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- 1. BEJ 30403 ELECTRONIC CIRCUIT ANALYSISAND DESIGN Chapter 2 Active Filter Part III Dr. Rahmat Sanudin Department of Electronic Engineering Faculty of Electrical and Electronic Engineering
- 2. Outline •Second Order LowPass Filter •Second Order High Pass Filter •Second Order Band-Pass Filter •Higher Order Filter Design
- 3. Second Order LPF •The second-order LPF consists of 2 pairs RC using Sallen & Key configuration • Sallen and Key configuration means an op-amp could accommodate up to 2 pairs of RC
- 4. Second Order LPF •At x node: 𝑉− = 𝑉𝑂 𝑅𝐴 𝑅𝐴 + 𝑅𝐹 • Amplifier’s Gain: 𝐴𝑉𝑜 = 𝑉𝑂 𝑉− = 1 + 𝑅𝐹 𝑅𝐴 V- V+
- 5. Second Order LPF •At y node: 𝑉𝑖 − 𝑉 𝑦 𝑅1 + 𝑉𝑂 − 𝑉 𝑦 𝑗𝜔𝐶1 = 𝑉 𝑦 − 𝑉− 𝑅2 • Substitute 𝑗𝜔 = 𝑠 𝑉𝑖 − 𝑉 𝑦 𝑅1 + 𝑉𝑂 − 𝑉 𝑦 𝑠𝐶1 = 𝑉 𝑦 − 𝑉− 𝑅2 V- Vo
- 6. Second Order LPF •At z node: 𝑉 𝑦 − 𝑉− 𝑅2 = 𝑉− ൗ 1 𝑗𝜔𝐶2 𝑉 𝑦 − 𝑉− 𝑅2 = 𝑗𝜔𝐶2𝑉− • Substitute 𝑗𝜔 = 𝑠 𝑉 𝑦 − 𝑉− 𝑅2 = 𝑠𝐶2𝑉− V- Vo
- 7. Second Order LPF IfR1 = R2 = R and C1 = C2 = C simplifying all those equations, the overall gain will be: 𝐴𝑉(𝑠) = 𝑉𝑂 𝑉𝑖 (𝑠) = 𝐴𝑉𝑂 (𝑅𝐶𝑠)2 + (3 − 𝐴𝑉𝑂)𝑅𝐶𝑠 + 1 𝐴𝑉𝑜 = 𝑉𝑂 𝑉− = 1 + 𝑅𝐹 𝑅𝐴 𝑉𝑖 − 𝑉 𝑦 𝑅1 + 𝑉𝑂 − 𝑉 𝑦 𝑠𝐶1 = 𝑉 𝑦 − 𝑉− 𝑅2 𝑉 𝑦 − 𝑉− 𝑅2 = 𝑠𝐶2𝑉−
- 8. Second Order LPF •With R1 = R2 = R and C1 = C2 = C, the overall gain : 𝐴𝑉(𝑠) = 𝑉𝑂 𝑉𝑖 (𝑠) = 𝐴𝑉𝑂 (𝑅𝐶𝑠)2 + (3 − 𝐴𝑉𝑂)𝑅𝐶𝑠 + 1 • The cut-off frequency: 𝑓𝑐 = 1 2𝜋𝑅𝐶 • If R1 ≠ R2 and C1 ≠ C2 : 𝑓𝑐 = 1 2𝜋 𝑅1𝐶1𝑅2𝐶2
- 9. Second Order LPF •The slope or roll-off is closer to the ideal frequency response
- 10. Second Order HPF •Similar to 2nd order LPF, 2nd order HPF also has 2 pairs of RC
- 11. Second Order HPF •At node x: 𝑉− = 𝑉 𝑜 𝑅𝐴 𝑅𝐴 + 𝑅𝐹 ⇒ 𝐴𝑉𝑜 = 𝑉 𝑜 𝑉− = 1 + 𝑅𝐹 𝑅𝐴 • At node y: 𝑉𝑖 − 𝑉 𝑦 ൗ 1 𝑠𝐶1 + 𝑉 𝑜 − 𝑉 𝑦 𝑅1 = 𝑉 𝑦 − 𝑉− ൗ 1 𝑠𝐶2 • At node z: 𝑉 𝑦 − 𝑉− ൗ 1 𝑠𝐶2 = 𝑉− 𝑅2
- 12. Second Order HPF •If 𝑅1 = 𝑅2 = 𝑅 and 𝐶1 = 𝐶2 = 𝐶 then the overall gain will be 𝐴𝑉 𝑠 = 𝑉 𝑜 𝑉𝑖 𝑠 = 𝐴𝑉𝑜 1 𝑠𝑅𝐶 2 + 3 − 𝐴𝑉𝑜 𝑠𝑅𝐶 + 1 • The cut-off frequency: 𝑓𝑐 = 1 2𝜋𝑅𝐶 • If 𝑅1 ≠ 𝑅2 and 𝐶1 ≠ 𝐶2, the cut-off frequency: 𝑓𝑐 = 1 2𝜋 𝑅1𝑅2𝐶1𝐶2
- 13. Second Order BPF •To build a second order BPF, it needs second order LPF and second order HPF 2nd order LPF 2nd order HPF
- 14. Second Order BPF •Frequency response second order BPF:
- 15. Second Order BPF •Finding the critical frequencies: 𝑓𝑐1 = 𝑓𝐿 = 1 2𝜋 𝑅𝐴1𝑅𝐵1𝐶𝐴1𝐶𝐵1 𝑓𝑐2 = 𝑓𝐻 = 1 2𝜋 𝑅𝐴2𝑅𝐵2𝐶𝐴2𝐶𝐵2
- 16. Second Order BPF •Centre frequency (f0): The geometric mean (average) of the cutoff frequencies • Quality factor (Q): The ratio of filter centre frequency (f0 = fc) to filter bandwidth (BW) 𝑄 = 𝑓0 𝐵𝑊 = 𝑓0 𝑓𝐻 − 𝑓𝐿 𝑓0 = 𝑓𝑐1𝑓𝑐2 = 𝑓𝐿𝑓𝐻
- 17. Higher Order FilterDesign @ 6 db/octave @ 12 db/octave @ 18 db/octave @ 24 db/octave
- 18. Higher Order FilterDesign The higher the order, the closer to the ideal response
- 19. Higher Order FilterDesign • When two (or more) filters are cascaded in series, the total transfer function is the product of the individual transfer functions • HTOTAL(s) = H1(s) H2(s) • The required transfer function must be factorised H1(s) H2(s) VIN(s) VIN(s) H1(s) VIN(s) H1(s) H2(s) 𝐻 𝑠 = 𝑉𝑂1 𝑉𝑖𝑛 × 𝑉𝑂 𝑉𝑂1 = 𝐻1 𝑠 × 𝐻2 𝑠 VO1 VO
- 20. Higher Order FilterDesign • In filter design, a term known “Butterworth Polynomials” has been formulated to aid the process of the design • These polynomials is useful to determine the required values of resistors related to filter gain • The list of “Butterworth Polynomials” is given in table in the next slide
- 21. Higher Order FilterDesign
- 22. Higher Order FilterDesign • In higher order filter design, the Sallen & Key configuration is normally used • The Sallen and Key configuration realises a two-pole filter section using a single op-amp • Synthesis by sections creates higher order filters by cascading first and second order sections • Under ideal assumptions, the order of the sections is irrelevant • Practically, dynamic range considerations decide the order of the sections
- 23. Higher Order FilterDesign • To get the highest maximum input signal range: • Sections with the largest damping ratio (lowest Q) should come first • Resonant sections with low damping ratios come last, possibly susceptible to noise • To get the lowest output noise: • Highest gain sections (i.e. most resonant) should come first • Subsequent sections attenuate noise • What is the best order ? • It depends on the required specification
- 24. Higher Order FilterDesign • Example 1: Design a 2nd order HPF with cut-off frequency of 200Hz • Solve for components related to frequency first. Let C1 = C2 = C and R1 = R2 = R • 𝑓 = 1 2𝜋𝑅𝐶 = 200 • Assume a value either R or C, then calculate another component. Let say assume 𝐶 = 2.2𝜇𝐹, thus • 𝑅 = 1 2𝜋 200 2.2𝜇 = 361 Ω
- 25. Higher Order FilterDesign • Find ratio of RF/RA by using Butterworth Polynomials Table 𝑅𝐹 𝑅𝐴 = 2 − 1.414 = 0.586 1.414 = 3 − 𝐴𝑉𝑜 = 3 − 1 + 𝑅𝐹 𝑅𝐴 Choose 𝑅𝐴 = 10𝑘Ω ∴ 𝑅𝐹= 0.586𝑅𝐴 = 5.86𝑘Ω
- 26. Higher Order FilterDesign • Example 2: Design the resistors 𝑅1 − 𝑅4 of a 4th order LPF as shown + + – – R R 1 3 R R 2 4 R R B1 B2 R R A1 A2 C C A1 A2 C C B1 B2 Vin Vout 22 kW 22 kW 22 kW 22 kW 4.7 nF 4.7 nF 4.7 nF 4.7 nF
- 27. Higher Order FilterDesign • Solution: Add an identical section except for the gain setting resistors. Choose R1-R4 based on the table for a 4th order design. + + – – R R 1 3 R R 2 4 R R B1 B2 R R A1 A2 C C A1 A2 C C B1 B2 Vin Vout 22 kW 22 kW 22 kW 22 kW 4.7 nF 4.7 nF 4.7 nF 4.7 nF The resistor ratio for the 1st section needs to be 0.152 (gain = 1.152); the 2nd section needs to be 1.235 (gain = 2.235). Use standard values if possible. 22 kW 3.3 kW 12 kW 15 kW
- 28. End of Chapter2 Part III