Active_filters_154986429212111725305c610d647dfcd.pdf

ACTIVE FILTERS

6.1 Low-pass Butterworth filter  Schematic diagram of a first –order low-pass Butterworth filter is shown below.

 The OPAMP is used in the non-inverting condition; hence it does not load down the RC network. 1 vin j2πfC  The voltage at the non-inverting terminal is v1  . vin  1 1  j2πfCR R j2πfC

 R  The output voltage is vo  1  F R1  AF vin Or, vo  1  j  f fH 

  RF  v in  v1  1   R1  1  j2πfCR  

where AF is the passband gain of the filter and fH is high cutoff frequency of the filter. vo AF   The magnitude and phase angle equations are and 2 vin 1   f fH  φ  tan1  f fH  .

 For f
2  0.707A F , at f > fH vo vin  AF .

 In designing low-pass active filter the value of C is generally taken less than 1 µF.  First –order low-pass Butterworth filter provides a 20 dB/decade roll-off. To get 40 dB/decade roll-off a second –order low-pass Butterworth filter should be used.  Schematic diagram of a first –order low-pas Butterworth filter is shown below. 1  The cut-off frequency is fH  . 2π R 2R 3C2C3

 The gain of the filter is

vo AF  . 4 vin 1   f fH 

 To design a second –order low-pas Butterworth filter we generally choose R2 = R3 1 = R, C2 = C3 = C and C  1 µF. Therefore, fH  . 2πRC  To get a Butterworth response the gain of the filter should be equal to 1.586. R Therefore 1  F  1.586 or, RF = 0.586R1. R1

 Higher order low-pass filters, such as third order, fourth-order low-ass filters are formed simply using first and second order low-pass filters, as shown.

Third order low-pass filter

Fourth order low-pass filter

Numerical 1. Design a low-pass filter at a cut-off frequency 1 kHz with a passband gain of 2. Assume C = 0.01 µF and RF = 10 kΩ. Redesign the same filter for cutoff frequency 1.6 kHz. 1 1 R   15.9 103 Ω  15.9 kΩ 2πCfH 2π  0.01 106 103 Since passband gain is 2, R1 = RF = 10 kΩ. R fH If R be the required resistance for cutoff frequency 1.6 kHz ( fH ) then  R fH Or, R  R

fH 1  15.9   9.94 kΩ . fH 1.6

2. Design a second order low-pass filter with cut-off frequency 1 kHz. Assume C = 0.0047 µF and R1 = 27kΩ. 1 1 R   33.86 103 Ω  33.86 kΩ . 2πCfH 2π  0.0047 106 103 R F  0.586R1  0.586  27  103  15.82 103 Ω  15.82 kΩ

6.2 High-pass Butterworth filter  Schematic diagrams of a first –order and a second-order high-pass Butterworth filters are shown below.  The filters can be achieved by interchanging the positions of the capacitors and resistors, as shown.  For the first –order high-pass Butterworth filter we have

j  f fL    R  R  j2πfCR v in v in  vo  1  F  v1  1  F  Or, vo  A F R1  R1  1  j2πfCR 1  j  f fL    where AF is the passband gain of the filter and fL is low cutoff frequency of the filter.  f fL  v  The gain of the first –order high-pass Butterworth filter is o  A F . 2 v in 1  f f  L

 Similarly, the gain of the second–order high-pass Butterworth filter is,  f fL  vo  AF . 4 v in 1   f fL 

First –order high-pass Butterworth filter

Second–order high-pass Butterworth filter

Numerical 3. Design a first order high-pass filter at a cut-off frequency 1 kHz with a passband gain of 2. Assume C = 0.01 µF and RF = 10 kΩ. Redesign the same filter for cutoff frequency 1.6 kHz. 1 1 R   15.9 103 Ω  15.9 kΩ 6 3 2πCfL 2π  0.01 10 10 Since passband gain is 2, R1 = RF = 10 kΩ. R fL If R be the required resistance for cutoff frequency 1.6 kHz ( fL ) then  R fL Or, R  R

fL 1  15.9   9.94 kΩ .  fL 1.6

4. Design a second order low-pass filter with cut-off frequency 1 kHz. Assume C = 0.0047 µF and R1 = 27kΩ. 1 1 R   33.86 103 Ω  33.86 kΩ . 6 3 2πCfL 2π  0.0047 10 10 R F  0.586R1  0.586  27  103  15.82 103 Ω  15.82 kΩ

6.3 Band-pass filters  A wideband band-pass filter can be assumed as a bandpass filter with Quality factor Q<10.  The filters with Quality factor Q>10 may be considered as a narrowband bandpass filter.  The quality factor (or figure of merit) of a filter can be defined as,

Q

f f fC fC   HL , BW fH  fL fH  fL

where fC, fL and fH are the center frequency, low cut-off and high cut-off frequencies. 6.3.1 Wideband band-pass filter  A wideband band-pass filter can be formed by cascading a high-pass and a lowpass filter.  For 20 dB roll-off a first order low-pass and a first order high-pass filters are cascaded (as shown) whereas for a 40 dB roll-off a second order low-pass and a second order high-pass filters are cascaded.  To realize a bandpass response fH must be higher than fL.

6.3.2 Narrowband band-pass filter  Schematic diagrams of a narrowband band-pass filter is shown below. It may be noted that the filter is in inverting-mode.

 A narrowband band-pass filter is generally designed for specific values of center frequency (fC) and Q or fC and bandwidth.  If we choose C1 = C2 = C then Q R1  2πfCCAF

R2 

R3 



Q

2πfCC 2Q2  A F



Q πfCC

R3 . The gain AF must satisfy the equation, A F  2Q2 . 2R1  The center frequency fC can be changed to a new frequency fC without changing where AF 

the gain and bandwidth by implementing, R2  R2  fC fC  . 2

Numerical 5. Design a wideband bandpass filter with fL = 200 Hz, fH = 1 kHz and passband gain 4. Assume C = 0.01 µF, C = 0.05 µF, R1 = R1 =10 kΩ. Also calculate the Q of the filter. 1 1 R   15.9 103 Ω  15.9 kΩ 6 3 2πCfH 2π  0.01 10 10

1 1   15.9 103 Ω  15.9 kΩ 6 2πCfL 2π  0.05 10  200

R 

Assuming the gain of both the stage is identical, i.e., 2 we have RF = RF =10 kΩ.

Q

fHfL 1000  200   0.56 fH  fL 1000  200

6. Design a narrowband band-pass filter with center frequency 1 kHz, Q = 12 and AF = 10. Redesign the circuit for center frequency 1.5 kHz, keeping A F and bandwidth constant. Assume C1 = C2 = C = 0.01 µF.

R1  R2 

R3 

Q 12   19.08 103 Ω  19.08 kΩ 3 6 2πfCCAF 2π 10  0.01 10 10



Q 2

2πfCC 2Q  A F





4



2π  10  0.01  10 6 2  32  10 3



 23.88  103 Ω  23.88 kΩ

Q 12   0.382 106 Ω  0.382 MΩ 3 πfCC π 10  0.01 106 2

2

f   1  R2  R2  C   23.88    10.61 kΩ  1.5   fC  6.5 Band-reject filters  Like band-pass filters, band-reject filters can also be wideband band-reject and narrowband band-reject type.

6.5.1 Wideband band-reject filter  Schematic diagrams of a wideband band-reject filter is shown below. It consists of a low-pass, a high-pass, and a summing amplifier.  To realize a bandpass response fL must be higher than fH.  Further, the gain of the low-pass and high-pass filter must be same.  The value of ROM is ROM  R2 R3 R4 .

Numerical 7. Design a wideband band-reject filter with fH = 200 Hz, fL = 1 kHz, gain of highpass and low-pass section is 2, and gain of summing amplifier is 1. Assume C = 0.01 µF, C = 0.05 µF, R1 = R1 = R4 = 10 kΩ.

R

1 1   15.9 103 Ω  15.9 kΩ 6 3 2πCfL 2π  0.01 10 10

R 

1 1   15.9 103 Ω  15.9 kΩ 2πCfH 2π  0.05 106  200

Since the gain of both the stage is identical, i.e., 2 we have RF = RF =10 kΩ. Furthermore the gain of the summing amplifier is 1. Therefore, R2 = R3 = R4 =10 kΩ.

Finally, R OM  R 2 R 3 R 4 

10  3.33 kΩ 3

6.5.2 Narrowband band-reject filter  Schematic diagrams of a Narrowband band-reject filter (or notch filter) is shown below. The job of the OPAMP is mainly to increase the Q of the filter.  Most common applications of notch filters are to reject a single frequency (such as 50 or 60 Hz power line frequency hum), in communications and biomedical instrumentation for eliminating undesired frequencies.  The value of the capacitor is generally chosen as C  1 μF .

Numerical 8. Design a 60 Hz active notch filter. Assume C = 0.068 µF.

R

1 1   39.01 103 Ω  39.01 kΩ 6 2πfNC 2π  60  0.068 10

6.6 All-pass filter  This filter is used to pass all the frequency components of the input signal without attenuation, while providing predictable phase shifts for different frequency components.

 All-pass filters find applications in telephone wires to compensate the phase change of different frequency components of the signals that they suffer during propagation through the line.  All-pass filters are also called delay equalizers or phase correctors.  Schematic diagrams of an all-pass filter is shown below.

 The output voltage of the filter can be found using superposition principle. For RF = R1 it is given by R   jXC   j2XC R1  vo    1  v in  v in 1   v in  v in  R  jXC  R1  R  jX C  R1 

2 2         2 j2πfC  j2πfC    v in  1    v in  1  Or, vo  v in  1   1  j2πfRC  1  j2πfRC  1     R    j2πfC  j2πfC    v 1  j2πfRC Or, o  vin 1  j2πfRC  Above equation implies the magnitude of the gain is unity but the phase difference between vo and vin is a function of input frequency f and is φ  2tan1  2πfRC  (in degree).  Above equation also implies that for fixed value of R and C, the phase angle (φ) varies from 0o to -180o as f varies from 0 to ∞.  If the value of R and C are interchanged in the figure, the phase shift between input and output becomes positive.

Numerical 9. Design an all-pass filter R1 = RF = 10 kΩ, R = 15.9 kΩ, C = 0.01 µF, f = 1 kHz. Calculate the phase shift.





φ  2tan1  2πfRC  2tan1 2π 103 15.9 103  0.01106  90o