CD output voltage too high for preamp: what to do?

Hi Chuck. I use the attenuators plugged into the amp inputs and on about 8db attenuation.That got rid of the gain noise I was experiencig between the pre and power. The Endlers are truely transparent, no real difference between the attenuators attached and not attached, except for the gain.
I was sceptical as well, but decided to try them and I am very pleased., The Rothwell attenuators are quite coloured and change the sound of the amp too much.

BTW, I don't use them anymore as I sold my VTL 7.5 and using a cj pre amp. The cj has the same amount of gain as the VTL, however there is no noise.

Thanks to everyone for their suggestions. I did set the Audible Illusions dip switches to minimum gain without enough of a change in volume. I will look into attenuators both at the pre inputs and inside the AI directly on the CD RCA plugs. Hifi57audio, I'd appreciate it if you'd let me know the formula to calculate the resistor values. I contacted Ayre: they won't adjust the CDP.
Illenema2 thought that the AI pre doesn't do well with attenuators at all. Does anyone have more thoughts on a change in preamp, although that solution is the most expensive!

Just for information, the CD inputs on the AI are not the most transparent. Try the aux. inputs. They're resistors on the inputs of the cd that are not on the other inputs. Art put them there to reduce glare on "Cheaper" CD players. Also, the output #2 has resistors also that, to me, degrade the sound. Always use output #1 for best sound. He did away with these in the L-2. He may have also done that with newer M3a's but I don't know that is the case. It did say in the owners manual.
I would locate the resistors on the main circuit board (I think they are near the dip switch) instead of the input resistors and replace them with the appropriate value. You do not want more in the signal path than absolutely required. Since these resistors are allready there, these would be the ones I personally would adjust.
Call Art Ferris at AI, he's a good guy and would probably help you or at least tell you if it's a bad idea.
The outboard controls would probably be the way to go honestly unless you returned it to him for adjustment.

The equation for voltage dividers is: (shunt/(series+shunt))=voltage reduction factor. VRF x Vin=Vout... (Vin/Vout)log20= db attenuation Put simply in your case, dropping from 2.25 to 1.5 is a voltage reduction of exactly 1/3. So, Series R would equal 1/3 of R total, and Shunt R would equal 2/3 of R total. R total =Series R + Shunt R. As concerns attenuation, R total may be any value you want. Obviously, if there are in fact resistors on the circuit board allowing further attenuation or gain reduction, use those. Hope this helps.

Almost forgot, as concerns input impedance; R total should be selected as to maintain yours. For instance if it's 50k and you plug a 50k attenuator into the inputs, your CD player will see a 37k load.( when set for 1/3 attenuation) and will vary over the range of the attenuator. And yes, if your source impedance is high enough, the upper frequences will be rolled off. Finding Rtotal is way too easy, it equals 150% of input impedance.(are you seeing that 1/3 relationship again?). Input impedance at audio frequences is almost purely resistive, so this can be measured with an ohm-meter. Alternatively, you could simply add a series resistor,to the CD input, of a value of 50% of input resistance and achieve the same voltage reduction. This of course would raise the input resistance by a third. All the above advice assumes a standard circuit design pratice used in almost all high end products. Mock it up first, with a set of male and female RCAs and resistors from Rat Shack, plug it in and listen, before taking the hammer and tongs to your preamp. O.K., now I'm done.