Are attenuators linear or logarithmic?

If you have an attenuator that scales in .5 db increments from -72db to -0db is it linear or logarithic?

Is -36db half the attenuation as -0db? At -36db on the scale are you killing half the power available? On a single frequency test tone would -36db on the scale be half the SPL of -0db?

Can't be that simple.

Jim S.
Attenuators are logarithmic because our hearing response to changes is SPL is logarithmic. Each step resistance is calculated to result in the voltage ratios being doubled for every 6 dB's. Even though each discrete step is linear, the combined steps when plotted resemble a logarithmic curve.

You are not affecting the power output of the amplifier with the preamp. The amplifier will still gain the signal from the preamp whether it's a -0.5 dB or - 72 dB signal. But the preamp signal from full to half attenuator will output about 2 percent of the max output signal voltage. This is assuming that it's trimmed in a linear way from zero to maximum, that is, each step resistor is doubled from full to (near) zero in 6 dB steps. Sometimes the attenuator will be trimmed to be less sensitive in the higher volume region because of the progressively small resistance needed.

What the power affect on the amplifier will be depends on the amplifer gain.

Thanks. I was just trying to estimate how much power the amps were pushing (or maybe better yet, the speakers were pulling) at certain levels of attenuation.

I recently went active with the speakers and was trying to guess at the change in speaker efficiency after removing the passive crossovers.

Kind of a backward way to go about it, I'm sure, but I am seriously electronics basics challenged.

Thanks again.
Jim S.
While we are on the subject - what is the relationship of potentiometer resistance to volume output?