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14th January 2003, 03:36 PM  #1 
diyAudio Member
Join Date: Mar 2002
Location: Athens+Ljubljana

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? 
14th January 2003, 03:48 PM  #2 
diyAudio Member

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 trueRMS 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. Jan Didden 
14th January 2003, 03:53 PM  #3 
diyAudio Member
Join Date: Nov 2002
Location: Florida

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. 
14th January 2003, 04:02 PM  #4  
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Join Date: Sep 2002
Location: Milton Ontario

Quote:
Vrms = .707 * Vpeak So if you had the peak voltage you would actually multiply the max power by .5 not .707. 

14th January 2003, 04:27 PM  #5 
diyAudio Member

True for sinewaves only
The above calculations are only true for sine signals.
\Jens 
27th January 2003, 07:06 AM  #6  
diyAudio Member

Quote:
One way to measure the actual current is to connect a known resistor, say 1ohm, 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 phaseangle 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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27th January 2003, 12:41 PM  #7 
diyAudio Member

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. 
27th January 2003, 01:44 PM  #8  
Electrons are yellow and more is better!
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Re: Calculation of RMS power output?
Quote:
All DVM's assumes sinus in AC measurements except for those with "true RMS". Most DVM's can handle frequencies to 12 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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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. 
27th January 2003, 03:06 PM  #10 
diyAudio Member
Join Date: Jan 2002
Location: Earth

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 nonzero and thus a product which is nonzero, even though the average power is actually zero since pure reactances do not dissipate energy. 
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