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Author Topic: Reactive power - Reactive Generator research from GotoLuc - discussion thread  (Read 266347 times)

Offline poynt99

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Luc,

This may be helpful in resolving the low impedance drive issue with your FG. It provides a few options, including Mark's buffer in an op-amp feedback loop to minimize cross-over distortion.

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Offline gotoluc

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Thanks for the reply poynt

I'm not sure I understand why the power going into the circuit cannot be measured in the standard way?

I looked at the circuit you propose and since electronics is not my strong side it doesn't help me understand either.

So here's what I propose, instead of building a circuit to avoid a possible problem with the 50 Ohm output resistor in the SG, how about showing me a simple circuit I can build that will make a higher voltage sine wave output then the SG can in the 700 to 800KHz range. This would eliminate the SG altogether and further test the circuit capabilities.

Thanks for your help

Luc





« Last Edit: April 21, 2014, 04:22:36 AM by gotoluc »

Offline gyulasun

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Hi Luc,

I would suggest to replace the bulb with a real metal film resistor of some ten Ohms and measure the AC voltage across it too. You can get acceptably low inductance load resistor from several metal film types connected in parallel like your 1 Ohm was assembled but no need for using higher than 4 or 5 pieces of them as a maximum.  Say you have 5 pieces of 220 Ohm metal film or carbon type (say 1/8 or 1/4 W rated each) and you parallel them so you get 44 Ohm and can use this in formula P=V*V/44 to get output AC power (V is in rms). If you have other resistor values in the some hundred Ohm range, just use them in parallel and check their value with a digital Ohm meter.
When using a bulb as a load, one can never know the output power dissipated in it if its instanteneous current is not measured together with the voltage across it, and a further problem is the bulb repesent a varying resistor as per its brightness changes, altering its loading effect to the circuit driving it, which may cause other unwanted effects in certain circuits (like changing the loaded Q of an LC tank, for instance)

All I mean with a real resistor instead of the bulb is that output power can also be measured, beside the MATH function of the scope.

Gyula

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Offline gotoluc

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Hi Luc,

I would suggest to replace the bulb with a real metal film resistor of some ten Ohms and measure the AC voltage across it too. You can get acceptably low inductance load resistor from several metal film types connected in parallel like your 1 Ohm was assembled but no need for using higher than 4 or 5 pieces of them as a maximum.  Say you have 5 pieces of 220 Ohm metal film or carbon type (say 1/8 or 1/4 W rated each) and you parallel them so you get 44 Ohm and can use this in formula P=V*V/44 to get output AC power (V is in rms). If you have other resistor values in the some hundred Ohm range, just use them in parallel and check their value with a digital Ohm meter.
When using a bulb as a load, one can never know the output power dissipated in it if its instanteneous current is not measured together with the voltage across it, and a further problem is the bulb repesent a varying resistor as per its brightness changes, altering its loading effect to the circuit driving it, which may cause other unwanted effects in certain circuits (like changing the loaded Q of an LC tank, for instance)

All I mean with a real resistor instead of the bulb is that output power can also be measured, beside the MATH function of the scope.

Gyula

Hi Gyula,

thanks for your post. 
I used the bulb just as a visual display to show the circuit outputs real current.  My first question was and still is,  when a circuits voltage and current is 180 degrees out of phase does this mean the input power is reflected back to the source?

Once that question is answered I would post the RMS voltage across a 10 Ohm 1% metal film resistor as load instead of the bulb.

I do know not to use bulbs to make power calculations but thanks for bringing it up.

Luc

Offline gyulasun

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Hi Luc,

To answer your question, yes, the output (not input) power seems to be reflected back to the source.

Here is an animation of Power Factor when the phase angle between the AC voltage and current changes between -90° and +90°.    http://demonstrations.wolfram.com/ACPowerFactorPrinciple/ 

You can read explanations in the Details section (under the 3 Snapshots) and the last sentence is: 

"If the phase angle were to be shifted to be greater than 90 degrees in either direction, the load would effectively become the source."

This means from the explanations that if the power curve is shifted in the negative direction (below the time axis as in your scope shot) then the load "delivers" power back to the generator.

Whether this is really happening in your circuit I do not know, the scope shows that to be the case. Further circuit explorations are needed.

Gyula
« Last Edit: April 21, 2014, 06:45:37 PM by gyulasun »

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Offline poynt99

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Any circuit element other than the pure source itself, is part of the load.

The RG of any function generator is actually part of the load, not the source. Therefore, if one does a source power measurement without somehow excluding RG (either mathematically, or with buffering), then one does not obtain an accurate source power nor polarity.

In this case, it appears only a fraction of total power is lost in RG, perhaps 18%, so the polarity should be correct. I am working on a sim that will hopefully help explain the apparent polarity reversal.
« Last Edit: April 21, 2014, 04:42:25 PM by poynt99 »

Offline poynt99

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OK Luc,

I've figured out why the phase of the current probe is reversed, but you'll need to try a test to see if we can get the circuit to operate close to the same without the MOT's. Please replace the MOTs with a small capacitor (CMM) as I am showing in the attached diagram. You may need to try a few different values, and obviously retune for resonance.

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Offline gotoluc

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Thanks poynt, I'll give it a try

Luc

Offline gotoluc

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OK Luc,

I've figured out why the phase of the current probe is reversed, but you'll need to try a test to see if we can get the circuit to operate close to the same without the MOT's. Please replace the MOTs with a small capacitor (CMM) as I am showing in the attached diagram. You may need to try a few different values, and obviously retune for resonance.

OK poynt,

I replaced the pile of MOT coils and cores with a variable air capacitor to ground like your in diagram. At 27pf I can get the same output 1.5v to the bulb and close to the same phase shift except with the MOT pile the mean is -82mv and with the air capacitor to ground the mean only goes to -65mv.

If this was just capacitance then I don't understand why there would be a difference, so I'm going to do other tests to try to get to the bottom of this.

Thanks

Luc

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Offline poynt99

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Luc,

I'm not sure where that wire going to the MOTs was actually connected once it got there, but you're right, there would be some inductance in series with that capacitor as well. How much, I'm not sure, but at least a few hundred nano-Henries from the connecting wire.

Not sure if that would make the difference or not.

Offline gotoluc

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Hi poynt,

none of the copper wires or coils are connected to anything. The single wire is only connected to the steel cores, so I wouldn't think there's any Inductance.

Please have a look at the next post for an update.

Thanks

Luc

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Offline gotoluc

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Hi everyone,

please find the test 2 video where I was able to Isolate and accurately calculate all power fed to the signal generator.

This test demonstrates to the best of my ability that power to the load is not coming from the signal generator.

Link to video: https://www.youtube.com/watch?v=SucknHDgE6E&feature=youtu.be

Luc

Offline gyulasun

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Hi Luc,

My first notice is that the signal generator continuously feeds the ferrite rod coil: the 1.263A-1.254A=0.009 Amper i.e. 9 mA change (no problem you mentioned 10 mA in the video) so this change in the input supply current of the sig gen shows that nicely when you remove then reconnect the BNC cable from the sig gen output (video time 6:40 or so) and the green cable to the transformer is unconnected.

And my second notice is that by connecting the green cable to the transformer input to complete its circuit and tuning the sig gen to the 180° out of phase V-I condition, you create an energy balance between the output of the sig gen and the energy picked up by the green cable in that while the sig gen continues to take up the same (or very nearly the same) 9-10 mA extra current (part of which surely continues to go into the ferrite coil like in the case of the unconnected green cable), and the green cable picks up the electromagnetic field of the ferrite coil and feeds it back as additional energy to the transformer input, this is why the green LED lites up. (When I mention the green cable I mean its associated (distributed) capacitance and small inductance what the cores of the MOTs and their hanging wires represent.)

I understand you mentioned removing the (yellow) transformer and substituting it with a Tesla coil what you would also feed with the single wire output of the center tap of the ferrite coil: in this case the green wire and its MOT enviroment would not be needed,  I accept this and I think this would confirm that your load  (now a LED and earlier a bulb, both coupled to the single wire output via the transformer)  is (or would be) fed by an Avramenko-plug-like circuit. In case of a Tesla coil (instead of the yellow transformer) the green wire (and its associated 'network' may not be needed, this would depend also on some factors.

My speculation that the green cable and its 'network' pick up and feed back energy to the transformer input, taken from the transmitted EM field of the ferrite rod coil, could be checked by wrapping up the ferrite coil (and its C core) in a metal box (Faraday cage) so that the ferrite coil should not be able to transmit much EM energy to its enviroment. Of course the enclosure surely would detune the sweet frequency spot so retuning the generator would be needed and notice that the green LED my not lite up at all in this case:  this would prove the picked up EM energy by the green wire in the un-enclosed case. In such metal enclosure, the ferrite cores would be positioned in the middle point of the inside volume and the three insulated coil wires would come out through tiny openings on 3 sidewalls (up, down and left side).

Of course I do not mean you build this enclosure, as I said I believe the Avramenko-plug-like energy transfer manifests in your setup and it is known for having high efficiency and a very little load back effect to the energy source (which I think is the generator through the center tap of the ferrite coil...).

Thanks for showing your findings. Further deductions on energy balance may be done by learning about the real power consumed by the load and comparing it to the output power of the generator. The transmitted energy in the form of EM field by the loopstick antenna would be difficult to estimate though.

Opinions are welcome.

Gyula

Offline gotoluc

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Thanks for your post Gyula,

here is test 3 with the LED on a Tesla coil and ferrite stick and yellow transformer in a Faraday cage.

Link to video: https://www.youtube.com/watch?v=j28oX7rYWHU&feature=youtu.be

Luc

Offline gyulasun

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Hi Luc,

When I mentioned the use of a Faraday cage I did not mean to use the Tesla coil, I did not mean to change your setup in any other respect, except the presence of an eclosure around the ferrite core coil because in that setup it was the only transmitting antenna.  Now you enclosed the ferrite rod coil, that is fine but created a new 'transmitting antenna'  in the shape of the air cored Tesla coil...  and I did not mean this, sorry.

I meant to enclose the loopstick, yes, and then trying to achieve the same result shown in the now last but one video by connecting the green wire to the yellow transformer to lite the LED again in the same way. Because I think that in that setup the green wire adds picked-up energy to the LED via the yellow transformer and the energy comes from the near field radiation of the loopstick.  I may be mistaken of course.

I will try to interpret your new test of course, I need some more time due to some other commitments.

Gyula

 

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