EE247 Lecture 4. For simplicity, will start with all pole ladder type filters. Convert to integrator based form- example shown
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1 EE247 Lecture 4 Active ldder type filters For simplicity, will strt with ll pole ldder type filters Convert to integrtor bsed form exmple shown Then will ttend to high order ldder type filters incorporting zeros Implement the sme 7th order elliptic filter in the form of ldder C with zeros Find level of sensitivity to component mismtch Compre with cscde of biquds Convert to integrtor bsed form utilizing SFG techniques Effect of integrtor nonidelities on filter frequency chrcteristics EECS 247 Lecture 4: Active Filters 2008 H.K. Pge C Ldder Filters Exmple: 5 th Order Lowpss Filter L2 L4 C C3 C5 Mde of resistors, inductors, nd cpcitors Doubly terminted or singly terminted (with or w/o R L ) Doubly terminted LC ldder filters Lowest sensitivity to component mismtch EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 2
2 LC Ldder Filters C L2 C3 L4 C5 First step in the design process is to find vlues for Ls nd Cs bsed on specifictions: Filter grphs & tbles found in: A. Zverev, Hndbook of filter synthesis, Wiley, 967. A. B. Willims nd F. J. Tylor, Electronic filter design, 3 rd edition, McGrw Hill, 995. CAD tools Mtlb Spice EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 3 LC Ldder Filter Design Exmple Design LPF with mximlly flt pssbnd: f3db = 0MHz, fstop = 20MHz fstop Mximlly flt pssbnd Butterworth Find minimum filter order : Here stndrd grphs from filter books re used fstop / f3db = 2 >27dB Minimum Filter Order 5th order Butterworth Pssbnd Attenution 3dB 2 Νοrmlized ω From: Willims nd Tylor, p. 237 Stopbnd Attenution 30dB EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 4
3 LC Ldder Filter Design Exmple Find vlues for L & C from Tble: Note L &C vlues normlized to ω 3dB = Denormliztion: Multiply ll L Norm, C Norm by: L r = R/ω 3dB C r = /(RXω 3dB ) R is the vlue of the source nd termintion resistor (choose both Ω for now) Then: L= L r xl Norm C= C r xc Norm From: Willims nd Tylor, p..3 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 5 LC Ldder Filter Design Exmple Find vlues for L & C from Tble: Normlized vlues: C Norm =C5 Norm =0.68 C3 Norm = 2.0 L2 Norm = L4 Norm =.68 Denormliztion: Since ω 3dB =2πx0MHz L r = R/ω 3dB = 5.9 nh C r = /(RXω 3dB )= 5.9 nf R = C=C5=9.836nF, C3=3.83nF L2=L4=25.75nH From: Willims nd Tylor, p..3 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 6
4 Lst Lecture: Exmple: 5 th Order Butterworth Filter Vin =Ω L2=25.75nH C 9.836nF L4=25.75nH C3 3.83nF C nF =Ω Specifictions: f3db = 0MHz, fstop = 20MHz >27dB Used filter tbles to obtin Ls & Cs Mgnitude (db) SPICE simultion Results 30dB 30 6 db pssbnd ttenution 40 due to double termintion Frequency [MHz] EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 7 LowPss C Ldder Filter Conversion to Integrtor Bsed Active Filter V V 2 V3 V4 V5 V6 I L2 I3 L4 I 5 C C3 C5 I 2 I4 I 6 I 7 To convert C ldder prototype to integrtor bsed filer: Use Signl Flowgrph technique Nme currents nd voltges for ll components Use KCL & KVL to derive equtions Mke sure rective elements expressed s /s term V(C ) =f(i) & I(L)=f(V) Use sttespce description to derive the SFG Modify & simply the SFG for implementtion with integrtors e.g. convert ll current nodes to voltge EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 8
5 LowPss C Ldder Filter Conversion to Integrtor Bsed Active Filter V V 2 V3 V4 V5 V6 I L2 I3 L4 I 5 C C3 C5 I 2 I4 I 6 I 7 Use KCL & KVL to derive equtions: I V = Vin V 2, V 2 2 =, V 3 = V 2 V sc 4 I I V =, V 5 = V 4 V 6, V 6 = V o = V 6 sc 3 sc 5 V V3 I =, I 2 = I I 3, I 3 = sl2 V5 V6 I 4 = I 3 I 5, I 5 =, I 6 = I 5 I 7, I7 = sl4 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 9 I LowPss C Ldder Filter Signl Flowgrph V = Vin V 2, I V 2 2 = sc, V 3 = V 2 V 4 I I V =, V 5 = V 4 V 6, V 6 = sc 3 sc 5 V o = V 6 V V3 I =, I 2 = I I 3, I3 = sl2 V5 V6 I 4 = I 3 I 5, I 5 =, I 6 = I 5 I 7, I7 = sl4 Vin V V 2 sc sl2 sc3 sl4 sc5 I 2 V3 V4 V5 V6 Vo I3 I4 I 5 I 6 I 7 SFG EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 0
6 LowPss C Ldder Filter Signl Flowgrph V V 2 V3 V4 V5 V6 I L2 I3 L4 I 5 C C3 C5 I 2 I4 I 6 I 7 Vin V V 2 sc V3 V4 V5 V6 sl2 sc3 sl4 sc5 Vo I I 2 I3 I4 I 5 I 6 I 7 SFG EECS 247 Lecture 4: Active Filters 2008 H.K. Pge Vin V V 2 sc I LowPss C Ldder Filter Normlize I 2 V3 V4 V5 V6 sl2 sc3 sl4 sc5 Vo I 3 I4 I 5 I 6 I 7 Vin V ' V * R V 2 V3 V4 V5 V6 * * R R sc * R sc * sl 3R sl sc * 2 4 5R V ' V ' 2 3 V ' 4 V ' 5 V ' 6 Vo * R ' V 7 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 2
7 Vin V ' V * R LowPss C Ldder Filter Synthesize V 2 V3 V4 V5 V6 * * R R sc * R sc * sl 3R sl sc * 2 4 5R V ' V ' 2 3 V ' 4 V ' 5 V ' 6 Vo * R ' V 7 * R V 2 V4 V6 sτ 2 sτ sτ 3 sτ 4 sτ 5 * R ' V 3 ' V 5 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 3 LowPss C Ldder Filter Integrtor Bsed Implementtion * R V 2 V4 V6 sτ 2 sτ sτ 3 sτ 4 sτ 5 * R ' V 3 * L2 * * L4 * * = C.R, 2 = = C.R, C.R, C.R, C.R * 2 3 = 3 4 = = * 4 5 = 5 R R τ τ τ τ τ ' V 5 Min building block: Integrtor Let us strt to build the filter with RC& Opmp type integrtor EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 4
8 OpmpRC Integrtor SingleEnded Differentil 2 R 2 R CI 2 2 CI CI Vo= Vin srci Vo Vo = ( Vin Vin ) srci Vin2 sr2ci ( Vin2 Vin2 ) sr2ci Note: Implementtion with singleended integrtor requires extr circuitry for sign inversion wheres in differentil cse both signl polrities re vilble R 2 R R R 2 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 5 Differentil Integrtor Bsed LP Ldder Filter Synthesize V 3 V 5 CI2 CI2 CI4 CI4 CI CI CI3 CI3 CI5 CI5 V 2 V 4 V O First itertion: All resistors re chosen=ω Vlues for τ x =R x CI x found from C nlysis Cpcitors: CI=CI5=9.836nF, CI2=CI4=25.45nF, CI3=3.83nF EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 6
9 Simulted Mgnitude Response 0.5 V ' V 3 V 4 ' V 5 0MHz 0MHz EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 7 Scle Node Voltges To mximize dynmic rnge scle node voltges Scle by fctor s EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 8
10 Differentil Integrtor Bsed LP Ldder Filter Node Scling V 3 X.2 V 5 X.8 X.2 X /.2 X.6/.2 X.2/.6 X.8/.6 X.6/.8 X 2/.8 X.8/2 V 2 V V O X 2 4 X.6 Second itertion: Nodes scled, note output node x2 Resistor vlues scled ccording to scling of nodes Cpcitors the sme : C=C5=9.836nF, C2=C4=25.45nF, C3=3.83nF EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 9 Mximizing Signl Hndling by Node Voltge Scling Before Node Scling After Node Scling Scle by fctor s EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 20
11 Filter Noise Integrted Noise Output noise voltge spectrl density.4μvrms Totl the output:.4 μv rms (noiseless opmps) Tht s excellent, but: Cpcitors too lrge for integrtion lrge Si re Resistors too smll high power dissiption Typicl pplictions llow higher noise, ssuming tolerble noise in the order of 40 μv rms EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 2 Scle to Meet Noise Trget Scle cpcitors nd resistors to meet noise objective 4μVrms s = 0 4 (V n /V n2 ) 2 Noise fter scling: 4 μv rms (noiseless opmps) EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 22
12 Differentil Integrtor Bsed LP Ldder Filter Finl Design R2 R C2 R6 R7 C2 C4 C R3 C R4 C3 C3 R5 R8 R9 C4 C5 R C5 R0 Finl itertion: Bsed on scled nodes nd noise considertions Cpcitors: C=C5=0.9836pF, C2=C4=2.545pF, C3=3.83pF Resistors: R=.77K, R2=9.677Κ, R3=0Κ, R4=2.82Κ, R5=8.493Κ, R6=.93Κ, R7=7.8Κ, R8=0.75Κ, R9=8.38Κ, R=0Κ, R=9.306K V O EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 23 C Ldder Filters Including Trnsmission Zeros All poles C L2 C3 L4 C5 Poles & Zeros C2 C4 C6 C L2 C3 L4 C5 L6 C7 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 24
13 C Ldder Filter Design Exmple Design bsebnd filter for CDMA IS95 cellulr phone receive pth with the following specs. Filter frequency msk shown on the next pge Allow enough mrgin for mnufcturing vritions Assume overll tolerble pssbnd mgnitude vrition of.8db Assume the 3dB frequency cn vry by 8% due to mnufcturing tolernces & circuit inccurcies Assume ny phse impirment cn be compensted in the digitl domin * Note this is the sme exmple s for cscde of biqud while the specifictions re given closer to rel product cse EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 25 C Ldder Filter Design Exmple CDMA IS95 Receive Filter Frequency Msk 0 Mgnitude (db) k 700k 900k.2M Frequency [Hz] EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 26
14 C Ldder Filter Design Exmple: CDMA IS95 Receive Filter Since phse impirment cn be corrected for, use filter type with mx. rolloff slope/pole Filter type Elliptic Design filter freq. response to fll well within the freq. msk Allow mrgin for component vritions & mismtches For the pssbnd ripple, llow enough mrgin for ripple chnge due to component & temperture vritions Design nominl pssbnd ripple of 0.2dB For stopbnd rejection dd few db mrgin 445=49dB Finl design specifictions: fpss = 650 khz Rpss = 0.2 db fstop = 750 khz top = 49 db Use Mtlb or filter tbles to decide the min. order for the filter (sme s cscded biqud exmple) 7 th Order Elliptic EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 27 C LowPss Ldder Filter Design Exmple: CDMA IS95 Receive Filter 7 th order Elliptic C2 C4 C6 C L2 C3 L4 C5 L6 C7 Use filter tbles & chrts to determine LC vlues EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 28
15 C Ldder Filter Design Exmple: CDMA IS95 Receive Filter Specifictions fpss = 650 khz Rpss = 0.2 db fstop = 750 khz top = 49 db Use filter tbles to determine LC vlues Tble from: A. Zverev, Hndbook of filter synthesis, Wiley, 967 Elliptic filters tbulted wrt reflection coeficient ρ Rpss= 0 log( ρ 2 ) Since Rpss=0.2dB ρ =20% Use tble ccordingly EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 29 C Ldder Filter Design Exmple: CDMA IS95 Receive Filter Tble from Zverev book pge #28 & 282: Since our spec. is Amin=44dB dd 5dB mrgin & design for Amin=49dB EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 30
16 Tble from Zverev pge #28 & 282: Normlized component vlues: C=.7677 C2= L2=.9467 C3=.534 C4=.0098 L4= C5= C6=0.72 L6= C7= EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 3 C Filter Frequency Response Component vlues denormlized Frequency response simulted Frequency msk superimposed Frequency response well within spec. Mgnitude (db) Frequency [khz] EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 32
17 Frequency Response Pssbnd Detil Pssbnd well within spec Mke sure enough mrgin is llowed for vritions due to process & temperture Mgnitude (db) Frequency [khz] EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 33 C Ldder Filter Sensitivity The design hs the sme specifictions s the previous exmple implemented with cscded biquds To compre the sensitivity of C ldder versus cscdedbiquds: Chnged ll Ls &Cs one by one by 2% in order to chnge the pole/zeros by % (similr test s for cscded biqud) Found frequency response most sensitive to L4 vritions Note tht by vrying L4 both poles & zeros re vried EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 34
18 RCL Ldder Filter Sensitivity Component mismtch in C filter: Increse L4 from its nominl vlue by 2% Decrese L4 by 2% Mgnitude (db) L4 nom L4 low L4 high Frequency [khz] EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 35 RCL Ldder Filter Sensitivity 5 0.2dB 5 Mgnitude (db) dB Frequency [khz] EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 36
19 Sensitivity of Cscde of Biquds Component mismtch in Biqud 4 (highest Q pole): Increse ω p4 by % Decrese ω z4 by % 2.2dB Mgnitude (db) dB kHz 600kHz Frequency [Hz] MHz High Q poles High sensitivity in Biqud reliztions EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 37 Sensitivity Comprison for CscdedBiquds versus C Ldder 7 th Order elliptic filter % chnge in pole & zero pir Pssbnd devition Stopbnd devition Cscded Biqud 2.2dB (29%) 3dB (40%) C Ldder 0.2dB (2%).7dB (2%) Doubly terminted LC ldder filters Significntly lower sensitivity compred to cscdedbiquds prticulrly within the pssbnd EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 38
20 7 th order Elliptic C Ldder Filter Design Exmple: CDMA IS95 Receive Filter C2 C4 C6 C L2 C3 L4 C5 L6 C7 Previously lerned to design integrtor bsed ldder filters without trnsmission zeros Question: o How do we implement the trnsmission zeros in the integrtorbsed version? o Preferred method no extr power dissiption no extr ctive elements EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 39 Integrtor Bsed Ldder Filters How Do to Implement Trnsmission zeros? C I C V I V V 2 3 L2 I 3 C I 2 I 4 V 4 C3 I 5 Use KCL & KVL to derive : I I I3 IC I 2 2 = I I3 I C, IC = ( V2 V4) s C, V 2 =, V sc 2 = sc Substituting for IC nd rerrnging : I I3 C V V 2 = s 4 ( C C ) C C EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 40
21 Integrtor Bsed Ldder Filters How Do to Implement Trnsmission zeros? C V I V V 2 3 L2 I 3 C I 2 I 4 V 4 C3 I 5 Use KCL & KVL to derive : I I C V 3 2 = V4 s( C C ) C C I3 I5 C V V 4 = s 2 ( C 3 C ) C 3 C Frequency independent constnts Cn be substituted by: VoltgeControlled Voltge Source EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 4 Integrtor Bsed Ldder Filters Trnsmission zeros C V I V V 2 3 L2 I 3 C C I 2 V 4 I 5 ( ) ( C 3 C ) C V 4 C C C I V 4 2 C 3 C Replce shunt cpcitors with voltge controlled voltge sources: I I3 C V V 2 = s 4 ( C C ) C C Exct sme expressions I3 I5 C V V 4 = s with C present s 2 ( C 3 C ) C 3 C EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 42
22 3 rd Order Lowpss Filter All Poles & No Zeros V I V V 2 3 L2 I 3 C I2 I 4 V 4 C 3 I 5 Vin V V 2 sc V3 V4 sl2 sc 3 Vo I I 2 I 3 I 4 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 43 Implementtion of Zeros in Active Ldder Filters Without Use of Active Elements V I V V 2 3 V 4 I 5 L2 I 3 ( C C ) ( C 3 C ) C V C 4 C C I 2 I V 2 4 C 3 C C C C C C 3 C Vin V V 2 V3 V4 s C C sl2 s( C3 C) ( ) I I 2 I 3 I 4 Vo EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 44
23 Integrtor Bsed Ldder Filters Higher Order Trnsmission zeros V 2 C V 4 C b V 6 Convert zero generting Cs in C loops to voltgecontrolled voltge sources C C 3 V2 V4 ( ) ( ) C C C 3 C C b C C V V 4 C C 2 C C 3 C V b 6 C 3 C b C 5 V 6 ( C 5 C b) C V b 4 C 3 C b EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 45 Vin V V 2 * R s( C C ) R * Higher Order Trnsmission zeros V V 2 V3 V 4 I V 5 5 V6 I 7 I L2 I 3 I L4 4 ( C C ) ( C 3 C C b) ( C 5 C b) C C V V 2 C 4 C C C C I 3 V b 4 C C 2 I 6 5 C b V b 6 C 3 C b C C C b C C C C 3 C C 3 C b b C 5 C b ' V V ' 2 V3 V4 V5 V6 * R * R sr * sl2 ( C3 C Cb) sl4 sr * C C ' V ' ' 3 V 4 V ' 5 V 6 ( 5 b ) EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 46 Vo * R ' V 7
24 Exmple: 5 th Order Chebyshev II Filter 5 th order Chebyshev II Tble from: Willims & Tylor book, p..2 50dB stopbnd ttenution f 3dB =0MHz EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 47 Trnsmission Zero Genertion OpmpRC Integrtor Cx 3 R f Vin Vin2 Vo Vo = s( C Cx ) R R2 Rf 2 R 2 R C C V x in3 C Cx EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 48
25 Differentil Integrtor Bsed LP Ldder Filter Finl Design 5 th Order AllPole R2 R C2 C R3 C R4 C3 C3 R5 R8 R6 R7 C2 C4 R9 C4 C5 R C5 R0 V O EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 49 Differentil Integrtor Bsed LP Ldder Filter Finl Design 5 th Order AllPole V 3 V 5 CI2 CI2 CI4 CI4 CI CI CI3 CI3 CI5 CI5 V 2 V 4 V O EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 50
26 Differentil 5 th Order Chebychev Lowpss Filter C2 C4 C C3 C5 C Cc Cb Cd All resistors Ω Cpcitors: C=36.nF, C2=4.05nF, C3=2.5nF,C4=5.344nF, C5=2.439nF Coupling cpcitors: C=.36nF, Cb=.36nF, Cc=.3nF, Cd=.3nF EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 5 5 th Order Chebyshev II Filter Simulted Frequency Response EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 52
27 7th Order Differentil Lowpss Filter Including Trnsmission Zeros Trnsmission zeros implemented with pir of coupling cpcitors EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 53 Effect of Integrtor NonIdelities on Filter Frequency Chrcteristics In the pssive filter design (C filters) section: Rective element (L & C) nonidelities expressed in the form of Qulity Fctor (Q) Filter impirments due to component nonidelities explined in terms of component Q In the context of ctive filter design (integrtorbsed filters) Integrtor nonidelities Trnsltes to the form of Qulity Fctor (Q) Filter impirments due to integrtor nonidelities explined in terms of integrtor Q EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 54
28 Effect of Integrtor NonIdelities on Filter Performnce Idel integrtor chrcteristics Rel integrtor chrcteristics: Effect of opmp finite DC gin Effect of integrtor nondominnt poles EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 55 Effect of Integrtor NonIdelities on Filter Performnce Idel Integrtor R C Idel opmp Idel Intg. log H ( ω) Idel Integrtor: DC gin= Single DC no nondominnt poles ω H(s) = o s ωo = / RC 0dB ψ Phse 90 o ω 0 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 56
29 Idel Integrtor Qulity Fctor Idel intg. trnsfer function: Since component Q is defined s:: ωo ωo H(s) = = = s jω ω j ωo H Q ( jω ) = R( ω) jx( ω) X ( ω) = R ( ω) Then: intg. Q idel = EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 57 Rel Integrtor NonIdelities Idel Intg. log H ( s) Rel Intg. log H ( s) ω 0 ω P = 0 ω 0 P2P3 ψ ψ 90 o 90 o ωo H(s) = H(s) s s s ( s o )( p2)( ω p3)... EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 58
30 Effect of Integrtor Finite DC Gin on Q log H ( s) ψ ω P = 0 ω ω o ω π ArctnP 2 ωo P Phse led@ ω ω o o (in rdin) 90 o Exmple: =00 P/ ω 0 = /00 phse error 0.5degree EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 59 Effect of Integrtor Finite DC Gin on Q Exmple: Lowpss Filter Idel intg Intg with finite DC gin Finite opmp DC gin Mgnitude (db) Droop in the pssbnd Phse ω 0 Droop in the pssbnd Normlized Frequency EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 60
31 Effect of Integrtor NonDominnt Poles log H ( s) ψ ω 0 P2P ω o ω ω π Arctn o 2 p i= 2 i ωo Phse ω p o i= 2 i (in rdin) 90 o Exmple: ω 0 /P2 =/00 phse error 0.5degree EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 6 Effect of Integrtor NonDominnt Poles Exmple: Lowpss Filter Idel intg Opmp with finite bndwidth Additionl poles due to opmp poles: Mgnitude (db) Peking in the pssbnd Phse ω 0 Peking in the pssbnd In extreme cses could result in oscilltion! Normlized Frequency EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 62
32 Effect of Integrtor NonDominnt Poles & Finite DC Gin on Q log H ( s ) ω P = 0 ψ 90 o ω 0 P2P3 ω o ω π Arctn 2 P ωo ω Arctn o i = 2 p i 90 P ω o ω o i = 2 pi Note tht the two terms hve different signs Cn cncel ech other s effect! EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 63 Integrtor Qulity Fctor Rel intg. trnsfer function: Bsed on the definition of Q nd ssuming tht: H(s) s... ( s s )( p2)( ω p3) o ωo << & >> p 2,3,... It cn be shown tht in the vicinity of unityginfrequency: Q intg. rel ω o i= 2 p i Phse ω 0 Phse ω 0 EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 64
33 Exmple: Effect of Integrtor Finite Q on Bndpss Filter Behvior 0.5 ο φ ω o intg 0.5 ο φ ω o intg Idel Idel Integrtor DC gin=00 Integrtor 00.ω o EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 65 Exmple: Effect of Integrtor Q on Filter Behvior ( 0.5 ο φ led 0.5 ο φ excess ω o intg φ ω o intg ~ 0 Idel Integrtor DC gin=00 & 00. ω ο EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 66
34 Summry Effect of Integrtor NonIdelities on Q Q intg. idel = Q intg. rel ω o p i= 2 i Amplifier finite DC gin reduces the overll Q in the sme mnner s series/prllel resistnce ssocited with pssive elements Amplifier poles locted bove integrtor unitygin frequency enhnce the Q! If nondominnt poles close to unitygin freq. Oscilltion Depending on the loction of unityginfrequency, the two terms cn cncel ech other out! EECS 247 Lecture 4: Active Filters 2008 H.K. Pge 67
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