Measure speaker/amplifier RMS power

[carrotSnack]

How can I measure the RMS power consumption of a speaker coil that is being fed an arbitrary periodic wave from an amplifier?
I know there are there multiple types of AC power. To be specific, how can I measure the RMS power consumption of the coil such that it is equal to the DC power used by a 100% efficient amplifier powering it?
I am using a car's class AB amplifier so measuring it's input DC power is not an accurate method.

 

[Nigel Goodwin]

Why are you wanting to do this? - it sounds a really pointless exercise?.

If you want to measure RMS power, then feed it from a sinewave (non-periodic waveforms are meaningless).

 

[Reloadron]

Read this link. No measuring the power to the amplifier is not an accurate method as it doesn't tell you what the actual power to the speaker coil is. There are losses along the way such as the heat generated by the amplifier and no amplifier is 100% efficient. The link provides a pretty good explenation as well as a good overview. This link is also pretty good and gets into the math a little.

Now if you actually want to measure the power to a speaker then this link explains the procedure and equipment requirements. There are also commercial products used in audio shops that are designed to measure the power but it can also be done with some test, measurement and diagnostic equipment.

<EDIT> Morning Nigel, I see you are up and about. Oh wait, much later on your side of the pond. </EDIT>

Ron

 

[carrotSnack]

Thank you for the links I'll be reading over them.

Why are you wanting to do this? - it sounds a really pointless exercise?.

Because I'm wanting to evaluate the performance of the coil, not the circuitry that drives it. I'm measuring other properties of the coil that are dependent on its power consumption.

 

[MikeMl]

Use a two-channel data-aquisition system to sample the voltage across the coil, and the current through the coil. Obviously, the sampling rate has to be (much?) higher than the frequencies present in the aperiodic waveforms you are interested in. Most DAQ systems come with analysis software which can do plotting, and math like power calculation and integration on the waveforms...

Some modern digital O'Scopes could do your instantaneous and RMS power plots with just the push of a (menu) button... Still would require capturing voltage on one trace, and current on the other.

 

[carrotSnack]

The first link gave the most insight into how this can be done for an arbitrary periodic wave.

What do you think about this:
Use a current sensor like this connected in series with the coil to measure the instantaneous current though it. It can measure up to 100kHz both +/- currents, more than fast enough.
Using an arduino, take samples of the current sensor output to get the instantaneous current through the coil.
Sample the instantaneous voltage over the coil with the arduino using a voltage divider to reduce the voltage to what the arduino can measure.
Use this equation to find the average power 1/(no. of samples) * SUM(v * i)

Though I wonder, should instead of SUM(v * i) should it be SUM(|v * i|) as sometimes v*i is negative? e.g. the very beginning of the graph

 

[KeepItSimpleStupid]

This **broken link removed** might help.

 

[carrotSnack]

So it seems it should be the average of v*i not |v*i| correct?

For a purely inductive load fed a sine wave, the voltage and current are 90 degrees out of phase and we know the average power to the load is 0.
Using v*i https://www.wolframalpha.com/input/?i=1/(2*pi)*integral(sin(x)*sin(x-pi/2), x, 0, 2*pi) gives an average of 0
Using |v*i| https://www.wolframalpha.com/input/?i=1/(2*pi)*integral(abs(sin(x)*sin(x-pi/2)),+x,+0,+2*pi) gives 0.32
Supporting v*i is correct.
Though there is no such thing as negative power, I suppose in this case when it is negative it means energy stored coil is being returned to the circuit correct?

 

[crutschow]

If you want an instantaneous indication of power, you can multiply the current times the voltage with an analog multiplier such as this.
You can average that signal with an RC low-pass filter to get the average power.

 

[dr pepper]

If its the amp your testing replace the speaker with a 8r power resistor.

A speakers resistance is complex, esp at resonance, if you test at a high enough freq away from resonance you'll get better measurements, and to make the speakers impedance flatter at higher freq's you can use a impedance compensation network (cap and resistor), an octave above resonance the impedance will be relatively similar to the characteristic (4r for car stuff), passive xovers often use impedance eq networks.

However as nige mentioned testing with a pulsed wave isnt a good idea, all kinds of other factors will affect things, even down to cable resistance.

 

[tomizett]

Hi carrotSnack. In the interests of trying to guide the discussion in a constructive direction, could you give us a little more guidance on what kind of solution you're aiming for - I'm just wondering, are you interested in building an instrument to make this measurement (ie, is that the project) or do you just need to design an experimental set-up using standard equipment (you just need to get on measuring your speakers)?

Also, do you have to perform your tests with real-world programme material (like music) or could you use a sine wave if it was easier to work with? Maybe you could use pink noise with a known crest factor - this seems to be quite common in measuring power amps and speakers.
I take it that you do need to know the real power? Just finding the apparent power would be quite easy with a true-RMS meter and a current shunt.

 

[carrotSnack]

I'm just wondering, are you interested in building an instrument to make this measurement (ie, is that the project) or do you just need to design an experimental set-up using standard equipment (you just need to get on measuring your speakers)?

The latter, though I'm willing to to build the measuring instrument myself if its not too hard and the alternative is expensive.

It's not a sine wave, that's why I was after a general method to use with any wave.
Yes, it is the real power I need.

I'm going to go with what I wrote in post #6. The current sensor is under $10 and I already have a capable microcontroller, it shouldn't be too difficult.
I don't see an alternative that wont cost me $100's.

 

[Nigel Goodwin]

My concern, as always in these incredibly vague threads, is what is he trying to achieve, and why? - if it's not something that is normally done, why does he feel a need to try and do it - what's the point?.

 

[tomizett]

With regard to digitising and processing your data, how about using a standard PC soundcard in place of the Ardino? It'll obviously cover the full audio bandwidth (if that's all you're interested in) - I don't know what the samplerate available on an Ardino is, but I'd not be supprised if it was not that great. If you put (say) voltage signal into the left channel and current into the right, then you should get a minimal and very repeatable phase lag between current and voltage readings. You then have massive computing power at your disposal to exract whatever data you want, and you could even use Matlab or similar to minimise your programming time.

Thinking about it, I would be tempted to use transformers (the kind of little audio matching transformers for DI boxes etc) to obtain isolation. Use a simple potential-divider for the voltage and a small shunt resistor, with the transformer primary in parallel, to sniff the current. The transformer secondaries can be fed directly into the soundcard.
You'll have to think about how much burden you can afford to insert with your current shunt (it'll spoil the daming factor, of course). It may be possible to hack your amplifier to include it within the feedback loop to elliminate this effect.

It all depends really on how accurate and "instrumentation-like" you need the system to be; what bandwidth, how flat, how accurate, how noisy, how linear. A solution like I've just suggested would undoubtedly be cheap-and-cheerful, but might at least help you to work out how good a "propper" solution would need to be.