A rough calculation would be output voltage squared divided by speaker resistance. If your meter gives an rms measurment your done if not, multiply by .707.
Use a dummy load and an adjustable input source for more accuracy. Input to amp should be around .775v to 1v. This method does not simulate a real world load but will give you a good idea of what you have. Good Luck
Use a dummy load and an adjustable input source for more accuracy. Input to amp should be around .775v to 1v. This method does not simulate a real world load but will give you a good idea of what you have. Good Luck
Unless you want to buy an Agilent or Bird wattmeter, this is simple and really works -- get a decent thermometer, a thermos of at least 1 litre capacity, fill with 1 litre of distilled water, place an 8 ohm resistor (insulate the leads with heat shrink tubing) of sufficient wattage to handle the load in the thermos.
Measure the temperature temperature at rest, crank up the amplifier and measure the temperature again. 1 calorie is the amount of energy needed to raise 1 liter of water 1 degree C -- you can convert to watt-hours by just figuring in the constants.
Yes, I did the graduate labs in physical chemistry --
Measure the temperature temperature at rest, crank up the amplifier and measure the temperature again. 1 calorie is the amount of energy needed to raise 1 liter of water 1 degree C -- you can convert to watt-hours by just figuring in the constants.
Yes, I did the graduate labs in physical chemistry --
That reminds me of the student who was asked how he could compute the height of a building with the aid of a barometer. Answer: climb to the top of the building, drop the barometer, time how long it takes until impact, then using the normal formulas for accelaration under 1 gravity, calculate the height of the building....
Jan Didden
Jan Didden
janneman said:
At least you won't have to purchase a bunch of polystyrene caps to integrate the measurements with your AD536 !
btw, at one time I did ask the people in the labs at CPC (they make salad and cooking oil) for the thermal transfer constants for this application on one of my ham band transmitters -- they actually know this stuff if you are diligent enough to look.
Or...
Go to the top of the building and tie the barometer to a length of string and lower it down until it is just suspended off the ground. Start the barometer swinging, and measure it's frequency. You can then compute the height from the mass of the pendulum, and the time it takes for a full cycle.
al/should probably stop now...
Go to the top of the building and tie the barometer to a length of string and lower it down until it is just suspended off the ground. Start the barometer swinging, and measure it's frequency. You can then compute the height from the mass of the pendulum, and the time it takes for a full cycle.
al/should probably stop now...
Go inside the building and find the senior building engineer. He will find the blueprints to the building and read off the height to you. It will be accurate to less than one foot. Give him/her the barometer as a gift for the favour.
Pros: No destruction. No dangerous swinging of pressure measurement devices. No need to know any other dimension.
ensen
Pros: No destruction. No dangerous swinging of pressure measurement devices. No need to know any other dimension.
ensen
Great, guys,
The only other one I know is: go to top of building, tie barometer to string, lower until it hits ground, pull up and measure length of string.
There is also a variation on the pendulum: measure the frequency at ground level and at the top. The difference will give you the difference in gravity, which in turn will give you the height. I confess I don't know how to do this one, but in principle it must be possible.
Jan Didden
The only other one I know is: go to top of building, tie barometer to string, lower until it hits ground, pull up and measure length of string.
There is also a variation on the pendulum: measure the frequency at ground level and at the top. The difference will give you the difference in gravity, which in turn will give you the height. I confess I don't know how to do this one, but in principle it must be possible.
Jan Didden
I think there's no such think as RMS output (wattage) in engineering standards. Wattage is simply P=V*I (or Prms=Vrms*Irms) or P=V2/R. RMS is always about AC. About 0.707 of the peak value is equivalent with DC voltage that would produce the same dissipation over a resistor.
Amplifier circuits are so complicated that "output" measurement number must be specified with measurement parameters or conditions in order that the number be meaningfull (even then only a few people can understand). And the front ends must also be able to deliver any current as requested by the amplifier.
Having said about class A and AB, I'm sure you have been concerned with power loss (heat). But if you measure the power right at the output, then it doesn't matter (much).
If it is about common practice (like speaker sensitivity measurement), I don't know, may be it is close to what PassFan had suggested. The load may be simplified into a dummy load (resistor), otherwise (due to speaker impedance nature on AC) there will be a "shift" between V and I, thus P = V * I * cos ("shift").
But if it is about personal needs, I will only concern about power before clipping. With 87dB speaker, I think my ears will be able to measure the output power (well, by listening to the "pop!" when the power is turned off )
Amplifier circuits are so complicated that "output" measurement number must be specified with measurement parameters or conditions in order that the number be meaningfull (even then only a few people can understand
). And the front ends must also be able to deliver any current as requested by the amplifier.Having said about class A and AB, I'm sure you have been concerned with power loss (heat). But if you measure the power right at the output, then it doesn't matter (much).
If it is about common practice (like speaker sensitivity measurement), I don't know, may be it is close to what PassFan had suggested. The load may be simplified into a dummy load (resistor), otherwise (due to speaker impedance nature on AC) there will be a "shift" between V and I, thus P = V * I * cos ("shift").
But if it is about personal needs, I will only concern about power before clipping. With 87dB speaker, I think my ears will be able to measure the output power (well, by listening to the "pop!" when the power is turned off
)
audio amp rms power output
to all the guys who answered my query on rms output of an audio power amp, thanx. you guys sure gave me right direction which i have been following eversince. "why bother about the power output, when whats important is that the amp is giving you the sound that you like" my sound preference is that it reproduces sound as it should sound. again thanx to all. 
to all the guys who answered my query on rms output of an audio power amp, thanx. you guys sure gave me right direction which i have been following eversince. "why bother about the power output, when whats important is that the amp is giving you the sound that you like" my sound preference is that it reproduces sound as it should sound. again thanx to all.
To measure the output power, connect a dummy load (say 8 ohms) connect an oscilliscope to the output and feed a sinewave into the amp at a suitable frequency... wind up the wick on the amp will the output waveform starts to clip, then wind it back a tad... measure the output voltage (either from the CRO or with a multimeter) and calculate the power.. (not sure how to do this, can't remember) make sure your dummy load can handle the current.. my piece of nichrome wire couldn't and got rather hot, then melted and twisted and burnt the table... heh.. anyway... this is my prefered method... cos its quite accurate and most (some) people have all the stuff required... lol heating water??
my piece of nichrome wire couldn't and got rather hot, then melted and twisted and burnt the table... heh.. anyway... this is my prefered method... cos its quite accurate and most (some) people have all the stuff required... lol heating water?? P=Vrms^2 / R
Vrms=V(peak of sine, or PSU voltage less a couple of volts) / sqrt(2) - this assumes split supplies, push pull amps and you only measure ground to rail rather than rail to rail.
It follows that:
Prms(for sine output)=Vpk/(2*R)
For bridged amps, multiply above result by 4.
Petter
PS: Since speakers are not typically resistive, this is not the power that is actually dissipated in the device, rather the power that would have been dissipated had the load been non-reactive. The proof point for this is that any small well insulated speaker typically does not burn even after playing on it a long time. We should thus refer to Power in the more correct term which would be "Power" to indicate that it is not in fact Power ...
Vrms=V(peak of sine, or PSU voltage less a couple of volts) / sqrt(2) - this assumes split supplies, push pull amps and you only measure ground to rail rather than rail to rail.
It follows that:
Prms(for sine output)=Vpk/(2*R)
For bridged amps, multiply above result by 4.
Petter
PS: Since speakers are not typically resistive, this is not the power that is actually dissipated in the device, rather the power that would have been dissipated had the load been non-reactive. The proof point for this is that any small well insulated speaker typically does not burn even after playing on it a long time. We should thus refer to Power in the more correct term which would be "Power" to indicate that it is not in fact Power ...
Jay said:I think there's no such think as RMS output (wattage) in engineering standards. Wattage is simply P=V*I (or Prms=Vrms*Irms) or P=V2/R. RMS is always about AC. About 0.707 of the peak value is equivalent with DC voltage that would produce the same dissipation over a resistor.[snip a lot]
Only for a sine wave is the .707 valid! The beauty about RMS is that it gives the correct power whatever the waveform.
Jan Didden
.707 is actually 1 / sqrt(2)
As far as I have been able to see, there has been some confusion in previous postings on this thread, but as of late, it has started to shape up (I even forgot to put the squared for Vpk myself). To recap:
Vrms=Vpk*1/sqrt(2)
Vrms^2=Vpk^2 * 1/2 = Vpk^2/2
"Power" is thus = Vpk^2/(2*R)
Petter
As far as I have been able to see, there has been some confusion in previous postings on this thread, but as of late, it has started to shape up (I even forgot to put the squared for Vpk myself). To recap:
Vrms=Vpk*1/sqrt(2)
Vrms^2=Vpk^2 * 1/2 = Vpk^2/2
"Power" is thus = Vpk^2/(2*R)
Petter
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