LC Ladder Filters

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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

=



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