PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2024-03-29, 06:15:15
News: A feature is available which provides a place all members can chat, either publicly or privately.
There is also a "Shout" feature on each page. Only available to members.

Pages: 1 2 3 4 5 6 [7] 8 9 10 11 12 13 14 15 16 17
Author Topic: Parametric Charging  (Read 60075 times)

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
Ok,i will up the sample rate on the screen,and give that a shot.

A few things--

This test was done on my second DUT,and the frequency is around 14.7KHz.

The resistors used on this DUT are the 1/2 watt carbon resistors..
On that note,the large gold ones were described as being carbon resistors by the seller,not the wire wound cement type,such as some of those white block ones i have,which is why i bought them-the gold ones.

I will redo the test tonight,post the results,and touch on a !hiccup! using this test method.

Brad

Sounds good Brad.

I hope those goldies are indeed low L. You could certainly measure their inductance, and compare to the 1/2W carbons.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4499


Buy me some coffee
Then we look deeper to verify the results.
For P(in) measurements, DMM's are excellent for finding the average on a signal/source that is/should be flat DC. Ainslie's circuit used a battery source for the input. The load was switched, therefore the current pulsed. So when measuring power back at the battery (the source), the voltage was a no-brainer, as it is simply the loaded battery voltage. The current on the other hand is pulsed, and this is where the DMM measurement shines above the rest, because it does an excellent job averaging the current from the fixed DC source. One simply multiplies the fixed loaded DC voltage times the measured average current, and you have the input power being delivered by the battery. In your case the source voltage is not fixed, as it is a pulsed DC output. So we have to use the tried and true method with the scope doing AVG[v(t) x i(t)]. BTW, this is an instantaneous measurement done by the scope, as it is taking many samples and performing the math on each sample across the full cycle. Doing this manually as you have been is not the same thing.
That's kinda the problem; there is other stuff happening in that wave form, and it all has to be accounted for. The best method we have available to achieve this is the one I've been asking you to use.

Well i must say that it is odd the manually calculating the average current using the scope values,using the scope to calculate the average current value,and the DMMs average current value all result in the same amount,while using the RMS value results in near twice the value--such as Itsu has also found using his current probe RMS value.

But anyway,if i use a battery,and switch the circuit on and off by way of the SG switching a mosfet at the same frequency and duty cycle,we can then trust our DMM to give us an accurate average current value-correct?

This being the case,and our scopes calculated average current value is the same as that of the DMM,and my manually calculated current value is also the same,then that would mean that my methods of calculating the P/in are correct-yes?

But how did you average the voltage in the Ainslie circuit,as it would be the supply voltage x duty cycle to get the average voltage over a full cycle. That is how the math works on a scope-is it not?,where V and I are averaged,and then multiplied together ?


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
I found another ferrit cored transformer:

prim.  6mH /  2 Ohm
sec.  2.5H / 30 Ohm

Using the 10V DC 20% duty cycle 28.8KHz signal without input diode.
I have a DC mA DMM in the red plus lead showing the (37.2mA rms? mean?) current going in, see picture.
(a 2th Fluke DMM across a temporary installed 1 Ohm csr shows 37mV).
Both of these will be AVG current.

Quote
The screenshot shows;

yellow: the voltage across the primary (still large spikes) 911mV rms
green:  the current from my current probe in the red FG supply lead 105mA rms
red:    the math function yellow x green presenting the average input power 20.8mW

So we see 20.8mW input.

A 3th fluke DMM shows the voltage across the 14860 Ohm resistor to be 11.2V
This represents a power out of 8.7mW indicating a cop = 0.42.
Did you subtract the CVR power from the 20.8mW? To obtain the CVR power, you need to know the RMS voltage across it.

Quote
Strange is the difference  between the input current DMM (37.2mA) backed up by the temp 37mV across a 1 Ohm resistor
and the current probe input current measurement of 105mA rms.
It isn't strange, because the DMM measures average current.

Quote
Are the DDM's unable to accuratly display the 28.8KHz current/voltage?  It looks like it.
The DMM displays the average very accurately. In general the only time we use the DMM for input power measurement is when the source is a fixed DC supply, and in that case it would be to measure the average current. I(avg) x Vdc = P(in). Since we are not using a fixed DC supply in this device, we don't use the DMM.

Quote
Using another methode of obtaining the input current (2 probes in differential setup across the temp. 1 Ohm csr)
shows a similar current signal as with the current probe, but due to the low level of signals the specific value
is hard to obtain, but i tend to believe the current probe is showing the corrent input current.

Regards Itsu
As I believe Brad suggested, you could increase your CVR to 10R rather than 1R. This will give you 10x the signal level to measure across the CVR. Set up an RMS measurement on the CVR voltage and you can then obtain the CVR power to subtract from the total (if your circuit configuration requires it).
   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
Well i must say that it is odd the manually calculating the average current using the scope values,using the scope to calculate the average current value,and the DMMs average current value all result in the same amount,while using the RMS value results in near twice the value--such as Itsu has also found using his current probe RMS value.
Indeed. Using the average current would result in a significant error, wouldn't you agree?

Quote
But anyway,if i use a battery,and switch the circuit on and off by way of the SG switching a mosfet at the same frequency and duty cycle,we can then trust our DMM to give us an accurate average current value-correct?
Yes, using the average DC input current is valid in this case. There is a consideration with this setup however: The device is only using milliwatts of power so there is a chance the SG driving the MOSFET gate could skew the P(in) measurement. In the RA case we were dealing with 30W or so of power, therefore not really a concern. You could insert a hefty series gate resistance to limit the effect, but that could slow down your switching as well.

Quote
This being the case,and our scopes calculated average current value is the same as that of the DMM,and my manually calculated current value is also the same,then that would mean that my methods of calculating the P/in are correct-yes?
Not quite. I suspect you are assuming you will be correcting the 12V battery (for example) by the duty cycle, when in fact that would not be correct. The duty cycle correction already takes place during the current measurement via the DMM's averaging. Since we are already using the DMM to obtain the average current, we don't need nor want to apply the duty cycle correction twice, which is what you would be doing if you corrected the voltage also.

Assuming minimal power loss in the wiring and the MOSFET, and assuming that the source power will then equal the power into the transfomer primary, the goal is to determine the power being supplied by the source, agreed? If a CVR is in series with the battery, the loss there can easily be subtracted out. Quick review: The power from a DC source is determined by the average voltage across it times the average current through it.

So, you will be measuring voltage across the battery, not the primary of the transformer. As such, you could use 3 DMMs and turn the scope off now that both your source and your load are DC.

Quote
But how did you average the voltage in the Ainslie circuit,as it would be the supply voltage x duty cycle to get the average voltage over a full cycle. That is how the math works on a scope-is it not?,where V and I are averaged,and then multiplied together ?
As per above, you don't correct for duty cycle when measuring the battery supply voltage, the DMM current measurement already automatically does the correction on the current (which is the parameter that is pulsing) for you. Performing the duty cycle correction twice (once on the battery voltage and once on the battery current) would be really un-cool.  8)
« Last Edit: 2018-09-04, 03:27:33 by poynt99 »
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4499


Buy me some coffee
Indeed. Using the average current would result in a significant error, wouldn't you agree?
Yes, using the average DC input current is valid in this case. There is a consideration with this setup however: The device is only using milliwatts of power so there is a chance the SG driving the MOSFET gate could skew the P(in) measurement. In the RA case we were dealing with 30W or so of power, therefore not really a concern. You could insert a hefty series gate resistance to limit the effect, but that could slow down your switching as well.
Not quite. I suspect you are assuming you will be correcting the 12V battery (for example) by the duty cycle, when in fact that would not be correct. The duty cycle correction already takes place during the current measurement via the DMM's averaging. Since we are already using the DMM to obtain the average current, we don't need nor want to apply the duty cycle correction twice, which is what you would be doing if you corrected the voltage also.

Assuming minimal power loss in the wiring and the MOSFET, and assuming that the source power will then equal the power into the transfomer primary, the goal is to determine the power being supplied by the source, agreed? If a CVR is in series with the battery, the loss there can easily be subtracted out. Quick review: The power from a DC source is determined by the average voltage across it times the average current through it.

So, you will be measuring voltage across the battery, not the primary of the transformer. As such, you could use 3 DMMs and turn the scope off now that both your source and your load are DC.
As per above, you don't correct for duty cycle when measuring the battery supply voltage, the DMM current measurement already automatically does the correction on the current (which is the parameter that is pulsing) for you. Performing the duty cycle correction twice would be really un-cool.  8)

Ok,i will get straight onto it when i get home from work.

Thanks for you time Poynt,great to have you here.

Hope you stick around till the end of this.


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4499


Buy me some coffee
Post removed--silly question  C.C
« Last Edit: 2018-09-04, 11:57:27 by TinMan »


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4499


Buy me some coffee
Sounds good Brad.

I hope those goldies are indeed low L. You could certainly measure their inductance, and compare to the 1/2W carbons.

I will indeed do that,just incase the ebay seller was telling a small fib  :-\

In the mean time,i will stick with the 1/2 watt carbon resistors until i have a look at the big ones.

Ok,well im glad you suggested increasing the point's,as it actually reduces the calculated measurement values,and not increase it. So my COP just went up  O0--see the difference in the two scope shots below.

Well,on with the math.
So we have the average of the math trace-804mW /10 ohms=80.4mW for total corrected P/in

RMS across 10 ohm CVR is 838mV
Power dissipated by CVR is 70.22mW
And a current value of 83.8mA

P/in for primary of transformer is there for a whopping 10.18 mW  O0

Are you happy with these calculations Poynt ?


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
I don't know if this helps but when making true duty cycle measurements, the input diode may skew results by not allowing the SG to actively  "pull down" the input.

In other words when the input "floats down" essentially disconnected from the SG by the diode, the duty cycle may not be accurate. It may act as a peak detector in a circuit with no inductance. With inductance, the primary reactance may alter input power measurements.

The reactive component cannot be absorbed by the generator as the diode blocks such action.

Of course i could be wrong about all this.

Regards


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4499


Buy me some coffee
I don't know if this helps but when making true duty cycle measurements, the input diode may skew results by not allowing the SG to actively  "pull down" the input.

In other words when the input "floats down" essentially disconnected from the SG by the diode, the duty cycle may not be accurate. It may act as a peak detector in a circuit with no inductance. With inductance, the primary reactance may alter input power measurements.

The reactive component cannot be absorbed by the generator as the diode blocks such action.

Of course i could be wrong about all this.

Regards

Hi ION

I dont think that is quite correct.
Once the on cycle has finished(duty cycle),we then have the flyback period.
If we look at the scope shot below,we can see that there is still current flowing through the CVR during this flyback period.

If we have say a 25% duty cycle,then for 25% of the cycle the FG is the source.
During the 75% off time,the inductor/transformer then becomes the source.
We can see the voltage inverts,and current continues to flow in the same direction.
This current is now flowing through the common(0 volts or ground as we call it),through the FG,and back into the circuit.

This is why i have a problem using the scopes math to work out P/in,as it will be accounting for power that is not only being delivered by the FG,but also the power being recycled by the transformer it self.
The scope shots clearly show an inductive kickback cycle during the off time,where the voltage is inverted,and current continues to flow in the same direction-->through the CVR on the ground rail.


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
Hi ION

I dont think that is quite correct.
Once the on cycle has finished(duty cycle),we then have the flyback period.
If we look at the scope shot below,we can see that there is still current flowing through the CVR during this flyback period.

If we have say a 25% duty cycle,then for 25% of the cycle the FG is the source.
During the 75% off time,the inductor/transformer then becomes the source.
We can see the voltage inverts,and current continues to flow in the same direction.
This current is now flowing through the common(0 volts or ground as we call it),through the FG,and back into the circuit.

This is why i have a problem using the scopes math to work out P/in,as it will be accounting for power that is not only being delivered by the FG,but also the power being recycled by the transformer it self.
The scope shots clearly show an inductive kickback cycle during the off time,where the voltage is inverted,and current continues to flow in the same direction-->through the CVR on the ground rail.


Brad

Yes, my thinking may not be worked out on this, although I do notice a float down on the yellow trace, although I'm not sure what the colored traces represent. Maybe they are the same as some earlier description, I will have to go back and check.

Can you try it without the diode just to be sure? But this could be dangerous to the SG if it is not internally clamped so maybe not a good idea..
« Last Edit: 2018-09-04, 18:39:08 by ion »


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
Well,on with the math.
So we have the average of the math trace-804mW /10 ohms=80.4mW for total corrected P/in

RMS across 10 ohm CVR is 838mV
Power dissipated by CVR is 70.22mW
And a current value of 83.8mA

P/in for primary of transformer is there for a whopping 10.18 mW  O0

Are you happy with these calculations Poynt ?

They appear correct.
   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
This is why i have a problem using the scopes math to work out P/in,as it will be accounting for power that is not only being delivered by the FG,but also the power being recycled by the transformer it self.
The scope shots clearly show an inductive kickback cycle during the off time,where the voltage is inverted,and current continues to flow in the same direction-->through the CVR on the ground rail.
Brad, this is in fact one of the beautiful things about this measurement method, it accounts for any energy going back to the source.

That means that during the kickback phase the power product reverses in polarity (notice the purple trace dips below zero?), and when this is averaged with the forward power, it actually decreases the over all forward power result.

So what you seem to think is working against you, is actually working for you in this case.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4499


Buy me some coffee
They appear correct.

Poynt

We need to be 99% + accurate here.
Is there anywhere mistakes could have been made here?

Is there another way to calculate P/in,so as to back up the shown results using your method?.

Looking at the circuit below,is there any way power can get to R1 without passing through CVR1 ?


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Experimentalist
Hero Member
*****

Posts: 2809


Buy me a beer
Brad

Maybe you have said before, but is your ground floating or real ground as shown in your diagram, if real ground then what is it connected to? apart from the scope ground.

Regards

Mike 8)


---------------------------
"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

As a general rule, the most successful person in life is the person that has the best information.
   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
Poynt

We need to be 99% + accurate here.
Is there anywhere mistakes could have been made here?

Is there another way to calculate P/in,so as to back up the shown results using your method?.

Looking at the circuit below,is there any way power can get to R1 without passing through CVR1 ?


Brad

I don't see any problems in the protocol, but your last question is a good one. Perhaps through capacitive coupling of the windings...

Remove your transformer and short the input leads, and output leads and treat it as a capacitor. See what your capacitor meter measures across the now two leaded component.

I created a simulation of the circuit, and just 100pF from pri-sec has a fairly significant impact on the CVR wave form.

Out of curiosity, have you tried driving this with a proper sine wave at about the same frequency?
   
Group: Experimentalist
Hero Member
*****

Posts: 1665
Interesting! 
PM writes: "Yesterday, I was trying to use some of the other mosfets using the newly measured capacitance change data and the results were terrible.  So, I thought I would replicate my previous circuit and data in post #9 for comparison and all attempts throughout the day to replicate the performance failed with all circuit values and measurement procedures identical.  The COPs were ~.8-.9 at best.  The weather here was overcast and rainy off and on all day during these test attempts! 

This morning, all the weather has moved on and we have a bright sunny day at the moment and when running the replication test at this moment in time, the COPs with gain have returned.  Any suggestions as to why this is happening? "

   Air humidity is clearly suspect as the culprit.  Possibly in your lab, but possibly humidity in the air outside...
   WHY humidity might matter so much, is another question. 
I would say it is very valuable to be able to turn the effect "off", and you have evidently found one way to do this, with increased air humidity.

Today is another very wet and rainy day and the COPs have again dropped!  For comparison, the two data tables below are of the same circuit tested in the same manner but the first was taken on yesterday 9-3-18 (sunny) and the second taken today 8-4-18 (rainy).  Obviously the permittivity of the outside air is affected by the humidity but locally at my bench, the relative humidity couldn't be changing by much due to being enclosed and air conditioned.

So, is the apparent energy gain due do an aetheric source or...?

Regards,
Pm
   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
PM, Is it a temperature related thing?

I would suggest running a calibration on your scope before taking your measurements, and then compare.

It is suspicious to me that in one case the P(in) are all over the map, both pos and neg, while the next time they are all pos. Suggests to me a change in the offset in the scope somewhere.
   

Group: Administrator
Hero Member
*****

Posts: 3866


Buy me some coffee
PM I noticed in an earlier post you were using a plug board to build on, these are well known to induce pf capacitance onto the traces, which would very likely be humidity responsive.
Although if air conditioned then maybe not, maybe affecting your house ground rod if it has one.
   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
In regards to testing another way, we could try the following as per the pic. This places the input CVR before the primary, which is ideally where it should be located. Without a current probe, it is a bit of a challenge, especially if we want to do further MATH on it.

If you can somehow multiply the MATH trace CH1-CH2 across the CVR x the CH3 probe, then this would be another way to measure the power input. I don't know if you can set up a MATH trace that already involves another MATH trace however.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4056
Itsu,

Well, the best way to tell is to do an experiment and although I'm using my Tek MDO, your TDS should measure in the same manner.

The test setup is a signal generator set at 500kHz sine wave driving a 100 ohm 1% non-inductive resistor with the current being measured with a Tek TCP0020 current probe.  I could have used a CSR and the results would have been more accurate on the second test.

The first scope pix shows the voltage and current waveforms with very little phase difference and the resulting mean Math channel calculation.   So, we multiply the rms values and get Pin = 4.415*42.47e-3 = 187.5mw. neglecting the small phase angle.  This compares favorably with the Math channel value of 187.3w.

In the next scope pix, we've added a ~100uh inductor to create some phase shift to move the power factor off unity so we'll be forced to use cosine correction.  Therefore now the Pin = 6.118*16.64e-3*cosine(72.48) = 30.65mw.  The Math shows 28.17mw for an error of 9% which I attribute to the fact that we are in the minimum current range of 10ma for the TCP0020 probe rather than the internal math calculations of the Tek scope.

Rather than repeat here, one can look at my post #88 for an explanation of how the "mean" or average measurements are taken on complex waveforms with the current Tek scopes.

Regards,
Pm   

PM,

thanks for the setup, i used this to, so this confirms that the x (multiply) function of the math is using
the rms values of the both channels used.

Now see what the + (add) and - (subtract) functions do with 2 signals, what values do they use?

Itsu
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4056
Hi Itsu.

What indications are you seeing that leads you to conclude the scope is performing any conversion (MEAN or RMS) on the v(t) x i(t) components?

When you set up the MATH channel/trace to perform CH1 x CH2 (for example), there is no processing before or after the result. You can see the result in Brad's scope shots (the purple trace), and it is just the raw output or result of the v(t) x i(t) process. When you set up a measurement on a channel/trace, then processing is performed and the result is indicated in the box labeled AVG or MEAN, or RMS etc.

In the case of the MATH trace set up for measuring power, the raw product is processed with the MEAN measurement function in order to take the instantaneous power trace and convert it to the average power.

Poynt99,

see post #146 from PM above, thats how i did it too, you can see that the math trace with x is using the rms values.

Its different when using the + or - functions, they seem to use the mean values, see screenshot below
where i subtract CH1 (yellow) from CH3 (purple) 3.28V - 1.65V = 1.63V like the red math trace shows.
If i would of have used the rms values of CH1 and CH3 to subtract them it would not match the red math trace value.

Itsu
   

Group: Administrator
Hero Member
*****

Posts: 3198
It's not as complicated as it may seem...
Itsu,

Did you miss my two posts to you? I think there is some confusion here regarding what the scope is doing when we utilize a AVG(CH1 * CH2) type measurement.

The scope is NOT using RMS nor any other process on the raw data being sampled on CH1 and CH2 in a "multiply" or any other MATH function trace.

What PM is saying is that the sampled AVG(CH1 * CH2) operation we set up for measuring average power is equivalent to multiplying the RMS values of the voltage and current, if there is no significant phase shift between the two. If there is significant phase shift between the voltage and current, then we can not simply multiply the two RMS values, unless we also incorporate the phase angle in the calculation. But since we are often dealing with non-sinusoidal wave forms, we can not do that, so we let the scope do all the work in determining the average power by setting up the CH1 * CH2 MATH trace. The result is RAW data, i.e. AVG[(sample1 vi product) + (sample2 vi product) + (sample3 vi product) +...]. The scope is sampling at a very high rate, as in 500Ms/s, and the average is a running average of these samples, because we apply a MEAN measurement to the samples.

So the samples are the RAW products of v(t) x i(t), nothing more.

I hope that helps.
   
Group: Experimentalist
Hero Member
*****

Posts: 1665
PM,

thanks for the setup, i used this to, so this confirms that the x (multiply) function of the math is using
the rms values of the both channels used.

Now see what the + (add) and - (subtract) functions do with 2 signals, what values do they use?

Itsu
Itsu,

Anytime you use the Math channel for a calculation let's say the "mean" or average, it uses the same method for all four math functions.  The instantaneous values at each sample along the horizontal time window of the waveforms you've selected are added or subtracted or multiplied or divided on at that moment in time and the result is stored in the scopes memory.  Then, depending on the cursor measurement mode you have selected say "between the vertical cursors", the scope knows that you wish to compute with the samples taken only between those cursors.  It then "integrates" or "sums" all those samples together to get one sum and then divides by the number of samples it took between your cursors.  This is of course for the "mean" measurement.  Long answer for your question but I hope it helps.

Pm 
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4056

Quote
Quote
Quote from: Itsu on 2018-09-02, 20:25:15

    I found another ferrit cored transformer:

    prim.  6mH /  2 Ohm
    sec.  2.5H / 30 Ohm

    Using the 10V DC 20% duty cycle 28.8KHz signal without input diode.
    I have a DC mA DMM in the red plus lead showing the (37.2mA rms? mean?) current going in, see picture.
    (a 2th Fluke DMM across a temporary installed 1 Ohm csr shows 37mV).

Both of these will be AVG current.

OK, good to know.

Quote
Quote
    The screenshot shows;

    yellow: the voltage across the primary (still large spikes) 911mV rms
    green:  the current from my current probe in the red FG supply lead 105mA rms
    red:    the math function yellow x green presenting the average input power 20.8mW

    So we see 20.8mW input.

    A 3th fluke DMM shows the voltage across the 14860 Ohm resistor to be 11.2V
    This represents a power out of 8.7mW indicating a cop = 0.42.

Did you subtract the CVR power from the 20.8mW? To obtain the CVR power, you need to know the RMS voltage across it.

I did not use a cvr in that setup, so nothing to subtract.  (i used momentarely a 1 Ohm cvr with a DC voltage DMM
across it to backup the current reading on the DC mA DMM).


Quote
Quote
    Strange is the difference  between the input current DMM (37.2mA) backed up by the temp 37mV across a 1 Ohm resistor
    and the current probe input current measurement of 105mA rms.
It isn't strange, because the DMM measures average current.

Right, it was strange to me as i thought that the DMM's would also show rms values, their not, thanks.

Quote
Quote
   Are the DDM's unable to accuratly display the 28.8KHz current/voltage?  It looks like it.

The DMM displays the average very accurately. In general the only time we use the DMM for input power measurement is when the source is a fixed DC supply, and in that case it would be to measure the average current. I(avg) x Vdc = P(in). Since we are not using a fixed DC supply in this device, we don't use the DMM.


Quote
 
Quote
  Using another methode of obtaining the input current (2 probes in differential setup across the temp. 1 Ohm csr)
    shows a similar current signal as with the current probe, but due to the low level of signals the specific value
    is hard to obtain, but i tend to believe the current probe is showing the corrent input current.

    Regards Itsu

As I believe Brad suggested, you could increase your CVR to 10R rather than 1R. This will give you 10x the signal level to measure across the CVR. Set up an RMS measurement on the CVR voltage and you can then obtain the CVR power to subtract from the total (if your circuit configuration requires it).


I did that in my latest setup showing a COP =0.81

Itsu
   
Group: Experimentalist
Hero Member
*****

Posts: 1665
Itsu,

Did you miss my two posts to you? I think there is some confusion here regarding what the scope is doing when we utilize a AVG(CH1 * CH2) type measurement.

The scope is NOT using RMS nor any other process on the raw data being sampled on CH1 and CH2 in a "multiply" or any other MATH function trace.

What PM is saying is that the sampled AVG(CH1 * CH2) operation we set up for measuring average power is equivalent to multiplying the RMS values of the voltage and current, if there is no significant phase shift between the two. If there is significant phase shift between the voltage and current, then we can not simply multiply the two RMS values, unless we also incorporate the phase angle in the calculation. But since we are often dealing with non-sinusoidal wave forms, we can not do that, so we let the scope do all the work in determining the average power by setting up the CH1 * CH2 MATH trace. The result is RAW data, i.e. AVG[(sample1 vi product) + (sample2 vi product) + (sample3 vi product) +...]. The scope is sampling at a very high rate, as in 500Ms/s, and the average is a running average of these samples, because we apply a MEAN measurement to the samples.

So the samples are the RAW products of v(t) x i(t), nothing more.

I hope that helps.

Yes I agree.  It seems there could be a fruitful discussion somewhere just on the use of the Math channel as all the newer scopes seem to  include this feature.  It's really cool when they allow you to create your own math expression! :)

Pm
   
Pages: 1 2 3 4 5 6 [7] 8 9 10 11 12 13 14 15 16 17
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-03-29, 06:15:15