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ESE319 Introduction to Microelectronics
1© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
Single-Amplifier-Biquad (SAB) Filter Sections
ESE319 Introduction to Microelectronics
2© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
3© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
[ 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:
ESE319 Introduction to Microelectronics
4© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
5© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
6© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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.
ESE319 Introduction to Microelectronics
7© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
8© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
9© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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:
ESE319 Introduction to Microelectronics
10© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
11© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
12© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
13© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
Sallen and Key LP Section with K = 1
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
ESE319 Introduction to Microelectronics
14© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
Sallen and Key LP Section with K = 11
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
ESE319 Introduction to Microelectronics
15© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
Butterworth Highpass SectionBy interchanging 2 R's and 2 C” on the LP schematic, we convert an LP topology to an HP one
HPLP
ESE319 Introduction to Microelectronics
16© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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:
ESE319 Introduction to Microelectronics
17© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
18© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
19© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
20© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
Filter ResponsePassbandresponse
Stopbandresponse
4
-34.14 dB
205.4 kHz
4.06 dB
1.98 kHz
ESE319 Introduction to Microelectronics
21© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
22© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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:
ESE319 Introduction to Microelectronics
23© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
24© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
25© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
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
ESE319 Introduction to Microelectronics
26© 2006 Philip V. Lopresti (edited 07Dec07 KRL)
Delyannis-Friend Bandpass Filter
C=10nF
Q=8T 0=10
0=104 rps⇒ f 0=
104
2=1.59 k Hz
Choose: