Interesting. Does the 1/8th rule apply to matching amplifiers with speakers as well, say for car audio?
My guess is that the 1/8 trick is for the typical characteristics of small-speaker headset. Big speakers may have different capacitance and inductance characteristics.
I don't know though. That's the first time I heard that approximation, and it doesn't seem to align with anything that I learned in electronics.
For a fixed Zout much higher than zero, if we want to suck the most power out of the amplifier, we should use a matching Zin indeed. E.g. if an amplifier has 16 ohms output impedance, then we will draw the most power out of it with a 16 ohm load.
Loads higher than 16 ohms will result in less current flowing.
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.
So from there, if we increase the impedance gradually, we obtain more and more power until we get to 16 ohms, and after that less and less power.
In amplifiers we care about efficiency, not with operating at the theoretical point where absolute maximum power is drawn. With a near zero ohm output, we ensure that all the voltage is dropped by the load rather than wasted in the source.
When we have a near zero ohm output impedance, Zin == Zout still applies! E.g. theoretically, the maximum power transfer from a 0.1 ohm amplifier would take place if we connect it to a 0.1 ohm load. And that reflects the trend in low-output-impedance audio amplifiers: the lower the speaker impedance you plug in, the power power you get: 16 ohms, 8 ohms, 4 ohms, ... You just can't go anywhere near 0.1, due to the practical limitations in the amplifier: ability to deliver current without frying itself.
But, so yes, Zin == Zout is relevant, but in the majority of the audio amplifiers built, which have very low output impedance, that theoretical point occurs at low value of Zout/Zin which usually out of reach of the absolute current delivery capability of the amplifier.
That is at odds with what the blog post says.
I'm now confused - for the best sound quality, are we to maximise efficiency of power usage or maximise accuracy of signal reproduction?
I didn't think you can do both.
In the case of a headphone amplifier on AC power, we don't have to think about efficiency, as long as it isn't ridiculous, like drawing 100W.
Class B amplifiers are only efficient when producing no output, or when producing output clipped to the power rails (like a square wave). Due to being almost fully off or fully on.
For a sinusoidal signal, the worst case efficiency occurs at about 40% of the output power delivered to the load, when the transistors will dissipate another 20%.
Source: https://eng.libretexts.org/Bookshelves/Electrical_Engineerin... Section 9.2.4: Class B Power Dissipation
(Yes, I checked a few youtube videos and blogs, but car audiophiles don't really get too scientific and it's hard to tell whether they really know what they are talking about).
As far as matching goes, you want amp power > speaker power handling. If the amp distorts before the speakers do, you're not getting the most out of the speakers, and clipping can be bad for some speakers too, especially tweeters.
Also, the THD figures quoted for amplifiers are usually at moderate output levels; they get worse with higher voltage swing. An amp that is getting close to clipping will generally have worse THD than one which is nowhere near clipping. (As well, THD gets worse with rising frequency.)
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.
> 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.
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.
That 1/8 rate is a property of our hearing, and so will apply to any system that creates sound. But it's a quite hard target to achieve in large bass speakers, so a lot of systems simply don't.
Power applications (like speakers or the power grid) are generally concerned with maximizing efficiency, which you get by lowering source impedance as much as possible.
There must be a mathematical description of these different situations but I can't be bothered to work it out. An important part of it is probably that an audio amplifier has feedback, an RF amplifier just emits a wave and the consumer must make itself compatible with it. Lower output impedance = more effective feedback control for an amplifier with feedback.
https://en.m.wikipedia.org/wiki/Smith_chart
You can put an amp 50ohm output into a 75 ohm cable or connectors but there will be negatives. The further away from output impedance the more negatives. Loss, distortion, and worse roll off, etc. Cables are fun too. for much higher frequencies, the less of the actual copper is used. The signal uses the surface of the wiring. For lower frequencies it uses the entire strand/solid copper wire.
RF mostly uses 50ohm but not always. Cable video (tv) is usually 75ohm.
If you are using a short piece of the cable to convey a voltage signal between two pieces of equipment (like the output of one amplifier into the input of another), it has no meaning. The voltage on the other end rises almost instantly and is reflected back.