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!

• It's supposed to be a filter, not an amplifier. That is the mid-band should have the same level on the output as the input. The final stage inverts the signal to provide negative feedback to the primary stage. Mar 10, 2017 at 20:57
• @Trevor but what does it filter if we take the output before it? What I do not understand most of all is why we nee feedback from it.
– Mu3
Mar 10, 2017 at 21:00
• Did you miss that class? ANyhoos.. read this. ece.uic.edu/~vahe/spring2012/ece412/biquad.pdf Mar 10, 2017 at 21:02
• Mar 10, 2017 at 21:05
• @Trevor thank you for the sources, I will read them. Do you know the name of the book that the first link was taken from?
– Mu3
Mar 10, 2017 at 21:07

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

• 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).

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

• Ah-hah, so it makes the life of a designer easier. But why the need for the third opamp in my case?
– Mu3
Mar 10, 2017 at 22:22
• Yeah, the op amps. The first opamp is a low pass filter the second a high pass and the last is just simply a gain stage. Mar 10, 2017 at 22:32
• Wait, in the picture that I provided first is bandpass, second is low-pass and third is the gain stage. However, it still escapes me - why is the gain stage even needed?
– Mu3
Mar 10, 2017 at 22:34
• Its to adjust the bandpass gain, if you didn't have the passband amplitude would not be adjustable. So you don't necessarily need it. If you designed a circuit and it had a gain of 1 then you could take it out and connect R4 to the input of the high pass stage. Mar 10, 2017 at 23:15

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.