Bandwidth of Tow-Thomas biquad

Thread Starter

Iamma

Joined May 13, 2022
8
I need to design a Tow-Thomas biquad in such a way to obtain a 7000 Hz Bandwidth. I've got the formulas for its design, 1705331471092.png and
1705331499100.pngand in course it says to make R2 = R4 and C1 = C2. The problem is that no matter how I change the values for these components I can't change the bandwidth how I want. In my knowledge the bandwidth is w0/Q. Can anybody help me understand how to select the values of the components?

This is the circuit I am working with:1705331747991.png
 

crutschow

Joined Mar 14, 2008
34,719
it says to make R2 = R4 and C1 = C2. The problem is that no matter how I change the values for these components I can't change the bandwidth how I want.
For those conditions, C1/C2 is a constant of 1, but 1/R2R4 is not.
Also R1 can be varied without affecting ω0.
So you adjust ω0 by your selection of R1R4C1C2 and Q by your selection of R1, R2, and R3.

Incidentally, that filter needs an input resistor as you can't connect the (-) input of an op-amp directly to a low impedance source.
 

MisterBill2

Joined Jan 23, 2018
19,091
Certainly "C" is correct about the input resistor. In fact, without an adequate input resistor the circuit will not function at all correctly. So it might be the source of your difficulty all along.
I am not familiar with the "Tow-Thomas biquad ", but I have used assorted op-amp circuits, and that input resistor to the summing junction is important indeed.
 

LvW

Joined Jun 13, 2013
1,766
I cannot fully understand your problem.
* For two equal capacitors C1=C2=C, the pole frequency (bandpass case: Center frequency wo) is given by wp=1/[C*SQRT(R2*R4)].
* The pole Q is (C1=C2=C) Qp=R1/SQRT(R2*R4)
* This means: You can select the pole data wp and Qp independently because you have tree tuning parameters: R1, C and R2*R4.
(and you can coose R2=R4 if you want)-
* As mentioned already, you need of course a resistor Rin between the signal source and the inverting input of the 1st opamp. This resistor appears in the numerator of the three transfer functions only and, thus, will deteremine the gain of the three flter functions.
* In the bandpass case, the numerator is N(s)=s*R2*R4*C2/Rin

Comment:
For a second-order bandpass, the pole Q (Qp) is identical to the quality factor Q (selectivity) which appears in the expression
BW=fo/Q in Hz (fo=wo/2Pi).
 
Last edited:

MisterBill2

Joined Jan 23, 2018
19,091
The value of the resistor in series with the input to the first op-amp must be at least as great as R4, or even R1+R4. Without a series resistor on the input it will certainly not perform as desired.
 

LvW

Joined Jun 13, 2013
1,766
The value of the resistor in series with the input to the first op-amp must be at least as great as R4, or even R1+R4. Without a series resistor on the input it will certainly not perform as desired.
Why do you think that this resistor "must be at least as great as R4" ?
I think, there is not such a restriction.

This resistor (in my above answer called "Rin") appears in the numerator only and determines the gain:

Bandpass numerator Nbp(s)=s*R2*R4*(C1+C2)/Rin
Lowpass numerator: Nlp(s)=R4/Rin
 

MisterBill2

Joined Jan 23, 2018
19,091
Consider the basic definition of a summing junction op-amp circuit. And I picked that value range because it would not interfere with the filter operation. Given that we have NO INFORMATION about the values of any resistors or capacitors it seems a reasonable starting point. It was nit defined as a mandatory value!!
 
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