Re: Impedance matching question

David Berlind
 

Thanks Craig... so, if I were to summarize what you wrote, at such short
distances, there's really no opportunity for a reflected signal to go out
of phase with the incident signal?

In watching the EEVblog videos, he's clearly using pretty high frequencies
(well out of the audio spectrum). So, your explanation is consistent with
that.

So, two questions remain.

1. why is an impedance match between output tubes and the output
transformer primary so important given the short physical differences. Or,
maybe the tube specs are not showing the actual impedance, but rather the
recommend Hi-Z on the load end to offer the optimal Low-Z to Hi-Z ratio?

2. Why is a Low-Z to Hi-Z ratio desired in audio applications vs. an
impedance match? I understand your point that it doesn't matter at low
distances, but Low-Z to Hi-Z appears to be an objective (iow, the objective
is to avoid a match, by orders of magnitude). Does the higher resultant
voltage (amplitude) at the load spread the signal out in a way that give
the amp more to work with from a fidelity POV?

On Tue, Mar 26, 2019 at 9:59 AM Craig Sawyers <c.sawyers@...>
wrote:

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.

Craig





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