Cut Off Frequency with Pot on Cmoy
Jun 16, 2014 at 2:18 PM Thread Starter Post #1 of 2

mendiola

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Could you help me to calculate the cut off frequency with the following schematic:
 
C2= 0.0000001 F (0.1uF)
R2= 22000 (22K)
fc= 72.34 Hz at -3dB
 
If I add a 10 K pot at middle position Do I have to make parallel 5K||5k = 2.5K and add it to 22K to get 24.5K ?
C2= 0.0000001 F (0.1uF)
R= 24500 (22k + 2.5k = 24.5K)
fc= 64.96 Hz at -3dB

Is correct the cutoff frequency including the pot?

 
Alfredo Mendiola Loyola
Lima, Peru
 
Jun 16, 2014 at 4:07 PM Post #2 of 2
Could you help me to calculate the cut off frequency with the following schematic:


 


C2= 0.0000001 F (0.1uF)

R2= 22000 (22K)

fc= 72.34 Hz at -3dB


 


If I add a 10 K pot at middle position Do I have to make parallel 5K||5k = 2.5K and add it to 22K to get 24.5K ?


C2= 0.0000001 F (0.1uF)

R= 24500 (22k + 2.5k = 24.5K)

fc= 64.96 Hz at -3dB




Is correct the cutoff frequency including the pot?





 


Alfredo Mendiola Loyola


Lima, Peru

 


It also applies to inverting input
Moving the wiper to the mid position and high pass filter??

Full-up, assuming low source impedance: bass cutoff is 0.1uFd against 100K or 15.92Hz.
way-down: bass cutoff is again 0.1uFd against 100K or 15.92Hz.
At "half loudness": bass cutoff is 0.1uFd against 100K+2K5 or 15.54Hz.
No difference except with careful measurement.
{http://groupdiy.com/index.php?topic=32778.0}
 

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