Electric Motor Secrets

[Robert49]

Hi Peter

Thanks for your response , it is greatly appreciated.
I've been following your work for years and Tom Bearden's ,John Bedini's and Joe Newman's.
I must say that the inspiration came from all of you.

I intend to try it at higher voltages as I did with my previous motor and will show the results. I have noticed that the higher the voltage, the higher is the percentage of power gain.

Thanks again

Robert

[Peter Lindemann]

Smaller Percentage....

Hi Peter

Thanks for your response , it is greatly appreciated.
I've been following your work for years and Tom Bearden's ,John Bedini's and Joe Newman's.
I must say that the inspiration came from all of you.

I intend to try it at higher voltages as I did with my previous motor and will show the results. I have noticed that the higher the voltage, the higher is the percentage of power gain.

Thanks again

Robert

Robert,

Yes, that is true. As the input voltage rises, the fixed voltage drops in the components become a smaller and smaller percentage of the total, therefore, the percentage gain goes up.

Out of curiosity, what is the air gap between your rotor and stator?

Peter

[Robert49]

Robert,

Yes, that is true. As the input voltage rises, the fixed voltage drops in the components become a smaller and smaller percentage of the total, therefore, the percentage gain goes up.

Out of curiosity, what is the air gap between your rotor and stator?

Peter

Peter

I will have to take the motor apart to measure the air gap but I can say it is approximatly 0.5mm

Robert

[gyula]

Hi Robert,

It is very good you continue your good 'habit' of testing such setups this year too, thanks for showing this latest 'wonder'.

My questions would be:

- the V2=196V voltage is an unloaded situation when the bulb is disconnected?
- when you try to load the motor shaft mechanically a little, how does the A1=1A input current draw changes?
- how 'clean' the output of your 100V DC supply is I wonder? You can trust in it because it is a well regulated and filtered DC source?

Thanks, Gyula

[Robert49]

Hi Gyula

V2 is loaded with bulb otherwise it would go up to over 2000volts.

The supply voltage is from a 15amp variac going through a 25amp bridge rectifier and filtered with a 220uf cap.

But I have to tell you that I found out my input ampmeter was bad so I have to redo the tests. I will test all my meters just in case.

One thing I forgot to say is the current going through the bulb is also going in the motor coils thus adding to the power of the motor.


Robert

[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

[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 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

[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

[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]

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,

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]

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]

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]

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