Gain pot -20dB

[Buffalo Tom]

Full clockwise 0dB unity gain. Full counterclockwise -20dB attenuation.
How do you do this? It's for the inputs on an input selector.

Thanks

[Dormammu]

Full clockwise 0dB unity gain. Full counterclockwise -20dB attenuation.
How do you do this? It's for the inputs on an input selector.

The manufacturer wanted to show off, so he marked it like that.
Without a point of reference, it is impossible to represent this claiming to be true.

[stallik]

Is it just a pot wired backwards sending (some) of the signal to ground?

[PRR]

It is a plain linear pot, normal potentiometer connection, except with a resistor added in the ground lead.  Say 100k pot and 10k resistor. (Ideally 100k pot and 11k resistor, but pots tend to run small, and this isn't laboratory work.)

If you work-out the mid-point: simple lin pot is -6dB at center, but pot standing on a 1/10 resistor is a part-dB higher, so -5dB.

Are you sure this is the taper you want? If input levels are not consistent and matched, it is "slow" at top and "abrupt" around -15.

[ElectricDruid]

+1 what PRR said. It's a linear pot with a resistor beneath it so it doesn't go all the way to zero  (or "-100dB" as it would be here). The strange scaling is what happens when you convert from linear to dB.

[Buffalo Tom]

It is a plain linear pot, normal potentiometer connection, except with a resistor added in the ground lead.  Say 100k pot and 10k resistor. (Ideally 100k pot and 11k resistor, but pots tend to run small, and this isn't laboratory work.)

If you work-out the mid-point: simple lin pot is -6dB at center, but pot standing on a 1/10 resistor is a part-dB higher, so -5dB.

Are you sure this is the taper you want? If input levels are not consistent and matched, it is "slow" at top and "abrupt" around -15.

Thanks PRR. I think the taper will be fine... Will build this and try. 

[MrStab]

Where Pt is the "top" half of the pot, Pb is bottom half, and Rf is your fixed resistor:

20log(1*(Pb+Rf)/(Pt+Pb+Rf))

e.g

20log(1*(5+3.3)/(5+5+3.3)) = -4.1dB if a 3k3 resistor is attached to a 10k linear pot which is set to midpoint.

Fully expect to have my maths brutally shot down. lol

Pot manufacturer datasheets will show you resistances at different points of travel for different tapers ("15A" is most common curve of log pot).

[Buffalo Tom]

Where Pt is the "top" half of the pot, Pb is bottom half, and Rf is your fixed resistor:

20log(1*(Pb+Rf)/(Pt+Pb+Rf))

e.g

20log(1*(5+3.3)/(5+5+3.3)) = -4.1dB if a 3k3 resistor is attached to a 10k linear pot which is set to midpoint.

Fully expect to have my maths brutally shot down. lol

Pot manufacturer datasheets will show you resistances at different points of travel for different tapers ("15A" is most common curve of log pot).

I believe your math is correct.. -5.3 dB with 100k pot and 10k resistor.