Do I really need a preamp?


I am considering another preamp to go with my CJ 2200. Am thinking about a CJ pre to go with it. However since my main source for listening is a cd player which has variable output, why do I need a good preamp. I have an Adcom GTP 450 tuner/preamp which will do the job for radio, tapes, etc. That said, how can I run cd and pre into the amp? As always, your help is appreciated.
artemus_5

Showing 2 responses by tom_nice

I second the opinion of Themusicshop, and have looked at the Welborne Labs kit (or assembled) unit he recommends. If you can do a kit, I'd ask Welborne how big the steps are, and negociate about getting different resistors for the kit, so as to make them no larger than 1 db, if they are larger than 1 db. (Typically they're about 2 db in the middle of the range.) If you arrange to have the control go down from 0 db in 1 db steps, you still have 23 db attenuation, which should be plenty for a CD player and halfway typical amp and speakers. You could get info from your manufacturers on how much attenuation you'd be likely to need. Reviewers almost always hate the variable output connection, and I presume with good reason. But if the output stage itself is decent, I disagree that an active preamp is needed for anything.
Welborne Labs informs me that their steps are 2 db, which I think are unfortunately large. Every commercial stepped attenuator I know of uses steps that large, presumably to be able to fit in with a wide range of power amps and sources. But each system needs only a small part of the huge range of attenuation allowed--literally from zero to infinite--and I would strongly recommend (1) getting a kit, so you can choose your own step size, which I'd make 1 db, (2) find out how much attenuation your system needs by rigging up a crude fixed attenuation network with female RCA connectors and resistors, no chassis needed, and (3) get your own resistors (which could be interspersed with some of the ones supplied with the kit). Don't know how to work out their values? The basic equation: attenuation = 20 log Rs/(Rs + Rg), where Rs is the series resistor and Rg is the resistor to ground.