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?And if you chose
C3 = 10*C4 ; R12 = 10*R1
Then
\(\omega _o = \frac{1}{10*R1*C4} * sqrt{2 + \frac{R1}{R2}}\)
\(Q = sqrt{ \frac{2*R2+R1}{9.6*R2}}\)
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 .
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?No, the equation is OK.
R12 = 10 * 100K = 1M, C3 = 10*C2 = 39nF
And then Fo = 281.121741Hz.
I wish, I had you knowledge to create those equations.But you could always use this full equations
http://images.elektroda.net/55_1291827003.png
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).
For optimizing the quality factor Q it is reasonable to use: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?
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.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.
Nie ma za co."Bardzo dziękuję za pomoc".
Mark
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.For optimize the quality factor Q it is reasonable to use:
R12 = 10 * POT and C3 = 10 * C4
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."Your" circuit does not fully meet all above requirements.
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.Well, all circuit have some drawbacks, your circuit for example has low Q factor.
Thread starter | Similar threads | Forum | Replies | Date |
---|---|---|---|---|
L | 10 band graphic equalizer | General Electronics Chat | 11 | |
S | Small screen to visualize sound (equalizer) | General Electronics Chat | 10 | |
R | Help with Constant Q Graphical Equalizer Design | Homework Help | 124 | |
S | graphic equalizer computations | Analog & Mixed-Signal Design | 5 | |
G | Graphic Equalizer Display Filter - MSGEQ7 | General Electronics Chat | 48 |
by Steve Arar
by Gary Elinoff
by Gary Elinoff