Author Topic: Accurate Measurements on pulsed system's harder than you think. (Read 84782 times)

EMJunkie · « **Reply #150 on:** December 13, 2015, 09:33:01 PM »

For a NON-Uniform, NON-Symmetrical Wave, taking the Peak Value, taking 0.707 of the Peak Value and calling this the true RMS Value just doesn't seem right to me. What is it I am missing that the scope is doing that I cant see?

I mean, we are measuring the total Area from the Scope Baseline, to Scope Trace, over time.

Chris Sykes
hyiq.org

verpies · « **Reply #151 on:** December 13, 2015, 09:40:59 PM »

Quote from: poynt99 on December 13, 2015, 05:17:12 PM

It should be noted that the following (from the above post, but abbreviated and generalized):
PRx = Vrms² / Rx
Holds for ANY wave form, with any duty cycle.

...but only when the load Rx is purely resistive.

EMJunkie · « **Reply #152 on:** December 13, 2015, 09:56:40 PM »

Quote from: verpies on December 13, 2015, 09:40:59 PM

...but only when the load Rx is purely resistive.

Yes, If I am not mistaken, RMS is supposed to be the DC Equivalent Figure given from a Sinusoidal Wave Form! So 0.707 of a Sinusoidal Wave would give the DC Value, of the Sinusoidal Wave Form, so the Area above the Wave, above the 0.707 would fill the missing Trace Area, giving a DC Value?

The wave Brad posted is far from Symmetrical.

See the Image below:

Chris Sykes
hyiq.org

verpies · « **Reply #153 on:** December 13, 2015, 10:31:59 PM »

Quote from: EMJunkie on December 13, 2015, 09:56:40 PM

Yes, If I am not mistaken, RMS is supposed to be the DC Equivalent Figure given from a Sinusoidal Wave Form!

Not only.
The RMS calculation can also be applied to other kinds of waveforms.

EMJunkie · « **Reply #154 on:** December 13, 2015, 10:36:02 PM »

Quote from: verpies on December 13, 2015, 10:31:59 PM

Not only.
The RMS calculation can also be applied to other kinds of waveforms.

Yes, yes, but only if they are symmetrical? I just used the Sinusoidal Wave Form as an example.

Is this right?

Chris Sykes
hyiq.org

EMJunkie · « **Reply #155 on:** December 13, 2015, 10:48:43 PM »

Its an interesting subject!!!

The equation Poynt gave us: P_Load = (V_rms x V_rms) / R_Load - I am not disagreeing with, its just that it wont hold true in all cases. Only because the RMS Equations outside this equation may not be sufficient to give correct values for NON-Symmetrical Waves! As far as I can see.

However, its a good generalized Equation, and holds true for Symmetrical Wave Forms.

As it may be the case that many out there may not understand what is actually being explained here, if I may try to elaborate some, and others please correct me if I am wrong!

The Equation Poynt99 gave us is from the Ohms Law Equation Set: P = V² / R - See Image below. Ohms Law is your friend!!! Use it!

What Poynt is saying here, is, that the Power P, in Watts, is the product of the Voltage V², (which is just Voltage x Voltage), taken in the form of RMS (Root Mean Square) value, divided by the Resistance R of the Load.

This equation gives you the Power P in Watts, consumed by the Load Element.

An Example:

Load Resistance R = 10 Ohms
Load Applied Voltage V_RMS = 10.5 Volts

Power P = 10.5² / 10 = 110.25 / 10 = 11.025 Watts

Chris Sykes
hyiq.org

verpies · « **Reply #156 on:** December 13, 2015, 11:07:29 PM »

I just got a question from a shy member in a PM:

Q: "So when should we use current RMS values to calculate power and when do we multiply the current by voltage ?"

A: When you have BOTH instantaneous current and instantaneous voltage available, then multiplying their magnitudes and averaging the resulting products is the surest universal method to calculate power.

However, sometimes you don't have BOTH voltage and current available. But in such case, all is not lost yet!
In your question you mention a case when only current measurement is available - "Is it possible to calculate power then?"

The answer is "yes", but with the following provisos:
The current must be in-phase with the voltage, or in other words the current must be proportional to the voltage.
This is tantamount to assuming a purely resistive load because current is proportional to voltage only when flowing through a pure resistor.

In such case you can use the RMS value of the current to calculate power using the formula P=I_RMS²R

In another, but similar case, mentioned by Poynt99 - if you only have the voltage measurement which is applied across a pure resistance:
- you can use the RMS value of the voltage to calculate power using the formula P=V_RMS²/R

You cannot use the average current (or average voltage) to calculate power because:
Sum of squares <> Square of the sum
or
Integral of squares <> Square of the integral.

An example:
Let's have a square current waveform flowing through a 1Ω resistor: 0A for 1ms and 10A for 1ms.
It is obvious that the average current of such current waveform is 5A because (0A+10A)/2=5A.
What about the power dissipated in the resistor?
It is also obvious that when 0A flows through the resistor then 0W is dissipated in that resistor.
However, when 10A flows through the resistor then 100W is dissipated in that resistor, because P=i²R which calculates to P=10A²*1Ω = 100W.
So we have a 0W pulse for 1ms and 100W pulse for another 1ms., or in other words: 100W half of the time. The average of that is 50W because (0W+100W)/2=50W.

If we used the average current (5A) to calculate power dissipated in that resistor, then we would obtain 25W because P=i²R and that calculates to P=5A²*1Ω = 25W ...which is very wrong.

Bonus Rant:
What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?

Answer:
Not with an arbitrary load (non-resistive) when the shape of the waveforms and the phase offset between them are not known.
For example, if you know that the shape of the current and voltage is sinusoidal and of the same frequency and the phase shift between them is 30º then you can use the formula P = i_RMS * V_RMS * cos(30º). ...but this formula will not work with other waveform shapes.

EMJunkie · « **Reply #157 on:** December 13, 2015, 11:29:40 PM »

Quote from: verpies on December 13, 2015, 11:07:29 PM

Bonus Rant:
What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?

Answer:
Not, with an arbitrary load (non-resistive) when the shape shape of the waveforms is not known and the phase offset between them.
For example, if you know that the shape of the current and voltage is sinusoidal and of the same frequency and the phase shift between them is 30º then you can use the formula P = i_RMS * V_RMS * cos(30º). ...but this formula will not work for other waveform shapes.

Fantastic Post, thanks for sharing Verpies!

Can you explain why the equation in the above quote will not work across other wave forms?

Thank You!

Chris Sykes
hyiq.org

verpies · « **Reply #158 on:** December 13, 2015, 11:56:56 PM »

Quote from: EMJunkie on December 13, 2015, 11:29:40 PM

Can you explain why the equation in the above quote will not work across other wave forms?

The first thing that comes to mind is the correction term cos(30º) which is related to the sinus shape.
For other shapes, a different correction term would have to be used.

It is important to keep in mind that the method that works for all shapes and I&V phases is the method that multiplies the instantaneous values of current and voltage and then averages the resulting products. You can't go wrong with that.

Addendum:
Err... you can go wrong with that, too ...but by making a conceptual mistake - not a mathematical one.
For example confusing the power dissipated by the light bulb in series with the power supply with the power delivered by the power supply.
It is possible for the bulb to dissipate 10W of power while transferring 1kW of power from the PS to the DUT. How you place your CSR and scope probes determines which power you measure...often inadvertently

In other words, the number of Watts will be correct but it will apply to a different power.

EMJunkie · « **Reply #159 on:** December 14, 2015, 12:05:39 AM »

Quote from: verpies on December 13, 2015, 11:56:56 PM

The first thing that comes to mind is the correction term cos(30º) which is related to the sinus shape.
For other shapes, a different correction term would have to be used.

Ah, of course!

because a Sinusoidal Wave is calculated: Asine(omega t)

The Cosine is the Covariance.

Chris Sykes
hyiq.org

verpies · « **Reply #160 on:** December 14, 2015, 12:18:50 AM »

Quote from: EMJunkie on December 13, 2015, 10:36:02 PM

Yes, yes, but only if they are symmetrical?

In my table, the sawtooth and the pulse train were not symmetrical.

tinman · « **Reply #161 on:** December 14, 2015, 12:31:57 AM »

Quote TK: You aren't doing it right !!!!!!!!

Quote Poynt: PRx = Vrms2 / Rx .Holds for ANY wave form, with any duty cycle.

Quote EMJ: @Poynt - So you're saying that the example TK gave is wrong? It appears so.

Question to Verpies-What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?
Quote Verpies: Not with an arbitrary load (non-resistive) when the shape of the waveforms and the phase offset between them are not known.

Question from TK--Why are you multiplying the Vdrop across the 100R by the "average current"?

I was not,and do not. Take for example the scope shot below. I use the instantaneous voltage across the resistor !not the supply voltage!,as there is no voltage across the resistor when Q1 is open,and the supply voltage is not always the voltage that is across the resistive load. The voltage to use is the actual voltage across the resistive load,and in the case of the screen shot below,that voltage is 12.2 volt's-not the open supply voltage of 12.4 volt's. If we know exactly what the value of the resistive load is,when can then use ohms law to calculate the dissipated power by that resistive load. As the value of the resistive load is exactly 51.2 ohm's,and the actual voltage is 12.2 volts,then the instantaneous dissipated power is 2.907 watts. As we also know that the exact duty cycle is 30%,then the average power dissipated by the resistive load is 872.1mW.

Then you will notice that (once again) the current from those calculations do not agree with the calculated instantaneous current the scope is showing across the CVR (blue channel).

But it's great to see you guys think it is so easy to measure the power in a pulsed system(see above quotes).

Brad.

poynt99 · « **Reply #162 on:** December 14, 2015, 01:53:14 AM »

Quote from: tinman on December 14, 2015, 12:31:57 AM

Quote TK: You aren't doing it right !!!!!!!!

Quote Poynt: PRx = Vrms2 / Rx .Holds for ANY wave form, with any duty cycle.

Quote EMJ: @Poynt - So you're saying that the example TK gave is wrong? It appears so.

Question to Verpies-What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?
Quote Verpies: Not with an arbitrary load (non-resistive) when the shape of the waveforms and the phase offset between them are not known.

Question from TK--Why are you multiplying the Vdrop across the 100R by the "average current"?

If you provide some examples of discrepancies (with the quoted numbers), I may be able to give you an answer.

Quote

Then you will notice that (once again) the current from those calculations do not agree with the calculated instantaneous current the scope is showing across the CVR (blue channel).

I don't know to what you are referring. Perhaps some actual numbers would be helpful.

Quote

But it's great to see you guys think it is so easy and to measure the power in a pulsed system(see above quotes).

For what you guys are doing here, yes it is easy and straight forward. Follow what is in that post of mine and you will be golden.

EMJunkie · « **Reply #163 on:** December 14, 2015, 02:08:35 AM »

Quote from: tinman on December 14, 2015, 12:31:57 AM

Quote TK: You aren't doing it right !!!!!!!!

Quote Poynt: PRx = Vrms2 / Rx .Holds for ANY wave form, with any duty cycle.

Quote EMJ: @Poynt - So you're saying that the example TK gave is wrong? It appears so.

Question to Verpies-What about the case when you have BOTH the RMS current and RMS voltage - is it possible to calculate average power from them?
Quote Verpies: Not with an arbitrary load (non-resistive) when the shape of the waveforms and the phase offset between them are not known.

Question from TK--Why are you multiplying the Vdrop across the 100R by the "average current"?

I was not,and do not. Take for example the scope shot below. I use the instantaneous voltage across the resistor !not the supply voltage!,as there is no voltage across the resistor when Q1 is open,and the supply voltage is not always the voltage that is across the resistive load. The voltage to use is the actual voltage across the resistive load,and in the case of the screen shot below,that voltage is 12.2 volt's-not the open supply voltage of 12.4 volt's. If we know exactly what the value of the resistive load is,when can then use ohms law to calculate the dissipated power by that resistive load. As the value of the resistive load is exactly 51.2 ohm's,and the actual voltage is 12.2 volts,then the instantaneous dissipated power is 2.907 watts. As we also know that the exact duty cycle is 30%,then the average power dissipated by the resistive load is 872.1mW.

Then you will notice that (once again) the current from those calculations do not agree with the calculated instantaneous current the scope is showing across the CVR (blue channel).

But it's great to see you guys think it is so easy to measure the power in a pulsed system(see above quotes).

Brad.

@Brad - Certainly I do not think that "it is so easy to measure the power in a pulsed system"!!!

I never said that, and to be honest, that's why I am here, to try and learn something!

Lets be honest, we are not even at measuring the Power Input to a Pulsed Circuit!!! We are only measuring the Power through a single Component!

Although there are problems with the solution, I think Poynt has suggested the best solution currently to measure a single Component in a Pulsed DC System. It should be the most accurate over all. TK's method is good, but there is, I believe a Complexity to it that will confuse most.

In saying this, I do like TK's method. He has shown that is can be pretty accurate also.

Resistance in any Resistor that is not purely Resistive is where part of this problem lays! As Verpies has pointed out!!!

Lets take an inductor for example. There are two Resistive components, it is typically written as such: Real Resistance+jImaginary Resistance in Ohms or Real Resistance-jImaginary Resistance in Ohms

+j is inductive ...(current lagging)
-j is capacitive... (current leading)

The Imaginary Resistance is X_L or X_C, which is the Inductive or Capacitive Reactance. If the Inductor is Resonant, then X_L and X_C cancel out and the DC Resistance remains. Which is the Real Resistance.

So, the Actual Resistance is again Changing with Time in a Pulsed System. If Resistance R Changes in the Equation: P = V²/R then this presents a massive Accuracy Error!!!

I should point out, this will present itself, typically as a Non-Symmetrical or Non-Uniform Wave Form!!!

Its a pretty complex subject! Not easy!

Chris Sykes
hyiq.org

digitalindustry · « **Reply #164 on:** December 14, 2015, 02:59:46 AM »

Quote from: poynt99 on December 14, 2015, 01:53:14 AM

If you provide some examples of discrepancies (with the quoted numbers), I may be able to give you an answer.
I don't know to what you are referring. Perhaps some actual numbers would be helpful.
For what you guys are doing here, yes it is easy and straight forward. Follow what is in that post of mine and you will be golden.

no offense but you are kind of coming across like a friend of mine i refer to as 'Homer Simpson'

he doesn't read anything , but then just assumes he is correct, is it because you have the word 'elite' in your user?

it's got a bit of that chess with a pigeon feel.

ha ha

: D

hey i'm sure you have ohms law covered at a higher understanding than i , i don't even know what a volt is measuring, but then the humble diode doesn't adhere to ohms law so as we seem to learn everything shifts as one learns more.

did you watch both the videos, also take into account the 'current' waveform in the original video?

News:

User Menu

Author Topic: Accurate Measurements on pulsed system's harder than you think. (Read 84782 times)

EMJunkie

Re: Accurate Measurements on pulsed system's harder than you think.

verpies

Re: Accurate Measurements on pulsed system's harder than you think.

EMJunkie

Re: Accurate Measurements on pulsed system's harder than you think.

verpies

Re: Accurate Measurements on pulsed system's harder than you think.

EMJunkie

Re: Accurate Measurements on pulsed system's harder than you think.

EMJunkie

Re: Accurate Measurements on pulsed system's harder than you think.

verpies

Re: Accurate Measurements on pulsed system's harder than you think.

EMJunkie

Re: Accurate Measurements on pulsed system's harder than you think.

verpies

Re: Accurate Measurements on pulsed system's harder than you think.

EMJunkie

Re: Accurate Measurements on pulsed system's harder than you think.

verpies

Re: Accurate Measurements on pulsed system's harder than you think.

tinman

Re: Accurate Measurements on pulsed system's harder than you think.

poynt99

Re: Accurate Measurements on pulsed system's harder than you think.

EMJunkie

Re: Accurate Measurements on pulsed system's harder than you think.

digitalindustry

Re: Accurate Measurements on pulsed system's harder than you think.