Attenuation

[nad]

Hello,

I've tried attenuating my cd input using a 10k restistor as a jumper and a 6.8k to ground, it attenuated the input but not enough.

Anyone know what values will double the above attenuation?


Thanks.

[PeteMcK]

L pad calculator - attenuation dB damping impedance decibel loudspeaker - sengpielaudio Sengpiel Berlin

[djoffe]

If we assume that the load driven by the attenuator is kind of high, then your current attenuator (10K series, 6.8K shunt) has 7.86 dB of attenuation. To get "twice" as much attenuation, psycho-acoustically, you add another 10 dB of attenuation. Your can do that by keeping the 10K series, and changing the shunt resistor from 6.8K to 1.5K. That attenuator, 10K series, 1.5 K shunt, gives 17.7 dB of attenuation, or about 10 dB more attenuation. This will sound about 1/2 as loud.

[nad]

Originally Posted by djoffe

If we assume that the load driven by the attenuator is kind of high, then your current attenuator (10K series, 6.8K shunt) has 7.86 dB of attenuation. To get "twice" as much attenuation, psycho-acoustically, you add another 10 dB of attenuation. Your can do that by keeping the 10K series, and changing the shunt resistor from 6.8K to 1.5K. That attenuator, 10K series, 1.5 K shunt, gives 17.7 dB of attenuation, or about 10 dB more attenuation. This will sound about 1/2 as loud.

Fantastic reply, much appreciated.

[jrockhead]

Originally Posted by nad

Fantastic reply, much appreciated.

In order for your CD player to sound its best you need to keep the players output impedance in mind. Specifically if its 47,000 ohms the attenuator should also have a 47,000 ohm impedance. With a resistor setup as you are playing with my guess its loading down the op amps and causing problems with distortion and excessive output current.

My elderly Sony CD player sounds it's best feeding into an attenuator that has a 100,000 ohm impedance. Even though it has LM4562's as the op amps and on paper they can drive a 600 ohm impedance.

Look on line for attenuator calculators where you input impedance along with attenuation, lots of them out there.

[cbdb]

Originally Posted by djoffe

If we assume that the load driven by the attenuator is kind of high, then your current attenuator (10K series, 6.8K shunt) has 7.86 dB of attenuation. To get "twice" as much attenuation, psycho-acoustically, you add another 10 dB of attenuation. Your can do that by keeping the 10K series, and changing the shunt resistor from 6.8K to 1.5K. That attenuator, 10K series, 1.5 K shunt, gives 17.7 dB of attenuation, or about 10 dB more attenuation. This will sound about 1/2 as loud.

As jrockhead said this might be to much load for the source, multiply the above resistors by around 5 to get you in the ballpark (or try a pot for the shunt to figure out the exact value and then replace it with a resistor).