Interconnect Inductance vs. Capacitance

No, as far as I am aware no one sells interconnects with capacitance high enough to "grossly" affect high end frequency response. But under extreme circumstances (high component output impedance, long cable length, high cable capacitance per unit length), it could become marginally significant. So in that sense it is potentially a system-level issue, that the consumer should be aware of.

For less than 6 foot IC's no worries but once you get to 12 feet or more then you would be surprised at how poorly some RCA line level gear will perform. I don't think the high end roll off is the most noticeable - to me there often seems to be a loss of dynamics when using very long runs - I'd avoid long RCA runs and go with XLR balanced if that is your need.

I agree with Vett's calculations, and the 12.5kHz answer. When I said that interconnect capacitance could become "marginally significant" under extreme circumstances, I was thinking of source components with active output stages. For passive preamps, or preamps with unbuffered resistive attenuators at their output, the effect can obviously be more than "marginal."

Audioquest4life, not sure where you are going wrong with your math, but for 100 ohm output impedance and 196pf capacitance, the 3db bandwidth would be:

1/(2*3.14*(100)*(196exp-12)) = 8,124,269 Hz (i.e., 8.1 MHz)

For Vett's example, it would be:

1/(2*3.14*(50000)*(255exp-12)) = 12,489 Hz

Although of course the 50K assumption is something of an oversimplification, and in practice I think the answer might not be quite that bad. The 50K output impedance assumes the control is set for 6db attenuation, and is the total impedance looking back into the output. But, first, I would think the control typically would be set for greater than 6db attenuation. Let's call it 12db, which would mean 25K between the output terminal and ground, and 75K between the output terminal and the preamp's internal voltage source which drives the attenuator. The high frequency rolloff would be determined, in this example, by the voltage divider ratio formed by the parallel combination of the 255 pf and the 25K, and the 75K. I'm not going to bother trying to figure that out, but my suspicion is that the net result would be a somewhat wider bandwidth than what would be provided by the 50K assumption. In any event, the 50K assumption does seem like a reasonable rough ballpark, which makes the point that the treble rolloff can be significant.

Regards,
-- Al

Shadorne -- Thanks for your observations. I just want to make sure it's clear to everyone that my previous post was written before I saw your last post, and was in response to the prior posts, not to your good observations.

Regards,
-- Al

Thanks Al for the comments. Let me further clarify two points.

If you think 6dB attenuation is not enough, a higher level of attenuation will actually lower the 3dB freq. A higher level of attenuation means that you will have a larger R value in series with the output. So at a lower sound volume, the highs will be rolled off even more and yields a narrower bandwidth!

The 2nd point is that my power amp needs 4V RMS to achieve full power, 100W/8ohm. The preamp has 12dB of gain. Most CD players have max outputs between 1V-2V. So the 6dB setting assumption is quite appropriate for my setup. I usually set the volume control knob anywhere between 10 to 3 o'clock, and I don't listen very loud....

Mathematics R us! lol...

Mathematics R us! lol...

LOL here too! As someone who also has multiple EE degrees, albeit one fewer than you do, this thread is definitely fun!

Agreed on the 6db, given your power amp's relatively low sensitivity. But I also chose 12db for my example in order to simplify my other comments (the references to 25K and 75K), which would have been harder to present if the impedances looking into the preamp output would have been 50K in both directions (to ground and to the signal source).

If you think 6dB attenuation is not enough, a higher level of attenuation will actually lower the 3dB freq. A higher level of attenuation means that you will have a larger R value in series with the output. So at a lower sound volume, the highs will be rolled off even more and yields a narrower bandwidth!

This would be true if the attenuator were simply a variable resistor in series with the output.

But I've been assuming (correct me if I'm wrong) that the end terminals of the attenuator are connected, respectively, to some signal source within the preamp (which itself is assumed to have negligible output impedance), and the preamp's ground. And that the preamp output is the wiper of the attenuator, with the output being referenced to preamp ground.

Given that, and using my 12 db example, the presence of the 25K in parallel with the cable capacitance makes for a very different situation than simply having some fraction of the 100K in series with the output. Without the capacitance, you get 12db at all frequencies. With the capacitance, you get a frequency-dependent voltage divider ratio equal to the combined impedance of the parallel combination of the 25K and 255pf (combined vectorially), divided by that figure plus 75K.

I'm not sure without doing some further analysis if that would result in greater bandwidth or less bandwidth than at a 6db attenuation setting (a 50K/50K setting on the attenuator, instead of 75K/25K). Note that in both cases, the capacitance is not being charged toward the source voltage. It is being charged toward some lower voltage, through an overall impedance which is not simply the resistance between the output terminal and the "top" end of the attenuator. In the 50K/50K case, the overall output impedance is 25K. In the 75K/25K case, the overall output impedance is only 18.75K. But of course the 25K is to ground, while the 75K is to the voltage source.

To use a wonderful expression I read in a completely different context a while back, my mind is becoming a bit too "pretzeled" by all of this to readily see the answer :)

Regards,
-- Al