Analog Circuits and Systems Prof. K Radhakrishna Rao Lecture 25: Active Filters 1 Review Inductor Simulation To convert RLC filters to Active RC filters Gyrator – Inductor Simulator (L=CR2) Active and Passive - Parameter Sensitivities in Active RC filters Effect of finite gain and finite gain bandwidth product on inductor simulated f0Q<<Gain Bandwidth Product 2 Review (contd.,) A0 Q of the inductor simulated = 2ω0 A0 ⎞ ⎛ ⎜1 − GB ⎟ ⎝ ⎠ Qa Q of the filter simulated = ⎛ Qa 2ω0Qa ⎞ − ⎜1 + ⎟ A0 GB ⎠ ⎝ 3 Q-enhancement- Sallen and Key The quality factor Qp of a second order passive RC filter is always less than 0.5 Qp < 0.5 is unacceptable for a general filter design Sallen and Key proposed use of negative and positive feedback, and active devices to enhance Q Several topologies similar to Sallen and Key filters are possible 4 Second Order Passive filter N (s ) Transfer function of second order passive filter = D (s) D(s) and N(s) are second order polynomials with D(s) having Q <0.5 5 Use of feedback to enhance Q -K ⎡⎣N ( s ) D ( s )⎤⎦ -KN ( s ) Vo = = Vi 1 + K ⎡⎣N ( s ) D ( s )⎤⎦ D ( s ) + KN ( s ) where K is the gain of the active device 6 Use of feedback to enhance Q (contd.,) For a general second-order passive RC/RL filter ⎡ s2 ⎤ s + p⎥ ⎢m 2 + n ωp N(s) ⎢⎣ ωp ⎥⎦ = D(s) ⎡ s2 ⎤ s + 1⎥ ⎢ 2 + ⎢⎣ ωp ωpQp ⎥⎦ where ωp is the natural frequency of the passive RC filter Qp is quality factor of the passive second-order RC/RL filter 7 Quality Factor Qp is always < 0.5 ⎡ s2 ⎤ s -K ⎢m 2 + n + p⎥ ωp Vo ⎢⎣ ωp ⎥⎦ = 2 Vi ⎡ ⎤ s s + (1 + pK)⎥ ⎢(1 + mK) 2 + (1 + nK) ωp ωpQp ⎢⎣ ⎥⎦ 1 + pK ω0 = ωp 1 + mK Qp Qa = (1 + pK) (1 + mK) (1 + nK) If K is positive, m and p are positive, and for all values of n it is a negative feedback system If K is negative m and p are positive, all positive values of n it is a positive feedback system 8 Enhancement of Qa Qa can be enhanced by increasing pK >0 with m = n =0, mK>0 with p=n=0, pK>0 and mK>0 with n =0. These make use of negative feedback. Qa can also be enhanced by making nK<0 and 0<|nK|<1. This constitutes using positive feedback. All types of filters can be designed using any of the Q-enhancement methods. 9 Second-order low-pass RC filter N(s) 1 = 2 D(s) (C1R1C2R2 ) s + (C1R1 + C2 (R1 + R2 )) s + 1 ωp = Qp = 1 R1C1R 2C2 1 ⎛ R1 ⎞ ⎜1 + ⎟ R2 ⎠ ⎝ with R1 = R 2 = R and C1 = C2 = C C1R1 + C2R 2 C2R 2 C1R1 1 1 ωp = and Qp = RC 3 10 Active Low Pass Filter m = 0, n = 0 and p = 1 Vo -K = Vi ⎡ s2 ⎤ s + (1 + K)⎥ ⎢ 2 + R1 = R2 = R and C1 = C2 = C ⎢⎣ ωp ωpQp ⎥⎦ The natural frequency of the active filter is now higher ω0 = ωp 1 + K ; Qa = Qp 1 + K Qa can be increased to the required value through suitable selection of K. Low-pass passive filter with amplifier gain -K and feedback 11 Structure of Active Low Pass filter Addition required between feedback signal and the input In order to get 2Vo the gain of VCVS will have to be made 2K A VCVS with gain -K can be realized by having buffer stage followed by inverting amplifier 12 Structure of Active Low Pass filter (contd.,) 13 Second order Butterworth low-pass filter Bandwidth = 40Hz, Qa = 1 2 . With R1 = R 2 = R and 1 C1 = C2 = C then Qp = 3 Qa = Qp 1 + K = 1+K 1 = 3 2 K = 3.5 and ω0 = 2π × 40 = 1+K 4.5 = RC RC 4.5 RC = 2π × 40 For R = 100kΩ; 4.5 C= µF = 84nF 2π × 40 14 Frequency Response of the Butterworth LP filter 15 Transient Response of the Butterworth LP filter 16 Frequency response of Low Pass Filter with Q =5 and f0 = 40 Hz R=100k; C=0.6mF; K=224 17 Frequency response of Low Pass Filter with Q =5 and f0 = 400 Hz C = 60 nF Finite GB of the Op.Amp Finite GB of the Op.Amp 18 Transient response of Low-pass Filter with Q =5 and f0 = 400 Hz C = 60 nF Q-Enhancement due to Finite GB 19 Low-pass Filter with Q =5 and f0 = 600 Hz C = 40 nF Q-Enhancement due to Finite GB 20 Observations Q increases from the specified value The natural frequency reduces slightly from the specified value At higher natural frequencies the transient responses are more oscillatory indicating Q enhancement Beyond a certain natural frequency the system becomes unstable and goes into oscillations at the natural frequency These deviations from the expected behavior are due to finite gain bandwidth product of the active devices used. 21 Effect of Gain Bandwidth Product of Op Amp Amplifier using a buffer and an inverting amplifier of gain K has a transfer function ⎛ 2 (1 + K ) s ⎞ Vo K = ; K ⎜1 ⎟ ⎜ ⎟ (1 + 2K)s ⎞ ⎛ s ⎞ Vi ⎛ GB ⎝ ⎠ 1 + 1 + ⎜ ⎟ ⎜ ⎟ GB GB ⎠ ⎝ ⎠⎝ Transfer function of the active low-pass filter ⎛ 2 (1 + K ) s ⎞ K ⎜1 ⎟ ⎜ ⎟ G B Vo ⎝ ⎠ = 2 Vi ⎛ 2 (1 + K ) s ⎞ s s + + 1 + K ⎜1 ⎟ 2 ⎜ ⎟ GB ωp ωpQp ⎝ ⎠ 22 Effect of GB Product of Op Amp (contd.,) Normalizing 2 (1 + K ) s ⎞ K ⎛ ⎜⎜1 ⎟⎟ 1+K⎝ GB ⎠ Vo = 2 Vi ⎛ 2K (1 + K ) s ⎞ s s + -⎜ +1 ⎟ 2 ω0 ω0Qp 1 + K ⎜⎝ GB(1 + K) ⎟⎠ Qa (due to GB) = Qp 1 + K ⎛ 2K (1 + K ) ω0Qp ⎞ ⎜⎜1 ⎟⎟ 1 + KGB ⎠ ⎝ GB should be large enough to make 2K (1 + K ) ω0Qp 1 + KGB 2Kω0Qa = = 1 GB 23 Examples Ex:1 Qa (specified) = 5 and f0 = 40Hz and K = 3.5 Qa (due to GB) = Qp 1 + K 2Kω0Qa ⎞ ⎛ ⎜1 - GB ⎟ ⎝ ⎠ = 5.23 Ex:2 Qa (specified)=5 and f0 = 400 Hz and K = 224 Qa (due to GB) = Qp 1 + K 2Kω0Qa ⎞ ⎛ ⎜1 - GB ⎟ ⎝ ⎠ = 48 24 Limitations of GB 2f0Qa = 1 for the filter to be stable in case inductance simulation GB 2Kf0Qa = 1 for the filter to be stable in case of filter using feedback GB 25 Fourth-order Butterworth Low-pass Filter 1 2 2 ⎡ ⎤ ⎡ s s s s ⎤ + 2 ⎥ ⎢1 + 1.848 + 2 ⎥ ⎢1 + 0.765 ω0 ω0 ⎥⎦ ⎢⎣ ω0 ω0 ⎥⎦ ⎢⎣ 26 Effect of finite GB Taking GB into account with f0 = 3.3 kHz (speech filter) Using 741 Op Amp having a GB of 1 MHz Q of the 2nd secondorder filter changes by 1% Q of the first secondorder filter changes by about 12.4% 27 High Pass Filter m = 1, n = 0 and p = 0 2 s -K 2 ωp Vo = 2 Vi ⎡ ⎤ s s + 1⎥ ⎢(1 + K ) 2 + ωp ωpQp ⎢⎣ ⎥⎦ Natural frequency of the active filter ωp ω0 = (1 + K ) Natural frequency of the active filter decreases by a factor 1 + K Qa = Qp 1 + K The quality factor Q of the active filter can be increased to the required value through suitable selection of K. 28 Passive HP filter 29 Active HP filter 30 Active HP Filter (contd.,) 2 s -K 2 Vo ω0 = 2 where ωp Vi s s + +1 2 ω0 ω0Q ( ) Qa = (C1R1 C2R 2 ) + ( 2 ωp 1 = ; ω0 = and C1R1C2R 2 1+K 1+K (C2R2 ) C1R1 ) (1 + (R1 R 2 ) ) = Qp 1 + K Required Qa is obtained by selecting K. ωp is determined for a specified ω0 and K. 31 Topology of active HP filter 32 Example If the lower cut off frequency is selected as 0.4 Hz. 1 Assuming C1 = C2 = C and R1 = R 2 = R; Qp = 3 1 Qa = for maximally flat response 2 1+K 1 Qa = Qp 1 + K = = ;K = 3.5 3 2 1 1 1 ω0 = 2π × 0.4 = = ;RC = RC 1 + K RC 4.5 2π × 0.4 4.5 1 For R = 100kΩ;C = = 1.8 µF 0.8π 4.5 33 Simulation 34 Single Op Amp Topology Buffer amplifier can be removed by suitable adjustment of the resistances R1 = R = R3//R4 and C1 = C2 = C 35 Q of active filters Q of the circuit gets enhanced by a factor of 1 + K in case of low-pass and high-pass active filters Natural frequency of the active low-pass filter ω0 = Natural frequency of the active high-pass filter ω0 = ( ( ) 1 + K ωp ωp 1+K ) 36 Effect of finite GB High Pass filter with f0 = 400 Hz and Q = 5 Q = 5 gives K = 224 For f0=400 Hz and R = 100 kW gives C = 265 pF 37 Effect of finite GB (contd.,) ⎡ 2 (1 + K ) s ⎤ K changes because of finite GB to K ⇒ K ⎢1 ⎥ GB ⎢⎣ ⎥⎦ ⎡ 2 (1 + K ) s ⎤ s2 -K ⎢1 ⎥ 2 GB ωp Vo ⎣ ⎦ = 2 Vi ⎡⎛ ⎤ ⎞ 2(1 + K)s s s ⎧ ⎫ + 1⎥ ⎢⎜1 + K ⎨1 ⎬⎟ 2 + GB ω Q ω ⎩ ⎭ ⎢⎝ ⎠ ⎥⎦ p p p ⎣ ⎡ 2 (1 + K ) s ⎤ s2 -K ⎢1 ⎥ 2 GB ωp Vo ⎣ ⎦ = Vi ⎡ ⎤ 2Kω0Qa ⎞ s2 s ⎛ 1+ + 1⎥ ⎢(1 + K ) 2 + ⎜ ⎟ GB ⎠ ωp ωpQp ⎝ ⎢⎣ ⎥⎦ Q of the high-pass filter simulated using 741 Op Amp having a GB of 1 MHz changes to 2.63 that is by 48% 38 Conclusion 39 Conclusion 40