Active Filters: realization of complex left hand plane poles (Complex Transfer Functions regarding the frequency) using only resistors and capacitors for the passive elements. Xavier Gago 1 Analogue Filter Characteristics Passive Filters: L, C and R Active Filters: A.O., C and R R2 L1 C1 XL = jLω 1 R1 C1 XC = jCω 2 L2 3 Vs C2 1 Vs Active Filters: advantages 1.- Full isolation between stages (full control of input and output impedances) - Easy design, regardless Z load - Easy multi-stage design 2.- It is possible to design complex filters without using inductances L (expensive) 3.- Fast filter design for a wide range of applications Active Filters: Disadvantages 1- They require power supply VCC (but circuit Vcc is available) 2- Limited bandwidth (due to the OPAMs). 3- Low power levels (impossible to use in speakers, mains filters,..) 2 1 Analogue Filter Characteristics Low Pass Filters types High Pass 10 10 5 5 0 0 -5 -5 -10 -10 -15 -15 -20 -20 modulo modulo -25 -25 -30 -30 -35 -35 -40 -40 -45 -45 0 1 2 3 4 5 0 1 2 3 4 5 10 10 10 10 10 10 10 10 10 10 10 10 f(Hz) f(Hz ) Band Pass Band Reject 10 10 0 0 -10 Example: Low-Pas -10 -20 -30 -20 modulo -40 modulo -30 -50 -40 -60 -50 -70 -80 0 1 2 3 4 5 6 -60 2 3 4 5 6 10 10 10 10 10 10 10 10 10 10 10 10 f(Hz) f(Hz) 3 Analogue Filter Characteristics Filter Order Filter order = Vi(s) VO(s) N(s) H(s) = Pole number = H(s) D()s D(s) order s = jω, complex variable, 10 0 VO ()s H(s) = -10 first order Vi ()s -20 -30 N order -40 -50 modulo -60 -70 fourth order Slope rate: 20N dB/dec -80 -90 -100 2 3 4 5 6 The higher ther order, the better the filter ! 10 10 10 10 10 f(Hz) 4 2 Filter Order and Attenuation The Gain H(s) for a low pass filter in the attenuation zone is |H| = k / ωn |H1| 20log H1 =20log k - 20n logω1 20log H2 =20log k - 20n logω2 |H2| 20log H1 - 20logH2 =20n log (ω2 - ω1 ) ω1 ω2 H (dB) - H (dB) 1 2 =20 n log ω2 - logω1 For a two frequencies ω 1 and ω 2 separated by a decade, H1(dB) - H2 (dB) =20 n dB / dec Useful to determinate the filter order for a pre-fixed attenuation 5 Analogue Filter Characteristics Filter Order First Order Filters: Transfer Function Low Pass High Pass ω s H(s) = G⋅ O H(s) = G⋅ s +ω O s +ω O Second Order Filters: Transfer Function Low Pass High Pass ω 2 s 2 H(s) = G⋅ O H(s) = G⋅ 2 2 2 2 s + a⋅ω O ⋅s +ω O s + a⋅ω O ⋅s +ω O Band Pass Band Reject s s 2 +ω 2 H(s) = G⋅ H(s) = G⋅ Z s 2 + a⋅ω ⋅s +ω 2 2 2 O O s + a⋅ω O ⋅s +ω O 6 3 Analogue Filter Characteristics Filter Response It determines the pass band characteristics 1- Butterworth 10 5 a- Moderate phase and amplitude response 0 b- Moderate oscillations for pulses -5 -10 Butterworth 2- Bessel -15 Bessel modulo -20 Chevyshev 1 dB a- Linear response -25 Chevyshev 2 dB b- Low oscillations in case of pulses -30 Chevysh3ev 1 dB c- Need high order -35 -40 2 3 4 5 10 10 10 10 3- Chebyshev f(Hz) a- Fast decrement of the gain b- Poor phase response c- Large oscillations in case of pulses d- NO monotone filter: ripple in the bandwidth 7 Response filter comparisons Butterworth Filter • Flat Response and Steep Attenuation • OK for Step Function Chebyshev Filter • Steeper than Butterworth • More Non-Linear than Butterworth 8 4 Analogue Filter Characteristics Chebyshev Filters Larger ripple Faster gain decrement 10 10 0 5 -10 0 -20 -5 -30 -40 -10 3dB Orden 3 -50 modulo modulo 1dB -15 -60 Orden 6 -70 -20 -80 -25 -90 -30 -100 2 3 4 5 2 3 4 5 6 10 10 10 10 10 10 10 10 10 f(Hz) f(Hz ) 9 First Order Active Filters First order low pass filter XBP1 1 1 VO ()s = ⋅ Vi ()s in out C ⋅s R ⋅C ⋅s +1 V ()s = 1 ⋅V ()s 1 1 O 1 i 2 +R1 ω O 1 R1 3 C1 ⋅s H(s) = ω O = 1 s +ω R C 1kohm O 1 1 1 C1 1uF C ⋅R 1 Vs H()s = 1 1 ω = ω = 1 C 3dB O s + R1C1 C1 ⋅R1 ω0 = pole frequency 10 0 0 -20 -10 -40 -20 -60 -30 fase modulo -40 -80 -50 -100 -60 -120 -70 -80 -140 0 1 2 3 4 5 6 0 1 2 3 4 5 6 10 10 10 10 10 10 10 10 10 10 10 10 10 10 f(Hz ) f(Hz) 10 5 n- order Active Filters Second order low pass filter 10 Type a kO Bessel 1.732 1.274 5 Butterworth 1.414 1 0 Chebyshev 1dB 1.045 0.863 -5 -10 Bessel Chebyshev 2dB 0.895 0.852 -15 Butterworth Chebyshev 3dB 0.767 0.841 modulo -20 Chevyshev 1 dB -25 Chevyshev 2 dB 2 -30 ω O Chevysh3ev 1 dB H(s) = G⋅ -35 2 2 s + a⋅ω O ⋅s +ω O -40 2 3 4 5 10 10 10 10 f(Hz) ω = ω ⋅k O C 3dB O Un-dumped natural frequency 11 n- order Active Filters Second order high pass filter Type a kO Bessel 1.732 1.274 10 Butterworth 1.414 1 5 0 Chebyshev 1dB 1.045 0.863 -5 Chebyshev 2dB 0.895 0.852 -10 Chebyshev 3dB 0.767 0.841 -15 Bessel modulo -20 Butterworth -25 Chevyshev 1 dB 2 s -30 H(s) = G⋅ Chevyshev 2 dB s 2 a s 2 -35 + ⋅ω O ⋅ +ω O -40 Chevysh3ev 1 dB 2 3 4 5 6 10 10 10 10 10 f(Hz) ω = ω k O C 3dB O Un-dumped natural frequency 12 6 n- order Active Filters Third order filter Second order section First order section Type a kO Type kO Bessel 1.447 1.274 Bessel 1.328 Butterworth 1.414 1 Butterworth 1 Chebyshev 1dB 0.496 0.911 Chebyshev 1dB 0.452 Chebyshev 2dB 0.402 0.913 Chebyshev 2dB 0.322 Chebyshev 3dB 0.326 0.916 Chebyshev 3dB 0.299 Fourth order filter Second order section Second order section Type a kO Type a kO Bessel 1.916 1.436 Bessel 1.241 1.610 Butterworth 1.848 1 Butterworth 0.765 1 Chebyshev 1dB 1.275 0.502 Chebyshev 1dB 0.281 0.943 Chebyshev 2dB 1.088 0.466 Chebyshev 2dB 0.224 0.946 Chebyshev 3dB 0.929 0.443 Chebyshev 3dB 0.179 0.950 13 n- order Active Filters Fifth order filter 2º order 2º order 1º order Type a kO akO kO Bessel 1.775 1.613 1.091 1.819 1.557 Butterworth 1.618 1 0.618 1 1 Chebyshev 1dB 0.714 0.634 0.180 0.961 0.280 Chebyshev 2dB 0.578 0.624 0.142 0.964 0.223 Chebyshev 3dB 0.468 0.614 0.113 0.967 0.178 Sixth order filter 2º order 2º order 2º order Type a kO akO akO Bessel 1.959 1.609 1.636 1.694 0.77 1.910 Butterworth 1.932 1 1.414 1 0.518 1 Chebyshev 1dB 1.314 0.347 0.455 0.733 0.125 0.977 Chebyshev 2dB 1.121 0.321 0.363 0.727 0.0989 0.976 Chebyshev 3dB 0.958 0.298 0.289 0.722 0.0782 0.975 14 7 Sallen – Key structure: low pass filter C1 1 2 R ⋅C1 ⋅C 2 R R U1 H()s = 2 1 1 s 2 + ⋅s + 3 2 Vs R⋅C R ⋅C ⋅C 2 1 1 2 2 C2 ω H(s) = K ⋅ O 2 2 s + a⋅ω O ⋅s +ω O In a Sallen-Key structure, K=1 2 1 a and ω from a⋅ω = ω 2 = O O R⋅C O 2 Filters Response data tables 1 R ⋅C1 ⋅C 2 1- kO is chosen as desired (ωO = kO ωC), then a is chosen in the table 2- C1 value is chosen 2 1 3- Others components are calculated as R = C = a C 2 2 2 ⋅ωO ⋅ 1 C1 ⋅R ⋅ωO 15 Sallen – Key structure: low pass filter Example: Second order low pass filter, Butterworth response, fC = 1 kHz and Gain=1 3 2 k0=1; ωO = 2 π 10 R = = 225 k By design, C1 = 1 nF 3 −9 a = 1.414 1.414 ⋅2⋅π ⋅10 ⋅10 Attention: it may be very high value C1 = 100 nF R = 2.25 kΩ 1 C = = 5⋅10 −8 = 50 nF 2 2 C = 50 nF 10 −7 ⋅()2.25 ⋅10 3 ⋅ 4⋅π 2 ⋅10 6 2 100nF 0 C1 XBP1 -20 2.25kohm 2.25kohm U1 in out R R -40 1 3 Vs -60 2 modulo 50nF C2 -80 -100 -120 2 3 4 5 6 10 10 10 10 10 f(Hz) 16 8 Sallen – Key structure: high pass filter R1 s 2 H()s = 2 1 C C U1 s 2 + ⋅s + 1 2 R 2 ⋅C C ⋅R1 ⋅R 2 3 Vs 2 R2 s 2 H(s) = K ⋅ 2 2 s + a⋅ω O ⋅s +ω O In a Sallen-Key structure, K=1 2 1 a and ω0 from a⋅ω = ω 2 = Filters Response data tables O R ⋅C O 2 2 C ⋅R1 ⋅R 2 1- kO is chosen as desired (ωO = kO ωC), then, a is chosen in the table 2- C value is chosen 2 1 R = R = 3- Components calculated as 2 1 2 2 a⋅ω O ⋅C R 2 ⋅C ⋅ω O 17 Sallen – Key structure: high pass filter Example: Second order high pass filter, Bessel response, fC = 5 kHz and Gain=1 4 kO = 1.274 ωO = ωC / kO = 2.466 10 2 a = 1.732 By design C = 10 nF R = = 4.68 k 2 1.732⋅2.466⋅10 4 ⋅10 −8 1 R = = 3.51k 1 2 2 4.68 ⋅10 3 ⋅()(10 -8 ⋅ 2.466 ⋅10 4 ) 10 3.51kohm 0 R1 -10 XBP1 10nF 10nF -20 U1 in out C C 1 -30 3 modulo Vs 2 -40 4.68kohm R2 -50 -60 -70 2 3 4 5 6 10 10 10 10 10 f(Hz) 18 9 Signal Conditioning: Summary Signal conditioning circuits amplify low-level signals and then isolate and filter them for more accurate measurements.