in this latter case input power was 130V*2.2A=286W, ok?

And when you put the bulb wire before A1 (i.e. to the left hand side of A1) then A1 showed 3.55A which is okay that it showed the sum of the earlier values A1 and A2 but you cannot use this summed current for calculating input power: this is what I do not think is correct. How would input power change from 286W to 461W by merely connecting the bulb wire before a current meter? A good current meter must be a short circuit for the current, it cannot cause such a huge change in power.
Is this okay? Maybe you meant something else with your question and I misunderstood.

Thanks, Gyula

[Robert49]

Gyula

"May I suggest you use a third Ammeter, say A3, in your above test setup for checking the input current which is actually drawn in the +V rail, inserting it to the most left hand side of the common point of the bulb and A1? So the bulb wire would go to the left hand side of A1 and and to this common point one wire of the A3 would be connected and the other wire of A3 would go to +V1 directly."

I just tried it and they both agree on 2.7amps

Thanks

Robert

[gyula]

Yes it was connected to the right hand side.

Robert

In the meantime I edited my post above please consider it. We posted in the same time

[Robert49]

Hi Robert,

Would like to clarify: in your above test the bulb wire was connected to the right or the left hand side of the Ampmeter?
Reference: In the schematic you showed in Reply #1574, the bulb was connected to the right hand side of the Ammeter A1.

Gyula

Yes it was connected to the right hand side.

Robert

[gyula]

Hi Robert,

Would like to clarify: in your above test the bulb wire was connected to the right or the left hand side of the Ampmeter?
Reference: In the schematic you showed in Reply #1574, the bulb was connected to the right hand side of the Ammeter A1.

EDIT: May I suggest you use a third Ammeter, say A3, in your above test setup for checking the input current which is actually drawn in the +V rail, inserting it to the most left hand side of the common point of the bulb and A1? So the bulb wire would go to the left hand side of A1 and and to this common point one wire of the A3 would be connected and the other wire of A3 would go to +V1 directly. This way all the guesswork would be over.

Gyula

[Robert49]

Gyula

"Regarding your last question on the 461 Watts input power: when you connect the bulb wire to the left hand side point of A1 (instead of the right hand side as shown in your schematic), the Ampmeter A1 measures not only the input current but the bulb current too . And I do not think it is correct to calculate the input power from this current (3.55A) measured that way and from the 130V input voltage. The input current must be measured in the input wire rail (either in +V or in -V) and no any branch wire should be connected to the left hand side (here in your case) of the meter."

I made a little test to show what I mean.
V1=50v
A1=1.8a
V2=94v
A2=0.87a
I put an ampmeter on one of the coils and it is showing 0.302a
0.302a x 9 coils =2.718a
Now add A1=1.8a to A2=0.87a ,you get 2.67a
The difference from 2.67a and 2.718 is fluctuations in measurements due to fluctuations in speed.

How does that sound to you?

Robert

[gyula]

Hi Robert,

Okay on your answers, thank you and I agree with them. Returning to my first question: V3 is indeed the difference between V2 and V1, and in my mind somehow I considered V3 as if one point of V3 had been connected to the negative rail instead of the positive rail, as you actually drew in the schematic for 100V input, my bad. So the V3=96V across the bulb back then in that setup was surely correct, in fact very nearly correct because only a 0.7-0.8V diode forward voltage drop must have been the small difference between the actual bulb voltage and the voltage difference of V2-V1 (assuming you used an Si diode, that is).

In this setup where you now use the 'beefier' capacitors, the input power is 286W and the bulb power is 135W [(230V-130V)*1.35A] because we have to consider the V2-V1 voltage difference for the bulb, right? While in your hand-loaded shaft case the input power went up to 520W and the bulb power also increased to 204W [266V-130V)*1.5A], it would be good to know the mechanical load in Watts you exerted on the shaft. This latter would call for a Proney brake test.

Regarding your last question on the 461 Watts input power: when you connect the bulb wire to the left hand side point of A1 (instead of the right hand side as shown in your schematic), the Ampmeter A1 measures not only the input current but the bulb current too . And I do not think it is correct to calculate the input power from this current (3.55A) measured that way and from the 130V input voltage. The input current must be measured in the input wire rail (either in +V or in -V) and no any branch wire should be connected to the left hand side (here in your case) of the meter.

I would have one more notice: it is interesting that in your setup where the bulb is connected to the -V rail (ground) instead of the +V rail, (post #1581 above) and the input voltage was 79V, the efficiency was better than it is now. It was 69% then and now it is 47.2% (no load was on the shaft in both cases). Of course I may have erred in my calculations above, please anyone speak up if I have.

I believe the efficiency can be improved when you have got a variable on time for the switches. And also by decreasing the air gap (which is now cca 0.5mm) between the rotor and stator as Peter said earlier for the attraction motor.

Greetings,
Gyula

[Robert49]

Hi Gyula.

I have changed the caps: 820uf on input and 3300uf on recuperation.
The setup is the same as in post #1574
I ran the motor at 4100rpm

V1=130v
A1=2.2a
V2=230v
A2=1.35a

As you can see, the A1 has dropped down considerably.
All measurements double checked with different meters.
I think the 3300uf cap bank made a big difference.Thanks a lot for the suggestion Gyula.
If I put a load on the shaft with my hand to half speed, A1=4a instead of 2.2a but a2 becomes 1.5a and V2=266v.
If I put the bulb wire before A1 , I get 3.55a on A1 which means the effective power going in the motor is 130v x 3.55a =461 watts ??

And by the way, the motor has a smoother sound now!!
Please comment Gyula and Peter

Robert

[Robert49]

Hi Gyula.

"I asked about the bulb load presence because I thought there was too big voltage difference between V2 and V3, it was 192V-96V = 96V. This may also indicate a problem in voltage measurement: there is only a series diode between A2 and one 'leg' of the bulb."

V3 is measured across the bulb and is the difference of potential between V2 and V1.

"OR the voltage meters showed correct levels but the problem could also be that the V2 voltage is not well filtered by the capacitors you indicated in the schematic between V2 rail and the negative ground. A scope shot would reveal how clean or smooth V2 is across the capacitor bank, if it contains big sawtooth-like waveforms, then the voltage difference could be much more understandable between V2 and V3 because then the waveforms would average out across the bulb via the series diode, especially when you place some 47 or 100 uF electrolytic capacitors across the bulb to filter further the recovered voltage."

I took a peek on the cap bank with the scope and Yes it needs more filtering.
So I will change the 66uf 2500v cap bank to 3300uf 450v and take a scope shot.

"On the input side the 220 uF puffer cap sounds good at first but perhaps still a low value due to the probably several amper peak current draw at each switching event. You may wish to use at least one more 220 uF in parallel, especially when you wish to increase the DC supply voltage above 100V."

I will change the cap on the supply to a 820uf.

"Just noticed your new measurements: I wonder what is the V3 amplitude across the bulb in this situation? (try to filter V3 too if you agree) You surely know that the bulb can directly draw current from the input DC voltage via any coil when the associated switch is off: what if you reconnect the bulb to the positive rail as before?"

The bulb draws current through the coils just for a short time and then when the potential of V2 equals or exceeds V1 , it only draws the current from the very high potential of the spikes. I have already verified that.

"I assume you adjusted the ON time for this situation too? Earlier you meantioned you still had a 5 degree dead time between pulses: was it so now too?"

The on time is not yet adjustable but it is coming soon.

Hope this answers your qustions
Thanks

Robert

[gyula]

Hi Robert,

Thanks for the answer and sorry to hear about the bad ampmeter. But it is very good you noticed it.

I asked about the bulb load presence because I thought there was too big voltage difference between V2 and V3, it was 192V-96V = 96V. This may also indicate a problem in voltage measurement: there is only a series diode between A2 and one 'leg' of the bulb. OR the voltage meters showed correct levels but the problem could also be that the V2 voltage is not well filtered by the capacitors you indicated in the schematic between V2 rail and the negative ground. A scope shot would reveal how clean or smooth V2 is across the capacitor bank, if it contains big sawtooth-like waveforms, then the voltage difference could be much more understandable between V2 and V3 because then the waveforms would average out across the bulb via the series diode, especially when you place some 47 or 100 uF electrolytic capacitors across the bulb to filter further the recovered voltage.
On the input side the 220 uF puffer cap sounds good at first but perhaps still a low value due to the probably several amper peak current draw at each switching event. You may wish to use at least one more 220 uF in parallel, especially when you wish to increase the DC supply voltage above 100V.

Just noticed your new measurements: I wonder what is the V3 amplitude across the bulb in this situation? (try to filter V3 too if you agree) You surely know that the bulb can directly draw current from the input DC voltage via any coil when the associated switch is off: what if you reconnect the bulb to the positive rail as before?

The hand load test shows the input puffer capacitor may indeed need some beefing-up, now that you know about the 3-4 Amper average current draw.

I assume you adjusted the ON time for this situation too? Earlier you meantioned you still had a 5 degree dead time between pulses: was it so now too?

Greetings,

Gyula

[Robert49]

Hi guys.

Here are the real measurements .
The bulb return is now to ground instead of +V

No load on shaft.
V1= 79v
A1= 3.7a
V2= 130v
A2= 1.55a
About 69% recovery??

Motor speed reduced to half with hand.
V1= 77v
A1= 4.44a
V2= 136v
A2= 1.57a
About 63%??
Not so amazing after all but still Beautifull!!!!
But I'm not giving up!

Robert