DIY XLR Attenuation (In-Built into XLR Cable)

Hi all,

Would appreciate all the experts help here to verify if my understanding is correct before I start building. Many thanks in advance.

As per mentioned (don't know if this had been discussed before, if so pardon me) - would like to DIY a pair of XLR inter-connect between the pre-amp to power amp with in-built attenuation (~12dB) at the power amp plug side.

I reference the article/formula at this site: http://wiki.slimdevices.com/index.ph...ve_attenuation.

The connection circuit I think should be something like this (in which I think both R1 & R2 values used have to be halve w.r.t. if it was unbalance), right ?


The formulas from the site are:-
Attenuation [dB] = 20 * Log( R1 / ( 1 / ( 1/R2 + 1/IMPload )))
New Source Impedance = 1 / ( 1 / (IMPsource + R1) + 1/R2 )
New Load Impedance = R1 + 1 / ( 1/R2 + 1/IMPload)

The values I gathered from my pre and power amp are:-
Power amp input impedance = IMPload = 104K-Ohms
Pre-amp output impedance = IMPsource = 500 Ohms

Using the following values:-
R1 = 82.4 K-Ohms - Or for Balance Line = 0.5*R1 = 41.2K-Ohms
R2 = 27.4 K-Ohms - Or for Balance Line = 0.5*R2 = 13.7 K-Ohms

The resulted values are:-
Overall attenuation ~ 11.6 dB
New Source Impedance ~ 20.6 K-Ohms
New Load Impedance ~ 104.1 K-Ohms

Are all my above assumptions + calculation/result correct ?

Regards, EH.

Attached Images