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Author Topic: Parametric Charging  (Read 36863 times)

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Good work indeed PM, TM

If I remember correctly Arie said that the power increases as the frequency increases.

I am wondering how the cop varies relative to drive voltage, does it scale linearly or squared.
   
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That last patent is where I am atm, but I'm using 3 phase same frequency, not two, the first phase is converted into a floating neutral and any positive wave is crossed over on the output.

Regards

Mike 8)

Hi Mike,

It is the Brasil patent BR0205594, right?   (Método para incrementar a energia cinética de elétrons utilizando uma combinação de campos magnéticos e eletromagnéticos permanentes)   

Maybe it cannot be retrived freely via the web I guess?  Did you have to pay for it?

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Good work indeed PM, TM

If I remember correctly Arie said that the power increases as the frequency increases.

I am wondering how the cop varies relative to drive voltage, does it scale linearly or squared.

Peter,

Thanks!

In my testing the output power does seem to decrease with frequency but I'm not sure if it is due to impedance mismatch or what Arie describes.

The COP tends to decrease slightly with an increase in supply voltage so it follows a more linear than square function IMO at this point.

Regards,
Pm
   

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Guys,

a word of caution, seeing those values of COP = 209% should trigger you to look for any (measurement) errors,
as normally this would not be possible.

Not sure if this any measurement error, calculation error or even the circuit itself that does not lend itself for
doing measurements with our equipment.

Tinman, could you use the math function on your scope to calculate the power in?
By the way, i miss your "first scope shot" in your last post.


Itsu

Hi Itsu

First,i have added the missing input pulse scope shot to my last post--Thanks for the reminder  O0

Regarding the measurements-
I have been very lenient toward the P/in,as in using higher than actual values to make the calculations.
If you look at the circuit i am using,you will see that i have also not accounted for the dissipated power of CVR two,and so this would also increase the COP calculations. In fact,the two CVRs them self dissipate more power than what the circuit consumes--!!apparently!!

My scope shots are quite clean and clear,and the frequency is very low.
We can use the scopes math function,due to the input not being an RMS value,nor can we use mean of average,due to the traces across the primary showing an AC type waveform,but where what you see across the primary is really power flowing through the secondary via capacitive coupling.

P/out is simple,a DC voltage across the 2200uF cap and 15k resistor combo.
P/dissipated by the two CVRs is also straight forward ohms law calculations.

But like you said--this shouldnt be happening,so we will continue on until an explanation is found.

Brad


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Pm and all here.

I believe we are looking at capacitive acceleration of electrons here--not sure how else to put it?

Seems to me that the electric field is causing a flow of electrons (current) and not the magnetic field here,and i believe i can show this.

More to come tonight.

Brad


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Good work indeed PM, TM

If I remember correctly Arie said that the power increases as the frequency increases.

I am wondering how the cop varies relative to drive voltage, does it scale linearly or squared.

Thanks Peter.

ATM i am concentrating on the power measurements i am getting,along with the !how! is this possible,or !where! is this extra energy coming from.

Im leaning toward an electric field effect due to capacitive coupling.

Much to do yet to confirm our findings.
Where is Darren (Poynt) when you need him ?


Brad

Brad
« Last Edit: 2018-08-31, 14:18:43 by TinMan »


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well
Darren is always around if you need him

??


Shall I ring him ?
   

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Hi Itsu

First,i have added the missing input pulse scope shot to my last post--Thanks for the reminder  O0

Regarding the measurements-
I have been very lenient toward the P/in,as in using higher than actual values to make the calculations.
If you look at the circuit i am using,you will see that i have also not accounted for the dissipated power of CVR two,and so this would also increase the COP calculations. In fact,the two CVRs them self dissipate more power than what the circuit consumes--!!apparently!!

My scope shots are quite clean and clear,and the frequency is very low.
We can use the scopes math function,due to the input not being an RMS value,nor can we use mean of average,due to the traces across the primary showing an AC type waveform,but where what you see across the primary is really power flowing through the secondary via capacitive coupling.

P/out is simple,a DC voltage across the 2200uF cap and 15k resistor combo.
P/dissipated by the two CVRs is also straight forward ohms law calculations.

But like you said--this shouldnt be happening,so we will continue on until an explanation is found.

Brad

Brad,


thanks for the screenshot, i was hoping it showed how you calculated the input power, but it does not make it clear to me.

It does not show any RMS or MEAN values (also i don't know where your probes are).
Not sure what you are saying above in bold, is it we can or we cannot?)

the screenshot shows an input square wave of 13.4V max DC (does your FG is capable of doing so?)

So no idea how you came to the mentioned:

Quote
So V/in is an average of 1.8v for the 20% on time.
I/in is 2.9v minus 1.8v= 1.1v across 10 ohms=110mA for our 20% on time.
So P/in for the 20% on time is 198mW

But no problem, it will pan out eventually, i guess.

Here a screenshot from a 13.4V (max) 28.9KHz 20% duty cycle all DC square wave signal (like yours) using
my MOSFET driver as my Rigol FG does not produce more the 10V DC square wave (20pp) signals.

The RMS value and the mean (average) are shown.



Regards Itsu

« Last Edit: 2018-08-31, 12:19:08 by Itsu »
   

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Mike,

The last patent in the list doesn't seem to be available.  Do you have a copy?

Regards,
Pm

Hi PM

I did have, but it seems it was on the computer that was taken when I was robbed. I had it in Dutch, not in English.

It was very much in the style of Jaños, The thing is with most of these patent applications, you have to read between the lines, not that there are things missing, just not in the order to make a replication. I'm talking about  how to get the negative or ground ref: on the output, and why I'm going for 3 phase in a "type of" wye configuration.

Regards

Mike 8)


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 author=Itsu link=topic=3655.msg69485#msg69485 date=1535705587]




Quote
thanks for the screenshot, i was hoping it showed how you calculated the input power, but it does not make it clear to me.

I thought the attached schematic would have made that clear--maybe not.
Here is the description from that post,the schematic and circuit are posted below.
The second scope shot shows the current and voltage trace,but where we now see a voltage across the coil by it self(yellow trace),and then the voltage across the coil and CVR series(blue trace).

Quote
Not sure what you are saying above in bold, is it we can or we cannot?)

Sorry-typo. Should say cant (cannot).

Quote
It does not show any RMS or MEAN values

As i said,you dont use RMS for measuring a DC current or voltage.
RMS is for AC only.
You will see in your scope shot that the mean and RMS values are very different.
When it is pulsed DC,mean or average should be used,and when it is AC,RMS should be used.

Now,in this setup,we are using a pulsed DC,but we can use mean or average due to the reverse current flow caused by the capacitive coupling to the secondary.
The only way to do it is manually,unless my scope can measure between cursors ?,which i dont know as of yet.
But non the less,the manual calculations are accurate.


Quote
(also i don't know where your probes are).

Sorry,i should get into that habit.
Schematic with scope probe placement below also.

Quote
the screenshot shows an input square wave of 13.4V max DC (does your FG is capable of doing so?)

Yes,open voltage. But when loaded,the max voltage depends on the impedance of the load,as my SG has a max current of 150mA at 1v.

Quote
So no idea how you came to the mentioned:

The same way we always measure power.
The average current during the 20% on time X the average voltage during the 20% on time,minus the power dissipated by the 10 ohm CVR during the 20% on time,and then divide the total by 5,or times the total by 20%.
This gives us an average P/in for continuous running.

Brad


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I've Pm'd Darren ;)

Cheers Peter


Brad


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well
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Hey Chet.

I see Peter has messaged him,so see how that go's.

I think we are on a winner here  O0


Brad


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Hi Itsu

First,i have added the missing input pulse scope shot to my last post--Thanks for the reminder  O0

Regarding the measurements-
I have been very lenient toward the P/in,as in using higher than actual values to make the calculations.
If you look at the circuit i am using,you will see that i have also not accounted for the dissipated power of CVR two,and so this would also increase the COP calculations. In fact,the two CVRs them self dissipate more power than what the circuit consumes--!!apparently!!

My scope shots are quite clean and clear,and the frequency is very low.
We can use the scopes math function,due to the input not being an RMS value,nor can we use mean of average,due to the traces across the primary showing an AC type waveform,but where what you see across the primary is really power flowing through the secondary via capacitive coupling.

P/out is simple,a DC voltage across the 2200uF cap and 15k resistor combo.
P/dissipated by the two CVRs is also straight forward ohms law calculations.

But like you said--this shouldnt be happening,so we will continue on until an explanation is found.

Brad

Brad,

In your highlighted statement above did you really mean "can't use the scopes math function" because it fits the context that follows?  If yes, actually we can use the scopes math function on these complex voltage and current wave forms to calculate the input power using the Math channel to display the 'mean' or average of say CH1(volts) x CH2(current).  If not, then ignore the following. :-[

Our digital sampling scopes take instantaneous amplitude measurements of each waveform many times per second so they can be stored and displayed later on the screen as you know.  This is perfect because it provides a means to perform calculations using these sampled values and in the case of power measurements, the voltage and current sampled values are multiplied times each other and the product is saved.  Then integration is performed on these product samples which simply means they are all added together or summed and then averaged by dividing by the number of samples over the measurement time period.  This is why it is important to do these measurements between the vertical cursors because they define the measurement time period.  The scope knows how many samples have been taken over that period so all can be calculated from there.

The results from this use of the Math channel is as accurate as the scope's vertical and horizontal resolution.  Most scopes are 8 bit on vertical which is 1 part in 256 with full screen deflection.  At half screen deflection, it is  1 part in 128 or 0.78% and so on which is why it is important to have the waveform being measured at the highest possible amplitude on the screen.  The horizontal sample rate determines the faithful reproduction of the wave being measured along with phase so it too is very important to have the highest sample rate possible for the measurement window.

IMO, at these low frequencies we are currently measuring, the scope math is reasonably accurate.

Regards,
Pm

Edit



 
   

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Pm--check this out  :o

Ok,i finally found another transformer that works much the same--in fact,even better.
Turns out i have 10 of them inside the battery operated CFL safety lights i scored from work when they changed to LEDs  O0



So,below is the scope shot of the new circuit,with the new transformer-->yes,the original circuit is still together,and will remain so  O0.
From this scope shot we get our values to make our P/in calculations.

The scope traces are very clean,and so accurate measurements are quite easy.
The blue trace (CH2)is across our 10 ohm CVR,and the yellow trace(CH1) is across the 10 ohm CVR and primary coil.
Our current is calculated using ohms law,where we have a clear 1.6 voltas across our 10 ohm CVR.
This gives us a current of 160mA

Now,our voltage across the coil is the value of CH1 minus the value of CH2
So our voltage across the coil is 1.9v(and you will see that i am using the max voltage here),minus 1.6v(voltage across the CVR),give us .3v across the coil.
At first i thought this could not be right,so i checked it using the reverse circuit(see circuit below,where the common rail is used as positive in). This lets me scope the coil voltage by it self,and it was indeed .3 volts(300mV) during the 25% on time.

So now we have 300mV X 160mA for our P/in during the 25% on time.
So during the 25% on time,our P/in is 48mW,where i have left the dissipated power by the CVR out.

Now we have our P/in to the transformer,which is 48mW for 25% of 1 full cycle.
Our average continuous P/in is 48/4,which is 12mW.

Our P/out is the voltage across R1(15k ohm),which for this test was 22 volts.
So our P/out is 32.2mW

Our COP is then 32.2/12x100 =268.3%


Brad


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Brad,

In your highlighted statement above did you really mean "can't use the scopes math function" because it fits the context that follows?  If yes, actually we can use the scopes math function on these complex voltage and current wave forms to calculate the input power using the Math channel to display the 'mean' or average of say CH1(volts) x CH2(current).  If not, then ignore the following. :-[

Our digital sampling scopes take instantaneous amplitude measurements of each waveform many times per second so they can be stored and displayed later on the screen as you know.  This is perfect because it provides a means to perform calculations using these sampled values and in the case of power measurements, the voltage and current sampled values are multiplied times each other and the product is saved.  Then integration is performed on these product samples which simply means they are all added together or summed and then averaged by dividing by the number of samples over the measurement time period.  This is why it is important to do these measurements between the vertical cursors because they define the measurement time period.  The scope knows how many samples have been taken over that period so all can be calculated from there.

The results from this use of the Math channel is as accurate as the scope's vertical and horizontal resolution.  Most scopes are 8 bit on vertical which is 1 part in 256 with full screen deflection.  At half screen deflection, it is  1 part in 128 or 0.78% and so on which is why it is important to have the waveform being measured at the highest possible amplitude on the screen.  The horizontal sample rate determines the faithful reproduction of the wave being measured along with phase so it too is very important to have the highest sample rate possible for the measurement window.

IMO, at these low frequencies we are currently measuring, the scope math is reasonably accurate.

Regards,
Pm

Edit



 

Hi Pm
Yes,i did mean !cant!,unless the math can be carried out between cursors.
We cant use RMS,because it is a DC pulse input,where the voltage never falls below the 0 volt line during that input pulse.
We cant use mean or average,because we have an AC component to the input voltage and current waveform,due to the capacitive coupling between primary and secondary.
 We only want to calculate the on time,while the scope will see a complete cycle,and try to calculate using the whole of the cycle.

My wave forms are very clean,and easy to get accurate voltage values from for the on time.

Now,i made a couple of errors earlier on,where i subtracted the dissipated power of the CVR from the calculated P/in,when i had already subtracted the CVR voltage from the total voltage to get the voltage across the coil,and used that voltage to make my calculations.

In my last post,the calculations where made where the dissipated power of the CVR was disregarded ,and not subtracted from the P/in value.
Even then,the result was a COP of 268.3%.

I am not seeing where i am going wrong here, !but!  ???

Maybe Poynt will see it straight away ?


Brad


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Pm--check this out  :o

Ok,i finally found another transformer that works much the same--in fact,even better.
Turns out i have 10 of them inside the battery operated CFL safety lights i scored from work when they changed to LEDs  O0



So,below is the scope shot of the new circuit,with the new transformer-->yes,the original circuit is still together,and will remain so  O0.
From this scope shot we get our values to make our P/in calculations.

The scope traces are very clean,and so accurate measurements are quite easy.
The blue trace (CH2)is across our 10 ohm CVR,and the yellow trace(CH1) is across the 10 ohm CVR and primary coil.
Our current is calculated using ohms law,where we have a clear 1.6 voltas across our 10 ohm CVR.
This gives us a current of 160mA

Now,our voltage across the coil is the value of CH1 minus the value of CH2
So our voltage across the coil is 1.9v(and you will see that i am using the max voltage here),minus 1.6v(voltage across the CVR),give us .3v across the coil.
At first i thought this could not be right,so i checked it using the reverse circuit(see circuit below,where the common rail is used as positive in). This lets me scope the coil voltage by it self,and it was indeed .3 volts(300mV) during the 25% on time.

So now we have 300mV X 160mA for our P/in during the 25% on time.
So during the 25% on time,our P/in is 48mW,where i have left the dissipated power by the CVR out.

Now we have our P/in to the transformer,which is 48mW for 25% of 1 full cycle.
Our average continuous P/in is 48/4,which is 12mW.

Our P/out is the voltage across R1(15k ohm),which for this test was 22 volts.
So our P/out is 32.2mW

Our COP is then 32.2/12x100 =268.3%


Brad
Brad,

At this point I seen nothing wrong with your calculations.  IMO, you have a different method of operation with your device but I could be wrong on that.  In any case, continue on! 8)

Pm
   

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It's not as complicated as it may seem...
Hey guys. Good to see you're still having fun with this stuff.

It is a bit of a shame however that apparently some important info I've offered in the past has seemingly been forgotten or just not utilized. Anyway, I would start with the following:

1) Replace the input side CVR scope measurement with a DC milli-ammeter measurement (DMM voltage across the CVR). This will give you an accurate average current going into the transformer: I(avg)=Vcvr(avg)/Rcvr

2) Using the scope, get an accurate peak and duty cycle measurement of the input voltage (after the diodes).

3) Calculate your P(in) like this: P(in)= (Vp x duty-cycle) x I(avg).

Is the P(in) measured this way still lower than your measured P(out)? If so, then there are other considerations we can look at. If not, then what might that indicate about some of the P(in) measurements being taken?

btw, When performed correctly, Input power measurements should come out Negative (for example when measured directly near a battery). The "Input Power" measurement being taken as per the schematics is not an input power measurement per se; it is a measure of the power dissipated in some of the components external to the source, and are actually part of the "load", as far as the source is concerned.

   
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All,

Errrr, hold the phone on my circuit as described in post #61 and all that follows!  The input current measurement taken on the low side of the primary is not an accurate representation of the current drawn from the signal generator as I have assumed.  Due to the tight coupling with the trifilar windings, current is conducted thru the primary winding capacitance to the secondaries and is not seen in the primary return line to ground.  Therefore the current waveform measured at the input of the primary winding is greater than the current measured at the return or low side of the primary.  See the two continuous running scope shots below to compare the input power levels with the different measurement points.

The first is with the current probe measuring the current in the input side of the primary and the second is with the probe in the low side or ground side of the primary.  As can be seen and calculated, the COP goes away using the high side current measurement in the pin calculation.  With the 3710 ohm load resistor across C1, the high side COP = 73.81^2/3710/2.03 = 1.468/2.03 = .723 and the low side COP = 73.32^2/3710/.807 = 1.45/.807 = 1.8 .  The high side is real and accurate.  :-[

When I changed the circuit topology for a supposed improvement, I was careless in my assumptions and this is the result.  This does not negate the performance of the original FWB parametric circuit but leaves the question, "can the FWB be improved topology wise?" .

Poynt,

Good to see you back!

Regards,
Pm
   

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Quote
Hey guys. Good to see you're still having fun with this stuff.

Great to see you still around Darren,and thanks for taking a look at this  O0

Quote
It is a bit of a shame however that apparently some important info I've offered in the past has seemingly been forgotten or just not utilized. Anyway, I would start with the following:

Never forgotten,and you will see i have also used the means you describe below--always do to confirm measurements.  O0

Quote
1) Replace the input side CVR scope measurement with a DC milli-ammeter measurement (DMM voltage across the CVR). This will give you an accurate average current going into the transformer: I(avg)=Vcvr(avg)/Rcvr
2) Using the scope, get an accurate peak and duty cycle measurement of the input voltage (after the diodes).
3) Calculate your P(in) like this: P(in)= (Vp x duty-cycle) x I(avg).

Indeed  O0
See video's below.

Quote
Is the P(in) measured this way still lower than your measured P(out)? If so, then there are other considerations we can look at.

Yes,way below.
Maybe time for those further considerations ?.

Quote
btw, When performed correctly, Input power measurements should come out Negative (for example when measured directly near a battery). The "Input Power" measurement being taken as per the schematics is not an input power measurement per se; it is a measure of the power dissipated in some of the components external to the source, and are actually part of the "load", as far as the source is concerned.

And when the total power dissipated by the circuit exceeds that supplied by the source ?.

The fist schematic below is for the first video.
The second schematic below is for the second video.

If you do not have time to watch both video's,then watch video 2,as that is the higher powered device.

https://www.youtube.com/watch?v=AnGD4BXsFyQ

https://www.youtube.com/watch?v=M5oYdyEGOAo

Both video's are unlisted


Brad


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And a third confirmation video.

https://www.youtube.com/watch?v=yOfgl9Uov3U


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It's not as complicated as it may seem...
Good job Brad triple checking the Pin measurements! It's good to make your measurements in more than one way to ensure one is not fooling one's self.

Along that train of thought, I went back through the thread and I couldn't find any Pin measurement you've done using the scope and the tried and true method of AVG[v(t) x i(t)]. I think you are familiar with this method, and I would encourage you to do this on one of your DUTs in question. I have confidence you will see quite different results.

The problem with the technique you have been using for Pin, is that assumptions are being made about the voltage and current used in the calculation. If the load driven by the signal generator was purely resistive, then the technique is valid. However, you are driving the primary of a transformer, and all bets are off in terms of peaks and duty cycles. Picking a Vp from the input voltage and ignoring everything else present on that wave form is not valid here. Using the tried and true AVG[v(t) x i(t)] method takes care of all the issues present with this type of scenario.

The Pin measurement for Rosemary's circuit is different in that the power supply "Input" is an assumed steady DC voltage. With that scenario, we can measure the average current with the DMM and multiply the result by the DC input voltage. So I led you astray somewhat in my last post, apologies. With your circuit one can see that the "Input Voltage" is anything but a steady DC state, and there is a reactance involved.

Have you yet purchased a decent meter that can do RMS measurements to say 100kHz? One cool thing that can be done with a true RMS meter is to determine the exact power being transferred to the secondary of your transformer. If you use a very low inductance CVR you can then determine the RMS current in the primary circuit. From this, you can now calculate the power dissipated1 in the primary of your transformer (in your case a primary resistance of 1.6R). Subtract that from the AVG[v(t) x i(t)] measurement, and the result is the remaining power (VA) going to the secondary. Of course you have the fancy scope that can do RMS of a trace, so buying a separate fancy true RMS meter is not necessary.

You have the scope and I would suggest you use it for these measurements, even as a double-check for those that seem like no-brainers.

1 This is the power dissipated (W) in the transformer primary, not the total power (W + VA) in the primary. P(W)(prim)= I(rms)2 x R(prim).
« Last Edit: 2018-09-01, 20:22:39 by poynt99 »
   
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All,

Before proceeding on with the parametric charger, I decided to profile the RFP14N05L mosfet I've been using and also some other devices I have on hand.  The schematic used for the testing the is attached including the math used to generate the traces to allow the determination of the capacitance values with increasing voltage.

Scope shots with samples of the measurements for the FRP14N05L, IRF740, and a 24mm piezo are also included for comparison plus a data table for the devices tested thus far.

Regards,
Pm

Edit 9/3/18-  Replaced data table with corrected info on 24mm piezo.
« Last Edit: 2018-09-03, 15:45:34 by partzman »
   

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Good job Brad triple checking the Pin measurements! It's good to make your measurements in more than one way to ensure one is not fooling one's self.

Along that train of thought, I went back through the thread and I couldn't find any Pin measurement you've done using the scope and the tried and true method of AVG[v(t) x i(t)]. I think you are familiar with this method, and I would encourage you to do this on one of your DUTs in question. I have confidence you will see quite different results.

The problem with the technique you have been using for Pin, is that assumptions are being made about the voltage and current used in the calculation. If the load driven by the signal generator was purely resistive, then the technique is valid. However, you are driving the primary of a transformer, and all bets are off in terms of peaks and duty cycles. Picking a Vp from the input voltage and ignoring everything else present on that wave form is not valid here. Using the tried and true AVG[v(t) x i(t)] method takes care of all the issues present with this type of scenario.

The Pin measurement for Rosemary's circuit is different in that the power supply "Input" is an assumed steady DC voltage. With that scenario, we can measure the average current with the DMM and multiply the result by the DC input voltage. So I led you astray somewhat in my last post, apologies. With your circuit one can see that the "Input Voltage" is anything but a steady DC state, and there is a reactance involved.

Have you yet purchased a decent meter that can do RMS measurements to say 100kHz? One cool thing that can be done with a true RMS meter is to determine the exact power being transferred to the secondary of your transformer. If you use a very low inductance CVR you can then determine the RMS current in the primary circuit. From this, you can now calculate the power dissipated1 in the primary of your transformer (in your case a primary resistance of 1.6R). Subtract that from the AVG[v(t) x i(t)] measurement, and the result is the remaining power (VA) going to the secondary. Of course you have the fancy scope that can do RMS of a trace, so buying a separate fancy true RMS meter is not necessary.

You have the scope and I would suggest you use it for these measurements, even as a double-check for those that seem like no-brainers.

1 This is the power dissipated (W) in the transformer primary, not the total power (W + VA) in the primary. P(W)(prim)= I(rms)2 x R(prim).

Hi Darren

Yes,i am a little confused here regarding this AVG[v(t) x i(t)] method you are talking about.
I thought that is what i had done all through the thread here,where the voltage during the 20% on time was averaged,and the current during the on time was averaged,and then the two multiplied to give our average P/in.

Should not the average P/in be calculated by only using the values during the 20% on time,as that is the only portion of the complete cycle the DUT is receiving energy from the SG. The rest of the waveforms through the remaining 80% of the cycle is recycled through the primary via the circuit as a whole.

If we look at the scope shot below,we see the on time where power is delivered to the transformer from the SG(between the red lines).
Between the green line's shows the inductive kickback cycle,where the voltage is inverted,but the current continues to flow in the same direction,and the last small portion before on time is C2 discharging

Now,the average voltage in that scope shot is the average voltage across both the CVR and primary of the transformer,not just the transformer it self,and that is why it cannot be used to calculate average power dissipated by the transformers primary only--the joys of having the scope and SG sharing a common ground  :(

So,when making my calculations during the 20% on time,i subtract the voltage across the CVR from the total voltage to get my voltage across the transformer. I confirmed this value by using the circuit below,where i could measure the voltage across the transformers primary only during the 20% on time portion of the cycle.
This showed me that my calculated voltage across the transformer in previous tests was correct.

Im not quite sure what you mean by this AVG[v(t) x i(t)] ,as i thought that was what i was doing throughout the whole thread here,but you say you dont see where i have done it ?.
If you mean average V over time X average current over time,then that is what i have done throughout the thread here,where i averaged the input  voltage across the coil during the 20% on time to the full cycle time,and the same for the current.

Now,if you mean using the average voltage X average current from the scope's measurements,then lets use the average values shown in the supplied scope shot.
Average voltage is 150mV,and average current is 316mV across 10 ohms for an average current of 31.6mA
Now our average P/in would be 4.74mW
Our P/out is 20v across 15kohm=26.66mW
We now have a COP of 562%

I am happy to test the DUT anyway you see fit,as all my testing so far gives COPs that should not be.


Brad


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It's not as complicated as it may seem...
Brad,

I'm no math guru, but what I am trying to convey with AVG[v(t) x i(t)] in words is:

Take the average (AVG) of the instantaneous voltage v(t) times instantaneous current i(t).

You do this by setting up one probe (CH1) measuring the input voltage, just as you have been, and one probe (CH2) measuring across the low-inductance (not wirewound power resistors) cvr for current. Set up a MATH trace for CH1xCH2, then set a measurement for the average value (or MEAN) of the MATH trace. You will have to scale the current based on the value of your CVR resistor. Some scopes allow you to scale, otherwise you do it in your head by dividing by 10 (for a 10R resistor).

That should now ring a bell with you...

This method has been discussed and recommended many times over the years, not only by myself, but by the true masters out there, TK, ION, etc. etc (apologies for those I have missed). It was used extensively during the Rosemary Ainslie saga as well. And I am fairly certain that we had you doing this as well on some test, or at least attempting to until we couldn't figure out how to get the AVG working on your MATH trace as I recall.

Anyway, give that a shot. It should be no problem with your new scope, and you very well may have already done it in recent times.
« Last Edit: 2018-09-02, 02:24:49 by poynt99 »
   
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