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How to determine power provided by a Class D amplifier — Parallax Forums

How to determine power provided by a Class D amplifier

Hi,

I have a number of different Class D amplifier boards. I noticed that they have different power outputs depending on the voltage applied to the amplifier, the impedance of the speaker, and apparently (?) the frequency of the signal provided to the amplifier. I am using a 4-Ohm speaker in a mono speaker setup.

How can I measure the Power that these boards supply to the audio speaker?

1) Is there some type of low cost Power meter that can be used?
2) Can I measure parameters with a standard VOM that will provide the Power value?

In advance, thank you.

Comments

  • kwinnkwinn Posts: 8,697
    You can get a rough idea of the power by measuring the average current to the speaker using a test signal, but getting an accurate measurement is a bit more complicated. The speaker impedance varies with frequency, and so does the meter response. Using an ADC to measure the AC current and voltage along with a uC to calculate power can provide a much better measurement.
  • Replace the speaker with a known resistor. Place that resistor in an insulated container filled with water at a known temperature. Run a sine wave test signal into the amp and turn up the gain until the output is just about to clip (or better yet, until you reach 1% THD). Leave it running for a known period of time. Measure the water temperature again. Do some maths.
  • Heater.Heater. Posts: 21,230
    Hint. You will need to know the mass and specific heat capacity of the water.
  • ErNaErNa Posts: 1,752
    How many horse power has a Rolls Royce? Answer: enough. When talking about audio power, many parameters have to be taken into account: most important: distortion. The "power" says nothing. The loudspeaker can have a certain efficiency in converting electrical power to sound pressure, but a common way to reduce distortion and to linearize frequency response it to dampen the eigen frequencies, so dissipating a big part of the power delivered. In your case you are speaking of 4Ohm speaker. Means: Power is U * I == U * U / R. The voltage of your power stage is the peak voltage, the power is the peak power the moment a sinus reaches maximun, average power is less. If e.g. your voltage is +- 10 V, the power stage has to drive 2.5A. But 4 Ohm is impedance, and this value is related to a frequency of 1kHz. The screenshot shows a 8 Ohm speaker, but you see, impedance is changing over frequency. For this reason: if it is loud enough and the sound is clear, power is enough ;-)
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