Baxandall tone control circuit

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Joined 2008
Hi all. I look for a Baxandall tone control circuit, which can control bass and treable on 20dB or more. My english is not good I know sorry

Hi

Your English is good enough::)
I hope this helps.

Max bass boost = (R1+R2) / R1

Max bass cut = R1 / (R1+R2)

Max treble boost = (R5+R1+2*R3) / R5

Max treble cut = R5 / (R5+R1+2*R3)

R4 shall be larger or equal to 10*(R5+R1+2*R3)

R2 shall be much larger than R1


In the example circuit:

Max bass boost = 20*log ((11k+100k)/11k) = 20dB

Max bass cut = 20*log (11k / (11k+100k)) = - 20dB

Max treble boost = 20*log ((3.6k+11k+2*11k) / 3.6k) = 20dB

Max treble cut = 20*log (3.6k / (3.6k+11k+2*11k)) = -20dB
 

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Disabled Account
Joined 2008
It look easier than it seem. Is possible use transistor on the way out, because operations amplifier and another integrate circuit is not allow for my poject.

Hi mckvak

Yes it is possible to use transistors instead of an opamp.

Maybe you should post the circuit you have made so far, that you would like to use the tone control with. If you have a gain stage maybe you should try to implement the tone control around that stage, or you could use a passive tone control.

stinius
 
There are diagrams where instead of R5 is a capacitor let's say about 1.8 or 3.3nF in series with the treble potentiometer and even if instead of C2 in some diagrams appears a fix rezistor. How can you calculate in this case? Another diagrams omit even R5 and the treble potentiometer (R4) is connected directly in paralel with R1-R2-R1. Which is the formula for this case? Thank you!
 
Ex-Moderator R.I.P.
Joined 2005
a while ago i played a bit with passive tone control simulator on Duncan site

it can cause really weird curves and bad effects if wrong values are chosen
I would be very careful with those things

are active controls equally tricky ?

btw, someone experienced with this stuff mentioned that passive controls cause much less bad side effects than active feedback controls

adding gain I believe is a different game, and does not make it active as such
 
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