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Old 14th January 2003, 03:36 PM   #1
protos is offline protos  Greece
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Default Calculation of RMS power output?

I was wondering how one can calculate an amplifiers power rms output at specific output volume into specific speakers with just a multimeter.
If , for example I use a 1khz tone from a test disc and I get let's say 4V output at the amp for 4ohm speakers then according to ohms law the power should be P=V^2/R so P=4x4/4 so P=4W.
But is this correct or should I multiply by .707 for RMS?
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Old 14th January 2003, 03:48 PM   #2
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Default RMS

Protos,

If your multimeter is indicating RMS, your calculation is correct, IF the speaker impedance is 4 Ohms. That may not be the case, irrespective of the nominal impedance. If you want you can also measure the speaker current and calculate power as Vrms X Irms.

A note: Your multimeter should be giving RMS values, but normally (unless it is a more expensive true-RMS meter) is calibrated to show the RMS value of a sine wave. Using this meter with a square wave or pulse signal for instance will not corretly indicate RMS values.

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Old 14th January 2003, 03:53 PM   #3
Jeff R is offline Jeff R  United States
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It isn't that simple. The actual impedance of a speaker is highly frequency dependent and may be several times higher or lower than the rating of the speaker. So, if you put in 1 KHz, the actual power will most likely not be what you calculate.

The only way to determine the power into an unknown load is to measure voltage and current and take the product.

You may be able to determine what the speaker impedance is at a particular frequency and then calculate the power by measuring the voltage at that one frequency.
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Old 14th January 2003, 04:02 PM   #4
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Quote:
But is this correct or should I multiply by .707 for RMS?
P=Vrms*Irms or =(Vrms)^2/R obviously

Vrms = .707 * Vpeak

So if you had the peak voltage you would actually multiply the max power by .5 not .707.
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Old 14th January 2003, 04:27 PM   #5
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Default True for sinewaves only

The above calculations are only true for sine signals.

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Old 27th January 2003, 07:06 AM   #6
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Quote:
Originally posted by Jeff R

The only way to determine the power into an unknown load is to measure voltage and current and take the product.

You may be able to determine what the speaker impedance is at a particular frequency and then calculate the power by measuring the voltage at that one frequency.
Yes, you'll need BOTH the current and the voltage.

One way to measure the actual current is to connect a known resistor, say 1-ohm, in series to the speaker and measure the drop across. But you'll have to add the power dissipated on this resistor also.

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.
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Old 27th January 2003, 12:41 PM   #7
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true RMS -- if you look at the expensive DVM's -- like the HP3478a (which are going for a song on EBay now) they use an Analog Devices AD536 chip for RMS detection. Chuck Hansen did an article on an RMS power meter for AudioXpress about 2 years ago using this chip. The AD737 is less expensive than the 536, but you have to limit the input to 200mV. Linear Technology made a chip with a thermal sensor for RMS measurement.

<p> Most of the less than $100 DVM's don't make very acurate RMS measurements over about 1kHz.
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Old 27th January 2003, 01:44 PM   #8
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Default Re: Calculation of RMS power output?

Quote:
Originally posted by protos
I was wondering how one can calculate an amplifiers power rms output at specific output volume into specific speakers with just a multimeter.
If , for example I use a 1khz tone from a test disc and I get let's say 4V output at the amp for 4ohm speakers then according to ohms law the power should be P=V^2/R so P=4x4/4 so P=4W.
But is this correct or should I multiply by .707 for RMS?
May I ask why you wonder? I think the guys here (including I) don't know what we should answer.

All DVM's assumes sinus in AC measurements except for those with "true RMS". Most DVM's can handle frequencies to 1-2 kHz except for the more advanced.

1 V rms sinus will be read 1 V on all DVM's, as long it's sinus!
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Old 27th January 2003, 02:34 PM   #9
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Yes, to get a result that is anywhere close to accurate, you will need to find both Vrms and Irms then:
Average Power = Vrms x Irms

Note the term 'Average Power', this is the correct term. You have RMS Voltage, RMS Current and Average Power.
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Old 27th January 2003, 03:06 PM   #10
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I agree with Audiofreak that average power is the product of Vrms and Irms provided that V and I are in phase. This is probably a reasonable approximation for many speakers at 1kHz.

Note that a pure reactance (capacitor or inductor) will give Vrms and Irms as non-zero and thus a product which is non-zero, even though the average power is actually zero since pure reactances do not dissipate energy.
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