# Equalizer filter formulae

#### Jony130

Joined Feb 17, 2009
5,092
Here you have

R2=R7

pot is 100k = R1

R12=R13

$$\omega_o = \sqrt{ \frac{2*R2 + R1}{ R1*R2*C3*C4*(R2+R12)}}$$

So

Fo = ωo / 2*∏

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#### Jony130

Joined Feb 17, 2009
5,092
And if you chose
C3 = 10*C4 ; R12 = 10*R1

Then

$$\omega _o = \frac{1}{10*R1*C4} * \sqrt{2 + \frac{R1}{R2}}$$

$$Q = \sqrt{ \frac{2*R2+R1}{9.6*R2}}$$

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#### MarkusBass

Joined Jul 31, 2010
8
Great - this exactly what I needed . Can you point me to any reference about the design? I need also Q-factor and gain/cut formulae.
I don't need to mention that the formula matches exactly measured values. Thank you very much.
EDIT: you were quicker (providing the Q-factor formula). Shouldn't there be R12 = 10* R2?

Mark

#### Jony130

Joined Feb 17, 2009
5,092
The only reference I know is

But a few months ago I with help of Mathematica software I analyzed this circuit: R1 = 10K
R2 = 3.3K
R3 = 1.8K
R4 = 100K
R5 = 100K
R6 = 470K
R7 = 10K
C1 = 47nF
C2 = 4.7nF
C3 = 22nF
C4 = 4.7nF

And for the bass boost F1 0.16/(R4 * C1)

F2
0.16/( R4||R1*C1)

Ku_max = (R1+R4)/R1

For mid boost (this circuit is the same as yours) The circuit matrix And the result  Last edited:

#### Jony130

Joined Feb 17, 2009
5,092
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• MarkusBass

#### MarkusBass

Joined Jul 31, 2010
8
And if you chose
C3 = 10*C4 ; R12 = 10*R1
Then

$$\omega _o = \frac{1}{10*R1*C4} * sqrt{2 + \frac{R1}{R2}}$$

$$Q = sqrt{ \frac{2*R2+R1}{9.6*R2}}$$
I think that there is a small mistake in the substitution. It should be R12 = 100*R2,which leads to a different formula. Can you look at it?
My formula looks like this:
$$\omega _o = \frac{1}{sqrt{1010}*R2*C4} * sqrt{1 + 2*\frac{R2}{R1}}$$
and it seems to yield the same results as the previous one.
Does it change the Q-factor formula?

Mark

#### MarkusBass

Joined Jul 31, 2010
8
The only reference I know is
But a few months ago,with help of Mathematica software, I analyzed this circuit: Thanks for all explanations. Accidentally I also analyze this curcuit (but with SPICE simulator and not Mathematica). Although it is similar to my curcuit, I did not like it. It's because the TREBLE pot has a great influence on the MID frequencies filter. And, as a result, it does not sound good .

Mark

#### Jony130

Joined Feb 17, 2009
5,092

#### MarkusBass

Joined Jul 31, 2010
8
No, the equation is OK.
R12 = 10 * 100K = 1M, C3 = 10*C2 = 39nF
And then Fo = 281.121741Hz.
Hmm, in my circuit POT (R1) = 100k, R12 = 220k and R2 = 2.2k. I'm confused where did you take 1M from. So R12 meets the condition: R12 = 100 * R2, but definitely R12 is not equal to 10 * R1 . And the resulting frequency is 594Hz (both measured and calculated from the first formula). Am I missing something?
But you could always use this full equations
http://images.elektroda.net/55_1291827003.png
I wish, I had you knowledge to create those equations.
And have you try this circuit http://www.analogzone.com/avt08063.pdf
No, I haven't tried it (I may check it). The article from TI explains why this circuit is not very good. R7 = 100R creates a high frequency roll-off (page 2) and 10k pot does not have expected logarithmic effect (page 5). Although they say that 10k low value pot solves this problem, but it doesn't (I checked it).
Somehow my circuit handles these 2 problems correctly . That's why I'm working on it.
PS: I wanted to send you a PM but it seems that new members are not allowed to send private messages. In that case: "Bardzo dziękuję za pomoc".

Mark

#### Jony130

Joined Feb 17, 2009
5,092
Hmm, in my circuit POT (R1) = 100k, R12 = 220k and R2 = 2.2k. I'm confused where did you take 1M from. So R12 meets the condition: R12 = 100 * R2, but definitely R12 is not equal to 10 * R1 . And the resulting frequency is 594Hz (both measured and calculated from the first formula). Am I missing something?
For optimizing the quality factor Q it is reasonable to use:
R12 = 10 * POT and C3 = 10 * C4
And this equation holds only with the above assumptions.

$$F_o = \frac{ \sqrt{2 + \frac{R1}{R2}}} {20*\pi*R1*C4}$$

$$Q = \sqrt{ \frac{2*R2+R1}{9.6*R2}}$$

"Your" circuit doesn't fully meet all the above requirements.

So you force to use these equations No, I haven't tried it (I may check it). The article from TI explains why this circuit is not very good. R7 = 100R creates a high frequency roll-off (page 2) and 10k pot does not have expected logarithmic effect (page 5). Although they say that 10k low value pot solves this problem, but it doesn't (I checked it).
Somehow my circuit handles these 2 problems correctly . That's why I'm working on it.
Well, all circuit have some drawbacks, your circuit for example has a low Q factor. So if you want to build more then six bands equalizer, you are forced to use Active inductor circuits.

And those circuits are much easier to design. And in professional equipment engineers also use gyrators .

"Bardzo dziękuję za pomoc".
Mark
Nie ma za co.
PS. To ci heca, dwóch Poloków rozmawia na forum po "ichnemu".

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#### MarkusBass

Joined Jul 31, 2010
8
For optimize the quality factor Q it is reasonable to use:
R12 = 10 * POT and C3 = 10 * C4
OK, I get it. But it means that several other components need to be changed too; R12 = 1M, R2 = 10k (otherwise gain is more than 20dB), capacitors C4= 900p, C3 = 9n (to keep the frequency at 600 Hz). As a result the Q-factor is even slightly lower than before.
Being frankly, I don't want to build a 31-band equalizer. I want to have just 3 bands (later maybe I will try 5 bands). And I want to have each filter to be independent from the other filters (as much as possible). I want to have gain/cut at 13 dB or less. It seems that this is quite a complex task since change of one component value influences several parameters.
"Your" circuit does not fully meet all above requirements.
Yes, I have to use the complex version of the formula. I still have some problems with Q-factor. I get values about 10% different from measured in SPICE but maybe I'm not measuring the bandwidth correctly.
Well, all circuit have some drawbacks, your circuit for example has low Q factor.
Yes, I know but with 3-bands EQ this shouldn't be a problem. I hope to be able to use this circuit with 5-bands too.
Jony can you send me PM - maybe this topic is interesting to others.

Mark

#### Jony130

Joined Feb 17, 2009
5,092
It is very strange but even I can not send PM to you.
So, maybe you check you Control Panel or ask administrator for help.

#### MarkusBass

Joined Jul 31, 2010
8
Can you try now? I changed permissions in my profile. If not, try Marek 2006 (bez spacji) on elektroda.

Mark

#### Jony130

Joined Feb 17, 2009
5,092
Well, unfortunately, PM is still not available.
It appears that you unlocked email only, not PM.
But i send PM on elektroda