Calculating Frequency Based on Pot Resistance

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lowatt

New member
Joined
Mar 15, 2022
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2
Location
Switzerland
Hi,
I have a question about how to calculate frequency in relation to the potentiometer resistance. The Calrec PQ11549 EQ has a dual-gang reverse log pot in each band circuit. What formula would be used to derive the frequency at a particular resistance setting? Let's take for example the High-Mid. I guess something that involves a pi*R*C?
Thanks in advance :)
PQ1549-HighMid.pngHigMidFrq.png
 
1 / (2 x Pi x R x C)

Will give you the cutoff (-3dB point) in Hz where R is the *overall* resistance (in ohms) seen on the output side of the capacitor - so not just the value of the pot setting, but the overall circuit value with the pot set at that point, and where C is the value of the capacitor in uF (or combined value if there are multiple capacitors).

It gets more complex if there are inductors as well as capacitors, and/or if some capacitors or inductors are strapped to ground while others are in the signal path. None of that seems to be going on here.
 
I guess that pesky 'u' in there was a case of fingers being faster than brain. Wouldn't be the first time! : o

Meanwhile, it's possibly worth adding that there are various free online 'calculators' that might be worth bookmarking.

A couple of decent ones are at:

https://www.digikey.com.au/en/resou...sion-calculator-low-pass-and-high-pass-filter

and

https://www.omnicalculator.com/physics/cutoff-frequency

both of which will (separately) calculate knee freq values for either inductance or capacitance.

Be aware, however, that there are others that appear to give erroneous results (order of magnitude issues - heh!). Case in point: http://www.learningaboutelectronics.com/Articles/Low-pass-filter-calculator.php#answer1 ... FWIW.
 
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