How do I find the actual wattage an amplifier is putting out

[jtvang125]

I'm curious as to how much power my audio amplifier is putting out. What do I need to measure to figure this out?

 

[f95toli]

It is actually quite tricky to measure this accurately.
The reason is that the power depends on the impedance of the load (i.e. the speaker) and the latter can be VERY frequency dependent for a real speaker.

Another problem is that the "maximum power" is not really that interesting, the distortion increases with power and at some point the amplifier will start clipping, this is not really big problem initially (at least not if the amplifier is clipping "gently") but sooner or later you will reach a point where the music sounds terrible and -worse- you risk destroying your speakers (an amplifier that is clipping outputs DC that heats up the voice coil).

Hence, measuring power of an amp feeding a speaker in "real time" is therefore somewhat tricky; the only way to do it is to measure both the voltage and the current from the amp but that is NOT something I would recommend unless you know what you are doing (voltage is not the problem, but measuring the current is).
The result will also depend on how long you are averaging (i.e.. the "time window") since peak power can be MUCH higher than the median power for real music.

Anyway, the standardized test is to measure the voltage across a 8 Ohm power resistor at 1% distortion and then calculate the power from P=V^2/R. This is what the wattage rating of an amp means.

 

[bobsmith1492]

I built a tester for our senior engineering project to do just that...

Several thousand lines of code later, I wouldn't recommend it.

An easy way would be to put a sine wave into a resistor and measure the voltage and resistance...

Put an oscilloscope across the output and turn it up until it just starts to clip - that would be your maximum continuous output power. If it's a cheap amp, though, you may burn it up so do it quickly!

 

[bobdole369]

Back in the day they measured and rated RF amplifiers (thats radio frequency) by measuring the power INPUT to the amplifier. f95toli is correct, however note that said resistor must be NON-INDUCTIVE - lest you "lose" some power due to reactance. Its very difficult to measure - The power output at 50hz might be completely different that the power output at 1000hz.

I still subscribe to the power input method - find you a big ammeter and stick it inline with the DC input of the amp itself. Create a sine wave as input and observe the output. Increase your input until clipping occurs and back off. Now view your ammeter and multiply by the voltage input.

Multiply by the known efficiency factor of that particular amps class and you get a ballpark.

 

[Martimus]

One of the first things I learned in Electronics class in college was that a simple transistor amplifier circuit draws the most power when it is amplifying an idle signal (In other words when it isn't doing anything).

Here is a pretty simple explanation for a common emitter amplifier: Common Emitter Amplifier
Page 7.4 shows the equation to calculate power. (Which shows that if you set Ve to 0, you have the highest power output)

I haven't built a transistor circuit in years though, so I am quite a bit rusty.

 

[Born2bwire]

Originally posted by: Martimus
One of the first things I learned in Electronics class in college was that a simple transistor amplifier circuit draws the most power when it is amplifying an idle signal (In other words when it isn't doing anything).

Here is a pretty simple explanation for a common emitter amplifier: Common Emitter Amplifier
Page 7.4 shows the equation to calculate power. (Which shows that if you set Ve to 0, you have the highest power output)

I haven't built a transistor circuit in years though, so I am quite a bit rusty.

That isn't correct, any time you are amplifying a signal you have to inject carriers into the output. You're reading the schematic wrong, the input to the circuit is on the base, not the emitter. In addition, that equation is for the large signal biasing power dissipation, so minimizing the difference between the common and the emitter (as you would expect) minimizes the bias power dissipation. Of course you can only minimize this so low until you fall out of saturation. But this equation is not taking into account the small signal power dissipation which arises due to the input. Your greatest power dissipation would occur, independent of you biasing as long as it is in saturation, when you have a DC input so large as to clip the amplifier.

 

[Modelworks]

The simplest way would be to use something like a killawatt meter on the wall plug.
Compare it with volume zero and then again at different volume levels and maybe graph the results.
Amplifiers are not 100% efficient so it is not going to be accurate, but it will give you a general idea.

 

[KIAman]

You can't measure accurately unless you establish a criteria for the measurement.

You need the resistance of the load, the frequency or frequency range, and time. Then you can calculate the "wattage" of your amp.

Considering you gave almost no information about the type of amplifier, if it is a home (AC) or car (DC) amp, manufacturer, speaker(s), impedance, I can't even begin to guess the rating.