ESE319 Introduction to Microelectronics Single-Amplifier-Biquad (SAB) Filter Sections

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1 Single-Amplifier-Biquad (SAB) Filter Sections

2 Sallen-Key Lowpass Section Analysis K R f R i Multiplying by R V j V p V j scr V j V o V j V p V i V p V j scrv p 0 and V o K V p Write 2 node equations V j V V i sc V R j V o V j V p j 0 R V p V V j p sc V R p 0 and V o K V p Eliminate V o using V o K V p, and combine terms V j V p V j scr V p 0 2

3 Sallen-Key Lowpass Section Analysis II V j V p 2 scr V j K scr V p V i V j scr V p 0 Eliminate V j (multiply V p Eq by (2 + scr)) V j 2 scr V j K scr V p V i V p 2 scr V j 2 scr scr V p 0 Add and solve for V p [ 2 scr scr K scr ]V p V i 3

4 Sallen-Key Lowpass Section Analysis III [ 2 scr scr K scr ]V p V i Multiplying to expand into a polynomial Collecting terms Dividing by (CR) 2 [ s2 [ scr 2 3sCR 2 KsCR ]V p V i [ s 2 CR 2 3 K scr ]V p V i 3 K CR s 2 CR ] V p 2 V CR i 4

5 Sallen-Key Lowpass Section Analysis IV Since V o K V p [ 3 K s2 CR s 2 ] CR V p 2 V CR i 2 K CR V o s 2 3 K CR s V 2 i CR 0 CR Note Q when K Q 3 K 5

6 Sallen-Key Lowpass Section Analysis V Identify filter parameters T s V o CR V i Design equations 2 K s 2 3 K CR s CR K s 2 0 Q s CR Q 3 K K 3! Note When K 3, T(s) oscillates or is unstable. 6

7 Sallen-Key 20 Khz Butterworth Section. Let's do the initial design with normalized R and C and R 0 k CR K / R T s n n C n 2 K s 2 n 3 K s R n C n s 2 n R n C n 2 n Q s n Normalized design equations R n C n where R n R R 0 and and Q 2 K R fn R in 3 Q C n 0 R 0 C K s n 2 2 s n Choose C n R n 7

8 Sallen-Key 20 Khz Butterworth Section C n R n C C n F F 25 0 R R 0 RR n R 0 R 0 k > R k where and F R 0 R 0 C7.958 nf F and R 0 k R fn R in.5857 R fn R in R in 0 R fn R 0 k > R in 0 k and R fn 5.86 k 8

9 Butterworth Sallen-Key Design Suppose we wanted to choose C 5. nf from the RCA Lab. 5. nf F R 0 R 0 Solving for R R k R.56 k C 5. nf Vi R.56 k Ohm Vj R.56 k Ohm C 5. nf Vp K Rf 5.86 k Ohm Ri 0 k Ohm Vo 9

10 Simulation Results K log 0 T j0 20 log 0 K4 db 4 db Pass-band Gain khz Stop-band Gain at 0x cutoff db khz 0

11 Butterworth Design Tables Choose RC for proper ω o and select K from the table NOTEAll Butterworth biquad stages have normalized 0n Filter Order K Values

12 Sallen and Key LP Section with K C 2 R R 2 K Recall when K > R C R 2 C 2 RC C When K R C R 2 C 2 T s V o V i s 2 C C 2 R R 2 C R C R 2 s C C 2 R R s 2 0 Q s 2 0 2

13 Sallen and Key LP Section with K Normalize frequency and impedance, i.e. s n s and Z n Z 0 R 0 T s n V o C n C 2n R n R 2n V i s 2 n C n R n C n R s n s 2 n 2n C n C 2n R n R 2n Q s n Design Formulas C n R n C n R 2n Q C n C 2n R n R 2n Design Formulas (Butterworth) 2 C n R n C n R 2n C n C 2n R n R 2n 3

14 Sallen and Key LP Section with K 2 C n R n C n R 2n C n C 2n R n R 2n Let C n and R 2n R n R n 2 2 R R n 2 n 2 2 Normalized Design C n C 2n 2 R n R 2n 2 C 2n R n 2 C 2n R n 2 2 Denormalized Design C C n 0 R 0 0 R 0 C R 0 R R 2 2 R 0 4

15 Butterworth Highpass Section By interchanging 2 R's and 2 C on the LP schematic, we convert an LP topology to an HP one LP HP 5

16 V j V p Sallen-Key Highpass Function Interchanging sc and /R in the LP nodal equations gives V j V i R V p V j R sc V j V o V j V p 0 sc V j V i V j V o sc V R R j V p 0 sc V p 0 sc V p V j V p R 0 Multiplying by R scrv j V j V o scr V j V p scrv i scr V p V j V p 0 Eliminating V o V o K V p 2sCR V j scr K V p scrv i scrv j scr V p 0 6

17 Sallen-Key Highpass Function II 2sCR V j scr K V p scrv i scrv j scr V p 0 Eliminate V j and multiplying by scr [ 2sCR scr scr scr K ]V p s 2 CR 2 V i Multiply out, divide by (CR) 2 and substitute V p K V o V o K s 2 s 2 3 K CR s CR 2 V i 7

18 Sallen-Key Highpass Function III T s V o V i K s 2 s 2 3 K CR s CR Note that the denominator polynomial (or poles) is the same as for the low pass section and the same design equations apply 2 0 CR Q 3 K T j K K 3! 8

19 20 Khz Butterworth Highpass filter R.56 k Ohm Vi C 5. nf Vj C 5. nf R.56 k Ohm Vp Ri 0 k Ohm K Vo Rf 5.86 k Ohm 9

20 Passband response 4.06 db Filter Response 4 Stopband response khz db.98 khz 20

21 Delyannis-Friend Bandpass Filter Node Equations V j V j V i R sc V j V o sc V j V n V j R 2 0 V n sc V n V j V n V o R 3 0 virtual ground open-loop op amp V o K V n Ideal op amp K V n 0 2

22 V n 0 ESE39 Introduction to Microelectronics Delyannis-Friend Bandpass Filter II V j sc V j V o 0 V R j V o V n 0 V n 3 scr 3 Substituting for V j V o V j V i R V i 2 V o sc V sc R 3 R R R o 3 Multiply by sc V o 2 sc R 3 R R 3 R 3 R R 2 2 sc R 3 sc V j V o sc V j V j R 2 0 V o sc R 3 R 2 0 V o s 2 C 2 V o V o R 3 R 2 sc R V i s 2 C 2 V o sc R V i 22

23 Delyannis-Friend Bandpass Filter III R 3 R R 2 2 sc R 3 s 2 C 2 V o sc R V i Dividing by C 2 [ s2 2 s C R 3 C 2 R 3 R R 2 ] V o s V C R i T s V o V i s C R s 2 2 s C R 3 C 2 R 3 R R 2 23

24 Delyannis-Friend Bandpass Filter IV T s V o V i s C R s 2 2 s C R 3 C 2 R 3 R R 2 0 Q T 0 s 2 0 Q s 2 0 Identify 2 0 C 2 R 3 R R 2 0 Q 2 C R 3 T V o 0 0 C R 3 R 3 V i 0 C R R 24

25 Delyannis-Friend Bandpass Filter V 2 0 C 2 R 3 R R 2 0 Q 2 C R 3 Given ω 0, Q and T 0, choose C and solve for the resistors R 3 2Q R C R 3 Q 0 2T 0 C 0 T 0 R R C 2 R C Q T 0 T j 0 R 3 2 R R 2 R 2 0 C Q R Normalize 0n and choose C R 3 2Q R Q n R T R 2 0 2Q R R 2 2Q R 25

26 Delyannis-Friend Bandpass Filter rps f k Hz 2 Q8 T 0 0 Choose C0 nf 26

27 Peak gain 9.98 db Simulation Results.66 khz High 3 db point 6.50 db.77 khz 27