Active Bandpass and Active Bandstop Filters Basedon Single Operational Amplifier(EE 221 Spring 2003 Midterm Project)Ramprasad Potluri∗April 20, 20031 Active Bandpass Filter (ABPF)1.1 Transfer functions of ABPFOne transfer function of an ABPF is the following:HBP=AvBss2+ Bs + ω20(1)Here, B is the bandwidth of the ABPF, ω0— the center frequency of the ABPF, and Av— the “gain” of the filter.B =ω0QEquation (1) can be transformed as follows:HBP=Avω0Qss2+ω0Qs + ω20=µAvQω0¶Ãss2ω20+1Qω0s + 1!(2)=G0ss2ω20+2ξω0s + 1(3)From Equations (2) and (3), the following can be seen:G0=AvQω0=AvBω20(4)ξ =12Q=B2ω0(5)∗Department of Electrical & Computer Engineering, University of Kentucky, Lexington, KY, USA.Figures created using Metagraf. Document prepared using LATEX.1andQ =12ξ(6)B = 2ω0ξ (7)Av=G0ω02ξ(8)1.2 One implementation of ABPFOne implementation of the ABPF is shown in Figure 1.ViVOR1R2R3C1C2−++−+−Figure 1: ABPF (From Encyclopedia of Elec-tronic Circuits, vol 7; Rudolf F. Graf andWilliam Sheets; McGraw-Hill; 1999; Page327.)ViVOR1R2R3C1C2−++−+−IinI1I2I4V1V−V+I3Figure 2: Analysis of ABPF.1.3 Analysis of the implementationThe following equations can be written based on Figure 2:V1R2= I3(9)Iin= I1+ I2+ I3(10)I4≈ I1(11)V−≈ V+(12)V1· jωC1= I1(13)(V1− VO)(jωC2) = I2(14)V−− VOR3= I4=⇒ −VOR3= I1(due to Equations (11) and (12)) (15)Vi− V1R1= Iin(16)21.4 Obtaining the transfer function of the implementationNote that equations (13), (14), (9), (16) define the components of Equation (10). So,Equation (10) can be written as follows:Vi− V1R1= V1· jωC1+ (V1− VO)jωC2+V1R2(17)To obtain a relationship between Viand VO, we need to eliminate V1from Equation (17).For this, we can use equations (13) and (15) thus:V1· jωC1= I1= −VOR3=⇒ V1= −VOR3· jωC1So, Equation (17) gives:Vi+VOjωC1R3R1= −VOR3+µ−VOR3jωC1− VO¶jωC2−VOR2R3jωC1=⇒ViR1= −½VOjωC1R1R3+VOR3+VOjωC1R3jωC2+ VOjωC2+VOR2R3jωC1¾=⇒ Vi= −VO½1jωC1R3+R1R3C2R1C1R3+ jωC2R1+R1jωC1R2R3¾=⇒VOVi=−11jωC1R3+R1R3+C2R1C1R3+ jωC2R1+R1jωC1R2R3=−jωC1R2R3R2+ jωC1R1R2+ jωC2R1R2+ (jω)2C1C2R1R2R3+ R1=−jωC1R2R3C1C2R1R2R3(jω)2+ (C1+ C2)R1R2jω + (R1+ R2)=−C1R2R3R1+R2jωC1C2R1R2R3R1+R2(jω)2+(C1+C2)R1R2R1+R2jω + 1The last equation can be written as follows (with the substitution s = jω and by takingC1C2R1R2R3R1+R2out of the denominator as the common factor):VO(s)Vi(s)=−1C2R1ss2+C1+C2C1C2R3s +R1+R2C1C2R1R2R3(18)Equation (18) is the transfer function of the filter from Vito VO.31.5 Determining the values of C’s and R’s for given Av, ω0, QBy comparing equations (1) and (18), the following can be written:ω0Q=C1+ C2C1C2R3(19)Avω0Q=1C2R1(20)ω20=R1+ R2C1C2R1R2R3(21)From equations (19) and (20), it can be seen that:Av=C1R3(C1+ C2)R1(22)1.5.1 First method of determining the values of C’s and R’sThis method works when 2Q2− Av> 0.LetC1= C2= CThen, Equation (19) gives:ω0Q=2CR3=⇒ R3=2Qω0C(23)Equation (20) gives:Avω0Q=1CR1R1=Qω0CAv(24)Equation (21) gives:ω20=µ1R1+1R2¶1C2R3=⇒ ω20=1C2R1R3+1C2R2R3=⇒ ω20−1C2R1R3=1C2R2R3=⇒ ω20C2R3−1R1=1R2=⇒ R2=R1ω20C2R1R3− 14Plugging in the expressions for R1and R3into this last expression gives:R2=Qω0CAvω20C2Qω0CAv2Qω0C− 1=Qω0CAv³2CQ2Av− 1´=⇒ R2=Qω0C(2Q2− Av)(25)Equation (22) gives:Av=R32R11.5.2 Second method of determining the values of C’s and R’sThis method can be used when 2Q2− Av< 0.LetR1= R2= RThen, equations (19), (20), (22) can be written as follows:ω0Q=C1+ C2C1C2R3=µ1C1+1C2¶1R3(26)Avω0Q=1C2R(27)ω20=2C1C2RR3(28)Equation (27) gives:C2=QAvω0R(29)Equation (26) gives:ω0R3Q−1C1=1C2(30)Equation (28) gives:1C1=RC2ω20R32(31)Use Equation (31) in Equation (30):ω0R3Q−RC2ω20R32=1C2=⇒µω0Q−C2ω20R2¶R3=1C25Into the last equation, plug in the expression for C2from Equation (29):µω0Q−QAvω0Rω20R2¶R3=Avω0RQ=⇒µω0Q−Qω02Av¶R3=Avω0RQ=⇒ R3=AvRQ11Q−Q2AvR3=2A2v2Av− Q2R (32)Plug in the expression for R3from Equation (32) and for C2from Equation (29) intoEquation (31) to obtain an expression for C1:C1=2RC2ω20R3=2Rω20(QAvω0R)(2A2v2Av−Q2)R=⇒ C1= (2Av− Q2AvQω0)1R(33)1.5.3 Algorithm to determine C’s and R’sINPUT: G0, ξ, ω0.OUTPUT: C1, C2, R1, R2, R3.1. Use equations (6), (7), (8) to obtain the corresponding values of Q, B, Av.2. Compute 2Q2− Av.IF 2Q2− Av> 0,(a) Choose C1= C2= C = some value.(b) Use equations (23), (24), (25) to determine R1, R2, R3.ELSE(a) Choose R1= R2= R = some value.(b) Use equations (29), (32), (33) to determine C1, C2, R3.END3. ENDRemark 1 : Note that when 2Q2− Av< 0 in Section 1.5.1, then:2Q2− Av< 0 =⇒ 2Q2< Av=⇒ 4Q2< 2Av=⇒ 3Q2< 2Av− Q2Thus, 2Av− Q2is positive when 2Q2− Avis negative. So, at least one of the two methodswill surely work.62 Active Bandstop filter (ABSF)2.1 One implementation of ABSFOne implementation of the ABSF is shown in Figure 3.ViVOR1R2R3C1C2−++−+−R4Figure 3: ABSF (From The Electrical Engineering Handbook, 2nd Ed.; Editor-in-Chief: RichardDorf; CRC Press 1997; Page 756.)2.2 Analysis of the implementationThe circuit of Figure 3 can be rearranged so that the ∆ portion in it is made obvious asshown in Figure 4 (note that we could have used this technique in the case of ABPF too).VOR2Z1=1jωC1Z2=1jωC2−++−R4ViR1R3+−Figure 4: Slight rearrangement of the circuit of Figure 3The ∆ in the circuit of Figure 4 can be transformed into a Y as shown in Figure 5.With respect to Figure 5, the following equations can be written:Za=Z1Z2Z1+ Z2+ R2=1jωC11jωC21jωC1+1jωC2+ R27VOZb=Z1R2Z1+Z2+R2Zc=Z2R2Z1+Z2+R2−++−R4ViR1R3+−Za=Z1Z2Z1+Z2+R2Figure 5: The circuit of Figure 4 after ∆ → Y transformation.Zb=Z1R2Z1+ Z2+ R2=1jωC1R21jωC1+1jωC2+ R2Zc=Z2R2Z1+ Z2+ R2=1jωC2R21jωC1+1jωC2+ R2The circuit of Figure 5 will be analyzed here. For convenience, this circuit has beenredrawn in Figure 6 to show some currents and voltages of interest.VO−++−R4ViZa+ R1R3+−ZbZci−i+v−v+V1Figure 6: The circuit of Figure 5 after slight modificationHere are two common approximations:1. i−≈ i+≈ 0, for a (near) ideal op-amp2. There exists a feedback in the circuit via Zb. So, v−≈ v+(virtual short circuit).From the first approximation, it follows thatV1− v−Zc= i−≈ 0 =⇒ V1≈ v−8Using the second approximation, it follows that V1≈ v+.But, by voltage division,v+=R4R3+ R4ViSo,V1≈R4R3+ R4Vi(34)Next, note that, since i−= 0,Vi− V1R1+ Za=V1− VOZbViR1+