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

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.

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

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

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

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.

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

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

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

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

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 -20dB/decade.

Frequency response of first order low pass filter

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

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

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

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.

Frequency response of first order High pass filter

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

Second order Butterworth Filters 1 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

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

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