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RE: How to calculate db variation on potentiometer

Looking at one channel, the potentiometer is a three terminal device, with a signal input, a signal output and a ground. There are, in effect, two variable resistors, R1 and R2, connected in series from the input to ground and the output taken from the connection between R1 and R2. The attenuation (voltage) is given by R2/(R1+R2).

When you connect a load, RL, the load appears in parallel to R2. The effective resistance of R2 is replaced by R3, as given by the formula
1/R3 = 1/R2 + 1/RL, or equivalently R3 = (R2 * RL) / (R2 + RL)
Then the attenuation (voltage) is given by R3/(R1+R3)

There is one more factor to consider. The load placed on the source will be R1 + R3. If the source impedance is R0, then the voltage input to the potentiometer will be reduced in the ratio (R1 + R3) / (R0 + R1 + R3).

The overall effect of these two steps is that the original output voltage will be reduced by the product of the two ratios, i.e.
[ (R1 + R3) / (R0 + R1 + R3) ] * [ R3 / (R1 + R3) ]

Take 20 log(base 10) of this value to get the dB.

You will need to know the source resistance, R0, and the load resistance, RL. Then you can measure the values of R1 and R2 at different settings of the potentiometer, getting the values for both channels.

Tony Lauck

"Diversity is the law of nature; no two entities in this universe are uniform." - P.R. Sarkar


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