Re: Impedance matching question

Craig Sawyers

That was a decent video, thank you. It doesn't seem like he had problems with the terminators
themselves. In fact, he uses them to address mismatches. I'd be curious to know (from anyone who
knows) why maximum power (matched impedance) is ideal in some situations, but a low-Z to high-Z
arrangement is the ideal in other situations (ie: guitar to amplifier or microphone to PA). I
realize the
outcome in the latter situation; you preserve the audio highs and lows. But need some schooling on
why to maximize voltage at the load (vs. maximizing transfer power) in those situations and why the
mismatch isn't destructive to the signal. And correspondingly, why the same isn't true of the
impedance match between output tubes and the primary of an output transformer since its a similar
audio application. Again, I understand the requirements and outcomes... but am confused about the
underlying physics.
It is to do with the frequency range. As soon as the length of the cable becomes a significant
fraction of the electrical wavelength in the cable, you need to impedance match. That is because
energy is reflected at an impedance discontinuity, so you end up with standing waves along the length
of the cable. For a 1 metre long coax the wavelength becomes a significant fraction of the cable
length by about 10MHz, so you need to impedance match.

With audio, the wavelength is so long (at 20kHz it is about 10km in a typical coax cable) you
absolutely do not need to match. Hence you guitar example.

Going way back long distance telegraph and telephone lines were significant fractions of an audio
wavelength in the cables, and they needed to impedance match.


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