www.Hobby-Electronics.info Twin T Notch Filter calculations
The diagram below shows a simple Twin T notch filter.
- I_{1} = (V_{in}-V_{1})/2R
- I_{2} = (V_{1}-AV_{2})2jpf2C = (V_{1}-AV_{2})4jpfC (AV_{2} is the output voltage of opamp U1B, and can be adjusted by R5)
- I_{3} = (V_{1}-V_{2})/2R = (V_{2}-V_{3})2jpfC (I_{3} flows through R2 and C4, since the input resistance of the opamp is infinite)
- I_{4} = (V_{in}-V_{3})2jpfC
- I_{5} = (V_{3}-AV_{2})/R
I_{2}+I_{3}=I_{1} => (V_{1}-AV_{2})4jpfC + (V_{1}-V_{2})/2R = (V_{in}-V_{1})/2R
Multiplying everything with 2R gives: (V_{1}-AV_{2})4jpfC2R + V_{1}-V_{2} = V_{in}-V_{1} =>
- (V_{1}-AV_{2})8jpfRC = V_{in}-2V_{1}+V_{2}
Equation 3. gives: V_{1}-V_{2}=(V_{2}-V_{3})4jpfRC =>
- V_{1}=(V_{2}-V_{3})4jpfRC+V_{2}
Substituting 7. in 6. gives: [(V_{2}-V_{3})4jpfRC + (1-A)V_{2}]8jpfRC = V_{in}-V_{2}-(V_{2}-V_{3})8jpfRC+V_{2} =>
- (1+4jpfRC)(V_{2}-V_{3})8jpfRC + (1-A)8jpfRCV_{2} = V_{1}-V_{2}
I_{3}+I_{4}=I_{5} => (V_{2}-V_{3})2jpfC + (V_{in}-V_{3})2jpfC = (V_{3}-AV_{2})/R => (V_{2}-V_{in}-2V_{3})2jpfC = (V_{3}-AV_{2})/R =>
(V_{3}-AV_{2})/2jpfRC = V_{2}-V_{in}-2V_{3} => (2 + 1/2jpfRC)V_{3} = V_{2}+V_{in}+(AV_{2})/2jpfRC =>
([4jpfRC+1]/2jpfRC)V_{3} = V_{2}+V_{in}+(AV_{2})/2jpfRC => V_{3} = (V_{2}2jpfRC + V_{in}2jpfRC + AV_{2})/(1+4jpfRC) =>
- V_{3} = (V_{in}2jpfRC + V_{2}[2jpfRC+A])/(1+4jpfRC)
=> V_{3}-V_{2} = (V_{in}2jpfRC + V_{2}[2jpfRC+A - 1 - 4jpfRC])/(1+4jpfRC) = (V_{in}2jpfRC + V_{2}[A - 1 - 2jpfRC])/(1+4jpfRC) =>
- V_{2}-V_{3} = -(V_{in}2jpfRC + V_{2}[A - 1 - 2jpfRC])/(1+4jpfRC)
Substituting 10. in 8. gives: -[V_{in}2jpfRC + V_{2}(A-1-2jpfRC)]8jpfRC = -(1-A)8jpfRCV_{2} + V_{in} - V_{2} =>
[V_{in}2jpfRC + V_{2}(A-1-2jpfRC)]8jpfRC = (1-A)8jpfRCV_{2} - V_{in} + V_{2}
Deviding everything by 8jpfRC gives: V_{in}2jpfRC + V_{2}(A-1-2jpfRC) = (1-A)V_{2} + (V_{in}-V_{2})/8jpfRC =>
V_{2}(A-1-2jpfRC) + V_{2}(A-1) - V_{2}/8jpfRC = -V_{in}2jpfRC - V_{in}/8jpfRC =>
And since opamp U1A is a unity gain amplifier, V_{out} = V_{2}. So the equation above also describes the output response.
V_{out}/V_{in} will be 0 if the nominator becomes 0. So 16(jpfRC)^{2} + 1 = 0 => 16(jpfRC)^{2} = -1 => j^{2}(4pfRC)^{2} = -1 =>
-(4pfRC)^{2} = -1 => f=1/(4pRC)
The V_{out}/V_{in} equation also shows that the width of the notch is determined by A. If A=1, the notch becomes so narrow that V_{out}/V_{in}=1. The notch becomes wider as A decreases.