> Loads lower than 16 ohms will cause power waste in the source impedance. For instance a zero ohm load will mean that all the power is dissipated in the 16 ohm source impedance: the zero ohm load draws current, but no power because it I^2R is zero.
Why are you even considering the case of a zero Ohm load with a 16 Ohm source impedance? Modern headphone amps (including really cheap ones) have source impedances in the range of 1 Ohm or less while headphones can range from 30-ish to 300 or more.
If you don’t understand the problem with your post, trying flipping the roles to match reality: Assume a 16 ohm load and a 0 (or realistically, 1 Ohm) source impedance: Now the “no power absorbed” side of this equation is the amplifier. All (or nearly all) power goes to the headphones. That’s exactly what we want.
> Loads higher than 16 ohms will result in less current flowing.
I think this is where you’re confused. Headphone amps are generally voltage limited sources. If you get to the point of current limiting then it’s going to distort intensely and people would turn the volume down because it’s so unpleasant. Any source impedance subtracts from the maximum voltage you can apply across the load, because the source impedance forms a resistive divider. These aren’t high frequency transmission lines where we’re trying to send GHz signals over impedance matched lines.
You don’t maximize power delivered to the headphone by adding an identical source impedance. You actually reduce it massively relative to a modern <1 Ohm source impedance amplifier.
If this still doesn’t make sense, consider why putting a 300 Ohm resistor in your headphone jack before connecting your 300 Ohm headphones isn’t going to improve anything. You’re just burning power in the source-side impedance and fighting against a fixed voltage limit.
I am not confused. You're engaged with the particulars, whereas my post is mostly about the Zout == Zin, and not specifically about headphones.
> Why are you even considering the case of a zero Ohm load with a 16 Ohm source impedance?
For completeness of the analysis. If we hold Zout constant, and likewise the output voltage of the voltage source, and consider a Zin of zero, in that case, power transfer is mimimal (zero). Likewise power transfer goes to zero for large Zin. In between those is the Zout = Zin point where the maximum transfer is obtained by the load from the source. This is part of explaining of what is Zout = Zin about; what is maximized.
For instance consider a R1 resistor in series with a battery. You may not change the battery or R1. For what value of R2 can you get R2 to dissipate the most power? The solution is R2 = R1.
It is useful to think about what happens if we make R2 zero; why wouldn't we consider it. A certain current will flow through a zero ohm R2, but it will not be dissipating any power. From there as we increase R1, the power dissipation curve rises. At R1 = R2, it turns around, and then converges to zero as R1 grows larger.
> Why are you even considering the case of a zero Ohm load with a 16 Ohm source impedance? Modern headphone amps (including really cheap ones) have source impedances in the range of 1 Ohm or less while headphones can range from 30-ish to 300 or more.
Modern tube amps (non hybrid) have significantly higher output impedance because they often have no overall negative feedback and a few special snowflake headphones have impedances as low as 0.1r and 2r.
The considerations are definitively relevant for some cases, as well as in general for analysis of the topic.
Why are you even considering the case of a zero Ohm load with a 16 Ohm source impedance? Modern headphone amps (including really cheap ones) have source impedances in the range of 1 Ohm or less while headphones can range from 30-ish to 300 or more.
If you don’t understand the problem with your post, trying flipping the roles to match reality: Assume a 16 ohm load and a 0 (or realistically, 1 Ohm) source impedance: Now the “no power absorbed” side of this equation is the amplifier. All (or nearly all) power goes to the headphones. That’s exactly what we want.
> Loads higher than 16 ohms will result in less current flowing.
I think this is where you’re confused. Headphone amps are generally voltage limited sources. If you get to the point of current limiting then it’s going to distort intensely and people would turn the volume down because it’s so unpleasant. Any source impedance subtracts from the maximum voltage you can apply across the load, because the source impedance forms a resistive divider. These aren’t high frequency transmission lines where we’re trying to send GHz signals over impedance matched lines.
You don’t maximize power delivered to the headphone by adding an identical source impedance. You actually reduce it massively relative to a modern <1 Ohm source impedance amplifier.
If this still doesn’t make sense, consider why putting a 300 Ohm resistor in your headphone jack before connecting your 300 Ohm headphones isn’t going to improve anything. You’re just burning power in the source-side impedance and fighting against a fixed voltage limit.