4 ms·

Digging deep into my memory here, but I always remember that when Zout==Zin that's what you want for maximum power transfer. My guess is that the 1/8 trick is

Digging deep into my memory here, but I always remember that when Zout==Zin that's what you want for maximum power transfer. 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.

marcosdumay 3y ago
Why would one optimize for maximum power transfer at all? Worse, why would one un-optimize the power source by adding impedance so that it works at maximum capacity? It's much better to just leave the extra capacity there. 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.
ahartmetz 3y ago
The radio stuff is about "wave impedance", the speaker or headphone stuff is more like a power supply situation at (in RF terms) very low frequencies. You want your power supply to have low impedance so it doesn't sag and keeps tight control over its output voltage. 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.
- screcth 3y ago
  Analysis becomes much simpler when frequencies are low enough. Wavelength is large compared to the dimensions of the circuit so you can ignore the effects of wave propagation.
willis936 3y ago
That's true for RF, but not for DC. Audio is "DC" because transmission lines are well approximated as ideal wires at frequencies below 100s of kHz.
- bgnn 3y ago
  nor fully correct. lumped circuit theory doesn't only depend on the frequency but also the cable length. EM waves travel at around 0.6C in solids (copper). wavelength = V/f where V=0.6C, and frequency is say 20kHz, which is 9km. If you have an audio cable of hundreds of meters you would need impedance matching.
  - BenjiWiebe 3y ago
    Rule of thumb for RF is that you can ignore impedance matching for <=1/10 WL. So for 20kHz you could ignore impedance matching if your cables are less than 900 meters long. How many audiophiles use cables 900m long? Or anyone else for that matter?
    - bgnn 3y ago
      There are certainly cases in big concert halls etc to run like 100m. 900m should be more a stadiums.
formerly_proven 3y ago
That's for maximizing the power transfer, which is important in radio applications because you want to maximize SNR (and because this is the condition where no power is reflected). 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.
kazinator 3y ago
Unlike what other commenters are saying, this Zout == Zin is applicable to audio. 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.
- Aurornis 3y ago
  > 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.

marcosdumay 3y ago

Why would one optimize for maximum power transfer at all? Worse, why would one un-optimize the power source by adding impedance so that it works at maximum capacity? It's much better to just leave the extra capacity there. 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.

ahartmetz 3y ago

The radio stuff is about "wave impedance", the speaker or headphone stuff is more like a power supply situation at (in RF terms) very low frequencies. You want your power supply to have low impedance so it doesn't sag and keeps tight control over its output voltage. 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.

screcth 3y ago
Analysis becomes much simpler when frequencies are low enough. Wavelength is large compared to the dimensions of the circuit so you can ignore the effects of wave propagation.

willis936 3y ago

That's true for RF, but not for DC. Audio is "DC" because transmission lines are well approximated as ideal wires at frequencies below 100s of kHz.

bgnn 3y ago
nor fully correct. lumped circuit theory doesn't only depend on the frequency but also the cable length. EM waves travel at around 0.6C in solids (copper). wavelength = V/f where V=0.6C, and frequency is say 20kHz, which is 9km. If you have an audio cable of hundreds of meters you would need impedance matching.
- BenjiWiebe 3y ago
  Rule of thumb for RF is that you can ignore impedance matching for <=1/10 WL. So for 20kHz you could ignore impedance matching if your cables are less than 900 meters long. How many audiophiles use cables 900m long? Or anyone else for that matter?
  - bgnn 3y ago
    There are certainly cases in big concert halls etc to run like 100m. 900m should be more a stadiums.

formerly_proven 3y ago

That's for maximizing the power transfer, which is important in radio applications because you want to maximize SNR (and because this is the condition where no power is reflected). 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.

kazinator 3y ago

Unlike what other commenters are saying, this Zout == Zin is applicable to audio. 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.

Aurornis 3y ago
> 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.