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Author Topic: Partnered Output Coils - Builders Group - Moderated!  (Read 17700 times)

Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #60 on: May 26, 2020, 02:38:52 AM »
We are solving for input power over a given time period in the example I gave, so we are at some point going to multiply voltage times current.  We will then convert the power to energy but let's just focus on the input power.  In my first example, the voltage and current is sampled at some frequency much higher than the waveforms of interest and each pair of these samples is multiplied together with the product is stored in a table.  At the end of the measurement period, these samples are added together and the total sum is divided by the number of samples to arrive at a true average of the input power.

In the second example, the same sampling method is used but only on one waveform so there is no product involved.  IOW, the current or voltage is sampled again at a frequency much higher than the frequency of interest and the magnitude of each sample is stored in a table.  The samples are then summed at the end of the measurement period and divided by the number of samples to arrive at the true average of that waveform.  However, if we now take the product of the individually averaged voltage and current, we will arrive at a different result as seen in the example.  The only time we can multiple the individual averages together and arrive at an accurate result is if the waveforms are linear.  For example, DC or ramped voltages and currents.  Any aberration in the waveform will cause errors to creep in.

These examples may at first seem the same but the difference is that the first takes the product of each current/voltage sample and then averages, while the second takes the average of the independent  current/voltage samples and then takes the product.

Scopes, simulators, and certain instruments like smart power analyzers, etc, use this sampling method as given in the first example.  Analog and most digital DMMs use various methods to achieve the second method of averaging the whole waveform.

No.  All the components used are linear.

Thanks Chris.

Regards,
Pm



Thank You Partzman!

That cleared up the dilemma of definition. I agree, of course, there is clearly a problem if the meter used is providing incorrect Calculated Results! Definitely a catch and something to be aware of!

I urge all, put this information into your Measurement Protocols! Its important that you know the simple, effective methods, if you are searching for the Holly Grail, then not knowing what the Holly Grail looks like, what to look for, you could zip right over it without knowing it! The most simple experiment being a success, but marking as a failure, when you have had bad data in your Protocol list all along!

Best wishes, stay safe and well,
   Chris Sykes

Free Energy | searching for free energy and discussing free energy


Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #61 on: May 26, 2020, 03:17:26 AM »
Hi Chris,

See my post, reply #48, in this thread.

Quote
RMS Power is a nonsensical number.

That is the last sentence in this article.
https://meettechniek.info/measurement/theory-definitions.html

Yet you use that nonsensical RMS Power calculation in your post and apparently think all RMS measurements lead to erroneous power and energy numbers. With 'True RMS' instruments, this is not the case. They in fact measure and display the RMS voltage and RMS current, but use proper algorithms to arrive at correct power and energy values. After all, RMS Energy makes no sense, however one can calculate it from RMS Voltage and RMS Current over the time interval

True RMS instruments can be used with non-linear loads. I thought the referenced article linked above did an excellent job explaining this. Below is another reference about non-linear loads and RMS. I realize it is written by a vendor (a well respected one, IMO).

https://www.fluke.com/en/learn/blog/electrical/true-measurements-of-non-linear-loads-require-a-true-rms-measurement-tool

Regards,
bi

Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #62 on: May 26, 2020, 03:28:11 AM »


For those serious researchers, now you know:

   1: The Goal is a Non-Linear Load on your DUT's Input!
   2: Proper and Correct Measurement Protocols to make sure you have greater chances of success.


RMS, if your Output is Sinusoidal and your Output Load is Linear, a Resistive Load, is fine to use as has been explained! There is a time and a place for the Protocols to be put in place, I believe I have made this very clear in recent posts.

Now, all you need do, is follow the very simple rules I have laid out in the first post, the goal: Making your DUT Input as Non-Linear as you can!

What does this mean?

It means, make your Input Coil, return as much Energy as you can! Your Input Coil moves from Input to an Output, returning Energy back to the Source! Thus the Circuit I gave, the Reverse Diode across Q1 is indication Power returning back to the Source. See image below again:

Perhaps the reason My Members have had so much success is because we have put the right protocols in place to make steps ahead, in the right direction. This is important, to know where you are going!

Best wishes, stay safe and well,
   Chris Sykes

Free Energy | searching for free energy and discussing free energy

Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #62 on: May 26, 2020, 03:28:11 AM »
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Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #63 on: May 26, 2020, 03:33:41 AM »
Hi Chris,

See my post, reply #48, in this thread.


Quote

RMS Power is a nonsensical number.



That is the last sentence in this article.
https://meettechniek.info/measurement/theory-definitions.html

Yet you use that nonsensical RMS Power calculation in your post and apparently think all RMS measurements lead to erroneous power and energy numbers. With 'True RMS' instruments, this is not the case. They in fact measure and display the RMS voltage and RMS current, but use proper algorithms to arrive at correct power and energy values. After all, RMS Energy makes no sense, however one can calculate it from RMS Voltage and RMS Current over the time interval

True RMS instruments can be used with non-linear loads. I thought the referenced article linked above did an excellent job explaining this. Below is another reference about non-linear loads and RMS. I realize it is written by a vendor (a well respected one, IMO).

https://www.fluke.com/en/learn/blog/electrical/true-measurements-of-non-linear-loads-require-a-true-rms-measurement-tool

Regards,
bi


@Partzman - Would you like to explain to Bistander?


Quote

nonsensical - having no meaning; making no sense.



Bistander, If you want to re-read the posts, they are very clear and concise if you read properly! Your first article does not ever mention the word Linear or Non-Linear? Therefore the article is quite simply invalid!

I have nothing more to say!

Best wishes, stay safe and well,
   Chris Sykes

Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #64 on: May 26, 2020, 05:12:25 AM »
Title of second article.

Quote
True measurements of non-linear loads require a true-rms measurement tool

I thought 'True RMS' was a more accurate RMS measurement which doesn't use hardware and software shortcuts employed in lesser expensive equipment intended for sinewave applications.

bi

ps. Graph attached is from first article. It shows the returned power and how it is integrated into the energy. True, for sinewave, but works for other waveforms also.

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #64 on: May 26, 2020, 05:12:25 AM »
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Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #65 on: May 26, 2020, 05:19:17 AM »
Title of that article.

I thought 'True RMS' was a more accurate RMS measurement which doesn't use hardware and software shortcuts employed in lesser expensive equipment intended for sinewave applications.

bi



Hi Bistander:

"Can the RMS value be negative? For example, the RMS for a DC signal is the DC value itself. If the DC signal is -ve, then will the RMS be equal to that -ve DC value?"

https://www.quora.com/Can-the-RMS-value-be-negative-For-example-the-RMS-for-a-DC-signal-is-the-DC-value-itself-If-the-DC-signal-is-ve-then-will-the-RMS-be-equal-to-that-ve-DC-value

Quote

No. It can’t. That is part of the reason for using RMS voltage for AC.

Because the mean value (average numerical value) of an AC voltage is zero it cannot be used to compute power. Squaring a negative number produces a positive. Finding the square root of a positive number is no problem. The mean value of this can be used to compute power.

RMS voltage calculations are made on the basis of a complete (360 degree) cycle.

The periodicity of the AC does not enter into the computation.

The mean value of a steady negative voltage, is negative. If you follow the RMS voltage calculating procedure, finding the mean, squaring it and then its square root, you get back where you started, but always with a positive number.

It is always worth considering extreme the extrapolation of values and procedures when exploring ideas that ‘are out of the box’. This is such a case.


Again using the Circuit I gave, using a purely DC Source, Not an AC Source, then the situation, a DC Source with a Non-Linear Load, RMS or True RMS is not correct! You need to know Used power, thus the Negative Sign, un-obtainable in RMS is necessary to know. What if you get back 10 watts and only used 1 watt?

Either method, True RMS or standard RMS give only positive numbers. If you run the Math you will find this to be true.

   double VSum = -3.0;
   double VSqu = Math.Pow(VSum, 2);
   double answer = Math.Sqrt(VSqu);
   answer = ?


Try it out, no matter how Negative the Sum of V is, its always comes out Positive! This is not Used Power in our case!

Another Analogy:

You pour 1 Litre of Water down a pipe. As you finish, the Pipe gushes back out 2 Liters of Water. How much total water was used?

Answer: 1 + -2 = -1


Best wishes, stay safe and well,
   Chris Sykes

Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #66 on: May 26, 2020, 06:28:56 AM »
OK. I guess what I'm seeing is that RMS Voltage and RMS Current are still valid for non-linear loads, those values can not be used for power and energy calculation. Then say for a device efficiency one needs average power which is derived from the energy which is calculated using the instantaneous voltage and current integrated over the time interval. I think what got me is the use of true RMS instruments.

Thanks,
bi

Free Energy | searching for free energy and discussing free energy

Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #66 on: May 26, 2020, 06:28:56 AM »
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Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #67 on: May 26, 2020, 07:51:51 AM »
OK. I guess what I'm seeing is that RMS Voltage and RMS Current are still valid for non-linear loads, those values can not be used for power and energy calculation. Then say for a device efficiency one needs average power which is derived from the energy which is calculated using the instantaneous voltage and current integrated over the time interval. I think what got me is the use of true RMS instruments.

Thanks,
bi


Hi Bistander, in some conditions, you can use RMS, but you do need to be aware of exactly what I have pointed out.

In the second article you pointed to, they say the same thing, in between the lines:

Quote from: "true-rms measurement tool" link=https://www.fluke.com/en/learn/blog/electrical/true-measurements-of-non-linear-loads-require-a-true-rms-measurement-tool


If you’re measuring a linear load—such as standard induction motors, resistance heaters, or incandescent lights—you can easily capture accurate rms measurements with an average responding measurement tool. However, if a nonlinear load is on that circuit, you need to use a true-rms measurement tool to get an accurate rms reading, or your measurements may read up to 40 % low.




You must remember, they are specifically talking about: "which calculates the effective value (or heating value) of any ac wave shape. In electrical term", they are expecting energy returned to the Source, thus Alternating, so this must be accounted for. Under DC Conditions, we normally would Not Expect Energy returned back to the Source.

You must remember, using RMS, you will not get any indication of Power coming back! I mean, for example, if anyone can show me a TRMS Meter indicating Power Returned to the Source? E.G: 1 + -2 = -1?

Best wishes, stay safe and well,
   Chris Sykes

Offline lancaIV

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #68 on: May 26, 2020, 09:04:22 AM »
What would be if you have an electric circuit with a C.O.P. ≤≥ 1 but getting a total process C.O.P.≥ 1 ?
Ambiental energy harvester and converter : to Gauss or TESLA units or PRESSURE !
Strong electromagnets and or PM/EM couples ,pumps/compressores,motors et cetera .... !

www.paxwater.com/biomimikry
 impeller as "electro-magnet"-centrifugal/centipetal movement demonstrator         

       Bionik is "Nature structure/behaviour mimikry/simulation/adaption and use !
DE3330899 Professor Dr.Ing. Ingo Rechenberg "Arrangement for increasing the speed of a gas or or liquid flow"             
                                SPEED ! And VELOCITY ? 8)

 The socio-economical Walras-law,the v.Weizsaecker "eternal tax law 'left side equals right side', is the Physics-"Entrophy"-definition !

But "seeding"-knowledge : you grave in 400 seemes and get 16000 seemes as gain per sqm !
                                    plants roots technology  ::) development ( turgor effect )

Sincerely

OCWL
p.s. : what you are doing all the time is like to analyze an animal or human by his/her/its EEG-/EKG-gram/graph         about their positive or negative influence ! Without natural sin !


Free Energy | searching for free energy and discussing free energy

Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #68 on: May 26, 2020, 09:04:22 AM »
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Offline partzman

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #69 on: May 26, 2020, 05:59:14 PM »
I think this has all been explained but here is a sim example that has a nonlinear input voltage which is also across a resistive load. 

We have a 100ma peak current sine source driving the nonlinear Coss or output capacitance of a mosfet with a 10k resistor in parallel.  The goal is to know what the power levels are for Pin and Pout across the load resistor.

Using the plot math which uses the averaging of the instantaneous products of the sampled voltages and currents, we see the input power I(I1)*V(VD1) = 320.6mw with an energy of 3.206uJ.  We also see using the same method that the power across R1 is V(VD1)*I(R1) = 312.26mw with an energy of 3.1226uJ.

Now we have an opportunity to compare the rms verses average power measurements for the nonlinear input and the load resistor.  First the resistor.  We see the voltage measurements across R1 as 55.88vrms and 41.523vavg respectively resulting in power levels of 55.88^2/10e3 = 312.26mw and 41.523^2/10e3 = 172.42mw respectively.  So even with a nonlinear voltage across a resistor, we must use the rms for an accurate power measurement while the average is inaccurate.  The ac rms power in a resistor will produce the same amount of heat as will the equivalent power in dc voltage and current.

We will now calculate the input power by using the rms values for voltage and current and multiply by the cosine of the phase angle between them.  To determine the phase angle, we will measure the time difference between the peaks of each waveform as the voltage is offset from zero.  So, the frequency of 200kHz has a period of 5us which results in 14ns for each degree.  We see the time difference between cursors 1&2 as 1.1575us for a phase lead of 1.1575e-6/14e-9 = 82.7 degrees.  Now using the rms values for input voltage and current we calculate the input power as 55.88*.0706*cos(82.7) = 501.3mw as compared to the true average power input of 320.6mw.  The only time we can use the rms to calculate the power of out-of-phase waveforms is if both waveforms are pure sinewave.  Any distortion will create errors.

We can also see that the using the average values for input voltage and current will result in an incorrect power input result.

Hopefully this will help with some of the confusion in power measurements.  I also hope that all my calcs are correct!

Regards,
Pm

Offline Thaelin

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #70 on: May 26, 2020, 10:55:37 PM »
  The way that this is going, I am wondering if any measurement that I make will be right? Baring of course having a gov lab to work in. I believe that there will be "no" perfect sine wave in which to measure. These devices live in harmonic hell the best I can see. Spectrum shows a real mess at times.


thay


Free Energy | searching for free energy and discussing free energy

Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #70 on: May 26, 2020, 10:55:37 PM »
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Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #71 on: May 27, 2020, 03:14:45 AM »
I think this has all been explained but here is a sim example that has a nonlinear input voltage which is also across a resistive load. 

We have a 100ma peak current sine source driving the nonlinear Coss or output capacitance of a mosfet with a 10k resistor in parallel.  The goal is to know what the power levels are for Pin and Pout across the load resistor.

Using the plot math which uses the averaging of the instantaneous products of the sampled voltages and currents, we see the input power I(I1)*V(VD1) = 320.6mw with an energy of 3.206uJ.  We also see using the same method that the power across R1 is V(VD1)*I(R1) = 312.26mw with an energy of 3.1226uJ.

Now we have an opportunity to compare the rms verses average power measurements for the nonlinear input and the load resistor.  First the resistor.  We see the voltage measurements across R1 as 55.88vrms and 41.523vavg respectively resulting in power levels of 55.88^2/10e3 = 312.26mw and 41.523^2/10e3 = 172.42mw respectively.  So even with a nonlinear voltage across a resistor, we must use the rms for an accurate power measurement while the average is inaccurate.  The ac rms power in a resistor will produce the same amount of heat as will the equivalent power in dc voltage and current.

We will now calculate the input power by using the rms values for voltage and current and multiply by the cosine of the phase angle between them.  To determine the phase angle, we will measure the time difference between the peaks of each waveform as the voltage is offset from zero.  So, the frequency of 200kHz has a period of 5us which results in 14ns for each degree.  We see the time difference between cursors 1&2 as 1.1575us for a phase lead of 1.1575e-6/14e-9 = 82.7 degrees.  Now using the rms values for input voltage and current we calculate the input power as 55.88*.0706*cos(82.7) = 501.3mw as compared to the true average power input of 320.6mw.  The only time we can use the rms to calculate the power of out-of-phase waveforms is if both waveforms are pure sinewave.  Any distortion will create errors.

We can also see that the using the average values for input voltage and current will result in an incorrect power input result.

Hopefully this will help with some of the confusion in power measurements.  I also hope that all my calcs are correct!

Regards,
Pm



Thank You Partzman, another excellent example.


Another Example:

My Below example is not as good as yours Partzman, but does show the situation I am deep diving into. The Circuit attached below:

   1: We have an Alternating Current Source, capable of 7 Volts.
   2: The simple Circuit, includes a Bulb, 6V @60ma, having a resistance of: 100 Ohms, modified to become highly Non-Linear
   3: R1 is a Current Shunt, or Current Sensing Resistor CSR, 1.0 Ohms.


We take the facts:

   1: The AC Source is only capable of 7 Volts, no more it is limited to 7 Volts only.

We can see, Peak to Peak the value is 10 volts, more than our 7 Volts.

   2: We have Positive Voltage and Negative Current, and Negative Voltage and Positive Current.

We should all know what this means! Positive Voltage and Negative Current or Negative Voltage and Positive Current should not occur in any AC System but here we see it does! Non-Linear operation, the Non-Linear Load is sending Current Back to the Source at the wrong stage of the Cycle.

Now, for most here, this situation should Scream Alert! Figuring out whats going on here is extremely simple, but very few have studied this and understand the true situation occurring here.

@Thay - This example should clear up the problems.

Best wishes, stay safe and well,
   Chris Sykes

Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #72 on: May 27, 2020, 04:32:03 AM »



I must ask, is it clear now, why we can not use RMS as an Input Measurement Method?

Best wishes, stay safe and well,
   Chris Sykes

Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #73 on: May 27, 2020, 05:41:41 AM »
Quote
   2: We have Positive Voltage and Negative Current, and Negative Voltage and Positive Current.

Chris,

Doesn't this occur in all AC circuits when power factor is not unity?

And for your last example, isn't the 7 volt figure RMS?
Also isn't the peak to peak 20V?

Regards,
bi

Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #74 on: May 27, 2020, 06:02:37 AM »
Chris,

Doesn't this occur in all AC circuits when power factor is not unity?

And for your last example, isn't the 7 volt figure RMS?
Also isn't the peak to peak 20V?

Regards,
bi



Hi Bistander,

My response is below, in between your quoted posts:


Doesn't this occur in all AC circuits when power factor is not unity?



No, observe the below wave forms, especially the resonance, Energy used and Energy returned, observe specifically the Energy Returned, Voltage and Current Polarity's are Opposite in Sign when Energy is returned to the Source.



And for your last example, isn't the 7 volt figure RMS?



No, please observe the AC Source ( V1 ) Specification, set and regulated to 7 V. EDIT: I may have misunderstood your question: 10 Volts Pk2Pk * 0.707 = 7.07 Volts RMS, if this is what you meant, then yes. Apologies if I did misunderstand you!



Also isn't the peak to peak 20V?



Where did you get 20 Volts Peak to Peak?

The Wave forms on the plot, the graph, reach peak 10 Volts Peak to Peak for the Red, Voltage Channel, and approximately 27 Volts peak to Peak for the Blue Channel, this is Current.

So no 20V pk anywhere there, not sure where you are reading this?

At all Readers, why is this not covered anywhere? Why has no one else covered this? How is it that I am covering this? How is it that I know all about this and no one else seems to? My example, given in post: #71 is very clear and very concise, yet no one else has ever pointed this out, why?

Are you mad yet? Been lied to for so long! Told to use RMS, when I have just proven to everyone it is Flawed and Nonsensical in some situations! No wonder so many fail, there is no way to succeed if you have the wrong information! Measuring Power on the Input as all used Power when there is a very good chance you have had Power coming back, returned power to power your Input, but accounting this as used not returned... Have I just opened up a bag of Worms here?

Best wishes, stay safe and well,
   Chris Sykes
« Last Edit: May 27, 2020, 09:44:55 AM by EMJunkie »

 

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