Calculation of RMS power output?

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#21
"Pure reactances do dissipate energy. "

This is exactly wrong. A pure reactance does not dissipate power regardless of what you feed into it.

Assuming one could make a pure reactance, one could get it to dissipate power if it was imperfect (such as exerted energy on something else, vibrated (phase is shifted, looks like real resistance), had real resistance or whatever), but then it is no longer a pure reactance.

The problem with definition of power (such as dB) is typically that one refers back to voltage and assumes in-phase current. Power is V^2/R by definition. If you assume that R is the same in both cases, you can suggest a power ratio in dB which would be

10LOG(P1/P2) (where P1=V1/R)
=10LOG(V1^2/V2^2)
=10LOG(V1/V2)^2
=20LOGV1/V2

The problem with the audio industry is that the definition of power is so loose that people can cheat. DIN and FTC norms alleviate that a little bit, but do you actually trust most manufacturers numbers?

Petter
 
#22
Traderbam,

Well even if we have a pure reactance (ie. zero DC resistance) it can still be used to dissipate energy. Take for example, a capacitor used in series with a AC voltage source to drop the voltage prior to an LED ... Ok, that's not a common usage and I've not explained it very well but i think you catch my drift.

x = 1 / (2 * pi * f * C)

If we ignore the constraint that an ideal capacitor / inductor would have infinite capacitance / inductance then since power is flowing thru the device, some amount of power is must be being dissipated by it.
 
#23
traderbam said:
"The power dissipated is a function of the AC voltage across the reactance and the frequency of said AC Voltage."

No. Capacitors and inductors store energy, they do not dissipate it. Otherwise when we buy them we would be concerned about their power dissipation rating, just like we are with resistors. Reactances store energy or power x time if you prefer, and this energy can be calculated by the integral of voltage across the reactive component x current flowing through it. These integrals equal the standard energy equations for Cs and Ls.

Power is a consideration in a sense when choosing certian Cs and Ls. The wire used in Ls is slightly resistive and is subject to a maximum current limitation like any other wire, and Ls that use ferrite cores suffer from core saturation if the current gets too high (which is not dissipative so doesn't waste power but does affect the storage capability). The metal film in capacitors is also slightly resisitive and will have a maximum current rating. The latter is mostly a concern in large psu electrolytics where a max ripple current rating is usually given.
Click to expand...

Exactly right. One consequence of the zero dissipation in pure reactances is the fact that the stored energy can be recovered. For instance, if you store charge in a cap by charging it, you can later connect a load across the cap to recover the energy. In fact, an L-C circuit that oscillates will continue to do so unless there is resistance in the circuit. There always is, so the oscillations die out.
That is how class C amps work: they put energy in an LC network that produces the sine swing by pulsing it.

Jan Didden
 
#24
OK, I'll take a stab at this.

A pure reactance (ideal C or ideal L) does not dissipate power. Power can be delivered into it, but that power can be removed from it without loss.

Consider a simple test case of an ideal AC voltage source connected across an impedance. The voltage across the impedance is:

V = A sin(wt)

If the impedance is a pure resistance R, then the current through the impedance is:

I = V/R = (A/R)sin(wt)

The instantaneous power delivered to the resistor is simply the product of the voltage and the current:

P = VI = (A^2/R)sin^2(wt)

(that's "A squared over R times sine squared omega t"). Simple trigonometry allows us to rewrite this equation as

P = (A^2/(2R)) (1 - cos(2wt))

Note the following:
(1) power into the resistor varies over time from zero to A^2/R
(2) the power varies with a frequency twice that of the input signal
(3) since the average value of a sinusoid is zero, then the average power delivered to the resistor by inspection is

Pav = A^2/(2R)

(also note that this average power is the square of the RMS voltage divided by the resistance)

Now consider the case where the impedance is an ideal capacitor. The current through the impedance is

I = C dV/dt = AC cos(wt)

and the instantaneous power delivered to the capacitor is

P = VI = A^2 C sin(wt)cos(wt)

again using a fundamental trig identity gives

P = (A^2 C/2) sin(2wt)

Note the following:
(1) Power into the capacitor varies from -(A^2 C/2) to +(A^2 C/2); this means that half the time the capacitor is receiving power from the source and half the time the capacitor is delivering power back to the source.
(2) The average power delivered to the capacitor is zero.
 
#25
This might clear up the whole "phase/power relationship" concept. Since the formula has not been written down yet

P= V(rms)*I(rms)*cos(Ov-Oi)

The "real" power is the projection of voltage onto the current. It is my understanding, as with others, that the reactive power loss is just a number to help us minimize real power loss.

Example, you are feeding an inductive load that causes the current to lag 30 degrees behind the voltage. This 30 degree lag means that you must supply more current to the load to produce a specific power (say 10KW). The more current you supply, the more real power loss in the lines. Hence, you need to reduce the reactive power loss (bank of caps) to minimize real power loss. This is less crutial when dealing with loudspeaker cables of course.
 
#26
"for example, a capacitor used in series with a AC voltage source to drop the voltage prior to an LED ..."

Now I see what you are saying, Audiofreak. You may be confusing a voltage dropped by a resistance and a voltage dropped by reactance. In the former the resistance dissipates heat and the AC voltage source has to provide the energy dissipated by the LED plus the energy lost by the resistor. However, with the capacitor used as the voltage dropper no energy is wasted so the AC source only provides the energy used by the LED and some energy that charges up the capacitor on one half of the AC cycle and is recovered by the AC source on the other half of the cycle.

The crux is that a voltage drop does not have to be achieved by dissipating energy. This is similar to how transformers work - they change voltage without wasting power (well ideally not).
 
#32
JensRasmussen said:
Just for fun I tried to find the actual RMS of a sinewave
I have made a zip with a word dok.
Hope it helps.
Click to expand...
Your math looks correct. You could have saved some effort after you solved the definite integral by taking the limit as T->infinity, which would have dropped the term containing sin(wT)cos(wT)/T immediately.
 
#33
johnthetweeker said:


Another thought, Jeff. Since a speaker's impedence isn't always resistive, wouldn't you need the phase-angle between the V & I for accurate measurement? (P = V * I * cos(phi)). So I think a simple DMM cannot be used for an accurate power measurement.
Click to expand...


In my approach to meaure power using a 4 quadrant multiplier which senses both voltage and current to the speakers, you are mulitiplying together the instantaneous values of V and I so, no, your equation is not valid. I guess we can rewrite P=VI as p(t) =i(t)*v(t). We can average p(t) to get average P using an integrator.

While both current and voltage will experience phase shifts in reactive loads compared to resistive loads, this does not apply to power, which is the product of those phase shifted voltages and currents.
 
#35
RobPhill33 said:
This, I think, would yeild the RMS value of the power function (using a sine wave). But, none of this really matters because everyone uses the average value and calls it RMS.
Click to expand...

(I'm doing this from memory, so forgive any mistakes I may make in the explanation) You can calculate the RMS value of any waveform. However, for audio a problem arises when trying to do the calculations on frequency varying waveform. If you take a sine wave, and start the RMS calcualtions, you get a sine squared term. When integrating the sine squared term, a 2*pi*F moves out in front sine function. This is the critical point. It means that the RMS power of a waveform is really frequency dependent. Yes, for a given peak to peak voltage, a sine wave of 10Khz has a higher RMS Power value than that of 1Khz.

What we can glean from this that the RMS value of power is really a useless number. "Average power" is calculated from the RMS voltage and currents, thus the term RMS power. It should really be called "Power calculated by the RMS method", which is really "Average Power". The problem is that no self respecting audio company would ever brag about a spec. called "Average Power"... ...sounds, well... ...average 🙂 🙂

I did a paper about 6 years ago on the calculation... ...it might be worth digging out again. It seems to me that the problem is in integrating a sin squared. If I remember right, I ended up doing it by tables. (The V squared term generates the sine squared...)

Anyone up for some calc?

-Dan
 
#37
AudioFreak said:


Pure reactances do dissipate energy.
For Example...
Assume DC resistance = 0
Assume an AC Voltage source
The power dissipated is a function of the AC voltage across the reactance and the frequency of said AC Voltage.
Click to expand...

Not really. The power 'dissipated' in a pure reactance is called 'immaginary' power.
AC power is NOT a function of frequency, but reactance is. The reactance formulae are:

X(L) = Reactance, inductive = 2*pi*f*L

X(C) = Reactance, capacitive = 1/(2*pi*f*C)

X(L) and X(C) can be used in place of R in Ohm's law to calculate the relation between AC volts and AC current.

Instantaneous AC power is a function of voltage, current AND the phase difference between the two:

P = V * I * cos (phi)
where phi is the phase difference between V & I.

Of course, V & I are related by the reactance. Note that if V & I are 90 degrees apart, cos(90) is zero, and therefore the power dissipated is zero, which would be the case of ideal inductances and ideal capacitances.

You can try these links for some fundas:

http://whatis.techtarget.com/definition/0,,sid9_gci213719,00.html


http://www.tpub.com/neets/book2/4.htm

Cheers!
 
#38
"for a given peak to peak voltage, a sine wave of 10Khz has a higher RMS Power value than that of 1Khz. "

No it shouldn't do. The integral is over the period T of the cycle, regardless of T, and is then divided by T. So the RMS value of a sinewave is only related to the peak value and is independent of frequency.
 
#40
traderbam said:

No it shouldn't do. The integral is over the period T of the cycle, regardless of T, and is then divided by T. So the RMS value of a sinewave is only related to the peak value and is independent of frequency.
Click to expand...

Hi,

Read carefully, I'm talking about the actual calculation of the RMS of a power waveform, not the calculation of RMS of a voltage waveform used to calculate Average power. By definition RMS power is the RMS of a power waveform. Try it for 1 volt sine wave, sin(wt), into a 1 ohm load... ...solving for the power waveform by P = V^2/R where V = sin(wt)

...calculate the RMS value of (sin(wt)*sin(wt)). I doing so, you end up taking the integral of a sine squared (wt) function...
(Note that 1 volt and into 1 ohms simplifies the math dramatically)

-Dan
 
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