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?
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?
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.
Jan Didden
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.
Jan Didden
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.
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.
Jeff R said:
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.
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.
<p> Most of the less than $100 DVM's don't make very acurate RMS measurements over about 1kHz.
protos said:
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!
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.
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.
Just my $.02 in support of traderbam:
Power does not really work in audio as most speakers are inductive, and some are capacitive. In practice, most people assume resistive load when giving up a figure.
Think about it, a well insulated enclosure dissipating 200W for a few minutes will roast. I can assure you the energy tranmitted to air is a fraction of this "200W" so the rest should have been heat and a fire hazard.
Petter
Power does not really work in audio as most speakers are inductive, and some are capacitive. In practice, most people assume resistive load when giving up a figure.
Think about it, a well insulated enclosure dissipating 200W for a few minutes will roast. I can assure you the energy tranmitted to air is a fraction of this "200W" so the rest should have been heat and a fire hazard.
Petter
I have thought, as have most, about the average/rms power thing. I believe that if you wanted to calculate actual RMS power you need a power function P(t). Plugging this function into the RMS function (S = integration sign)
P(rms) = (S( ( (P(t) )^2) ))^(1/2)
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.
P(rms) = (S( ( (P(t) )^2) ))^(1/2)
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.
"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.
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.
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