Interconnect Inductance vs. Capacitance
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 |
The interactions between human and machine are always fascinating to me. A lot of them are beyond textbooks. Shadorne's example of dynamic and lively sound is a good one. I am not sure what electric characteristics can be used to describe it. I also think that a system is only as good as its weakest point. In my example, if I get a pair of ICs that have higher capacitance, it can never sound good in my system. This is, of course, assuming that I still have good hearing.... lol... A fellow Dude preamp owner likes the top line model from Audio Horizons the best. AH's website states the higher end models have higher capacitance. This seems odd to me. He is testing Blue Jeans Cables' ICs now. It will be interesting to see what he thinks.... |
Vett93 -- It's been too long since I studied Thevenin's Theorem. It looks like I was right, and bandwidth will be greater when the attenuator is set for 12db attenuation, compared to when it is set for 6db attenuation. The Thevenin equivalent circuit for the preamp output with the attenuator set for 6db, at the mid-point of its resistance range (what I've referred to as "50K/50K"), is a voltage source equal to one-half of the voltage being applied to the attenuator, in series with 25K. The Thevenin equivalent circuit for the preamp output with the attenuator set for 12 db attenuation (what I've referred to as "75K/25K"), is a voltage source equal to one-quarter of the voltage being applied to the attenuator, in series with 18.75K. Therefore the higher attenuation setting will result in a lower source impedance, resulting in a smaller RC time constant and a wider bandwidth. -- Al |
