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How to relate dB and percentage of pot travel?

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AnalogNewb

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Hi Everyone,

I am making a front panel for a mixer and I need to know where to put the dB hash marks. I have a fader and an ohmmeter but I don't have a easily accessible audio circuit, nor do I have an accurate dB meter on hand.

How do I calculate the resistance corresponding to each 5 dB of drop, assuming that the top of the fader is zero dB drop and the bottom is infinite drop?

Can I just use this formula, normally used to calculate volts from dB and instead use it to calculate percentage of fader resistance, assuming the top of the faders is 1 volt or 100%?

V = Vref * 10 ** (dB/20)

Where Vref is 1, V represents my percentage, and dB is -5, -10, etc?

Thanks.
 

How about this:
if you assume, say, 60 dB total power range (100%) and 1 Mohm total resitance, you get a 30 dB voltage (resistance) range i.e., a 1000:1 resitance ratio.

100% 30 dB 1 M
80% 24 dB 500 k
60% 18 dB 250 k
40 % 12 dB 125 k
20 % 6 dB 62.5 k
0% 0dB 31.25 k
or you can use -dBs if you reverese the scale ..

From here you can easily calculate voltage drop with reference to your 1V input ..

IanP
:wink:
 
Assuming your pot is linear, then you can turn the formula in respect of Vref.

To power som enumber of ten, use the hat sign "^". Eg 10^2 = 100.

You can read more about decibel here.
 
I should have mentioned that the pot is set up as a voltage divider. max resistance has the pot at ground, thus infinite dB reduction.

Also, it's an audio taper pot.

I mapped out my fader using the formula I posted and the hash marks "look" right so I'm hoping for the best...
 

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