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Stereo to mono is simple to apply to a signal that came from audio jack. But I've encountered problems with the Butterworth filter.

There is an example circuit what I've done, first, the two channel were summed into one, then is applied the filter. enter image description here The filter type should be an high pass, the cut should be at 106Hz 12dB/oct.

  • R1 & R2: 1kOhm
  • C1: 220nF
  • R3: 68kOhm

Only the filter: works correctly.

Only the mono: works correcty.

The problem: mono + filter is only (almost) 1/6 of the maximum volume. Why? Please correct me if I've done a mistake.

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  • \$\begingroup\$ Would you mind showing your calculation for \$f_c\$? \$\endgroup\$
    – jonk
    Jul 24 '17 at 21:39
  • \$\begingroup\$ @jonk I haven't calcuated nothing, i just picked the scheme online, please help me \$\endgroup\$
    – Northumber
    Jul 24 '17 at 21:52
  • \$\begingroup\$ Can you link to the online scheme? \$\endgroup\$
    – jonk
    Jul 24 '17 at 21:55
  • \$\begingroup\$ Here the scheme: electronics-tutorials.ws/filter/filter_6.html And here scheme plus (at the end) the value table: caraudiowiki.it/costruire-un-crossover-attivo @jonk \$\endgroup\$
    – Northumber
    Jul 24 '17 at 22:01
  • \$\begingroup\$ Are you using two pairs of capacitors and resistors for your RC filter portion? \$\endgroup\$
    – jonk
    Jul 24 '17 at 22:13
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R1 and R2 load each other's output stages. For example, if the instantaneous voltage to the left of R1 is +1.0 V and the instantaneous voltage to the left of R2 is -1.0 V, the voltage to the C-R filter is 0 V. If the left channel is at full volume and the right channel is at zero volume, the voltage to the filter is -6 dB relative to the inputs because the two resistors form a voltage divider.

Also, since there is only one reactive element (C1), the filter performance is 6 dB / octave, not 12 dB / octave.

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