Jagau's successful ZPM replication

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  • Last Post 30 June 2023
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Jagau posted this 09 December 2022

My replication of Figther's ZPM. As my two coils were already mounted on an AMCC320 I was able to replicate quickly.
In the first image the 2 coils mounted on the AMCC320

 

In the second image when the power supply is turned on at 24 volts DC he sees an input power of 4.32 watts on 24 volts and 180 ma when my TL494 oscillator is supplied to both coils through an IGBT pulsed at 38.2% on the high side. Channel1 is IGBT pulse 

 

Image of the oscilloscope with the current probe on the PS

You can notice that the oscilloscope show on a my TEK current probe (green) 292mv =292 ma  (rectangular form) and 38.2% DTC

Last image the 118 volt 4 Watts lamp lit with 34 ma x118.8v = therefore 4.0392 watts of power produce at the output on a large capacitor and 4 diodes in parallel in order to produce an almost pure DC which can be read by the 2 DDMs

Calculation of input power in DCM mode with reactive component.

As indicated by the power supply we have an instantaneous pulse power of 4.32 watts, this is the instantaneous power not the average power over time.

V and I is what PS provided a DM as wrong lecture for input due to high frequency so to calc PIn we take well known formula for inductor.

Irms on scope is 0.292 Amp, DC voltage is 24volts

Pin= I*V*sqrt(Duty_Cycle)/sqrt(3)


Pin= 0.292 Amp x 24 Vdc x sqrt 0.382 / sqrt 3 = 2.533 watts

So for  Pout of 4.032 watts  i need only 2.533 watts at input, Still a pretty good COP
Jagau

beyond-unity zpm
Fighter posted this 22 December 2022

Hello friends,

Jagau asked me to lock his threads until he returns.

He needs to take care of some personal things, he will give more details when he returns.

Regards,

Fighter

"If you want to find the secrets of the universe, think in terms of energy, frequency and vibration."
Nikola Tesla
Jagau posted this 31 March 2023

Thanks Fighter for unblocking the thread.
For those who have questions I would like you to ask them here
THANKS

Jagau

Jagau posted this 31 March 2023

Hi Atti
Could you tell me with which circuit you took to made this interesting oscilloscope shot? 

and explained it a bit more.


THANKS
Jagau

Atti posted this 31 March 2023

Hi Jagau.

 

You can read and read the details here. Watch Cap4 video.   https://www.beyondunity.org/thread/atti-s-successful-zpm-replication/?p=3
 The power source is not a laboratory power supply but a battery.12V+18V .  
The current and voltage values taken from the battery are also shown.  The phase relationships are also shown.  

 

Atti.

AlteredUnity posted this 31 March 2023

Hey Jagau, I have plenty of components at my disposal and seeing the promising results(even in my experiment though the coils are wound opposite.) I am interested in your diode(s) and output capacitor configuration. I haven't took the time too hook up my siglent for readings, and apply your formula, but think it could really simplify my measurments. Ps. Last time I tried tuning a 120v load it tripped my plugs gfi, and can't remember but may have tripped circuit breaker too.

Jagau posted this 01 April 2023

Hi
So with the schematic that Fighter provided on his thread you seem to have done well, when you have the time and the opportunity show us that. Fighter and I will be happy to help you
Jagau

AlteredUnity posted this 28 April 2023

Quoting:Jagau

My replication of Figther's ZPM. As indicated by the power supply we have an instantaneous pulse power of 4.32 watts, this is the instantaneous power not the average power over time.

V and I is what PS provided a DM as wrong lecture for input due to high frequency so to calc PIn we take well known formula for inductor.

Irms on scope is 0.292 Amp, DC voltage is 24volts

Pin= I*V*sqrt(Duty_Cycle)/sqrt(3)


Pin= 0.292 Amp x 24 Vdc x sqrt 0.382 / sqrt 3 = 2.533 watts

So for  Pout of 4.032 watts  i need only 2.533 watts at input, Still a pretty good COP
Jagau

 

You are correct Jagau in your calculations, your power supply does only display instantaneous power, it doesn't take into account duty cycle nor is it made to measure a pulsed dc signal

I believe my siglent does integrate the values and show the true power measurement, minus *cos(theta)(so doesn't account for phase shift) so im excited to see results when I set back up.

Though from basic formula I came across: Average Power = Instantaneous Power * Duty Cycle. V = 24 I = .292 InstantaneousPower = (24*.292)=7.008w AveragePower = instantaneous-power * dutycycle

So actual input power would be (7.008) * .382 = 2.677W. I believe the formula your using is for a 3phase power system. Correct me if I'm wrong, I'm just posting as I learn.Either way it's still Beyond Unity.

Jagau posted this 28 April 2023

Hi


One thing that needs to be clarified and has been discussed a lot here is instant power. Teledyde Lecroy which is a large manufacturer of very expensive laboratory oscilloscopes said this:


Oscilloscopes, whether analog or digital, are voltage responding instruments. Current is measured using a suitable transducer, usually a current probe or resistive shunt. The oscilloscope display is the instantaneous function of voltage or current vs. time. The product of these quantities is instantaneous power.

And more for those who still doubt they said this:

The product of the effective (rms) current and effective (rms) voltage is called the apparent power. Apparent power is represented by the symbol S and is measured in units of Volt-Amps (VA).

You can find the original article here:


 https://teledynelecroy.com/doc/power-real-and-apparent

In addition for educational purposes another excellent YT from another Keysight oscilloscope manufacturer,
Listen carefully to what he says at 1 minute 12 sec. Very important.


You may have learned in your classes that power is voltage times current
but if you take the rms voltage times the rms current that doesn't produce what's called real power that's apparent power and not watts.


These things are not obvious explained but with these two examples I believe that it is clearer.

Jagau

Jagau posted this 19 June 2023

Hi all

Maybe this oscilloscope maker article Lecroy,

 can help you understand how to find power with an oscilloscope

https://archive.eetasia.com/www.eetasia.com/ART_8800701576_765245_TA_7f8ada44.HTM#

Jagau

Jagau posted this 30 June 2023

Another great article on taking measurements with Pulse Width Modulation PWM, that I used in my calculations on the ZPM and a reference that can help you to explain my result.

 https://lygte-info.dk/info/DMMPWM%20UK.html

Power

When working with pulsed current or PWM you cannot just multiply voltage and current to get power, here the RMS  values can be useful.

Example:
0V to 10V with 10% duty cycle will delivering 0A at 0V and 1A 10V. It is easy to calculate that it will deliver 10 watt when (Voltage x Current -> 10 x 1) on and with 10% duty cycle the power is 1 Watt.

But using the DC values that is 1V (10% of 10V) and 0.1A (10% of 1) gives 0.1 Watt, i.e. very wrong.


What about RMS values, that is 3.16V and 0.316A and gives the correct 1 Watt. 

But there is a caveat on that, the RMS part will include current running both ways, this is fine for dimensioning the cable and calculating power for resistors, but when the circuit has a inductor or capacitor that returns some current each cycle, the above power calculations will fail!

With pure resistance without PWM, MEAN and RMS are equal. This is different with PWM especially when using inductive and capacitive elements, ie with a return to the source of a certain power that you must take into account in the calculations in this case MEAN and RMS are different.
  Ex: a power supply (in the presence of an inductive and capacitive component) gives the power it is consumed to the right, it does not give what is returned to the source at left. In your calculations you must take into account the power returned to the source, this should be done with an oscilloscope.

Multiplying both current and voltage waveforms while accounting for phase angle gives a reliable response.  Note, if you multiply the two numerical values by hand it will only give the apparent power.

We could even go so far as to say the electric companies that supply us with electricity charge us for what counter sees at the moment,  not what we return, but to us as a researcher we can calculate it in our true power used

Jagau

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