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If an audio power amplifier would power a 8 Ohm speaker through its proper(?) output and if the volume knob is set such that the amplifier outputs 8Vrms 440Hz sine wave across the speaker leads the power delivered to the speaker can be found roughly as if Im not wrong:

P = V^2/R = 64/8 = 8 Watt

But now if we add a parallel speaker the equivalent resistance will be 4 Ohm.

In this case will the amplifier act as a current source which will still output 8Vrms but double the current it sources? How about the typical effect of the mosfet power amplifier’s output impedance?

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Your scenario should separate linear model from a model that includes limiting values of voltage and/or current. The amplifier is more voltage source than a current source within its linear range...
If the amplifier pumps twice the current into a 4 ohm load compared to an 8 ohm load, then it is a voltage source. A high-quality amplifier will have a much lower output resistance, compared to the resistance of the load (speaker).

When you turn up the volume, you run into limits: a voltage limit (mostly set by the DC supply for the power amp) and a current limit. A voltage source has difficulty supplying enough peak current into a small load: when you add speakers in parallel, more current must flow.
An example audio power amplifier:
An amplifier has a damping factor of 180, which basically means that its output resistance is 180 times smaller than its 8-ohm load. Such tiny output resistance is achieved with a great deal of negative feedback in the amplifier. Within the voltage and current limits, the amplifier runs linearly and the model is a simple voltage source, with 44.4 milli-ohms Thevenin output resistance.
But the power supply likely sags when more current flows - DC supplies are often not regulated.
And the high-power output stage is not perfectly efficient when more current flows, so the peak current limit decreases for small load resistors (like 4 ohms or 2 ohms). Feedback cannot correct these voltage and current limits.
When voltage or current limits are reached, distortion increases greatly. So a power amplifier is specified for maximum power at some (small) distortion. The example power amp is tested for maximum power where distortion has increased to 1% distortion (THD):

  • 250W... 8 ohm
  • 410W... 4 ohm
  • 600W... 2 ohm

Note that halving the load from 8 ohms to 4 ohms doesn't double the power from 250W to 500W, and similarly at 2 ohm load, power is less than 1000W. This is a result of limiting of voltage and/or current.

When operated within its linear range, you can assume the simple voltage-source model:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Amazing and neat answer with circuit model example thanks! \$\endgroup\$ – user1999 Oct 6 '18 at 17:41
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"If an audio power amplifier"

Modern Audio Power Amplifiers are designed to have very low output impedance. What you are asking depends on the specifications of that particular power amplifier.

If the power amp can supply the amount of current that the load requires, it doesn't matter what the actual load impedance is. In other words, if the amp can supply 100 Watts into an 8 Ohm load (about 3.5 Amps) and you connect a 4 Ohm load at 8 Vrms output, the total load current is 2 Amps. This is well within the current and power ratings of that particular amplifier.

In general, the output voltage of an Audio Power Amplifier remains fairly constant as the load changes. The changes in output voltage as the load is varied occur for several reasons: resistance between where the negative-feedback point is in the amplifier to the output terminals, how much negative-feedback is actually used inside the amplifier, some other (minor) factors.

My experience with a large variety of Audio Power Amplifiers is that the change in output voltage as the load is a very low percentage. This is assuming, of course, that the input to the amplifier is constant and that the amplifier is not into current limit or power limit.

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  • \$\begingroup\$ Okay let me ask this way; imagine the audio power amp has zero output impedance and the volume is set to constant. And the 8 Ohm speaker terminals show 8V. What happens to the voltage if we remove the speaker and connecting 4 Ohm? Assuming we didnt move the amplification or volume \$\endgroup\$ – user1999 Oct 6 '18 at 17:04
  • \$\begingroup\$ Assuming a perfect amplifier with zero output impedance, the output voltage will remain constant regardless of the load impedance. \$\endgroup\$ – Dwayne Reid Oct 6 '18 at 17:19
  • \$\begingroup\$ I see which means it will source twice current in that case \$\endgroup\$ – user1999 Oct 6 '18 at 17:21
  • \$\begingroup\$ Exactly correct! \$\endgroup\$ – Dwayne Reid Oct 6 '18 at 17:22
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An audio amplifier is not a constant current or constant power device.
The Volume control sets the voltage output, and you can see this by simply disconnecting your 8 Ohm load ...the output drive voltage should remain constant. However, all amplifiers have an output impedance (of more than just the FET) and the output voltage will droop slightly with increasing load (lower impedance). How much load you can apply and how much droop you experience depends on the amplifier.

As you lower the load impedance, you increase the current requirements for the amplifier. At some point it cannot supply the required current and the amplifier will become non-linear, and eventually potentially blow a fuse or shutdown due to overtemperature.

The output impedance of the amplifier should be constant (and not load dependent) as long as the amplifier is still linear. Once it becomes non-linear, it would be design dependent what happens to the output impedance. You would expect that non-linear includes an increase in distortion, severe clipping and loss of gain (Vout).

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In this case will the amplifier act as a current source which will still output 8Vrms but double the current it sources?

Only if the amplifier is designed for that load.

Usually, amplifiers are designed for a specific load, and feeding twice the load with it will not lead to twice the power delivered.

So, no.

See: Impedance matching for maximum power transfer.

How about the typical effect of the mosfet power amplifier’s output impedance?

These impedances are typically very low with given minimum load impedance, and within operational bounds.

Amplifiers aren't perfect nor linear: They might be very linear (i.e. work like a low-impedance voltage or current source) up to a specific current (or voltage), but then simply break down or run into saturation.

So, modelling an amplifier as an ideal current or voltage source plus a source impedance usually doesn't cut it.

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  • \$\begingroup\$ Do need impedance matching for such amplifiers. I thought that were for 600 Ohm old sytems \$\endgroup\$ – user1999 Oct 6 '18 at 16:32
  • \$\begingroup\$ 600Ω?! are you sure? \$\endgroup\$ – Marcus Müller Oct 6 '18 at 16:53
  • \$\begingroup\$ Yes 600 Ohm was too big so power matching was necessary as far as I read Im not into the subject btw \$\endgroup\$ – user1999 Oct 6 '18 at 16:54
  • \$\begingroup\$ 600Ω output impedance doesn't sound like an amplifier at all; that's a gigantic output impedance and only allows for extremely little power to be delivered (if the load is matched, 600Ω, and your amplifier can generate a whole whopping 100V, then the power delivered is just U²/R = 10⁴/600 = 18.3W; that's nearly nothing for an amplifier with a 100V output voltage capability). There's something wrong with your numbers. \$\endgroup\$ – Marcus Müller Oct 6 '18 at 16:55
  • \$\begingroup\$ 600 Ohm was the old standard which required power matching ; modern amplifiers do not need power matching afaik \$\endgroup\$ – user1999 Oct 6 '18 at 17:00

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