Twin T active notch filter [closed]

Question

So I know that this a band reject filter with the quality factor of Q

I have two questions:

1-Why in one T the resistance is half the other resistances in the second T.

2- How would adding a voltage follower increases Q?

Answer 1

The double-T RC circuit has a third-order transfer function. However, if the grounded C (lets call it C3) is identical to (C1+C2) and the grounded R (lets call it R3) is R3=(R1||R2) we have a so-called pole-zero cancellation which results in second-order filter with the desired bandstop characteristic:

wp=wz=1/RC and H(w=0)=H(w infinite).

Note that the mentioned conditions can be fulfilled for a variety of R and C combinations.

Examples:

C1=C and C2=C/kc and C3=C(1+1/kc)

R1=R and R2=kr and R3=R * kr/(1+kr).

For a "tuned" double-T network we require kr=kc=k. For this case, the corresponding transfer function reveals that Qp=k/[2(1+k)].

We can show that Qp,max=0.5 for very large k values; for the simple case (as in the given circuit) with k=1 we have Qp=0.25.

This a very bad value (bad selectivity). For a much larger Q we can modify the circuit and connect R3 not to ground but to the output of a fixed gain positive amplifier having a gain A.

That means: We use positive feedback for increasing the selectivity (larger Q). Just adding a voltage follower does not have any positive influence on Q. It only provides decoupling between the filter and the load.

EDIT: The most simple solution for increasing Qp is to select A=1 (voltage follower) and to use different k factors: kc=1 and kr>1.

In this case: wp=wz=1/[RC*SQRT(kr)] and Qp=0.5*SQRT (kr)

Answer 2

1) because if you would R instead of R/2 in the top branch, the RC timeconstant would be different from the bottom branch. Same as there is 2 x C in the top branch but 2 C in the bottom branch.

3) Because without the follower the impedance Rl would load the filter and change its behaviour. Example: what would happen without the voltage follower so Rl is directly at the output and Rl has a value of R/10 or even R/10000. What would happen ?

Answer 3

1) The voltage at positive and negative terminals are almost same. In this case it is vout. Assume the input source vin=0. Now analyse the upper arm T circuit. The net capacitance seen by the resistance R/2 is , 2C (both capacitors are parallel). Time constant if this upper T part is (R/2*2C) = RC.

Consider the lower T. Net resistance seen by the capacitor with capacitance 2C is R/2 ( Same reason as said above - the resistors are in parallel). Again time constant is (2C*R/2)=RC.

Both upper and lower T has same time constant. Symmetrical contribution in charging and discharging rate.

2) voltage follower is not just where vout = vin. Remember that voltage follower has large input impedance. If you directly connect Rl(load resistance), you will have lot changes. You may try that by yourself

Answer 4

Other folk are explaining the 2C thing so I'll concentrate on: -

How would adding a voltage follower increases Q

I believe you may be thinking of the bootstrapped twin T notch filter. This can improve Q dramatically: -

Concentrate on where the output of U1b connects - it now connects to the junction of the "2 C" capacitor and the "0.5 R" resistor (formerly connected to ground). Increasing Q has this effect on the frequency response: -

This type of filter is commonly used to notch out 50 or 60 Hz AC pick-up on audio signals. It has minimal effects on the audio outside of a few hertz either side of the notch frequency. Here's a 60 Hz design taken from an old National semiconductor data sheet by the looks of it: -