michiganpat
Super Member
Hey everyone, I've got all the pieces for a custom speaker project based on some $10 AR2ax cabinets I picked up. one question I have concerns the speaker level pots, which I have thoroughly cleaned.
the two potentiometers that are used for mid and high range attenuation are a 0-16 ohm pot, with 2 legs and a common ground, so one leg is X ohms, the other 16-X. measuring it all the way one way or the other, it's acutally ~.3-15.7 ohm according to my multimeter. The way they are originally wired in, one leg is in series, the other in parallel to the driver. in it's "high" position, with the parallel leg at 15.7 ohm, the series leg at .3, it should provide ~0.7db of attenuation. turning all the way down essentially will turn off the driver as the parallel leg goes to 0 if I understand it correctly.
with an L pad circuit, which is the main "influence" of performance, the series leg or the parallel leg? playing with an online L-pad calculator, it looks like as the series resistor approaches the driver's impedence, attenuation approaches infinity....consequently, as the parallel leg approaches 0, the attenuation approaches infinity. what if you have what I have, a nominal 4 ohm impedence driver, where say, the series leg is 4 ohms, but the parallel leg is ~11.5 ohm? will it have infinite attenuation, or ~2.6db (depending on which value I look at)?
the two potentiometers that are used for mid and high range attenuation are a 0-16 ohm pot, with 2 legs and a common ground, so one leg is X ohms, the other 16-X. measuring it all the way one way or the other, it's acutally ~.3-15.7 ohm according to my multimeter. The way they are originally wired in, one leg is in series, the other in parallel to the driver. in it's "high" position, with the parallel leg at 15.7 ohm, the series leg at .3, it should provide ~0.7db of attenuation. turning all the way down essentially will turn off the driver as the parallel leg goes to 0 if I understand it correctly.
with an L pad circuit, which is the main "influence" of performance, the series leg or the parallel leg? playing with an online L-pad calculator, it looks like as the series resistor approaches the driver's impedence, attenuation approaches infinity....consequently, as the parallel leg approaches 0, the attenuation approaches infinity. what if you have what I have, a nominal 4 ohm impedence driver, where say, the series leg is 4 ohms, but the parallel leg is ~11.5 ohm? will it have infinite attenuation, or ~2.6db (depending on which value I look at)?