Single-Amplifier-Biquad (SAB) Filter Sectionsese319/Lecture_Notes/Lec_24_SAB_Biquads_0… · ESE319...

ESE319 Introduction to Microelectronics

1© 2006 Philip V. Lopresti (edited 07Dec07 KRL)

Single-Amplifier-Biquad (SAB) Filter Sections



Sallen-Key Lowpass Section Analysis

V j−V i

RsC V j−V o

V j−V p

R=0

V p−V j

RsCV p=0

Write 2 node equations:

K=1R fRi

Vj:

Vp:

and

V o=K V p

V p−V jsCRV p=0

Eliminate Vo using Vo = K Vp,

and combine terms:

V jsCR V j−V o V j−V p=V i

and

Vj:

Vp:

Vj:

Multiplying by R:

Vp: −V j 1sCR V p=0V o=K V p



[ 2sCR 1sCR − 1K sCR ]V p=V i

2sCR V j− 1K sCR V p=V i

Sallen-Key Lowpass Section Analysis II

Eliminate: Vj (multiply Vp Eq by (2 + sCR))

−V j1sCR V p=0

2sCR V j− 1K sCR V p=V i

Add and solve for Vp:

−2sCR V j2sCR 1sCR V p=0

Vj:

Vp:

Vj:

Vp:



Sallen-Key Lowpass Section Analysis III[ 2sCR 1sCR − 1K sCR ]V p=V i

Multiplying to expand into a polynomial

[ sCR 23sCR2−1−KsCR ]V p=V i

Collecting terms:

[ s2 CR 2 3−K sCR1 ]V p=V i

Dividing by (CR)2:

[s2 3−K CR

s 1CR 2]V p= 1CR

2

V i



Sallen-Key Lowpass Section Analysis IV

[s2 3−K CR

s 1CR 2]V p= 1CR

2

V i

Since Vo = K V

p:

V o= 1CR

2

K

s2 3−K CR

s 1CR 2 V i

0=1CR

Q= 13−K

Note: Q=12≠ 22when K=1



T s=V o

V i=

1CR 2

K

s2 3−K CR

s 1CR 2=

02K

s20

Qs0

2

Sallen-Key Lowpass Section Analysis VIdentify filter parameters:

Design equations:

0=1CR Q= 1

3−K K3!

Note: When K ≥ 3, T(s) oscillates or is unstable.



Sallen-Key 20 Khz Butterworth Section

0=0=1CR

=220⋅103

Choose:

Q= 12

K=1R fnRin

=3− 1Q=3−2=1.5857

Rn=RR0

Cn=0R0C

1. Let's do the initial design with normalized R and C.R0=1 kand

Normalized design equations:1

RnCn=1 and

T sn=K /RnC n

2

sn2 3−K

RnC nsn

1RnC n

2

= K

sn2 1Qsn1

= Ksn22 sn1

where and

Cn=1⇒ Rn=1



Sallen-Key 20 Khz Butterworth SectionCn=1⇒ Rn=1

C=C n

0R0F= 1

220⋅1031R0F= 25

1R0

F=7.958R0

F

R=RnR0

0=0=220⋅103

R0=1 k => R=1 k C=7.958nFand

R0=1 kwhere and

1R fnRin

=1.5857⇒ R fn=0.5857 Rin Rin=10⇒ R fn=5.857

R0=1 k => Rin=10 k and R fn=5.86 k



Butterworth Sallen-Key Design

C5.1 nF

C5.1 nF

R1.56 k Ohm

R1.56 k Ohm

Ri10 k Ohm

Rf5.86 k Ohm

ViVo

K

Vj Vp

Suppose we wanted to choose C = 5.1 nF from the RCA Lab.

5.1nF=7.958R0

F⇒5.1⋅10−9=7.958⋅10−6

R0R0=

7.958⋅10−6

5.1⋅10−9=1.56 k⇒R=1.56 kSolving for R0:



Simulation Results

Pass-band Gain

Stop-band Gain at 10x cutoff

K=1.5857⇒20log10∣T j0 ∣=20 log10K=4dB

1 kHz

4 dB

205.4 kHz

-36.45 dB



Butterworth Design Tables

Filter Order K Values2 1.586

4 1.1522.235

6 1.0681.5862.483

8 1.0381.3371.8892.610

Choose RC for proper ωo and select K from the table:

NOTE:All Butterworth biquad stages have normalized 0n=1



Sallen and Key LP Section with K = 1

K = 1

C1

C2

R1

R2

T s=V o

V i=

1C1C2R1R2

s2 1C1R1

1C1R2 s 1

C1C2R1R2

=02

s20

Qs0

2

Recall when K > 1: R

1C

1 = R

2C

2 = RC

When K = 1: R

1C

1 ≠ R

2C

2




T sn=V o

V i=

1C1nC2n R1nR2n

sn2 1

C1n R1n 1C1n R2n sn 1

C1nC 2nR1n R2n

= 1

sn2 1Qsn1

Normalize frequency and impedance, i.e. sn=s0

Z n=ZR0

and

1C1n R1n

1C1n R2n

= 1Q

1C1nC 2n R1n R2n

=1

Design Formulas Design Formulas (Butterworth)1

C1n R1n 1C1n R2n

=2

1C1nC 2n R1n R2n

=1




C1nR1n 1C1n R2n

=2 1C1nC 2nR1nR2n

=1

Let C1n

= 1 and R2n

= R1n

= Rn:

2Rn

=2⇒Rn=22

=21

C2n Rn2=1⇒C2n=

1Rn2=12

Normalized Design:

C1n=1

C2n=12

R1n=R2n=2

C1=C1n0R0

= 10R0

R1=R2=2 R0

Denormalized Design:

C2=1

20R0



Butterworth Highpass SectionBy interchanging 2 R's and 2 C” on the LP schematic, we convert an LP topology to an HP one

HPLP



Sallen-Key Highpass FunctionInterchanging sC and 1/R in the LP nodal equations gives:

sCRV jV j−V o sCR V j−V p =sCRV i

sCR V p−V j V p=0

V o=K V p

2sCR1 V j− sCRK V p=sCRV i

−sCRV j sCR1 V p=0

Multiplying by R:

V j−V i

RsC V j−V o

V j−V p

R=0 sC V j−V i

V j−V o

RsC V j−V p =0Vj:

Vp:V p−V j

RsC V p=0 sC V p−V j

V p

R=0

Eliminating Vo:



Sallen-Key Highpass Function II

−sCRV j sCR1 V p=0

Eliminate Vj and multiplying by sCR:

[ 2sCR1 sCR1 −sCR sCRK ]V p=s2 CR 2V i

Multiply out, divide by (CR)2 and substitute

V o=K s2

s2 3−K CR

s 1CR 2 V i

2sCR1 V j− sCRK V p=sCRV i

V p=1KV o



Sallen-Key Highpass Function III

T s=V o

V i= K s2

s2 3−K CR

s 1CR 2

Note that the denominator polynomial (or poles) is the same asfor the low pass section and the same design equations apply:

0=1CR Q= 1

3−KK3 !

∣T j∞∣=K



20 Khz Butterworth Highpass filter

VjVi

C5.1 nF

C5.1 nF

Vp

R1.56 k Ohm

R1.56 k Ohm

VoK

Rf5.86 k Ohm

Ri10 k Ohm



Filter ResponsePassbandresponse

Stopbandresponse

4

-34.14 dB

205.4 kHz

4.06 dB

1.98 kHz



Delyannis-Friend Bandpass FilterNode Equations:

V j−V i

R1sC V j−V osC V j−V n

V j

R2=0

sC V n−V jV n−V o

R3=0

V o=−K V n

Vj:

Vn:

Ideal op amp: K∞⇒V n0open-loop op amp

virtual ground



Delyannis-Friend Bandpass Filter II

−V o

sC R3 R1−V iR1

−2V o

R3−sCV o−

V o

sC R3R2=0

−sCV j−V o

R3=0⇒V j=−

V osCR3

V o

R3R12 sCR3

V os2C 2V o

V o

R3R2=− sC

R1V i

1R3

1R1

1R2

2 sCR3

s2C2V o=− sCR1V i

Vj:V n0V j−V i

R1sC V j−V osC V j

V j

R2=0

V n0 Vn:

Substituting for Vj:

Multiply by sC:



Delyannis-Friend Bandpass Filter III

[s2 2C R3

s 1C2R3⋅R1∥R2 ]V o=−

sC R1

V i

Dividing by C2:

1R3

1R1

1R2

2 sCR3

s2C2V o=− sCR1V i

T s=V o

V i=

− sC R1

s2 2C R3

s 1C 2R3⋅R1∥R2



Delyannis-Friend Bandpass Filter IV

Identify:02= 1C2 R3⋅R1∥R2

0

Q= 2C R3

T 0=∣V o

V i∣=0

=0

C R1C R320

=R32 R1

T s=V o

V i=

− sC R1

s2 2C R3

s 1C 2R3⋅R1∥R2

=−

0

QT 0

s20

Q s0

2



Delyannis-Friend Bandpass Filter V

02= 1C2R3⋅R1∥R2

0

Q= 2C R3

T 0=∣T j0∣=R32R1

Given ω0, Q and T

0, choose C and

solve for the resistors:

R3=2QC0

R1=R32T 0

= QC0T 0

R1∥R2=1

02C2R3

= 120CQ

R2=R1

20CQ R1−1

R3=2Q R1=QT 0

R1∥R2=12Q

R2=R1

2Q R1−1

Normalize and choose Cn = 1:0n=1



Delyannis-Friend Bandpass Filter

C=10nF

Q=8T 0=10

0=104 rps⇒ f 0=

104

2=1.59 k Hz

Choose:



Simulation ResultsPeak gain

High 3 dB point

1.66 kHz

1.77 kHz

19.98 dB

16.50 dB

Single-Amplifier-Biquad (SAB) Filter Sectionsese319/Lecture_Notes/Lec_24_SAB_Biquads_0… · ESE319...

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