measuring power out of an audio amp

[hummusdude]

I'm using an 8 ohm resistive load hooked up to the speaker terminals with a 1 volt sine wave at 1 kHz input. On my scope I measure 18.6V peak on the speaker terminals. Converting to rms gives 18.6 * 0.707 = 13.15 volts rms. To get power I use (13.15^2)/8 = 21.15 watts.

My speakers are 6 ohms. So if I measure 18.6 volts peak at 8 ohms, does the voltage stay the same when I connect 6 ohm speakers? The power supply is pushing +-25VDC from a 250 VA transformer so I know there's still some headroom there. And I can increase the input a bit and get 21.2 volts peak at the speaker terminals (still with 8 ohms resistive) before I start to see clipping. I think the clipping is likely from the power supply because it's not rated to go any higher and the amp can handle a bit more.

It seems like the voltage would not droop if the power supply can supply enough current, which I think it can in this case. I sure wish I'd bought 6 ohm power resistors!

[james_s]

Yes the voltage will stay the same and lower impedance speakers will draw more power from the amplifier resulting in (all else being equal) louder sound. Be aware that the amplifier will also run hotter, the load on the power supply will be greater and not all amplifiers can safely handle lower impedance speakers, although 6 ohms is probably close enough that no reasonable quality amp will have trouble with it.

[DavidAlfa]

Well, that should be tested, the amplifier might be unable to handle the extra current, clipping the signal.
But in theory, yes, as long as the voltage stays the same.

[radiolistener]

On my scope I measure 18.6V peak on the speaker terminals. Converting to rms gives 18.6 * 0.707 = 13.15 volts rms. To get power I use (13.15^2)/8 = 21.15 watts.

That's correct. If you have Vpk amplitude, you can just divide square by double load:

18.6^2 / (8 * 2) = 345.96 / 16 = 21.6225 W

My speakers are 6 ohms. So if I measure 18.6 volts peak at 8 ohms, does the voltage stay the same when I connect 6 ohm speakers?

No, it depends on your amplifier output impedance. You can check it by measure output amplitude for two different known load resistors. More difference for resistance leads to better precision. But at the same time very high or very low load can be out of working range for your amplifier, in such case the result will be distorted. Resistors should NOT be inductive.

For example, for exactly the same input:
At 8 Ω load your amplifier produce 18.6 Vpk amplitude.
At 4 Ω load your amplifier produce 17 Vpk amplitude.

Then you calculate amplifier output impedance:

Z = (R1 - R1*(U1/U2)) / (U1/U2 - R1/R2) = (8 - 8*(18.6/17)) / ((18.6/17) - (8/4)) = -0.75296 / -0.90588 = 0.8311 Ω

When you know amplifier output impedance, you can calculate output voltage for specific load (simple voltage divider). And when you know voltage and load resistance you can calculate power for a new load: P = Upk^2 / (2 * Rload)

But since most of audio amplifiers usually have pretty low output impedance, the result for 6 Ω load should be close enough to 8 Ω load.

[Doctorandus_P]

On my scope I measure 18.6V peak on the speaker terminals.

And what is the "zero" reference during that 18.6V peak measurement?
If that reference is anywhere different from the zero-crossings of the sinewave, then that number does not mean anything.

Using loudspeakers with a lower impedance will put more current through them, and thus more power, but there are limits. One limit is the current that the output transistors can deliver. Another limit is that your power supply voltage sags a bit under these higher currents, and therefore the the amplifier may clip at a slightly lower output voltage.

[Terry Bites]

BTW the speaker is far from being a resistor, even at 1kHz and comparative results using reistors may not give you the answer you are looking for. The actual sound pressure level produced by the speaker will depend very much on its mechanical design and enclosure. Speaker effciencies vary widely. see https://en.wikipedia.org/wiki/Electrical_characteristics_of_dynamic_loudspeakers

Power amplifiers have output resistances in the milliohms at mid frequncies.

[TimFox]

The very low output impedance of typical analog solid-state power amplifiers is the direct result of voltage feedback from the output terminals.  This will try to keep the output voltage constant (for a given input voltage and variable load resistance), but there are other limits including the sag of the DC power supply at high output current that the feedback cannot overcome (since the current has already hit a limit).  When operating at relatively low power, the output voltage should be quite independent of load resistance over a wide range, since the output current is well below the limits imposed by the output devices and the power supply.
There was a very long thread a while ago where the original poster misunderstood the "maximum power transfer theorem" and insisted that the maximum output power from a power amplifier was when the load impedance equaled the source impedance, which is true for many circuits but generally not relevant to the "optimum load resistance" for a power amplifier.

[hummusdude]

I put the scope ground lead on one side of the 8 ohm resistor and the probe on the other. I'm not sure how the zero reference would be anything other than where the sine wave crosses the y axis. Do you mean on my scope?

[hummusdude]

The very low output impedance of typical analog solid-state power amplifiers is the direct result of voltage feedback from the output terminals.  This will try to keep the output voltage constant (for a given input voltage and variable load resistance), but there are other limits including the sag of the DC power supply at high output current that the feedback cannot overcome (since the current has already hit a limit).  When operating at relatively low power, the output voltage should be quite independent of load resistance over a wide range, since the output current is well below the limits imposed by the output devices and the power supply.
There was a very long thread a while ago where the original poster misunderstood the "maximum power transfer theorem" and insisted that the maximum output power from a power amplifier was when the load impedance equaled the source impedance, which is true for many circuits but generally not relevant to the "optimum load resistance" for a power amplifier.

um...I think that was me that was very insistent about max power transfer (chuckles).

[TimFox]

The maximum power output theorem is relevant to maximizing gain (below power limit), but not necessarily power output.

[Terry Bites]

The maximum power transfer theorem?

[TimFox]

I worded it that in quotes the first time, but improvised (not in quotes) the second time.