How important is the pre-amp?

Georgie in product protection mode.

No, that some else here, all that info and maths above is correct

And just to let you know, your out of touch, and need to do your homework.
I don’t own Lightspeed Attenuator, for at least 6mts now.

Cheers George

It was such a seismic shock to the audio world that no one noticed.

This is false.

To explain to members with the maths of it.
As a 10kohm passive has an outuput impedance of around 2.5kohm, the "cheapest interconnect cables" can be <100pf capacitace per ft and less. It would if be use you 2mts!! of it, still be only 300pf!!!
And this 300pf capacitance with the passives 2.5kohm output impedance, will still only be only -3db down at 212khz!!!! in hf rolloff, a dog would have trouble hearing this.

A 10K passive may well have an output impedance of nearly 10K, depending on the position of the volume control. This is in series with the source impedance, which often includes an output coupling capacitor. When you put a capacitor and a resistance in series, you create a simple filter. Filters can and do color the sound. This is easily heard. The most common complaint leveled at passive controls is a loss of bass and overall impact when the volume is anything less than full. This complaint has been mentioned by others earlier on this thread.


Its simple physics. It is possible for a thing to be too simple, and a volume control by itself is an example.


Another example of the sort of problem passives can cause is actually laid out in George’s example above. Let’s go with that 10K value. What that means is that the source (perhaps a CD player or DAC) is driving that 10K control. Many sources simply aren’t going to be happy with a load like that and will have higher distortion and often a loss of bass (since the 10K load interacts with the output coupling cap, creating a low frequency roll off). The formula for the cutoff frequency thus generated is:
f = 1,000,000/R x C x 2Pi where f is frequency in Hertz
R is resistance in Ohms (in this case 10K)
C is capacitance in microFarads
2Pi is 6.28.
(Normally this equation is seen with a ’1’ instead of 1,000,000 but algebra allows us to do this since normally C is would otherwise be Farads rather than microFarads. Farads are not a useful value in audio applications.)
What this formula shows is that as R goes down, C must be increased. You want the cutoff frequency to be less than 2Hz so that phase shift is not introduced at 20Hz. This means that the minimum value for C is going to be about 6.8uf. Its a simple fact that due to inductance created when a capacitor is rolled that a cap that big will introduce coloration simply on account of the inductance. So many manufacturers choose to avoid values like that on that account.


George will (and has in the past) point out that a direct coupled output from the source will not have this frequency pole. But a load like that will affect its distortion- and limit the kinds of sources with which such a control will be successful! The obvious thing to do is increase the value of the control; by doing so you can run into Miller Effect problems with the input stage of the amplifier being driven. Miller Effect is caused by the input capacitance of the amplifier and is governed by the same formula, only now were talking about a high frequency effect. To avoid this problem, usually the control is buffered by some sort of active circuit.
The simple fact is that passives will only work right in limited situations. This is why you see so much conversation about this, because different people have different systems. So we are often talking about apples to oranges when it comes to passive controls.

Georgie in product protection mode.

As I said no, but Ralph most definitely is.