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


I have an Audible Illusions preamp which is designed for a 1.5v input signal. I am told that this is standard. My Ayre CX-7e puts out 2.25v. They play beautifully together but I have too little control over volume: I cannot use the preamp vol controls beyond 9 o'clock because it is way too loud. Likewise it is hard to get just the right volume in the limited range available. This is worse with some CDs for reasons I do not know.
Audible Illusions will change the preamp attenuation board for $275, but before I do so I wonder how others have dealt with the problem of newer CD players with higher outputs than their preamps were designed for.
gmargo
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.
Hifi57audio, I was wondering about exactly the same point concerning imput impedances, when inserting external shunt type attenuators:

Why do the EVS or Endler attenuators typically use a much lower impedance attenuator?

For example from the Endler website:

These are 4k or 20k shunt type attenuators which place only one resistor in series with the signal. Series resitance values of 1k, 2k, 4k and 20k were evaluated with 4k sounding the best. I use a proprietary technique on the switched side of the circuit to eliminate the possibility of an open circuit in case the contacts bounce. This helps keep things ultra quiet even in the presence of dc.

And from the EVS site:

Attenuators are available in a low impedance (1K) and a high impedance (10K) version.

Hifi57audio, any clarification on these low impedances for external stepped attenuator would great!

Thanks,

Rene
I ordered the Endler attenuators in the 4k 24 step version. I am going to use them on the amp with my preamp in the chain feeding them. I did read in the PF review that used alone with a source feeding it, while the attenuators were very musical and transparent, there was a tradeoff in reduced soundtage and imaging (more 2 dimensional sound is what the end result was). Has anyone experienced this? Might this be different using the preamp in the chain?

My preamp is a Joule Electra LA-100 MkIII, with input impedance of 20k ohms and output impedance of 300 ohms. Currently it is set at 11db of gain.