LC Ladder Filter Design Example. Design a LPF with maximally flat passband: f-
3dB = 10MHz, fstop = 20MHz. Rs >27dB. From: Williams and Taylor, p. 2-37.
EE247 Lecture 4 • Ladder type filters –For simplicity, will start with all pole ladder type filters • Convert to integrator based form- example shown
–Then will attend to high order ladder type filters incorporating zeros • Implement the same 7th order elliptic filter in the form of ladder RLC with zeros – Find level of sensitivity to component mismatch – Compare with cascade of biquads
• Convert to integrator based form utilizing SFG techniques
–Effect of Integrator Non-Idealities on Filter Frequency Characteristics EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 1
LC Ladder Filters Vin
C1
Vo
L4
L2
Rs
C3
C5
RL
• Design: – Filter tables • A. Zverev, Handbook of filter synthesis, Wiley, 1967. • A. B. Williams and F. J. Taylor, Electronic filter design, 3rd edition, McGraw-Hill, 1995.
– CAD tools • Matlab • Spice EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 2
LC Ladder Filter Design Example Design a LPF with maximally flat passband:
f-3dB = 10MHz, fstop = 20MHz Rs >27dB
Stopband Attenuation dB
• Maximally flat passband F Butterworth • Find minimum filter order : - Use of Matlab - or Tables • Here tables used -3dB
fstop / f-3dB = 2 Rs >27dB Minimum Filter Order F5th order Butterworth EECS 247
1
2
Νοrmalized ω From: Williams and Taylor, p. 2-37
Lecture 4: Active Filters
© 2007 H.K. Page 3
LC Ladder Filter Design Example Find values for L & C from Table: Note L &C values normalized to
ω-3dB =1 Denormalization: Multiply all LNorm, CNorm by: Lr = R/ ω-3dB Cr = 1/(RX ω-3dB ) R is the value of the source and termination resistor (choose both 1Ω for now) Then: L= Lr xLNorm C= Cr xCNorm EECS 247
Lecture 4: Active Filters
From: Williams and Taylor, p. 11.3 © 2007 H.K. Page 4
LC Ladder Filter Design Example Find values for L & C from Table: Normalized values: C1Norm =C5Norm =0.618 C3Norm = 2.0 L2Norm = L4Norm =1.618
Denormalization: Since ω-3dB =2πx10MHz Lr = R/ω-3dB = 15.9 nH Cr = 1/(RXω-3dB )= 15.9 nF R =1
FC1=C5=9.836nF, C3=31.83nF FL2=L4=25.75nH EECS 247
From: Williams and Taylor, p. 11.3
Lecture 4: Active Filters
Example:
© 2007 H.K. Page 5
Last Lecture: Order Butterworth Filter
5th
L2=25.75nH
L4=25.75nH
Vo
Rs=1Ohm C1 9.836nF
Specifications: f-3dB = 10MHz,
fstop = 20MHz Rs >27dB Used filter tables to obtain Ls & Cs
C3 31.83nF
Magnitude (dB)
Vin
0 -5 -10 -20
RL=1Ohm
SPICE simulation Results
30dB
-30 -40 -50 0
EECS 247
C5 9.836nF
Lecture 4: Active Filters
10
20
Frequency [MHz]
30
© 2007 H.K. Page 6
Low-Pass RLC Ladder Filter Conversion to Integrator Based Active Filter
Vin
•
+ V3 −
+ V1 − V2 Rs I 1
I3
L2
C3
I4
I7
V6
Vo
I5
L4
C1
I2
+ V5 −
V4
C5
I6
RL
Use KCL & KVL to derive equations: I V1 = Vin − V2 , V2 = 2 sC1 I V4 = 4 , V5 = V4 − V6 sC3 V I1 = 1 Rs
, I 2 = I1 − I 3
I4 = I3 − I5
, I5 =
EECS 247
, V3 = V2 − V4 , V6 =
V5 sL4
I6
Vo =V6
sC5 V3
,
I3 =
,
I 6 = I 5 − I7
sL2
V I7 = 6 RL
,
Lecture 4: Active Filters
© 2007 H.K. Page 7
Low-Pass RLC Ladder Filter Signal Flowgraph I V1 = Vin − V2 , V2 = 2 sC1 I V4 = 4 , V5 = V4 − V6 sC3
Vin
V I1 = 1 Rs
, I 2 = I1 − I 3
I4 = I3 − I5
, I5 =
1 V1
−1
1 Rs I1
EECS 247
1
V2
1
V5 sL4
−1
V3
, V3 = V2 − V4 , V6 =
,
I 6 = I 5 − I7
V4
1
1 sC3
1
I4 SFG
Lecture 4: Active Filters
sL2
1
s L2
I3
V3
I3 =
1
−1
Vo =V6
sC5
,
sC1 I2
I6
−1
V I7 = 6 RL
,
−1
V5
V6
1
1
1
s L4
sC5
I5
1
I6 − 1
Vo
1 RL
I7
© 2007 H.K. Page 8
Low-Pass RLC Ladder Filter Signal Flowgraph
Vin
Vin
+ V3 −
V2 + V1 − I Rs 1
−1
1 V1
V2
1
1
−1
V3
V4
1 1
sC1
s L2
sC3
1
I3
−1
I4
C5
I6
1
−1
I5
C3
1
I2
I7
V6
L4
I4
1 Rs I1
I3
L2 C1
I2
+ V5 −
V4
V6
−1
V5
RL
1
1
1
s L4
sC5
I6 − 1
1
I5
Vo
1 RL
I7
SFG EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 9
Low-Pass RLC Ladder Filter Normalize Vin
1 V1
−1
V2
1 Rs
Vin
I1
1
1 V1
−1
EECS 247
V4
1
−1
V5
V6
1
1
1
1
1
1
sC1
s L2
sC3
s L4
sC5
−1
I2
V2
1
V2'
1
I3
V3
sC1R* 1
−1
V3
1
R* Rs
V1'
1
− 1 V3'
−1
I4 − 1
V4
1
I5
1
V5
−1
R*
1
R*
sC3R*
s L2
1
V4'
Lecture 4: Active Filters
−1
V6
1 1
sC5 R*
s L4
V5'
I6 − 1
1
V6' − 1
Vo
1 RL
I7
Vo
R* RL
V7'
© 2007 H.K. Page 10
Low-Pass RLC Ladder Filter Synthesize V2
V2'
− 1 V3'
V2 +
+
−1
-
1
sτ 2
sC5 R
V6' − 1
1
V6
1
+
sτ 3
-
V3' EECS 247
1
sτ 4
R* RL
*
s L4 1 V5'
Vo
1
V4
1
s1τ 1
V4'
-
Vin − R* Rs
sC3R
1
1
R*
*
s L2
V6
−1
V5
1 1
R*
sC1R* 1
V4
−1
V3
1
R* Rs
V1'
1
+
−1
1 V1
-
Vin
Vo R* R
1
sτ 5
+
V7'
-
L
V5'
Lecture 4: Active Filters
© 2007 H.K. Page 11
Low-Pass RLC Ladder Filter Integrator Based Implementation V2 +
-
-
+
1
sτ 2
+
1
s1τ 1
V6
V4 1
+
sτ 3
-
V3'
τ 1 = C1 .R*
L , τ 2 = 2 = C2 .R* R*
sτ 4
Vo R* R
1
sτ 5
+
-
L
V5' , τ 3 = C3 .R*
Building Block: RC Integrator
EECS 247
1
-
Vin − R* Rs
Lecture 4: Active Filters
L4 , τ4 = = C4 .R* R*
, τ 5 = C5 .R*
V2 =− 1 V1 sR C
© 2007 H.K. Page 12
Negative Resistors V1Vo+
V2+
V1V2-
Vo+
V2+
Vo-
V1+ EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 13
Integrator Based Implementation of LP Ladder Filter Synthesize V2
V3' V4 V5'
Vo EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 14
Frequency Response 1
1
0.5 V2
V3'
0.1 0.5
V4
Vo
10MHz EECS 247
V5'
10MHz Lecture 4: Active Filters
© 2007 H.K. Page 15
Scale Node Voltages Scale Vo by factor “s”
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 16
Node Scaling V2
X 1.2 V3’ X 1.2
X 1/1.2 X 1.6/1.2
X 1.2/1.6
V4 X 1.6
X 1.8/1.6
X 1.6/1.8
V5’ X 1.8
X 2/1.8
X1.8/2
VO X 2 EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 17
Maximizing Signal Handling by Node Voltage Scaling Before Node Scaling
After Node Scaling
Scale Vo by factor “s”
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 18
Filter Noise Total noise @ the output: 1.4 μV rms (noiseless opamps) That’s excellent, but the capacitors are very large (and the resistors small Æ high power dissipation). Not possible to integrate. Suppose our application allows higher noise in the order of 140 μV rms … EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 19
Scale to Meet Noise Target Scale capacitors and resistors to meet noise objective
s = 10-4 Noise: 141 μV rms (noiseless opamps) EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 20
Completed Design V2
V3'
V4
5th order ladder filter Final design utilizing: -Node scaling V5' -Final R & C scaling based on noise considerations EECS 247
Vo
Lecture 4: Active Filters
© 2007 H.K. Page 21
Sensitivity • C1 made (arbitrarily) 50% (!) larger than its nominal value • 0.5 dB error at band edge • 3.5 dB error in stopband • Looks like very low sensitivity
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 22
Differential 5th Order Lowpass Filter
+
-
+
+
+
+
+
+
-
-
+
-
+
-
-
-
-
+
Vin
-
Vo • Since each signal and its inverse readily available, eliminates the need for negative resistors! • Differential design has the advantage of even order harmonic distortion components and common mode spurious pickup automatically cancels • Disadvantage: Double resistor and capacitor area! EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 23
RLC Ladder Filters Including Transmission Zeros All poles
Vin
C1
Poles & Zeros
Vo
L4
L2
Rs
C5
C3
C6
C4
C2
RL
Vo Rs
Vin
EECS 247
L4
L2 C1
C3
Lecture 4: Active Filters
L6 C5
C7
RL
© 2007 H.K. Page 24
RLC Ladder Filter Design Example • Design a baseband filter for CDMA IS95 cellular phone receive path with the following specs. – Filter frequency mask shown on the next page – Allow enough margin for manufacturing variations • Assume pass-band magnitude variation of 1.8dB • Assume the -3dB frequency can vary by +-8% due to manufacturing tolerances & circuit inaccuracies
– Assume any phase impairment can be compensated in the digital domain * Note this is the same example as for cascade of biquad while the specifications are given closer to a real product case EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 25
RLC Ladder Filter Design Example CDMA IS95 Receive Filter Frequency Mask
Magnitude (dB)
+1 0 -1
-44 -46 600k 700k 900k 1.2M Frequency [Hz] EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 26
RLC Ladder Filter Design Example: CDMA IS95 Receive Filter • Since phase impairment can be corrected for, use filter type with max. roll-off slope/pole Æ Filter type Æ Elliptic • Design filter freq. response to fall well within the freq. mask – Allow margin for component variations & mismatches • For the passband ripple, allow enough margin for ripple change due to component & temperature variations Æ Design nominal passband ripple of 0.2dB • For stopband rejection add a few dB margin 44+5=49dB • Final design specifications: – fpass = 650 kHz Rpass = 0.2 dB – fstop = 750 kHz Rstop = 49 dB • Use Matlab or filter tables to decide the min. order for the filter (same as cascaded biquad example) – 7th Order Elliptic EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 27
RLC Low-Pass Ladder Filter Design Example: CDMA IS95 Receive Filter 7th order Elliptic
C6
C4
C2
Vo Vin
L4
L2
Rs C1
C3
L6 C5
C7
RL
• Use filter tables to determine LC values
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 28
RLC Ladder Filter Design Example: CDMA IS95 Receive Filter • Specifications – fpass = 650 kHz Rpass = 0.2 dB – fstop = 750 kHz Rstop = 49 dB • Use filter tables to determine LC values – Table from: A. Zverev, Handbook of filter synthesis, Wiley, 1967 – Elliptic filters tabulated wrt “reflection coeficient ρ”
(
Rpass = − 10 × log 1 − ρ 2
)
– Since Rpass=0.2dB Æ ρ =20% – Use table accordingly EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 29
RLC Ladder Filter Design Example: CDMA IS95 Receive Filter • •
Table from Zverev book page #281 & 282: Since our spec. is Amin=44dB add 5dB margin & design for Amin=49dB
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 30
• Table from Zverev page #281 & 282: • Normalized component values: C1=1.17677 C2=0.19393 L2=1.19467 C3=1.51134 C4=1.01098 L4=0.72398 C5=1.27776 C6=0.71211 L6=0.80165 C7=0.83597
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 31
RLC Filter Frequency Response • Frequency mask superimposed Magnitude (dB)
• Frequency response well within spec.
-5
-15
-25
-35
-45
-55
-65 200
300
400
500
600
700
800
900
1000
1100
1200
Frequency [kHz]
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 32
Passband Detail -5
• Passband well within spec. Magnitude (dB)
-5.5
-6
-6.5
-7
-7.5 200
300
400
500
600
700
800
Frequency [kHz] EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 33
RLC Ladder Filter Sensitivity • The design has the same specifications as the previous example implemented with cascaded biquads • To compare the sensitivity of RLC ladder versus cascaded-biquads: – Changed all Ls &Cs one by one by 2% in order to change the pole/zeros by 1% (similar test as for cascaded biquad) – Found frequency response Æ most sensitive to L4 variations – Note that by varying L4 both poles & zeros are varied
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 34
RCL Ladder Filter Sensitivity Component mismatch in RLC filter: – Increase L4 from its nominal value by 2% – Decrease L4 by 2% -5
L4 nom L4 low L4 high
Magnitude (dB)
-15 -25 -35 -45 -55 -65 200
300
400
500
600
700
800
900
1000
1100
1200
Frequency [kHz] EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 35
RCL Ladder Filter Sensitivity -5.7
-5.9
-6.1
-5
0.2dB
-6.3
Magnitude (dB)
-15
-6.5 200
-25
300
400
500
600
700
-50
1.7dB
-35 -55
-45 -60
-55 -65 600
700
800
900
1000
1100
1200
-65 200
300
400
500
600
700
800
900
1000 1100 1200
Frequency [kHz]
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 36
Sensitivity of Cascade of Biquads Component mismatch in Biquad 4 (highest Q pole): – Increase ωp4 by 1%
2.2dB
– Decrease ωz4 by 1%
Magnitude (dB)
0 -10 3dB
-20 -30 -40 -50 200kHz
EECS 247
600kHz Frequency [Hz]
1MHz
High Q poles Æ High sensitivity in Biquad realizations
Lecture 4: Active Filters
© 2007 H.K. Page 37
Sensitivity Comparison for Cascaded-Biquads versus RLC Ladder • 7th Order elliptic filter – 1% change in pole & zero pair Cascaded RLC Ladder Biquad Passband deviation
2.2dB (29%)
0.2dB (2%)
Stopband deviation
3dB (40%)
1.7dB (21%)
Doubly terminated LC ladder filters B Significantly lower sensitivity compared to cascaded-biquads particularly within the passband EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 38
RLC Ladder Filter Design Example: CDMA IS95 Receive Filter 7th order Elliptic
C6
C4
C2
Vo Vin
L4
L2
Rs C1
L6 C5
C3
C7
RL
• Previously learned to design integrator based ladder filters without transmission zeros Æ Question: o How do we implement the transmission zeros in the integratorbased version? o Preferred method Æ no extra power dissipation Æ no active elements EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 39
Integrator Based Ladder Filters How Do to Implement Transmission zeros? Ca
+ V1 − I 1
V2
L2
Rs
Vin
•
+ V3 −
I2
C1
Use KCL & KVL to derive :
I 2 = I1 − I 3 − I C , a
V4
I3
I C = (V2 − V4 ) sCa , a
I4
C3
I V2 = 2 , sC1
Vo
I5
RL
V2 =
I1 − I 3 − I C a sC1
Substituting for IC and rearranging : a Ca I1 − I 3 V2 = + V4 × s ( C1 + Ca ) C1 + Ca EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 40
Integrator Based Ladder Filters How Do to Implement Transmission zeros? Ca
+ V1 − I 1
V2
Vin
V4
C1
I2
Vo
I3
L2
Rs
•
+ V3 −
I4
I5
C3
RL
Use KCL & KVL to derive : Ca I1 − I 3 V2 = + V4 × s ( C1 + Ca ) C1 + Ca
Frequency independent constants Can be substituted by: Voltage-Controlled Voltage Source
Ca I3 − I5 V4 = + V2 × s ( C3 + Ca ) C3 + Ca EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 41
Integrator Based Ladder Filters Transmission zeros Ca
+ V1 − I 1
+ V3 −
V2
L2
Rs
(C1 + Ca )
Vin
I2
+ -
Ca V4 C1 + Ca
V4
I3
I5
(C3 + Ca )
RL
+ -
Ca I 4 V2 C + C 3 a
• Replace shunt capacitors with voltage controlled voltage sources: Ca I1 − I 3 +V V2 = s ( C1 + Ca ) 4 C1 + Ca
Ca I3 − I5 +V V4 = s ( C3 + Ca ) 2 C3 + Ca EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 42
3rd Order Lowpass Filter All Poles & No Zeros + V1 − I 1
+ V3 −
V2
Vin
C1
I2 Vin
1 V1 1 Rs
1
I1
EECS 247
V3 − 1
1
I2
RL
V4
1
1
1
1
s C1
s L2
sC3
−1
Vo
C3
I4
− 1 V2
I5
V4
I3
L2
Rs
1
I3
Vo
1 RL −1
I4
Lecture 4: Active Filters
© 2007 H.K. Page 43
Transmission Zero Implementation W/O Use of Active Elements + V1 − I 1
+ V3 −
V2
(C1 + Ca )
Vin
C + V - 4 C aC 1+ a
1 Rs I1 EECS 247
Ca C3 + Ca 1 V3 − 1
− 1 V2
1
V4
1
1
1
1
s (C1 + Ca )
s L2
s (C3 + Ca )
I2
−1
I3
Lecture 4: Active Filters
RL
Ca + - V2 C I4 3 + Ca
Ca C1 + Ca 1 V1
Vo
(C3 + Ca )
I2
Vin
I5
V4
I3
L2
Rs
1
I4
Vo
1 RL
−1 © 2007 H.K. Page 44
Integrator Based Ladder Filters Higher Order Transmission zeros Ca
V2
V6 C5
C3
C1
Convert zero generating Cs in C loops to voltagecontrolled voltage sources
Cb
V4
V6
V2
V4
(C1 + Ca )
(C3 + Ca + Cb )
(C5 + Cb )
C
C + V - 4 C aC 1+ a
a + V2 C3 + Ca Cb + - V6
+ -
Cb V4 C3 + Cb
C3 + Cb
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 45
Higher Order Transmission zeros + V1 − I1
Rs
Vin
+ V3 −
V2
I3
L2
(C1 + Ca )
I4
+ + -
Ca + V4 C1 + Ca
I2
1 V1
V1' EECS 247
1 V3 − 1
− 1 V2
R* Rs
1
(C3 + Ca + Cb )
V4
1
R*
1
s (C1 + Ca ) R*
s L2
s R* (C3 + Ca + Cb )
−1
V3'
1
V4'
Lecture 4: Active Filters
−1
I7 Vo
(C5 + Cb )
RL
Cb + V4 C5 + Cb
I6
Cb C3 + Cb
1
V2'
V6
L4
Ca V2 C3 + Ca Cb V6 C3 + Cb
Ca C3 + Ca
Ca C1 + Ca Vin
I 5 + V5 −
V4
Cb C5 + Cb V5
−1 R* s L4
V5'
V6
1
Vo
1 * s R ( C5 + Cb ) 1 V' 6
−1
R* RL
V7'
© 2007 H.K. Page 46
Example: 5th Order Chebyshev II Filter
EECS 247
Lecture 4: Active Filters
•
5th order Chebyshev II
•
Table from: Williams & Taylor book, p. 11.112
•
50dB stopband attenuation
•
f-3dB =10MHz
© 2007 H.K. Page 47
Realization with Integrator V1 =
1 ⎡Vi −V1 − V2 ⎤ + Ca V Ca + C1 3 s (Ca + C1 ) ⎢⎣ Rs R* ⎥⎦
-Rs
Rs
R*
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 48
5th Order Butterworth Filter
V2
From: Lecture 4 page 14
V3' V4 V5'
Vo EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 49
5th Order Chebyshev II Filter Opamp-RC Simulation V2
V3' V4
V5' Vo EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 50
+
+
+
+
+
+
-
+
+ +
+
+
-
-
+
+
-
+
-
-
-
-
-
-
-
Vin
7th Order Differential Lowpass Filter Including Transmission Zeros
-
Vo
Transmission zeros implemented with pair of coupling capacitors EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 51
Effect of Integrator Non-Idealities on Filter Frequency Characteristics • In the passive filter design (RLC filters) section: – Reactive element (L & C) non-idealities Æ expressed in the form of Quality Factor (Q) – Filter impairments due to component non-idealities explained in terms of component Q
• In the context of active filter design (integrator-based filters) – Integrator non-idealities Æ Translated to have form of Quality Factor (Q) – Filter impairments due to integrator non-idealities explained in terms of integrator Q EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 52
Effect of Integrator Non-Idealities on Filter Performance • Ideal integrator characteristics • Real integrator characteristics: – Effect of opamp finite DC gain – Effect of integrator non-dominant poles
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 53
Effect of Integrator Non-Idealities on Filter Performance Ideal Integrator Ideal Intg.Intg. Ideal l o g H (s)
C
R
Vin
+
Vo 0dB
I d e al o pa m p D C ga i n = ∞ Sing le po le @ D C → n o no n- d om in an t po le s −ωo H( s )= s ωo = 1/ RC EECS 247
Lecture 4: Active Filters
ω0
ψ -90o
© 2007 H.K. Page 54
Ideal Integrator Quality Factor Ideal intg. transfer function:
H( s )=
Since component Q is defined as::
s
H ( jω ) = Q=
Then:
−ωo
−ωo
=
jω
1
=− j
1 R (ω ) + j X (ω )
ω ωo
X (ω ) R (ω )
t g. = ∞ Qiidneal
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 55
Real Integrator Non-Idealities Ideal Intg. l o g H (s)
a
ω0
P1 =
ψ
ψ
-90o
-90o
H( s )= EECS 247
−ωo s
Real Intg. l o g H (s)
H( s )≈ Lecture 4: Active Filters
(
ω0 a
ω0
P2P3
−a
1+ s a
ωo
)(1 + p2s )(1 + p3s ) .. . © 2007 H.K. Page 56
Effect of Integrator Finite DC Gain on Q l o g H (s) a
ωo
∠ −π
ω0 P1 =
ω 2
+ Arct an P1
ωo
ω0 a
-89.5
ψ
-90
P1
ωo
→ P h a s e l ea d @ ωo (in radian )
-90o
Example: P1/ ω0 = 1/100 Æ phase error ≅ +0.5degree EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 57
Effect of Integrator Finite DC Gain on Q
Magnitude (dB)
Ideal intg Intg with finite DC gain
• Phase lead @ ω0
ÆDroop in the passband
Droop in the passband
1
Normalized Frequency EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 58
Effect of Integrator Non-Dominant Poles l o g H (s)
ωo
ω0
∠ −π 2
P2P3 -90
ψ
ω
∞ ω
− Ar ct an ∑ po i =2 i
∞ ω o → Phase l ag @ ω o p i =2 i (in radian )
∑
-90.5
-90o
Example: ω0 /P2 =1/100 Æ phase error ≅ −0.5degree EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 59
Effect of Integrator Non-Dominant Poles
Magnitude (dB)
Ideal intg Opamp with finite bandwidth
Peaking in the passband
• Phase lag @ ω0
ÆPeaking in the passband In extreme cases could result in oscillation!
1 Normalized Frequency EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 60
Effect of Integrator Non-Dominant Poles & Finite DC Gain on Q l o g H (s) a
ωo P1 =
ω0 a
ω0
∠ −π
P2P3
ψ
2
-90
ω
+ Arctan P1
ωo ∞ ω − Arctan ∑ o i = 2 pi
-90o Note that the two terms have different signs Æ Can cancel each other’s effect! EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 61
Integrator Quality Factor Real intg. transfer function:
H( s )≈
(
−a
1+ s a
Based on the definition of Q and assuming that:
ωo
ωo > 1
1 ∞ 1 −ω ∑ 1 o a i = 2 pi
ad @ ω
0
Phase
l ag @
ω0
© 2007 H.K. Page 62
Example: Effect of Integrator Finite Q on Bandpass Filter Behavior 0.5ο φexcess @ ωointg
0.5ο φlead @ ωointg
Ideal
Ideal
Integrator P2 @ 100.ωo
Integrator DC gain=100
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 63
Example: Effect of Integrator Q on Filter Behavior ( 0.5ο φlead −0.5ο Æ φerror @
φexcess )
@
ωointg
ωointg ~ 0 Ideal
Integrator DC gain=100 & P2 @ 100. ωο EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 64
Summary Effect of Integrator Non-Idealities on Q int g. = ∞ Qideal int g. ≈ Qreal
• • •
1 1 a − ωo
∞
1 p i =2 i
∑
Amplifier DC gain reduces the overall Q in the same manner as series/parallel resistance associated with passive elements Amplifier poles located above integrator unity-gain frequency enhance the Q! – If non-dominant poles close to unity-gain freq. Æ Oscillation Depending on the location of unity-gain-frequency, the two terms can cancel each other out!
EECS 247
Lecture 4: Active Filters
© 2007 H.K. Page 65