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Old 24th March 2007, 09:52 PM   #1
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Lightbulb Input / Output impedance, Resistor network Vs. Dynamics

Hello !

As all of us know, a good audio design is not only made of good components, but also depends of its implementation. Nothing new.

But, following this rule, I was wondering about our I/O impedance, transistors and opamps implementation, how it can affect the Dynamics of an audio system.

A basic rule is, that a resistive link will be less conductive as the current trought it increases. All of us know it.

So, let's suppose that we have a low impedance audio source (100Ohms), and a very high impedance audio input to drive (10M Ohms). So, if I connect it with a hard wire, or with a 1K resistor, or even 10k, 100k, I'll not perceive any difference in sound, just because a 100k resistor or a 0ohm hard wire are conductive enough to drive the input, without have a perceivable sound attenuation.

Music is a great AC signal, with attacs and decays. For example,imagine a song starting with sweet flute, and few moments later, a very strong drum is played. So, if we measure, the AC signal of the music will become stronger (more voltage pump) when the drums are playing.

So, imagine that we have an mp3 player (low impedance source, capable of drive headphones) driving directly an amplifier that it's input impedance is 47k.

Now imagine that we have the same mp3 player and the same amplifier, but now, the mp3 player's signal is passing trought a 100:1 attenuation network (100k link resistor + 1k resistor to the ground), and after this atenuation, an audio pre amp that amplifies 100x the sound again, and it's finally connected to the amplifier, that it's input impedance is 47k.

So, theoretically, the result of both configurations should be the same.

In the first configuration, the mp3 player have a low enough output impedance to drive the amplifier without loss, so the music will be played 100%.

In the second configuration, the mp3 player's sound is divided 100 times, and then multiplied 100 times by the pre amp, and the sound will be 100% again. After, the pre amp is connected to the amplifier, and this pre amp can drive the amplifier without any problems too. (let's assume that it's output impedance and driving capabilities are exactly the same as the mp3 player.)

If we put an sinewave, of any kind of continuous AC signal to play in both configurations, they will behave exactly the same way, and there will be NO sound difference between them.

But if we put that music, that starts with a sweet flute and suddenly a very strong drum starts to be played, probably the drum will sound compressed in the second configuration.

Why?

Because in the second configuration, when the drums are playing, there will be more current flowing through the 100K resistor, because of the more voltage pump, and the low impedance at the resistor's output (1k). So at the drums moment, the 100K resistor will not be conductive enough to have no loss.

Is it right ?!



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Old 24th March 2007, 10:16 PM   #2
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resistors don't change with the signal, they have the same conductivity for small, large, slow or fast signals
Of course they have thermal properties, but this is not important here.
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Old 24th March 2007, 10:27 PM   #3
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ehheehhe I could imagine
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Old 25th March 2007, 12:19 AM   #4
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It's an interesting subject, but I doubt that the resistors are the problem, as noted above.

There are may other things going on, true output impedance (see this page) and his very clever 40Hz square wave tests of portable mp3 players. I've started to use this test on other equipment like amps and preamps.

Then there is the subject of gain structure. You mention it above, that the signal would be attenuated then re-amplified. This happens all over the place and to a very large degree. Most signal paths have far too much gain, which means attenuating the signal at various points along the path then re-amplifying, etc.
This can't be the best sounding approach in most cases, so keeping the gain structure more even seems like a good goal.

In some cases tho, you do want a nice strong signal that gets attenuated at the receiving end because it helps overcome noise.
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