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Author Topic: Reactive power - Reactive Generator research from GotoLuc - discussion thread  (Read 253436 times)

Offline poynt99

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I must clarify something, because I was working on the notion of a pure LRC circuit. In this case, yes I would say the circuit is highly inductive and a signal inversion is taking place somewhere.

However, Luc is using a transformer as an inductor, and this changes things when the secondary is loaded or shorted.

What I found is this:

When the secondary is shorted and therefore left with the secondary's DC winding resistance as a load, that load is reflected back to the primary, effectively shorting it to a degree as well. As an example, let's say the primary inductance is 280mH and the secondary is 5H, with a DC winding resistance of 100 Ohms. When the secondary is shorted, the 100 Ohm winding resistance is reflected back across the primary as a ratio of the secondary to primary inductance, in this case 5H/.28H = 17.86. So this would result in the equivalent of a 5.6 Ohm resistor being placed across the primary winding.

When Luc says that the "magic" is in shorting the MOT secondary, what he is really doing is essentially eliminating the MOT inductance, making the circuit mostly capacitive, hence the "i leading v" result we see in the scope shot. The voltage swing across the primary will change from 350V to a few hundred mV when the secondary is shorted.

Now, there's still a problem; the capacitive reactance of the 25uF cap at 60 Hz is not enough to cause a 90º phase shift across the network, and we are back to what I said before about the circuit having to be purely capacitive (or mostly) with a capacitor value of about 1uF, not 25uF. I would like to see Luc measure his capacitor stack to see how close it really is to 25uF.

I'd also like to see Luc remove the MOT and test the circuit again. He should get a similar 90º result.

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Offline poynt99

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Luc just emailed me this:

what is relevant is the zero effect to the generator and delivering over 20 Watts to the load!... this is what needs to be understood.
Actually, you've not proven that the resistor is dissipating 20W. Why? Because that 1k resistor is going to be inductive. The only way you are going to be 100% certain is to either measure its temperature and try to match it with a DC supply, or you use the scope to perform AVG[v(t) x i(t)] in order to get the average power of the resistor.

Before you get too excited about any measurement, especially a power measurement, you must confirm that measurement by at least one other means. This mantra has been stated endless times in this forum but it seems to always fall on deaf ears. And to date, no one here has made a single watt of OU power, which proves that they've been wrong 100% of the time they make a claim based on their botched power measurements.

Offline tim123

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In order to prove there's real power being developed, Luc could presumably replace the resistive load with a FWBR. Then put the same resistive load across the DC terminals (+ smoothing cap) and measure the DC voltage...

The trouble with the AC measurements is they have to include the power factor. I really don't think you can calculate AC power in the way that Luc was doing, but then he hasn't set out exactly how he's calculating these values (such as 10-20w out and zero in) in a post, so I'm not sure.

I have tried again - this time with the circuit from this post (i.e. with the load on the secondary):
http://www.overunity.com/14013/reactive-generator-research-for-everyone-to-share/msg377250/#msg377250

I found that if I calculated the power across the load *assuming* a power factor of 1 - then it appears to be OU.

All the calcs are complicated, and so I did what I suggested above - and used a FWBR. This time the output was sensibly UU...

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Offline poynt99

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You can in fact calculate the real power in the resistor by measuring the rms voltage across it as Luc has done, but ONLY if the resistor itself is non-inductive, AND the two meter leads are right tight against the body of the resistor. But in this case since we are dealing with 60Hz, the latter requirement is not necessary, within reason.

Sure, Luc could use a FWBR on the output of the secondary as you suggested and tested yourself. This is yet another possible check that he has most likely NOT performed.

Are you using a resistor similar to his?

Offline tim123

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I was using a standard low-power resistor - 150ohms - probably wire wound - not sure...
As the instruction is to use only 1-2 watts - i thought it'd be ok. It was.
I didn't use the variac this time.

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Offline conradelektro

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Here you can see how Jean Louis Naudin made a faulty measurement (measured too much power output) because the resistor also was inductive (not only resistive):

http://jnaudin.free.fr/rvproject/html/ugentest22.htm (NOTE TECHNIQUE du 26 octobre 2012)

He not only had to deal with the induction of the resistor but also with the so called "reflected HF power" which falsified his measurement http://jnaudin.free.fr/rvproject/html/ugentest23.htm

The subject of Naudin's test is not relevant for this thread. I cite his pages because it shows the difficulty of correctly measuring HF output from a coil.

The pitfalls are "induction in the measurement resistor" (too much power is measured) and "reflective power" or "missmatch loss" (too little power is measured) http://en.wikipedia.org/wiki/Mismatch_loss (more pronounced at higher frequencies).

May be it helps.

Greetings, Conrad

Offline poynt99

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I have tried again - this time with the circuit from this post (i.e. with the load on the secondary):
http://www.overunity.com/14013/reactive-generator-research-for-everyone-to-share/msg377250/#msg377250

I found that if I calculated the power across the load *assuming* a power factor of 1 - then it appears to be OU.

How did you do it exactly? Mind posting your results?

Measuring the rms voltage across a non-inductive load resistor should yield the correct power when using: v^2/R.

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Offline poynt99

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All the calcs are complicated, and so I did what I suggested above - and used a FWBR. This time the output was sensibly UU...
Remember that you will lose some power to the diodes.

Offline poynt99

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I was using a standard low-power resistor - 150ohms - probably wire wound - not sure...
As the instruction is to use only 1-2 watts - i thought it'd be ok. It was.
I didn't use the variac this time.
What are the specs for your MOT?
What value cap did you end up using?

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Offline tim123

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How did you do it exactly? Mind posting your results?...

Ok, I'll run thru the test again later, and fully document it...

Thx for the other info too... Will try to include it in my tiny brain...

:), Tim

Offline tim123

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Ok, hopefully I've done this right...

Attached is a schematic of the test setup.
 - Mains is 250v - measured
 - MOT is UK 230/250v 600w (I think).
 - Resistor is standard 1/4 watt carbon film - 148 Ohms - According to my DMM

Measurement:

 - I used the wattmeter to read overall power in
 - The DMM (I presume) is showing RMS volts AC,
 - The 'scope is using just CH1 to show RMS Volts AC

Note 1: The scope is battery-powered (DSO203) - so I can connect the two ends of the probe directly over the load.

Note 2: The scope is not calibrated... Note the DMM & Scope show different values.

Note 3: I cannot connect my mains 'scope probe's earth leads to the Neutral line - even at the end of the circuit. It still trips the RCDs...

Results:

5 uF:
 - Watts : 1 W (approx - it wanders within 0.3 for all the tests...)
 - DMM   : 6.47 V
 - Scope : 5.8 V
 - V2/R  : 0.23 W

6.5 uF:
 - Watts : 2 W
 - DMM   : 8.1 V
 - Scope : 7.3 V
 - V2/R  : 0.36 W

7.5 uF:
 - Watts : 3 W
 - DMM   : 10 V
 - Scope : 8.4 V
 - V2/R  : 0.48 W

Note: Power factor remained at 0 throughout - according to the wattmeter.

Also attached is
 a) A pic of the test bench.
 b) A scope shot - showing the waveform - at 7.5uF

Another note: I put a solenoid coil around one of the secondary's wires - as a basic current probe. I couldn't get much amplitude with the small currents, but it did seem to work. The signal from the solenoid coil & the voltage probe were out of phase... I'll test that some more if folks are interested...

Let me know if I've done anything wrong, or need to do more...

Regards, Tim :)

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Offline poynt99

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Thanks Tim.

Wow, you've got some harmonics going on there. Is your grid power clean?

Can you measure the resistance and inductance of both the primary and secondary of your MOT?

The difference between the scope and DMM could be from the harmonic noise we see there. The scope should be the more accurate of the two.

Also, it occurred to me that these power meters were not really designed to be that accurate down around the 1W range. I am amazed that yours even has tenths of a watt!

Anyway, it would appear in all cases that your output power is below the input power, even using the higher DMM measurement.

Obviously the power factor computation by the meter is incorrect (i.e. it can not be 0) if the resistor etc. is dissipating power. You can confirm the actual phase angle (hence PF) using the second channel of your scope. See my attached diagram. NOTE: You will also be able to get the true input power with this probe arrangement, provided your scope can do the required math.

Offline tim123

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Wow, you've got some harmonics going on there. Is your grid power clean?

The harmonics go away with higher capacitance / power throughput. See pic below... I don't know what causes them (harmonic reactance form small caps?). (EDIT: They only go away on the DSO scope - the other one still shows them)

Quote
Can you measure the resistance and inductance of both the primary and secondary of your MOT?

Primary:
 - 2.4 Ohms
 - 0.27 Henries
 - with 2ndary shorted = 0.05 H

Secondary:
 - 133.5 Ohms
 - over 20 Henries
 - with 1ary shorted = 3.8 H

Quote
The difference between the scope and DMM could be from the harmonic noise we see there. The scope should be the more accurate of the two...

I've done another test. Pic attached. A few changes this time:

 - 10 Ohm, wire wound, variable power resistor (100w) as the load (on 2ndary), instead of the 148 Ohm...
   I wanted to try a lower resistance and see what happened...

 - I included my mains scope - to take the same measurement - so I now have 3 things all reading the voltage across the load.
   I wanted to have another source for comparison.

Test run is depicted - using 12.5uF.
 - Watts : About 5 W
 - DMM : 1.116 V
 - DSO Scope : 2.515 V Pk, 0.899 V RMS

 - Mains Scope : About 2.5V Peak, Probably... This takes a little explaining - and I have a potential problem...

  1) To read the voltage using this scope, I have to put probe A on one side of the resistor, and probe B on the other. I then have to use MATH A-B to get the true voltage. BUT my scope doesn't display the calculated wave's peak or RMS - so it has to be judged by eye. lol.

Yellow = Chan A, Blue = Chan B, Red = MATH A-B
 
 2) For some reason my scope is displaying 10 times more than it should...
 - Probes are set to 10x. (Max scale is 5v per div on scope)
 - BUT Scope has to be set to 'Probe 1x' to match the other 2. If I set it to 10x - it's 10x too big!

I don't understand this..... Will engage tiny brain, and see if I can figure it out... I may be some time... :)

Offline nilrehob

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I just did a first test:
http://www.youtube.com/watch?v=xc056vnuIYU

/Hob

Offline poynt99

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The harmonics go away with higher capacitance / power throughput. See pic below... I don't know what causes them (harmonic reactance form small caps?).
Well it makes sense actually. We are high-pass filtering the grid voltage by using a relatively small capacitance. So we are seeing the noise present in the grid voltage. As you increase the cap value, the voltage increases and there is more of the 60Hz signal drowning out the "noise".

Quote
Primary:
 - 2.4 Ohms
 - 0.27 Henries
 - with 2ndary shorted = 0.05 H

Secondary:
 - 133.5 Ohms
 - over 20 Henries
 - with 1ary shorted = 3.8 H
Thanks for the specs.

Quote
Test run is depicted - using 12.5uF.
 - Watts : About 5 W
 - DMM : 1.116 V
 - DSO Scope : 2.515 V Pk, 0.899 V RMS
The signal is still a bit noisy so the DMM may be giving an error here.

Quote
- Mains Scope : About 2.5V Peak, Probably... This takes a little explaining - and I have a potential problem...

  1) To read the voltage using this scope, I have to put probe A on one side of the resistor, and probe B on the other. I then have to use MATH A-B to get the true voltage. BUT my scope doesn't display the calculated wave's peak or RMS - so it has to be judged by eye. lol.
Sounds like you are doing a differential measurement A-B. That's fine. It's not recommended, but you could try lifting the earth gnd temporarily on your grid scope to see if that would permit use of your probe gnds. But be careful, your scope chassis could become line voltage...potential danger!

Try the phase and input power test with your DSO203. It will be interesting to see how it compares to the watt meter PF indication and power.

 

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