18 dB self-made attenuator pad for line-level signals
How to calculate the resistors
Calculating resistors for other attenuations is really not difficult. Firstly, you need to determine the attenuation factor (F) from the dampening value (D). These two formulas may come in handy:
D = 20 x log F
F = 10 ^ (D/20)
In the example above, the attenuation factor is the impedance on the input side (2600 + 680 + 2600 = 5880) over the impedance on the output side (680): F = 5880 / 680 = 8.65. This leads to a dampening of D = 20 x log 8.65 = 18.7 dB.
Now the other way around:
Say you want to create a pad with 9 dB attenuation. The attenuation factor is F = 10 ^ (9/20) = 2.81
If we keep the output resistance at 680 ohms (it could be lower - a couple of hundred ohms will do) then the input impedance should be 680 x 2.81 = 1910 Ohms. So the two resistances on top and bottom of the circuit need to be: (1910-680) / 2 = 615 Ohms. You may not be able to find exactly these values but a close approximation will not generate much error.
For example, say you have two 600 ohm resistors and combine them with the 680 ohm resistor in the middle, the dampening becomes:
F = (600 + 680 + 600) / 680 = 2.76
D = 20 x log 2.76 = 8.8 dB
The result is not really far apart from the intended 9 dB and should be fine.