Bandpass filter help

[fbchurch2009]

I was going to order this kit from jameco but I had some questions about the filtering!

http://makezineblog.files.wordpress.com/2010/10/ledcolororgan-schem_r2.gif

I wanted to pretty much keep everything the same but I wanted to change where the bandpass occurs (and maybe slightly change the bandwidth). I noticed that the only thing that changes for each filters is the three capacitor values. How would I go about choosing the values for the filter components? I have done an "online filter calculator" but all of the values are not as rounded as the one in this schematic...

To me, they look like third order filters: The first two stages is an active filter where there is multiple feedback. Then the third stage is the passive high pass filter that is connected to the output of the op amp. I don't know how to go about fitting it for my need. Thanks!

 

[LvW]

As far as I understand your problem you ask how to design the bandpass stages. OK?
Here is a basic set of formulas (original: R1=1k, R5=560k) assuming two identical cacpacitors C:

* Select two equal capacitors suitable for the desired center frequency
* Center angular frequency wc=1/[C*SQRT(R1R5)]
* Quality factor (Q=fc/BW): R5/R1=4Q^2
Thus, you have two equations for the two unknown resistors.

 

[fbchurch2009]

Thanks, that helps! A couple of more quick questions... I ran the center angular frequency equation with the filter at the top and got an answer of 900Hz. Their schematic says 3000Hz though?

What about the 680 ohm resistor? I'm guessing it just sets the center frequency... (but then why isn't it apart of the formula?)

Finally, I am not too certain about the circuitry after the op amp and before the transistor. Their video said line conditioning. After the signal has been filtered, why can it be just connected to the base of the transistor without anything else?

 

[LvW]

Thanks, that helps! A couple of more quick questions... I ran the center angular frequency equation with the filter at the top and got an answer of 900Hz. Their schematic says 3000Hz though?
What about the 680 ohm resistor? I'm guessing it just sets the center frequency... (but then why isn't it apart of the formula?)

Sorry - but I have to apologize. I have overlooked the 680 ohm resistor because in most cases this resistor is set to infinity and all the formulas I have given apply to such a design only.
For a finite value other formulas apply. Are you still intersted in a revised set of formulas?

 

[fbchurch2009]

Yes please!

 

[LvW]

OK - let`s designate the grounded 680 Ohm as R2 (R1 and R5 as before).
Again, we can set C1=C2=C
* R5=2Q/(wc*C)
* 4Q^2=R5[(1/R2) + (1/R1)]

Thus, you can select a value for C as well as for R1 or R2.

 

[fbchurch2009]

Thanks! Is there anyway to see how the passive high pass filter after the op amp will affect this calculation or is it just something that needs to be simulated?

 

[LvW]

Thanks! Is there anyway to see how the passive high pass filter after the op amp will affect this calculation or is it just something that needs to be simulated?

I do not see a high pass filter. Rather I see a rectification circuit (function of the diode).

 

[Laplace]

Tried to get a Bode plot of the three bandpass filters in the referenced circuit. Assumed that the amplitude control is at the midpoint, giving 6K in parallel with 680Ω as the input resistance.

 

[fbchurch2009]

Thanks for all of the help! I think I got the filtering part down. The only thing left I don't understand is the 2 caps, 2 resistors and a diode after the op amp. To me, I thought the 2.2uF would isolate the transistor circuit and the op amp circuit, so I don't understand the need for a diode? Also, since I want to have a bandpass filters of higher frequencies, how would I go about choosing these values? Are they frequency dependent?

 

[Laplace]

First, you do realize that the bass & treble labels in the circuit are interchanged? As for the two capacitors and two resistors, the 2.2 uF is for AC coupling to block the DC bias point of the op-amp, while the 39K resistor is to set the DC bias point of the capacitor. The diode rectifies the AC signal, passing the positive peaks to charge the following capacitor and drive the base of the transistor to switch the LEDs. I would assume that the purpose of the 100K resistor is to discharge the base capacitor to assure complete turn-off of the LEDs. Not sure why the transistor base capacitors have such different values.