Probably the choices of the impedance values involved.
Amount of attenuation is pretty much straight calculation via resistor-divider rule
Or in English, if you put a 1000ohm in-line with a 10ohm load it will attenuate by around 40dB (slightly more, can't be bothered to calculate 10/1010). But if you put 1000ohm in-line with a 50k load it isn't going to do jack. (0.17dB)
Amount of attenuation is pretty much straight calculation via resistor-divider rule
Or in English, if you put a 1000ohm in-line with a 10ohm load it will attenuate by around 40dB (slightly more, can't be bothered to calculate 10/1010). But if you put 1000ohm in-line with a 50k load it isn't going to do jack. (0.17dB)
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