ECEN 622 TAMU ACTIVE RC FILTERS - Texas A&M …s-sanchez/622 Lecture 2 Active... · ACTIVE RC...

ACTIVE RC FILTERS

By Edgar Sánchez-Sinencio 1

1. Basic Building Blocks

• First-Order Filters

• Second-Order Filters, using multiple VCVS

• Second-Order Filter, using one VCVS ( Op Amp)

• State-Variable Biquad

2. Non-Ideal Active – RC Filters

• Using VCVS ( Op Amp) vs. VCCS ( transconductance Amp)

• Second-Order Non-idealities

• Fully Differential Versions

• Fully Balanced, Fully Symmetric Balance Circuits

3. Introduction to Matlab and Simulink for filter Design and

filter approximation techniques

ECEN 622 TAMU

622 (ESS) A1

ACTIVE - RC FILTERS

The basic building block is illustrated below

1

2

1

2

i

o

Z

Z1

A

11

Z

Z

sV

sVsH

cases particularconsider Next we

Z

ZsH

then, A that assume usLet

1

2

rentiator Diffe CRssH

Integrator CsR

1sH

2

21

Vi

Z1 Z2

A Vo

Vi

R1

C2

Vo

Vi

R2C1

Vo


3

Z1

or

ZF = Z2

Can be:

EXAMPLE: Let

FF

FF

11

11

CsR1

RZ,

CsR1

RZ

Assuming ideal op amp A . Then using (1)

p

zn

FF

111F

1

011

s1

s1K

)CsR1(

CsR1RR

V

VH

(4)

1H

1

F

R

R

z p

zp 1

F

R

R

F

1

C

C

p z

zp

R

sC

1

R

C

R C

sC

sRC1

C

R sRC1

R

By Edgar Sánchez-Sinencio

4

Particular cases are easily derived from (3) and (4)

Integrator: F1 R,0C

1FFF1

F1

RsC

1

CsR

1

R

RH

Differentiator ; 0C,R F1

1F111

F1 CsRCsR

R

RH

Low-Pass: 0C1

FF

1

F

1CsR1

R

R

H

High-Pass: 1R

FF

11

1

F1

CsR1

CsR

R

RH


5

One pole and one zero

What are the key differences between Eqs. (4) and (5)?

(5) CsR1

CsR1

C

C

CsR1

sC

sC

CsR1

V

V

11

FF

F

1

11

1

F

FF

i

o

Exercise 1. Obtain the transfer function of the following circuit.

A Vo

R1 RF1 RF2

CF

Vi

C1

Vi

R1 RF CFC1


6

Second-Order Filters Based on a Two-Integrator Loop.

• We can design a second-order filter by cascading two inverters. i.e.

)6(

1RCRCsRCRCs

R

R

R

R

RsC1RsC1

R

R

R

R

v

V

2F2F1F1F2F2F1F1F2

2

2F

1

1F

2F2F1F1F

2

2F

1

1F

i

o

What are the locations of the poles?

2F2F1F1F

2F2F1F1F2

2F2F1F1F2F2F1F1Fp

RCRC2

RCRC4RCRCRCRCs

2,1

ViR1

RF1RF2

CF2CF1

R2Vo


7

To have complex poles it requires that

?0RCRC2RCRC 2F2F1F1F2

2F2F2

1F1F

Which it is impossible to satisfy. Therefore, cascading two first-order filter yield a

second-order filter with only real poles.

The general form of the second order two-integrator loop has the following topology.

(7a)

Z

Z

Z

Z1

Z

Z

Z

Z

H

2

2F

1

1F

2

2F

3

1F

-1

Vi

Vo

Z3 ZF1

Z2

ZF2

Z1


8

Note the similarity of Eq. (7a) with (2). Also observe that A “-1” needs to be inserted

before or after the second inverter to yield a negative feedback loop.

Let us consider the following filter where

2F2F

2F2F

1F1F221133

RsC1

RZ,

sC

1Z,RZ,RZ,RZ

Thus Eq. (7a) yields:

2o

o2

21o

22F11F2F2F

2

22F31F

2F2F

22F

11F

2F2F

22F

31F

sQ

sRCRC

1

RC

ss

RCRC

1

H

RsC1

RR

RsC

11

RsC1

RR

RsC

1

H


9

By injecting in different current summing nodes a general biquad filter can be obtained.

22F11F2F2F

2

3113211F222F31F

3

o

2F2F

22F

11F

2F2F

12F

2F3

2F2F

2

2F2

2F2F

RR

31F3

o

RCRC

1

RC

ss

KRsCVKRsCVRCRC

1V

V

RsC1

RR

RsC

11

RsC1

VR

RK

RsC1

R

RK

RsC1RsC

1V

V

22F

R1CF1 r

r

RF2

CF2

Vo

R2

R3VLP=V3

VBP=V2

-VBP=V1

r/K2

RF2/K3

Vo2Vo1


10

Exercise 2. Obtain the expressions of Vo1 and Vo2.

More general biquad expressions and topologies can be obtained by adding a summer.

o301o20in10oT VKVKVKV

Exercise 3. Draw an active-RC topology of the block diagram show above.

Exercise 4 a) For only obtain Vo and Vo1 when instead of the resistor RF2/K3 a

capacitor K4 CF2 is used. b) For only obtain Vo1 when the resistor R3 is replaced

by a capacitor KHPCF1.

0V1

0V3

S

Vin

Vo1

Vo

-K10

-K20

-K30


11

By using also the positive input of the op amp other useful filters can be obtained.

V2

V1

Z1 Z2

ZR

ZC

1

2

1

22

CR

R1

1

2

o

Z

Z1

A

11

Z

Z1V

ZZ

ZV

Z

Z

V

Example. Phase shifter Z2=R2=R1, Z1=R1, ZR=R 21C VV withA and sC

1Z

sRC1

sRC1

sRC1

sRC2sRC12

sRC1

sRC1

V

V

1

o


12

Sallen and Key Bandpass Filter

K

Vi

Vo

C1

R2

C2R1

R3

K is a non-inverting amplifier

Using Nodal Analysis

(3) V

(2) 0

2

122 V-

(1) 11

2o

21

1322

31211

KV

RsCVsC

R

V

R

VVsC

RRCCsV io

21231

31

132

3

1

3

22

2

11

11

CCRRR

RR

CR

s

R

R

R

RK

CRs

CR

sK

sV

sVsH

i

o


13

A particular case is for R1=R2 =R3=R, C1=C2=C

Then

Q

24K and

2RC

Q and givena for or

K4

2Q;

RC

2

o

o

2

2o

Exercise 5. Prove the transfer function is a BP filter of the following circuit

212122121

212

11

i

o

CCRR1K

1s

CR

1

CRR

RR

1K

1s

CR

s

1K

K

sV

sVsH

-K

Vi

R1 C2C1

Vo

R2


A4In the past before IC fabrication, active filters implementation preferred one op amp

structure. One very popular type is the Sallen and Key unity gain implementations.

QRC

CQC

CC

RRR

sQ

s

sH

o

oo

oLP

21

4 2

2

1

21

22

2

One also popular topology is the Rauch Filter

22113211

2

1322

CRCR

1

R

1

R

1

R

1

C

ss

CRCR

1

sH

LP Rauch Filer

LP Sallen - Key

ViVo

R2 R1

C1

C2

1

ViVo

C2

C1

R3

R1

R2


15

Another technique for analysis and design based on state-variable uses building blocks.

CIRCUIT REPRESENTATION

V1

V2

R2

R1C1

Vo1

S 1/sV1

V2

22CR

1

11CR

1

Vo1

Note: V1 and V2 can take

any value including Vo.

S

V1

V2

V3

V4

V5

K4

Vo2

-K1

-K2

K3

K5

V1

V2

V3

V4

V5

R1

R2

R3

R4

RF

Vo2

R5


16

Let us apply to a two-integrator loop plus Mason’s Rule.

For Second-topology

S 1/s 1/s1oVinV 3oV

-Ko1-1

-KQ

Ko2

2oV

-K1

BP KKsKs

sVK

s

KK

s

K1

Vs

K

V

HPKKsKs

VsK

s

KK

s

K1

VKV

2o1oQ2

in1

22o1oQ

in1

o2

2o1oQ2

in2

1

22o1oQ

in11o


17

Next we show that we can go from an Active-RC representation into a

block diagram or vice versa.

KHN Biquad Filter

V1

R4

R1

R5 R6 R7

R2

C1 C2

VHP VBP VLP

S 1/s 1/sVi

3

5

R

R

-R5/R4

VLPVHP VBP

16CR

1

27CR

1

Q43

5

21

1 KR||R

R1

RR

R


18

2716

45

16

Q2

2

3

5

22716

45

16

Q

3

5

i

HP

CRCR

RRs

CR

Ks

sR

R

sCRCR

RR

s

1

CR

K1

R

R

V

V


ACTIVE RC FILTERS


1. Basic Building Blocks

• First-Order Filters

• Second-Order Filters, using multiple VCVS

• Second-Order Filter, using one VCVS ( Op Amp)

• State-Variable Biquad

2. Non-Ideal Active – RC Filters

• Using VCVS ( Op Amp) vs. VCCS ( transconductance Amp)

• Second-Order Non-idealities

• Fully Differential Versions

• Fully Balanced, Fully Symmetric Balance Circuits

3. Introduction to Matlab and Simulink for filter Design and

filter approximation techniques

ECEN 622 TAMU

ELEN 622 (ESS)

Op Amp Non-Idealities

A

iV1Z

2Z

A

V0

oV

21

1

1

2

1

2

1

2

where

11

11

ZZ

Z

A

Z

Z

A

Z

Z

A

Z

Z

sH

Integrator

Case 1 sAs

GB;

sCZ;RZ

2

211

1

11

1

1

1

GBRC;

GBssRCGB

RCs

GB

sRCsH

o

GBtanjH;

RCjGB

RCjH

90

2

1 1

2

Non-Ideal Active-RC Integrators


o.

GB;

jAtan

GBtan

75

10

1i.e.

1 o11

M

o

o

oo

o

GB

j

GB

jH

1

1

11

2

2

error 5

10 i.e.

1

11

2

2

2

2

0

%~

.GB

GB

GB

M

o

o

M


It follows that the ideal -6 dB/octave roll-off expected from an ideal integrator

changes to -12 dB/octave at the frequency of the parasitic pole given by

CRs tp

1

which may be approximated by,

CRs tt

1 for

dB

A

o

0

-6 dB/octave

1/AoC

R

1/CR

-12 dB/octave

t

22By Edgar Sánchez-Sinencio

In general

jXR

jT

1

then we define the integrator Q-factor by

jAQ

R

XQ

tI

I

GBQ t

L


Making an analogy of QL of an inductor

L

oL

R

LQ

RLL

Lossy Part

For an integrator one can obtain

jAGB

GBGB

RC

RCQI

112

Miller Integrator

C

Rvi

voS

GB


How can we compensate this degradation of performance?

a)

CGBRC

1

jAQI

C

R

RC

vi vo

s

GBsA

If we make

Ideally we obtain

RCsV

sV

i

o 1

This integrator yields a positiveC

R

R1

R2

A1

A2

vivo


ACTIVE – RC INTEGRATOR: Pole Shift and Predistortion

622 (ESS)

26

(1b)

1sRCsA

1sRC

1

V

V

(1a) sRC

1

sRC

11

sA

11

sRC

1

V

V

i

o

i

o

)s(A

+

-Vo

Vi

R C

A(s)

; where Ao is the DC gain and 3dB the dominant pole in

open loop.

Then (1b) becomes

(2)

RCsAs

RC

A

RCRC

1Ass

RC

A

V

V

dB3dB3o

2

dB3o

dB3dB3dB3o

2

dB3o

i

o

Let GB=Ao3dB

G. Daryanani, “Principles of Active Network Synthesis and Design,” John Wiley and Sons, 1976.

dB3

dB3o

s

A)s(ALet


27

The roots of the denominator are

(3a)

RCGB

41

2

GB

2

GBP

2/1

2dB3

2,1

Using the approximation 1–X)1/2≅ 1 −X/2 for X<<1, then

)b3(

RCGB

21

2

GB

2

GBP

2dB3

2,1

Thus the roots yield

GBRCA

1GBP

RCA

1P

o2

o1


28

The Bode Plot Looks Like


29

PREDISTORTION; FREQUENCY COMPENSATION

In order to relax the bandwidth op amp requirement one can use a RC or CC on

the Miller Integrator. That is

B. Wu and Y. Chiu, “A 40nm CMOS Derivative-Free IF Active-RC BPF with Programmable Bandwidth and Center Frequency Achieving Over 30dBm IIP3”,

IEEE JSSC, Vol. 50, No. 8, pp 1772-1784, August 2015.

GB

1R Cor

GB

1C R Use CC

Equivalent

Vi

RC

R

C

VoA(s)

Vi

R RCC

Vo

Vi

R

C

Vo

Vi R

C

Vo

CC

CC


30

Vi

Vi

Z1

Z1 Z2

ZF

A

Zo ZL

ZL

Vo

Vo

Gm

Using VCVS vs. VCIS in Active-RC Filters

The motivation is to use OTA (VCIS) instead of more power hungry Op Amp (VCVS)

1

2

i

o

L1

21

1m

2m

m

L1

21

1

2m

1

2

i

o

1

F

i

o

1

F

1

F

i

o

oo

Z

Z

V

V

ZZ

ZZ

Z

1g

Z

1g

g

ZZ

ZZ

Z

1

1

Zg

11

Z

Z

V

V

Z

Z

V

V

then,A If

Z

Z1

A

11

Z

Z

V

V

0ZRFor


Using VCVS

𝑉𝑖 − 𝑉𝑥𝑍1

+𝑉𝑜 − 𝑉𝑥𝑍𝐹

= 0

𝑉𝑜 = −𝐴𝑉𝑥

Vx

Signal flow graph:

VxVi Vo−𝐴

𝑉𝑥 = 𝑉𝑖𝑍𝐹

𝑍1 + 𝑍𝐹+ 𝑉𝑜

𝑍1𝑍1 + 𝑍𝐹

β

𝑍𝐹𝑍1 + 𝑍𝐹

𝑍1𝑍1 + 𝑍𝐹

Using Mason’s rule:

𝑉𝑜𝑉𝑖=

−𝐴𝑍𝐹

𝑍1 + 𝑍𝐹

1 + 𝐴𝑍1

𝑍1 + 𝑍𝐹

=−𝑍𝐹/𝑍1

1 +1𝐴

1 +𝑍𝐹𝑍1

Thus as 𝐴 → ∞ the gain becomes −𝑍𝐹/𝑍1


Using VCCS

𝑉𝑖 − 𝑉𝑥𝑍1

+𝑉𝑜 − 𝑉𝑥𝑍2

= 0

𝑉𝑜 − 𝑉𝑥𝑍2

+𝑉𝑜𝑍𝐿

= −𝐺𝑚𝑉𝑥

Vx

Signal flow graph:

VxVi Vo

1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿

𝑉𝑥 = 𝑉𝑖𝑍2

𝑍1 + 𝑍2+ 𝑉𝑜

𝑍1𝑍1 + 𝑍2

β

𝑍2𝑍1 + 𝑍2

𝑍1𝑍1 + 𝑍2

Using Mason’s rule:

𝑉𝑜𝑉𝑖=

𝑍2𝑍1 + 𝑍2

1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿

1 −1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿

𝑍1𝑍1 + 𝑍2

=−𝑍2/𝑍1

1 +1

𝐺𝑚𝑍2 − 11 +

𝑍2𝑍1

1 +𝑍2𝑍𝐿

What conditions do we need to impose on𝐺𝑚 for proper operation?

𝑉𝑜 =1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿

𝑉𝑥


Using VCCS: Acceptable 𝐺𝑚 Range

Vx

Signal flow graph:

VxVi Vo

1 − 𝐺𝑚𝑍21 + 𝑍2/𝑍𝐿

𝑍2𝑍1 + 𝑍2

𝑍1𝑍1 + 𝑍2

𝑉𝑜𝑉𝑖=

−𝑍2/𝑍1

1 +1

𝐺𝑚𝑍2 − 11 +

𝑍2𝑍1

1 +𝑍2𝑍𝐿

Note from the signal flow graph that having a negative feedback loop requires 𝐺𝑚𝑍2 > 1

For the gain to approach the ideal gain of −𝑍2/𝑍1, we need

𝐺𝑚𝑍2 − 1 ≫ 1 +𝑍2

𝑍11 +

𝑍2

𝑍𝐿𝐺𝑚𝑍2 ≫ 1 + 1 +

𝑍2

𝑍11 +

𝑍2

𝑍𝐿

Thus, guaranteeing this second condition automatically guarantees the negative feedback condition


Using VCCS: Practical Considerations

Vx

In practice, 𝑍𝐿 = 𝑅𝑜|| 𝑍𝐿𝑜𝑎𝑑 where 𝑅𝑜 is the OTA’s output resistance and 𝑍𝐿𝑜𝑎𝑑 is the external load impedance

𝐺𝑚𝑍2 ≫ 1 + 1 +𝑍2𝑍1

1 +𝑍2𝑍𝐿

One should note that 𝐺𝑚 and 𝑅𝑜 are not independent since increasing current to increase 𝐺𝑚 will reduce 𝑅𝑜.To a first order, one can consider 𝐴 = 𝐺𝑚𝑅𝑜 to be constant.

Finally, in cascaded filter designs, the load of one stage is the input resistor of the next one. We can thus assume 𝑍𝐿𝑜𝑎𝑑 = 𝑍1 as a realistic condition (𝑍𝐿𝑜𝑎𝑑 = ∞ places a looser constraint on 𝐺𝑚)

Substituting for 𝑍𝐿 as described above yields the following constraint on 𝐺𝑚:

𝐺𝑚 ≫1

𝑍1

2 1 + 𝑍1/𝑍2 + 𝑍2/𝑍1

1 −1𝐴1 + 𝑍2/𝑍1


Numerical Example

• Ideal VCVS and VCCS components from Cadence were used to simulate the above circuits.

• Two configurations were tested:1. Unity gain inverting amplifier (𝑍2 = 𝑍1 = 𝑅)2. Lossy integrator with corner frequency 1 MHz (𝑍2 = 𝑅||𝐶)

• To have a fair comparison, the value of 𝐴 was fixed to 30 for both the VCVS and VCCS implementations (thus the VCCS had 𝑅𝑜 = 30/𝐺𝑚). This is a typical value for the voltage gain of a single stage amplifier.

• In all tests, 𝑅 = 100𝑘Ω, the output resistance of the VCVS is set to 1 𝑘Ω and a load capacitance is added to the output of the amplifier to give an output pole at 10 MHz.

• With these numbers, the constraint on 𝐺𝑚 is 𝐺𝑚 ≫ 54 𝜇𝑆

• The value of 𝐺𝑚was swept from 30 𝜇𝑆 to 600 𝜇𝑆 and the simulation results are shown in the following slides.


Simulation Results: Inverting Amplifier

VCVS Response

Increasing 𝐺𝑚

VCVS Response

Increasing 𝐺𝑚


Simulation Results: Lossy Integrator

VCVS Response

Increasing 𝐺𝑚

VCVS Response

Increasing 𝐺𝑚


Conclusions

• It is possible to use VCCS (OTA) instead of VCVS (Opamp) in active-RC filters in order to avoid using costly buffer stages.

• Proper performance requirements place a lower limit on the transconductance of the OTA used.

• Using a transconductance of 10-15x the minimum requirement yields a comparable performance to a design employing an Opampimplementation.


It can be shown that for equal s

GBsA , the Tow-Thomas filter has the following

deviations

GB

k

GBQ;GB

Q

GBQQQ

o

o

o

oooo

oo

oa

2

2

and

414

or

41

1

Vi

QR R

R/k

A1A2

A3R

VBP

CCC1

C2

VL

r

r

-VL

C

rr

VBP -VL

Improved integrator version with positive QI

2

2

1

22

11

GB

kQ

GB

GBkQQ

ooo

o

oa

Improved version by replacing noninverting integrator:


Single Ended

Fully-Differential Version

How to generate Fully-Differential Filters based on Single-Ended

Version?

CRsV

V

i

o 1

+

-

iVR

oV

C

+

-iV

oV

R C

+

-

iV R

oV

C

+

-

-

+R

C

R

iV

iV

oVoV

C


Particular Case. Assume no is Available.vi-

Symmetric conditions

Read fully balanced - fully

symmetric circuits from 607.

XQ RRRRR

11111

2303

+

-

-

++

-

C

C

R

R R

RoV

iV

oV

iV

-

++

--

+

R2

R3

R01

RQ

RX

R03

R02

C1 C

2

VBPV

HPV

LP

iV

State -Variable

Filter

+

-

iV

oV

R

R-

iV

C

C

oV

iV

-+


KHN State Variable Two-Integrator Filter

Use Mason’s Rule:

Next we consider the fully-differential version of the KHN filter.

,1

20202

CRK

,03

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How can we take advantage of improved combination of ± QI in fully differential versions?

QRR

R/k

A1

C1

+

-A2

+

-

C2R

RR/k C1

QRR

C2

inV

inV

oV

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A1

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-A2

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C2R

RR/k C1

QRR

C2

inV

inV

oV

oV

SAME!


622 (ESS)

Effects of Non-Ideal Op Amps on the Tow-Thomas Biquad

)3,2,1i(A When i

2121

21221

32o

2121

2121o2

AA

1

A

1

A

11

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1

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A

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A

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AA

1Q3

A

Q1

A

1Q21

QsssD

are finite, the denominator becomes of the transfer function yields:

R

( )

( )

( )

( )

( )

( )A1=∞

A2=∞A3=∞

Vi

R/K

QR

C

R

VBPVLP

-VLP

C r

r


3,2,1i,s

GBALet i

i

Furthermore assume the range of interest .1Q,1GB

and A

GB

i

o

oi

i

Then D(s) becomes:

3

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o

1

o

o

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a

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2

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or

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ECEN 622 (ESS)

TAMUKEY FILTER PARAMETERS IN ACTIVE-RC FILTERS

• Dynamic Range

• Signal-To-Noise Ratio

• Total Output Noise

• Noise Power Spectral Density

• Total Area

Resistor and Capacitors can be expressed as:

resistor R of terminals theinput to thefrom function transfer theis fH Where

1 r

fH

R2

V

R2

HVfP

dsinput yiel sinusoidala for ndissipatiopower resistor The

ues.filter val normalized theare c and r where

CcC,RrR

i

2i

2i

2ii

R

Reference. L. oth et all, “General Results for Resistive Noise in Active RC and MOSFET-C Filters”, IEEE Trans on Circuits

and Systems II, Vol. 42, No. 12, pp. 785-793, December 1995.


Focusing on the noise resistor, the power spectral density is given by

(2) fHrkTR4fHkTR4fS2

o2

oR

The definition of fH o is pictorially shown below:

Thus, the total output noise (mean squared value) due to the resistors become

In practice the upper limit of the integration is limited to a useful practical value.

o RR dffSN

+

-Vi

VR

+

-

fHV o


The signal-to-ratio for a given Vi and frequency f is given by

2cc

12c

1

BP

BP

2c

c2

c

BP

2

R

f2

maxi

max,R

max,RRf

R

2i

fjffQjf

jffQfH

yieldssH above thenotation following theUsing

sQ

s

sQ

sH

examplefiler order BP-seconda consider usLet

(4) N2

fHmaxVDR

Then resistors. thein ndissipatiopower specified maximum theis P where

PfPmax

and

(3) N2

fHVSNR


12

icR Qaa

R

VfP

For the biquad shown below

and

a

a/Q1

C

kT2NR

OBPR VaNa limited by linearity and by resistor power dissipation which is

proportional to (a)2.

R

CC

RQR/a

QR/a 1/a

VOBP

-1Vi


Fully Differential Fully Balanced Circuits

What is the problem with single-input / single-output?

A

o

o

FZ1Z

iVnV

oVo

A

0

A

VVV

ZZ

ZVV

on1

F1

1on

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icmid1

Fo

icmidi

No elimination of

common-mode signal.

How to solve this problem?

o

FZ1Z

oVo

1Z

o

2V

1V

FZ

)VV(Z

ZV 21

1

Fo

No common-mode output.

2

)VV()VV(VVVFor 21

21icmidi


How to obtain a fully differential circuit?

We will discuss two potential approaches

Approach 1

2PV

FR

o2V o

1R

1R 1oV

FR

o1V o

1R

1R 2oV

FR

2nV

2P1P VV

1PV1nV

Remark:More robust to reject

common-mode signals

Approach 2

FR

o

1V o

1R

)VV(R

R

)VV(R

RV

211

F

121

F1o

1PV

1nV

FR

o

o

2V o

o

)VV(R

RV 12

1

F2o

2PV

2nV

FR

o

o

)VVVV(R

RVV 1221

1

F2o1o

)VV(R

R2V 21

1

FoD

conditions 1P1n2P2n VV;VV

Remark: sensitive to CM signals

FR


First-Order FB Low Pass with Op Amp

*

.subckt opamp non inv out

rin non inv 100K

egain 1 0 (non, inv) 200K

ropen 1 2 2K

copen 2 0 15.9155u

eout 3 0 (2, 0) 1

rout 3 out 50

.ends

*vin 3 31 ac 1.0

vin 31 0 ac 1.0

x1 4 1 2 opamp

x2 4 11 22 opamp

R1 3 1 1K

R11 3 4 1K

R1B 31 4 1K

R1BB 31 11 1K

RF1 2 1 1K

RF1B 22 11 1K

RF11 4 0 1K

RF11B 4 0 1K

C1 2 1 0.159155u

C1B 22 11 0.159155u

C1A 4 0 0.159155u

C11B 4 0 0.159155u

rdummy 3 31 1

.ac dec 10 10Hz 10KHz

.probe

.end

1C

1FR

oiV o

1R

11R11FR

A1C

B11C

o

iV o

B1R

BB1R

B1C

B1FR

oV

B11FR

oV

22

1131

4

23 1


Fully Balanced T-T Active-RC Implementation

1C

2C

1oRR

R

2oR

o oV

QR

o

iV

o

R

KR

KR1oR

QR

2oR

1C

QR

2oR

1C

R

R

2C

2C

1oR

R

o

o

iV

o R

KR

KR

oV

1oR1C

QR

R

2C

2oR


Introduction to Matlab and

Simulink For Filter Design

622 Active Filters


Texas A&M University


Example 1: Ideal Integrator

57

R = 1K C = 0.159mF


Bode Plot: Ideal Integrator (Matlab)

s=tf(‘s’);

R=1e3; %Resistor Value

C=0.159e-3; %Capacitor Value

hs=1/(R*C*s); %hs= Vo(s)/Vi(s)

figure(1)

bode(hs) %Create Bode Plot

grid minor %Add grid to plot

H= gcr; %change X-axis

units

h.AxesGrid.Xunits = ‘Hz’; %Set units to

Hz

pole(hs); %calculates hs poles

zero(hs); %calculates hs zeros

58

-20

-10

0

10

20

30

Magnitu

de (

dB

)

10-1

100

101

-91

-90.5

-90

-89.5

-89

Phase (

deg)

Bode Diagram

Frequency (Hz)


59

2) Go to: Tools => Control Design=> Linear Analysis 3) Then press: Linearize model

1) Create Model using Gain, Integrator, and In/Out blocks

Bode Plot: Ideal Integrator (Simulink)


Tow-Thomas Biquad (Simulink)


Output Waveform (Scope)


Integrator Non-ideal amplifier

62

clear clcs=tf('s'); R=1; %Resistor ValueC=0.159e-3; %Capacitor Valuehs1=-1/(R*C*s); %hs= Vo(s)/Vi(s)figure(1) bodemag(hs1)hold onf=1e3;for i=1:5;GBW=2*pi*f;A=GBW/s;Beta=R/(R+1/(s*C));hs2=-1/(R*C*s)*1/(1+1/(A*Beta));hold onbodemag(hs2,{2*pi*1,2*pi*1e5}) f=10*f;endgrid minor %Add grid to ploth= gcr; %change X-axis unitsh.AxesGrid.Xunits = 'Hz'; %Set units to Hzlegend('ideal', 'GBW=1kHz','GBW=10kHz', 'GBW=100kHz','GBW=1MHz', 'GBW=10MHz',1)

100

101

102

103

104

105

-80

-60

-40

-20

0

20

40

60

80

Magnitu

de (

dB

)

Bode Diagram

Frequency (Hz)

ideal

GBW=1kHz

GBW=10kHz

GBW=100kHz

GBW=1MHz

GBW=10MHz


Filter Approximation: Low-Pass Butterworth

63

The squared magnitude of a low-pass butterworth filter is given by:


Pole-zero plot

64

Pole-Zero Map

Real Axis

Imagin

ary

Axis

-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

10.080.170.280.380.50.64

0.8

0.94

0.080.170.280.380.50.64

0.8

0.94

0.2

0.4

0.6

0.8

1

0.2

0.4

0.6

0.8

1


Bode Plot

65

-120

-100

-80

-60

-40

-20

0M

agnitu

de (

dB

)

10-2

10-1

100

101

102

-270

-180

-90

0

Phase (

deg)

Bode Diagram

Frequency (rad/sec)


Low-pass Chebyshev Filter

• Use the Matlab cheb1ap function to design a second order

Type I Chebyshev low-pass filter with 3dB ripple in the pass

band

w=0:0.05:400; % Define range to plot

[z,p,k]=cheb1ap(2,3);

[b,a]=zp2tf(z,p,k); % Convert zeros and poles of G(s) to polynomial form

bode(b,a)

grid minor;



67

-80

-60

-40

-20

0

Magnitu

de (

dB

)

10-2

10-1

100

101

102

-180

-135

-90

-45

0

Phase (

deg)

Bode Diagram

Frequency (rad/sec)



w=0:0.01:10;


[b,a]=zp2tf(z,p,k);

Gs=freqs(b,a,w);

xlabel('Frequency in rad/s');

ylabel('Magnitude of G(s)');

semilogx(w,abs(Gs));

title('Type 1 Chebyshev Low-Pass Filter');

Grid;

68

% Another way to write the code!



69

10-2

10-1

100

101

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Type 1 Chebyshev Low-Pass Filter


Inverse Chebyshev• Using the Matlab cheb2ap function, design a third order

Type II Chebyshev analog filter with 3dB ripple in the stop band.

w=0:0.01:1000;


[b,a]=zp2tf(z,p,k); Gs=freqs(b,a,w);

semilogx(w,abs(Gs));

xlabel('Frequency in rad/sec');

ylabel('Magnitude of G(s)');

title('Type 2 Chebyshev Low-Pass Filter, k=3, 3 dB ripple in stop band');

grid


Inverse Chebyshev

71

10-2

10-1

100

101

102

103

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency in rad/sec

Magnitude o

f G

(s)

Type 2 Chebyshev Low-Pass Filter, k=3, 3 dB ripple in stop band


Elliptic Low-Pass Filter

• Use Matlab to design a four pole elliptic analog low-pass filterwith 0.5dB maximum ripple in the pass-band and 20dBminimum attenuation in the stop-band with cutoff frequency at200 rad/s.

w=0: 0.05: 500;

[z,p,k]=ellip(4, 0.5, 20, 200, 's');

[b,a]=zp2tf(z,p,k);

Gs=freqs(b,a,w);

plot(w,abs(Gs))

title('4-pole Elliptic Low Pass Filter');

grid


Elliptic Low-Pass Filter

73

0 50 100 150 200 250 300 350 400 450 5000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

14-pole Elliptic Low Pass Filter


Transformation Methods

• Transformation methods have been developed where a low

pass filter can be converted to another type of filter by simply

transforming the complex variable s.

• Matlab lp2lp, lp2hp, lp2bp, and lp2bs functions can be used to

transform a low pass filter with normalized cutoff frequency,

to another low-pass filter with any other specified frequency,

or to a high pass filter, or to a band-pass filter, or to a band

elimination filter, respectively.


LPF with normalized cutoff frequency, to

another LPF with any other specified frequency

• Use the MATLAB buttap and lp2lp functions to find the transfer function of a third-order Butterworth low-pass filter with cutoff frequency fc=2kHz.

% Design 3 pole Butterworth low-pass filter (wcn=1 rad/s)

[z,p,k]=buttap(3);

[b,a]=zp2tf(z,p,k); % Compute num, den coefficients of this filter (wcn=1rad/s)

f=1000:1500/50:10000; % Define frequency range to plot

w=2*pi*f; % Convert to rads/sec

fc=2000; % Define actual cutoff frequency at 2 KHz

wc=2*pi*fc; % Convert desired cutoff frequency to rads/sec

[bn,an]=lp2lp(b,a,wc); % Compute num, den of filter with fc = 2 kHz

Gsn=freqs(bn,an,w); % Compute transfer function of filter with fc = 2 kHz

semilogx(w,abs(Gsn));

grid;

xlabel('Radian Frequency w (rad/sec)')

ylabel('Magnitude of Transfer Function')

title('3-pole Butterworth low-pass filter with fc=2 kHz or wc = 12.57 kr/s')


LPF with normalized cutoff frequency, to

another LPF with any other specified frequency

76

103

104

105

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Radian Frequency w (rad/sec)

Magnitude o

f T

ransfe

r F

unction

3-pole Butterworth low-pass filter with fc=2 kHz or wc = 12.57 kr/s


• Use the MATLAB commands cheb1ap and lp2hp to find the transfer function ofa 3-pole Chebyshev high-pass analog filter with cutoff frequency fc = 5KHz.

% Design 3 pole Type 1 Chebyshev low-pass filter, wcn=1 rad/s


[b,a]=zp2tf(z,p,k); % Compute num, den coef. with wcn=1 rad/s

f=1000:100:100000; % Define frequency range to plot

fc=5000; % Define actual cutoff frequency at 5 KHz

wc=2*pi*fc; % Convert desired cutoff frequency to rads/sec

[bn,an]=lp2hp(b,a,wc); % Compute num, den of high-pass filter with fc =5KHz

Gsn=freqs(bn,an,2*pi*f); % Compute and plot transfer function of filter with fc = 5 KHz

semilogx(f,abs(Gsn));

grid;

xlabel('Frequency (Hz)');

ylabel('Magnitude of Transfer Function')

title('3-pole Type 1 Chebyshev high-pass filter with fc=5 KHz ')

77

High-Pass Filter


High-Pass Filter

78

103

104

105

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency (Hz)

Magnitude o

f T

ransfe

r F

unction

3-pole Chebyshev high-pass filter with fc=5 KHz


Band-Pass Filter

• Use the MATLAB functions buttap and lp2bp to find the transfer function of a 3-pole Butterworth analog band-pass filter with the pass band frequency centered at fo = 4kHz , and bandwidth BW =2KHz.

[z,p,k]=buttap(3); % Design 3 pole Butterworth low-pass filter with wcn=1 rad/s

[b,a]=zp2tf(z,p,k); % Compute numerator and denominator coefficients for wcn=1 rad/s


f0=4000; % Define centered frequency at 4 KHz

W0=2*pi*f0; % Convert desired centered frequency to rads/s

fbw=2000; % Define bandwidth

Bw=2*pi*fbw; % Convert desired bandwidth to rads/s

[bn,an]=lp2bp(b,a,W0,Bw); % Compute num, den of band-pass filter

% Compute and plot the magnitude of the transfer function of the band-pass filter

Gsn=freqs(bn,an,2*pi*f);


grid;

xlabel('Frequency f (Hz)');

ylabel('Magnitude of Transfer Function');

title('3-pole Butterworth band-pass filter with f0 = 4 KHz, BW = 2KHz')


Band-Pass Filter

80

102

103

104

105

0

0.2

0.4

0.6

0.8

1

1.2

1.4

Frequency f (Hz)

Magnitude o

f T

ransfe

r F

unction

3-pole Butterworth band-pass filter with f0 = 4 KHz, BW = 2KHz


Band-Elimination (band-stop) Filter

• Use the MATLAB functions buttap and lp2bs to find the transfer function of a 3-pole Butterworth band-elimination (band-stop) filter with the stop band frequency centered at fo = 5 kHz , and bandwidth BW = 2kHz.

[z,p,k]=buttap(3); % Design 3-pole Butterworth low-pass filter, wcn = 1 r/s

[b,a]=zp2tf(z,p,k); % Compute num, den coefficients of this filter, wcn=1 r/s


f0=5000; % Define centered frequency at 5 kHz

W0=2*pi*f0; % Convert centered frequency to r/s

fbw=2000; % Define bandwidth

Bw=2*pi*fbw; % Convert bandwidth to r/s

% Compute numerator and denominator coefficients of desired band stop filter

[bn,an]=lp2bs(b,a,W0,Bw);

% Compute and plot magnitude of the transfer function of the band stop filter

Gsn=freqs(bn,an,2*pi*f);


grid;

xlabel('Frequency in Hz'); ylabel('Magnitude of Transfer Function');

title('3-pole Butterworth band-elimination filter with f0=5 KHz, BW = 2 KHz')


Band-Elimination (band-stop) Filter

82

102

103

104

105

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency in Hz

Magnitude o

f T

ransfe

r F

unction

3-pole Butterworth band-elimination filter with f0=5 KHz, BW = 2 KHz


How to find the minimum order to meet

the filter specifications ?

The following functions in Matlab can help you to find the

minimum order required to meet the filter specifications:

• Buttord for butterworth

• Cheb1ord for chebyshev

• Ellipord for elliptic

• Cheb2ord for inverse chebyshev


Calculating the order and cutoff frequency

of a inverse chebyshev filter

• Design a 4MHz Inverse Chebyshev approximation with Ap gain at passband corner.

The stop band is 5.75MHz with -50dB gain at stop band.

clear all;

Fp = 4e6; Wp=2*pi*Fp;

Fs=1.4375*Fp; Ws=2*pi*Fs;

Fplot = 20*Fs;

f = 1e6:Fplot/2e3:Fplot ;

w = 2*pi*f;

Ap = 1;

As = 50;

% Cheb2ord helps you find the order and wn (n and Wn) that

%you can pass to cheby2 command.

[n, Wn] = cheb2ord(Wp, Ws, Ap, As, 's');

[z, p, k] = cheby2(n, As, Wn, 'low', 's');

[num, den] = cheby2(n, As, Wn, 'low', 's');

bode(num, den)


Bode Plot

85

-200

-150

-100

-50

0

Magnitu

de (

dB

)

105

106

107

108

109

1010

540

720

900

1080

1260

Phase (

deg)

Bode Diagram

Frequency (rad/sec)


References

[1] S. T. Karris, “Signals and Systems with Matlab

Computing and Simulink Modeling,” Fifth Edition.

Orchard Publications

[2] Matlab Help Files


ECEN 622 TAMU ACTIVE RC FILTERS - Texas A&M …s-sanchez/622 Lecture 2 Active... · ACTIVE RC...

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Transcript of ECEN 622 TAMU ACTIVE RC FILTERS - Texas A&M …s-sanchez/622 Lecture 2 Active... · ACTIVE RC...