Swend
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This power value is impossible for real circuits.(IIR) P = -3.731 W
Hi,
Your three forumulas apply for
* DC voltage and DC current
* instantaneous voltages and current (V and I at the same time)
* and some other cases
But they don´t necessarily apply for pulsed or any other waveforms.
This power value is impossible for real circuits.
* you don´t have negative resistance
* and for any value of current the result can never be negative. (I x I is always positive)
Your diagram:
Voltage and current have opposite direction, this is impossible on a real resistor. Check the polarity / sign.
Also with a resistive load: V = I x R, but this seemns d´not to be the case with your circuit.
It seems you messure V and I.
But how do you measure them?
* Instantaneous
* RMS (over which time?)
* average (over wihich time?)
All these values may be correct or false depending what you want to measure/calculate. You have to choose the correct calculation scheme.
The picture shows "DC pulse".
I expect a square wave with a fixed ON time and ON voltage (non fluctuating) and a fixed OFF time and OFF voltage (non fluctuating)
But this seems not to be that case with your circuit. Please clarify. Show the voltage waveform at the voltage source.
In addition to what klaus has said, P is not UUR. P is UU/R.
The waveorm is one problem.my question to "Which formulas should I apply to the measured waveform"?
Impossible .... unless your "load" is an energy generator.Yes for IIR, but for UI the result is negative i.e. IF the current really is negative.
Let's say you have an ideal 1M resistor --> your DMM easures 1M, I can hold it in my hand and measure the resistance with a DMM
It is a current measurement device, thus the arrow shows the technical current direction (from positive voltage to negative voltage)the Polarity Arrow points towards positive.
For non ohmic loads you need to use: P(t) = V(t) x I (t)I would like to calculate the instantaneous power, which calculation scheme do you suggest?
For me it does not seem to be a DC pulse...but we don't see the source voltage at all...I wrote "DC pulse" because the traces looks like a DC pulse to me.
...
"UUR" is just a more meaningful name than if I named them I, II and III.
As you can see in my OP "(UUR) P = U² / R"
Your informations do not match.
* your load is no "R"
* it behaves like a "complex energy source".
--> correct the sign.
Let's say you have an ideal 1M resistor --> your DMM easures 1M
Now add a parallel capacitor --> your DMM still measures 1M (but it will behave totally different at a pulsed source)
Now add a series inductance --> and still your DMM measures 1M, (but it will behave even worse at a pulsed source)
Because it measures with DC, nit AC. It is no RLC meter!
It is a current measurement device, thus the arrow shows the technical current direction (from positive voltage to negative voltage)
For non ohmic loads you need to use: P(t) = V(t) x I (t)
For me it does not seem to be a DC pulse...
but we don't see the source voltage at all...
You can calculate instantaneous power at a single point in time - then integrate, then average over larger portions of time
but the simple formulae you listed only apply when the input components are non time varying - or single phase AC all in phase ....
If I may ask, why are you measuring voltage at that node?
If your load is a resistor, then you cannot have current leaving the resistor when the voltage across the resistor is +ve.
Also, whatever you are using to measure the current, make sure that it is properly connected.
You talk about R, an ohmic load. But an ohmic load has a constant relationship of V and I.Could you please elaborate on that claim?
As already explained... Mabe it is 1M OHms at DC. But your pulse measurement show other values.It is a physical 1 Mega ohm standard high voltage glass resistor. So physically it is a resistor.
I (just for me speaking) see no need for further discussion. V(R) x I(R) is always positive on a true resistor.No, the arrow indicates electron flow i.e. from negative to positive, check the definition of the "polarity Arrow" that I linked to.
What "load" do you need to measure the voltage of an AC outlet? ... or a car battery?I can't show you something that I don't have. The load completes the circuit, if there is no load, there is no circuit, if there is no circuit, there is no current, if there is no current there is no voltage. That's the nature of the source.
In your measurement results I can not see a true R. I see a complex load. Thus the results will depend on waveform. You are free to ignore this.Then forget that I said "DC pulse", you can call it what you prefer, I will refer to it as "the trace".
You talk about R, an ohmic load.
But an ohmic load has a constant relationship of V and I.
But in your case it is not constant, it varies with time.
You have measured V and I.
You calculated V x I to get P.
So please calculate V/I of your measured values. To me it seems the result is not constant "1M Ohms".
Thus it is no ohmic load and can't be treated (calculated) according Ohm´s law.
Use a DMM and it shows 1MOhms.
Use an RLC measuremnt tool and it shows something different.
I (just for me speaking) see no need for further discussion. V(R) x I(R) is always positive on a true resistor.
What "load" do you need to measure the voltage of an AC outlet? ... or a car battery?
--> I´m very sure I did measure the battery voltage without connecting a dedicated load.
In your measurement results I can not see a true R. I see a complex load. Thus the results will depend on waveform. You are free to ignore this.
Mathematics:
P = V x I
and R = V / I (rearranged: V = R x I, I = V / R)
Now take: P = V x I .... and replace "I" with "V/R" according above formula "I = V/R".
The result is P = V x (V / R) and solve this: P = V x V / R = V^2 / R.
This means P = V x I and P = V^2 / R
In your case this is not true.
There are just three possible reasons:
* The measurement of I is wrong
* The measurement of V is wrong
* or "R" is not a true ohmic "R". Maybe it is non linear, maybe it is complex...
I don´t know which reason generates the problem. I can only guess.
What is the essnce of C1?
Looking at the waveforms, I'd guess that at least two of the three points apply. Respectively it's impossible to determine what's actually happening when only looking at the waveforms. It's necessary to validate the measurements and the resistor impedance independently.There are just three possible reasons:
* The measurement of I is wrong
* The measurement of V is wrong
* or "R" is not a true ohmic "R". Maybe it is non linear, maybe it is complex...
Looking at the waveforms, I'd guess that at least two of the three points apply.
It's necessary to validate the measurements and the resistor impedance independently.
A point that makes me completely distrust the waveforms in post #1 is that they start "out of nothing".
So we e.g. can't determine the As integral applied to the Pearson probe.
According to datasheet, it's typically saturated at 0.0006 As, which is apparently exceeded in the experiment. Depending on the preceding current waveform, the falling current measurement may be simply transducer saturation. If (unexpectedly) no saturation happens, there will be still a current droop due to 140 Hz lower cut off frequency.
Pearson probes are fine instruments, but you need to understand their limitations.
Similar problems might occur with other measurements too.
In stead of expressing your personal distrust maybe you should start by posing me a question.
You want others to do something....Is this really the ob of forum members to work for you - for free?no one offered to validate my measurements
You want us to belive in all you say and in all your measurements.So the only novel thing here is that I can demonstrate true negative resistance on command, since I can't really see any applications for it - it's pretty useless.
Hi,
Then the only conclusion can be (and don't treat this as joke or offense): You found the perpetuum mobile. Even better: you solved the energy problems of our world.
--> Your invention is far away from being useless. It´s the most important invention ever.
Obviously the traces don't start "out of nothing" I just chose to start at that particular point when load is applied, prior to that point both traces are 0V complete flatliners.
"Prior to that point" isn't quite exact, you have actually cut the rising edge and we can't know how long is lasts. The voltage and current transfer functions may involve either non-linearity (e.g. core saturation) or a impulse response of some time duration, probably both. In this case, the hidden rising edge echoes in the visible response.
I agree with KlausST that the current measurement channel seems to be inverted. I' d think about possible alternative explanations if I see the pulse response form the start.
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