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

[EMJunkie]

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]

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]

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]

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]

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