The attenuation is needed to get a digital equalizer working at peak performance between a preamp and the monoblocks.

Many thanks.

If your attenuators have affected the tonal balance, then perhaps the input impedance is reactive rather than resistive. Sometimes a feedback loop has to be considered too. I can provide you a simple diagram for a 12dB resistive attenuator based on a specific input resistance but if the input is reactive or complex, then it will affect the tonal balance, as you've described.

Can you provide more info, a schematic perhaps?

Wayne

Resistive and reactive? Don't know ... guess it is time to pull out an electronics textbook.

The current attenuators (20 db) do the same thing to every system they have been installed into (1 tubed, 2 solid state). They attenuate the signal but the bass and dynamics of the system get squashed.

There are no schematics of the amps. Is that what would help? The digital eq has an output impedance of 120 ohms.

Try this. Try a couple of attenuators, with different impedance levels. Try hooking them up as a series resistor followed by a shunt resistor. Pick one of the two decibel levels and try both attenuators shown. Let me know what you hear.

For 12dB:
36k series, 12k shunt
- or -
150k series, 47k shunt

For 20dB:
43k series, 4.7k shunt
- or -
430k series, 47k shunt

Many thanks!

The input impedance of the amps measure about 100k. So that would put the total input impedance of 100K + 47K = 147k?

I've built 4 attenuators of various values so far using cheap Radio Shsck resistors. It seemed a reasonable assumption that cheap resistors would sound better in the circuit than cheap pots.

Yet every one of the attenuators (on tube and solid state amps) seems to have an overly strong attenuation in the treble. Could the cheap parts be the issue or is it more likely that I still haven't gotten the impedance issues ironed out?

If your amplifier's input impedance is 100k, then putting 47k in shunt will reduce it to 32k. The formula for parallel resistors is:

RT = 1 / (1 /R1 + R2...)

For two resistors, you can also calculate with this formula:

R1 · R2 / R1 + R2

At any rate, the total amount would be 32k. That's probably OK, but I might try values that shifted it less. Maybe use a divider that has a shunt value equal to that of your input impedance.

Since the shunt resistor is paralelled with the input impedance, that value becomes lower than you might think. So in this case, if you went with a 100k ohm shunt resistor, you'd effectively have 50k input impedance. The series resistance you'd want for 12dB attenuation would be about 150k. So instead of having resistor values of 150k series and 47k shunt, you'd use 150k series and 100k shunt. That would make the circuit function as 150k series value and 50k shunt, providing 12dB attenuation. From the resistor values, it would aoppear to be an 8dB attenuator. But because of input impedance, the attenuation value would be 12dB.

!!

The latest and greatest attenuator circuit was built last night. Early listening in the basement system sounded ok with a tad of rolloff of the high frequencies. It is in the basement system cooking for the day and we'll see what it can do tonight in the main system.

Thanks for all the help!