What are the formulas to work out the insertion loss and then the return loss? I would like to work out what the cross talk amount would be for NYD's schematic that I posted earlier.

Referring to the schematic you posted earlier, the bus loss is given by Rshunt / (Rshunt +20K)

Rshunt is typically 234 ohms which gives a bus loss of 38.74dB.

Return loss is harder to workout but usually the worst case is with the oan centred. Assuming VR2 is 20K then we need to work out the voltage at the junction of R1 and R2.

From slider to ground is 10K so the voltage on the slider is roughly the bus voltage times 10K / (20K + R2) or ~ 10dB

down

From slider to the top is 1) in parallel with R2 i.e 5K. This forms the top arm of another divider. The bottom arm of this is:

1. R1 in series with whatever the pot VR1 resistance is. assuming the source has a very low impedance then the worst case is with VR1 at 5K + 5K which is 2K5 . So the total is 4K7 +2K5 = 7K2

2. The impedance to ground of the bottom pan pot and R3 and R5. The bottom half of this pot is 10K and the top half is 10K in parallel with R3 i.e 5K so the total is 15K. This is in parallel with R5 (20K) which makes the combination about 15 *20 / (15 + 20) which is 8.6K.

1. in parallel with 2. is 7.2 * 8.6 / (7.2 * 8.6) = 3.9K

So the second divider is 3.9 / (3.9 + 5) = 7dB.

So the total return loss is 10dB + 7 dB = 17dB.

Bus loss is 38.74 so total crosstalk is 38.74 + 17 = 55.74 dB.

Not up to real pro standards but definitely good enough for normal stereo.

The reason the return loss is relatively poor is the use of R1. Without it the crosstalk would drop another 4dB to nearly 60dB.

These figures are not 100% accurate because the network is quite complex and I have made some assumptions in order to simplify it. The best way to get the exact answer is with a simple simulation.

Cheers

ian