Open discussion for projects on this forum.

[Ufopolitics]

You can determine the internal resistance/output impedance of the PSU through making a simple voltage divider on the output. When the measured voltage across the load resistance is exactly half the output voltage, then the load resistance is equal to the PSU output resistance.





Voltage divider - Wikipedia

But on the subject of PSUs, I find it cheaper, easier and more convenient and reliable to get some transformers and regulators and make my own. And if you blow it up then it costs all of 50p to get a new regulator, which may have built in thermal and short circuit protection anyway. Then there's no trouble of unknown circuitry and switching frequencies and what not. I don't like the idea that I can light up 90V neon bulbs off a single wire from a 12V DC switched mode power supply...

Thanks, I will look into all that!!




Ufopolitics

[Ufopolitics]

UFOpolitics, you scrambles my text in YOUR Quote 2 from MY Quote 1 to your favor, as you think. But, you are only making yourself foolish!
The point of my post was NOT to say that my contraption is better. My intention was to make YOUR contraption BETTER!! The same with the shortcutted G-part too!
( I'm not good in English but you are ...... ! )

And the signal goes through zero because it comes from the secondary! (NOT a Field signal)

How should you achieve OU whether you seems to misinterpret most things??

PS: Run Amps =8 ("Yellow")
regards / Arne

How did you come up with such output at secondaries being that yellow shaped signal?...which happens to be identical to fields signal?


Are you assuming this or is it the software response?

There is one thing you both (Bistander and You) are missing here...

The Input Signal to Primaries only effects a Spatial Fluctuation of Field on Secondaries, which is soft and smooth fluctuation...and Secondaries output signal is absolutely not the same as primaries!!!

Your simulation software (in case resulting wave is due to it) is just "assuming" this transfer would be done through the same iron core, or by your input, assigning software a full, common core just like a transformer with field signal as input signal...and not like in reality it takes place, through Space.

No Sim Software is designed/programmed to reproduce exact Figuera's Full Circuit Conditions

Unless you are a Sim Software Advanced Programmer Engineer...and honestly I don't think so.

This conversation is completely nonsense, and I see why MM don't want all this noise there at the building thread...

Anyways many thanks for trying to make my system better...I appreciate it...but I have my own ways to go for it.


Ufopolitics

[dR-Green]

You can determine the internal resistance/output impedance of the PSU through making a simple voltage divider on the output. When the measured voltage across the load resistance is exactly half the output voltage, then the load resistance is equal to the PSU output resistance.





Voltage divider - Wikipedia

But on the subject of PSUs, I find it cheaper, easier and more convenient and reliable to get some transformers and regulators and make my own. And if you blow it up then it costs all of 50p to get a new regulator, which may have built in thermal and short circuit protection anyway. Then there's no trouble of unknown circuitry and switching frequencies and what not. I don't like the idea that I can light up 90V neon bulbs off a single wire from a 12V DC switched mode power supply...

[seaad]

UFOpolitics, you scrambles my text in YOUR Quote 2 from MY Quote 1 to your favor, as you think. But, you are only making yourself foolish!
The point of my post was NOT to say that my contraption is better. My intention was to make YOUR contraption BETTER!! The same with the shortcutted G-part too!
( I'm not good in English but you are ...... ! )

And the signal goes through zero because it comes from the secondary! (NOT a Field signal)

How should you achieve OU whether you seems to misinterpret most things??

PS: Run Amps =8 ("Yellow")
regards / Arne

[Ufopolitics]

Ufo,

It appears Mr. Seaad's graph showing AC voltage (+ & - swing) is taken from the secondary output, which will be AC. It is basically a transformer secondary. Transformers only work on AC or changing DC. So just the AC component on the primary current is transformed to the secondary.

Bistander,

I can see clearly Two Signals there...one Yellow as the Primaries stepping signal, plus a Red AC sinewave ...and so I was only referring to the stepped yellow signal, not the AC Sine.

You all are patting yourselves on your backs thinking you've found free energy. I see no evidence of that. You have only gotten a signal from your partG by a modification which seaad and I were pointing to a month ago.

Be interested to see data from the complete circuit (with secondary and primary).

Regards,

bi

Bistander, done that and tested...prior with resistor tests (not yet with part g).

Wake up!!, main point here is that Virtual Fields Fluctuations Displacement DO Generate an Induction on Secondaries...I did that rough test and got same light bulbs lit up hooked directly to secondary coil. And at low frequencies.

Spatial Virtual Fluctuations from Static Primaries Electromagnets Fields DO generate an EMF Induction on Secondaries.

[bistander]

Ac/dc

...

You keep doing your virtual simulations with Field Signal passing the zero zone.

How many more times it is required to tell you it must be positive (above zero) at all times??!!
...

Ufo,

It appears Mr. Seaad's graph showing AC voltage (+ & - swing) is taken from the secondary output, which will be AC. It is basically a transformer secondary. Transformers only work on AC or changing DC. So just the AC component on the primary current is transformed to the secondary.

You all are patting yourselves on your backs thinking you've found free energy. I see no evidence of that. You have only gotten a signal from your partG by a modification which seaad and I were pointing to a month ago.

Be interested to see data from the complete circuit (with secondary and primary).

Regards,

bi

[Ufopolitics]

Hi Ufo,

I guess it would be clearer if they stated the specifications like this:
60Volts or 10 Amperes (whichever comes first on the particular load) and up to 600 Watts maixmum.
So:

Using resistive loads for simple examples.

6 Ohms will max out current at 10A and voltage at 60V and be at max power of 600W.

Say you have a 4 Ohm load. It will draw 10A from the PSU which will regulate at 10A and set the voltage accordingly, 40V in this case. If the load heats up and goes to 5 Ohms, the PSU will maintain that regulated 10A and automatically raise to 50V. (If you have the voltage adjustment set above 50.) In this example the 600W rating is not achieved.

Say you have a 20 Ohm load. You could set the PSU to supply anywhere 0 to 60V. Current will be determined by Ohm's Law but would never exceed 3A. If you had the current dial on the PSU set lower than 3A, say at 2A, then the voltage will be limited to 40V such that 2A is not exceeded.

PSU = power supply unit.

You have 10A max or a lesser setting and/or 60V max or a lesser setting and the PSU will regulate at which value it encounters first.

I hope that is understandable.

bi

Yes, thanks for your time Bistander!!

Yes, it seems there is something very different from this PSU and the Extech I had for years...on the Extech I can set a CL (Current Limit) to Max, then just dial voltage, or viceversa...as both potentiometers dialing travel distance from min to max is the same...Not so for the newer one I have ...the Current (A) Pot is kind of normal dial distance...but Voltage Pot seems like endless travel to reach Max gain...not the same type at all...and this set me completely off when attempting to reach a proper VA adjustment.

Unfortunately the Extech is only 18V and 3Amps Max output.

I measured the resistance from both terminals of PSU when it was OFF...and gave me 1.140 Kohms...and when it is ON (no power dial of course) it goes down to 1.132 Kohms...have no idea if that would help...or mean any real value...but just a travel of meter signal through electronic circuit of PSU...(It takes very long for meter to read final, steady reading)

Resuming here...so you believe it will give me close to Full Output at 6 ohms within circuit?

I actually do not attempt to reach the full 600 Watts...but around 50V and like 4 or 5 Amps Max...which is about less than 50% from total output wattage (200 to 250 W).

I will see by testing with some resistors directly...and observing increase ability as I increase R.

Problem is that all Linear feed PSU's are untouchable because of such high prices!

I have a new small 70 A Arc Welder, which I may start playing with it in order to convert it to a nice and robust Linear Output PSU...It has a massive Transformer!!...But it don't have Pots to regulate but Min-Max current switch...

Problem is to find some decent priced high wattage pots...as at the same token...I have too many projects going on at once...so, I wouldn't like to add another one..

Hope this resistance solves issue.


Thanks again Bistander!


Ufopolitics

[Ufopolitics]

You could say that the effective internal resistance of the PSU is 6 ohms, so the PSU won't be able to supply full output power, or maximum rated power, to any load below 6 ohms

Voltage drop - Wikipedia

Internal resistance - Wikipedia

Thanks,

My PSU is a switched output and not linear...I only wished it would be linear...

About Voltage drop...yes...it won't let me dial above certain value...is like doing nothing when I dial it up.


Ufopolitics

[Ufopolitics]

UFO, some pics from my 50Hz sim. Probably much faster then in your test. The higher freq gives higher XL values in sim than in your testbed= lower current. And have mostly higher fixed milli Henrys than yours. Supply voltage 24V !! I have 20 steps but only 1 + 1 time segment fully ON you have 2 + 2. Primarys 20 + 20mH secondary 1200mH. Observe the high start current! Run Amps =8 (yellow). This sim is not fully otimised! / Arne

I can only read from above:

..."Mine is Much Faster"..."Mine is Much Higher"..."Mine is Better..." than yours...

Only one big difference...your whole thing is Virtual...mine is VERY REAL.

Believe me I can accelerate that small Motor and Assembly up to 10,000 RPM's...and also Increase V and A Higher than that on video...but I will just blow my bulbs...and am not to do that stupidity.

You keep doing your virtual simulations with Field Signal passing the zero zone.

How many more times it is required to tell you it must be positive (above zero) at all times??!!

Idk really why I am spending time here with you...when I have so many more important things to write and do here...


Ufopolitics

[dR-Green]

You could say that the effective internal resistance of the PSU is 6 ohms, so the PSU won't be able to supply full output power, or maximum rated power, to any load below 6 ohms

Voltage drop - Wikipedia

Internal resistance - Wikipedia

[seaad]

UFO, some pics from my 50Hz sim. Probably much faster then in your test. The higher freq gives higher XL values in sim than in your testbed= lower current. And have mostly higher fixed milli Henrys than yours. Supply voltage 24V !! I have 20 steps but only 1 + 1 time segment fully ON you have 2 + 2. Primarys 20 + 20mH secondary 1200mH. Observe the high start current! Run Amps =8 (yellow). This sim is not fully otimised! / Arne

[bistander]

Regulated power supplies

Hello Guys,


As you may know I recently bought a brand new PSU, which is the new kind based on Switching Mode, a piece of crap, I know...but my budget is low at this point...and the reason I got it was to run my Part G and Primaries...BUT needless to say it would not even add V when G was closed looped...Now it works...but it is still constrained as I am using very low ohms at all components...so can not "dial up" Voltage to its max potential output.

My question is:

Are these units "Regulated Switching Controls" or a "Protection" built in which is constrained directly based on Ohms or Kirchhoff Laws equations?

The reason why I am concluding this way...is because when I tested Figuera Device with resistors it did render its full output which is 600W (60VX10A)...all the way till resistors started smoking....and PSU started its automatic fan motor...

Now that I am using the low ohms Toroid winding...plus low Primaries as well...it only allows me to drive it up to a certain voltage which is far from Max V...while amps do reach max or 10A.

If it would be so...then I would have to just maintain on my circuit the required minimal resistance values...like for a real example:

My PSU is 60V and 10 Amps...which based on ohms law in order to fully drive it to max Wattage it must have at least a resistance of 6 ohms or above to play within its safety zone?

Calculating from Ohm's Law: R=V/I so 60V/10A = 6 Ohms

If it is so...then is just a piece of cake to wind my primaries and part g in order that they add up these resistance values...

I wish I would be correct...but am I?

Please let me know...and thanks in advance!!


Ufopolitics

Hi Ufo,

I guess it would be clearer if they stated the specifications like this:
60Volts or 10 Amperes (whichever comes first on the particular load) and up to 600 Watts maixmum.
So:

Using resistive loads for simple examples.

6 Ohms will max out current at 10A and voltage at 60V and be at max power of 600W.

Say you have a 4 Ohm load. It will draw 10A from the PSU which will regulate at 10A and set the voltage accordingly, 40V in this case. If the load heats up and goes to 5 Ohms, the PSU will maintain that regulated 10A and automatically raise to 50V. (If you have the voltage adjustment set above 50.) In this example the 600W rating is not achieved.

Say you have a 20 Ohm load. You could set the PSU to supply anywhere 0 to 60V. Current will be determined by Ohm's Law but would never exceed 3A. If you had the current dial on the PSU set lower than 3A, say at 2A, then the voltage will be limited to 40V such that 2A is not exceeded.

PSU = power supply unit.

You have 10A max or a lesser setting and/or 60V max or a lesser setting and the PSU will regulate at which value it encounters first.

I hope that is understandable.

bi

[Ufopolitics]

A Question for Bistander or Citfta...

Hello Guys,


As you may know I recently bought a brand new PSU, which is the new kind based on Switching Mode, a piece of crap, I know...but my budget is low at this point...and the reason I got it was to run my Part G and Primaries...BUT needless to say it would not even add V when G was closed looped...Now it works...but it is still constrained as I am using very low ohms at all components...so can not "dial up" Voltage to its max potential output.

My question is:

Are these units "Regulated Switching Controls" or a "Protection" built in which is constrained directly based on Ohms or Kirchhoff Laws equations?

The reason why I am concluding this way...is because when I tested Figuera Device with resistors it did render its full output which is 600W (60VX10A)...all the way till resistors started smoking....and PSU started its automatic fan motor...

Now that I am using the low ohms Toroid winding...plus low Primaries as well...it only allows me to drive it up to a certain voltage which is far from Max V...while amps do reach max or 10A.

If it would be so...then I would have to just maintain on my circuit the required minimal resistance values...like for a real example:

My PSU is 60V and 10 Amps...which based on ohms law in order to fully drive it to max Wattage it must have at least a resistance of 6 ohms or above to play within its safety zone?

Calculating from Ohm's Law: R=V/I so 60V/10A = 6 Ohms

If it is so...then is just a piece of cake to wind my primaries and part g in order that they add up these resistance values...

I wish I would be correct...but am I?

Please let me know...and thanks in advance!!


Ufopolitics

[bistander]

And a few hours later

Hi UFO, marathonman, and all, ....

Remember when I first confirmed that the g core works I also said "although my windings are a little bit different in design"

I have also tested your method marathonman but so far found that it does not work.


Now to make it work this is what you do -
...

You may of already wound a continious coil around the toroid with the load connections at opposing sides.
Disconnect these two load connections.
Make a splite in the continious winding wherever you like, you will now have a beginning and end to the torroid winding.
Connect the load connections to each side of the this split.

Connect the wires from you commutator around the G core at intervals.
It will now work.
...

netica

Over on the forbidden thread, member netica posts the above which would appear (on the surface) to support my contentions about partG. His fix to the continuous wound toroid avoids the opposing Ampere-turns which I was saying would cause little or no flux in the toroid core essentially destroying winding inductance.

bi

[bistander]

More part G

Hello Bistander,

Yes, I got it now, thanks very much...(about time right?...)

I did read that page related to Kirchhoff 1 & 2nd Laws (KCL, KVL)

Interestingly there is a Closed Circuit (on the EXAMPLES part) based on Three Resistors (R1,R2,R3) and Two Batteries (E1,E2), this generates Three Currents, (I1, I2, I3)...And if you look again at that particular circuit, but take off Battery or Power Source E2...We will have exactly Part G as R2 (Center) plus the Two Primaries properly connected there as R1, R3...

By taking off E2 still I3 would be a negative current, connected to negative end of E1...and so, we should still have both Loops (closed circuits) S1, S2...correct?

So, further on, could we apply KCL to that new circuit without E2, since it is identical to all three Components (Part G, Primary 1, Primary 2) from Figuera?

Hi Ufo,

I've been meaning to reply. I guess now is good since you have started testing your partG on the forbidden (to me) thread.

What you say above is true at certain points in the rotation of G. At other points, not so simple because R2 needs split as there will be two unequal paths in the toroid winding. However the similarity holds to my circuit analysis used for post#1322. Pasted below for easy reference.

Bistander,
...

Now about your comment above about Part G...

Can I ask You if you are considering -at all- the fact that Part G is not connected directly to negative supply terminal?

Sure Ufo, I considered the whole primary circuit including the supply. I'll post two of MM's images for easy reference.

But instead, its negative connection derives from both sets of primaries coils in series?

Wouldn't the fact that Part G component is connected right between the two sets of coils, which also have a resistance as an impedance?

PartG is used as a voltage (or current) divider to proportion the current from the source to the primary coils N and S. It is desired to have a certain min/max relationship of current through N and S, where N is max when S is min and vice versa.

At 12:00 (point 8) on the toroid is where primary N connects, so when the brush (being at supply positive) is at point 8, maximum current goes to N. It is desired that at this point, minimum current will flow to primary S which is connected to point 1 on the toroid.

Current must flow through the toroid winding between point 8 and point 1 to have continuity to supply positive via the brush. There are two paths in the toroid winding between 8 and 1, the right side and the left side, which are in parallel with each other and of equal lengths and turns. So to figure the impedance of the toroid winding (8 to 1) we have 2 impedances in parallel, each with equal resistance and inductive reactance. Because the inductance of the 2 coils is linked, the mutual inductance must be considered. Because the current direction is opposite in the 2 halves of the toroid winding, that mutual inductance essentially cancels the inductive reactance for the equivalent impedance.

Therefore, the potential difference between point 8 and point 1 will for all practical purposes be zero. This means the current will not be a minimum in primary S as desired, but essentially maximum, the same as in primary N.

I only looked at this one position (brush at point 8) because this case has the 2 toroid winding halves equal and the parallel combination of 2 equal inductance coils with mutual inductance is easily calculated. Obviously the same is true for brush at point 1. All the other points in between I am unsure and not about to attempt those calculations.

I see the same results looking at it another way using the flux to figure inductance. I mentioned this to you before. When you have half of the toroid mmf CW and the other half equal and CCW, resulting flux is zero and therefore the inductance of the coils on the toroid is zero. As the brush rotates around the toroid, the potential changes and mmf (Ampere-turns) are unlikely to be equal and opposite so there may be inductance in the coils and impedance causing unequal currents in primary N and primary S.

My take anyway,

bi

edit: Weird thought. What if you wound the left half of the toroid backwards from the right side half? Mutual inductance would add instead of subtract.

Hope this helps some.

bi