1. Active filters
A. V. Gokhale.
Y.C.C.E, Nagpur

2. Advantages of active over passive filters
Active filters do not use inductors which are large, heavy and costly elements at audio frequency range.
The op-amp used in non-inverting configuration gives high input impedance and low output impedance avoiding loading of source as well as load.
In active filters as inductors are not used the quality factor extend upto few hundred.
Can be fabricated on chip due to inductor less design.
The output in passive filters is always less than input, but in case of active filters the output in pass band is gain times input.
Limitations: The high frequency response is restricted by bandwidth and slew rate of op-amp.

3. Classification / Type / Order
Based on frequency available in output as
Low pass filter
High pass filter
Band pass filter
Band stop or reject filter.

4. Classification / Type / Order
Type of Active filter:
Butterworth filter
Chebychev filter
Caur filter

5. Classification / Type / Order
Order of Active Filter:
Slope of the stop band decides the order of the active filter.
Slope given by
dB/decade where f2=10xf1
dB/octave where f2=2xf1
dB/decade
dB/octave
order of filter 20 6 1 40 12 2 60 18 3 80 24 4

6. Transfer function of active filter
An active filter is specified by the voltage transfer function,
Under steady state condition i.e. at s = j,
where is magnitude or gain of the transfer function and is phase of the transfer function.

7. Normalized Butterworth Polynomial
The order of the filter is given by the normalized denominator of the transfer function known as normalized Butterworth polynomial.

8. First Order Butterwoth Low Pass Filter
The transfer function of the circuit is given by
The voltage at non-inverting terminal V1 is given by

9. Transfer function / cut-off frequency
The closed loop gain of the non-inverting amplifier is
Thus the transfer function is given by
Let

10. Transfer function / cut off frequency
For steady state response substitute
where
is higher cut-off frequency.
The magnitude of the transfer function is given by
and the phase angle is given by

11. Frequency Response f =100fH f =10fH f >> fH f = fH f << fH
From the magnitude of transfer function
f =100fH
f =10fH
f >> fH
f = fH
f << fH
Magnitude of TF
Frequency
The magnitude decreases by 20dB (20x ln 10) each time frequency is increased 10 times. Thus the slope in the stop band will be dB/decade.

12. Frequency response of first order low pass filter

13. First Order Butterwoth High Pass Filter
The transfer function of the circuit is given by
The voltage at non-inverting terminal V1 is given by

14. Transfer function / cut-off frequency
The closed loop gain of the non-inverting amplifier is
Thus the transfer function is given by
Let

15. Transfer function / cut off frequency
For steady state response substitute
where
is higher cut-off frequency.
The magnitude of the transfer function is given by
and the phase angle is given by

16. Frequency Response f =fL/100 f =fL/10 f >> fL f = fL
From the magnitude of transfer function
f =fL/100
f =fL/10
f >> fL
f = fL
f << fH
Magnitude of TF
Frequency
The magnitude increases by 20dB (20x ln 10) each time frequency is increased 10 times. Thus the slope in the stop band will be 20dB/decade.

17. Frequency response of first order High pass filter

18. Design Steps for First Order Butterworth Filter Low Pass / High Pass
Design for given cut-off frequency and pass band gain i.e. fL or fH and AF
Select value of capacitor less than or equal to 1µF.
C  1µF
Calculate value of resistor using formula of cut –off frequency
For given value of pass band gain calculate value of R1 & RF assuming value of R1
For better performance use RF  100 KΩ

19. Second order Butterworth Filters1 y1 y3 y4 y2
Vin VA VB
Applying KCL at node VB
Eq 2
Substitute value of VA in EQ.1
The gain of non-inverting amplifier is given by
Applying KCL at node VA
Eq 1

20. Generalized Transfer Function of Second Order Filter
Second Order Low Pass
Second Order High Pass

21. Second Order Butterworth Low Pass Filter
The generalised transfer function of second order filter is
From the figure the admittances are
y1=1/R1
y2=1/R2
y3=sC3
y4=sC4