Tow-Thomas biquad filter design

Question

I am a student of EE, and recently in my book I read about the Tow-Thomas biquad filter.

The book simply states that "R4 and R4' should be of the same values, as well as R3 and R3'", but does not give any explanation as to why. I could not find any explanation on the Internet.

In my understanding, the third op-amp provides a voltage gain of -1. So here is question: why is that amplifier even needed and why is important to have specific resistances of the same value?

Thanks for your time and responses!

Answer 1

"...why is that amplifier even needed and why is important to have specific resistances of the same value?"

Some answers to your questions:

• No, it is not necessary that the inverter stage has a gain of "-1". However, it is allowed and convenient - so, why not?

• Yes, such an additional inverting stage is necessary because a fixed and stable DC operating point needs negative DC feedback (1 or 3 or 5 ... inverting stages within a feedback loop).

• More than that, it is also not necessary that other parts (as R3) are chosen to be equal. You are free to select all the parts - as long as the time constants and parts ratios in the formulas for realizing the desired filter parameters have the correct values. However, in many cases it is simply convenient to have the chance for using the same parts values.

• Comment: The last two stages form a non-inverting integrator stage (inverting integrator with succeeding phase inversion). These two stages can be replaced by a single opamp which can operate as a non-inv. integrator (phase-lead integrator or Deboo-integrator).

Answer 2

It makes up for easier analysis, I'm not going to analyze this for you because it's a homework question, I'll show you a different example so I don't deprive you of valuable learning. I you look at an instrumentation amplifier equation it helps to have resistors equal each other for the overall gain (and common mode).

$$\frac{V_{out}}{\Delta V_{in}} = \bigg(1+\frac{2R_2}{R_1}\bigg)\bigg(\frac{R_6 (R_3+R_4)}{R_3 (R_5+R_6)}+\frac{R_4}{R_3}\bigg)$$

If we let R3=R5 and R4=R6 you get this:

$$\frac{V_{out}}{\Delta V_{in}} = 2\frac{R_4}{R_5} $$

So the answer is in a round about way is it makes things nice for analysis and circuit design.

To solve this find the transfer function of each block and solve it open loop (by treating the first stage as an summing amplifier and breaking the loop between R3 and R4). Then set the output of the first stage equal to the other disconnected input of the summing amplifier

Answer 3

You need that 3rd stage for 180* phase-shift, needed to loop back around to the input of stage-1, the "gain" answer, while one may think that, is not correct. Notice that since you have fixed-value resistors, any 3rd stage 'gain' is fixed, and the goal using this type of circuit is generally with unity gain or less whilst signal conditioning is performed.

In real-world applications, one could use a variable resistor to actually go from unity-gain to say a gain factor of 10x. This type circuit was often used in high dollar analog audio processing circuits during the last 25-years of the 20th century.