eces 352 winter 2007 ch. 12 active filters part 2 1 second order active filters based on bridged-t...

Ch. 12 Active Filters Part 2 1ECES 352 Winter 2007

Second Order Active Filters Based on Bridged-T Networks* Develop second order active filters based on op

amps. Assume ideal op amps, i.e.

No input current Negligible voltage drop between the

inputs, i.e. infinite gain* Use feedback network with R’s and C’s

* General form of the filter function is:

* Consider case where we use a particular RC combination called the bridged-T network.

Gives well recognized filters, e.g. low pass, high pass, bandpass, notch, etc.

Other RC combinations possible, but give less useful (more complicated frequency dependent) filter functions.

* Will analyze bridged-T networks to show they give second order filters.

* Can get different filters based on where input is connected into the bridged-T network.

+A

Bridged – T Network

V-≈ 0

Vi

Vo

I-≈ 0

Bridged – T network

20

02

12

2

)(

)()(

sQ

s

asasa

sV

sVsT o

i

o


Injecting the Input Signal* Where do we inject the input signal ?

* Need to inject the signal at a point where it does not change the poles (expression for t(s) ).

* We will inject the signal at the bottom of R4 which was at ground.

* Since we find the poles by turning off the signal source when we use the Gray-Searle technique, this means that we will not be modifying the poles (as long as we take into account the source resistance Rs).

* Take the output off the op amp output.

* Where we inject the signal determines what type of filter we get !

Rs

4321413231

2

4321321

2

1111

1111

)(

RRCCRCRCRCss

RRCCRCCss

V

Vst

b

a

Vi

Vo

VS


* When injecting the input signal, we need to take into account the finite source resistance.

* Can divide R4 into two parallel resistors. For any α > 0 and < 1, we always get

* We can easily correct for Rs by subtrac-ting Rs from the needed value for R4/

* This approach provides more flexibility in taking Rs into account, especially when R4 is small.

* If we don’t take Rs into account, the poles (expression for t(s) ) will be altered and the filter design will not meet the specifications.

Injecting the Input Signal – Source Resistance Problem

4321413231

2

4321321

2

1111

1111

)(

RRCCRCRCRCss

RRCCRCCss

V

Vst

b

a

Rs

Vi

Vo

VS

Rs

VS

Vi

Vo

4

444

44

1

1

1

1

1R

RRR

RR


* Analysis to do: Find T(s) = Vo(s)/Vi(s)

* Assume op amp is ideal: Input currents are virtually zero. Gain is very large so voltage

difference between inputs is nearly zero.

Inverting terminal (-) grounded since noninverting (+) is grounded.

Injecting the Input Signal - Bandpass Filter

2143213

2

14

44

44

3321

3240

33201

324214

3244

44

321

0144110

32011

32

0

1

1

2

2

1

122110

23

3

1111

grearranginafter get we

and for ngsubstituti so1

. of in termsget tosubstitute

and currents theof in terms express Now

111

11

11

1

1

grearranginso1

11

1

11

grearranginso

amp), op (ideal 0 and 0 Since

. of in terms currents allGet :Objective

CCRRCCRss

sCR

V

VT(s)

IIR

IIR

V

V

V

RRsCsC

RsCRV

R

V

RsCVsC

RsC

V

RIIII

RsC

V

RI

RI

RsCsC

VsCRIZIV

RsCVsCI

RsC

V

sC

I

sC

I

sC

IZIZIV

IR

VIV I

V

i

o

iii

o

i

ooi

o

C

CC

o

o

I3

I1I2

I4Ii

I-= 0

V-= 0

Filter’s transfer function


Second Order Filter has Bandpass Form* Circuit analysis gives

* Recall the form of second order filter is

* So our filter is a bandpass filter since a2

= 0, a0 = 0 where

* Finally, recall form of transfer function for bridged-T network was t(s)

* So the numerator of t(s) is in fact the same as the denominator of T(s), so the poles of T(s) are the same as the zeros of t(s)!

2143213

2

14

1111CCRRCCR

ss

sCR

V

VT(s)

i

o

20

02

12

2

)(

)()(

sQ

s

asasa

sV

sVsT o

i

o

141 CR

a

4321413231

2

4321213

2

1111

1111

)(

RRCCRCRCRCss

RRCCCCRss

V

Vst

b

a

T(dB)

0

0 dB

-3 dB

Qo


* Transfer function for bandpass filter is

* Given the filter specifications (0 and 0 /Q), we can determine the R’s and C’s.

* We can also use the desired transmission at 0 to determine the size of .

We can get gain [T(0))>1] at 0 if so desired.

How to Design a Specific Bandpass Filter

4

321

21

4321

213141

20

02

1

20

02

12

2

1

1

111sin

)(

)()(

R

RCC

CCQ

RRCC

CCRQceand

CRa

where

sQ

s

sa

sQ

s

asasa

sV

sVsT

o

o

o

i

o

T(dB)

0

0 dB

-3 dB

Qo

21

2

4

3)(CC

C

R

RT o


Example - Bandpass Filter Design* R and C calculations

Three specifications Five elements to specify: R3 R4 C1 C2 α

Conventional approach Set two equal, e.g. C1 = C2

Pick a convenient size for them C1 = C2 = 5 nF

Define ratio m = R3/R4 ; R3 = R

Simplify equations for o and Q

Solve for two parameters: R & m Calculate R3 & R4.

T(dB)

0

0 dB

-3 dBQ

o

1)(

)(/101

/101

40

50

oT

passbandsradxQ

sradx

oo

o

QmRC

soRC

m

RRCC

mR

R

C

C

R

RCC

CCQ

2

1

2

1

2

1

4321

4

3

4

321

21

Filter specifications:

KK

m

RR

Qsradx

sradx

QQm

RKnF

xRsox

sradx

QRC

o

o

o

1.0400

40

and

10 so and400/101

/1014

/44

405

sec102sec102

/101

22

34

2

4

522

3

44

4

Solution


Bandpass Filter Example

00constant

01

1111 21

2

4

3

2

2143213

2

14

sass

jsasCC

C

R

R

sasss

s

CCRRCCRss

sCR

V

VT(s) o

i

o

T(dB)

0

0 dB

-3 dBQ

o

From previous analysis we know C1 = C2 = 5 nFand R3 = 40 kΩ and R4 = 0.1 kΩ so peak value of T(ωo) is given by

20055

5

1.0

40

21

2

4

3

nFnF

nF

K

K

CC

C

R

R

To get unity transmission (0 dB) at 0, we set T(ωo) = 1.

KKR

andKKR

so

so

1005.0005.01

1.0

120

005.0

1.0

005.0200

11200

44

So if Rs = 5K, we can account for it by making R4/α = 15 K


Bandpass Filter Example - Poles* Transfer function for bandpass filter

sradxsradx

Qso

sradxnFKRC

m

RRCC

mR

R

C

CQ

and

sradnFKCR

a

where

sQ

s

sa

sV

sVsT

o

o

i

o

/10110

/101

/101)5(40

4001

104002

1

2

1

2

/10)5)(1.0(

005.0

)(

)()(

45

5

4321

4

3

4

141

20

02

1

sradxjsradx

sradxj

sradx

QQ

jQ

pp

sQ

sCCRRCCR

ss

PP

/105/105

2

/101

2

/101

4122

,,

01111

33

44

202121

20

02

2143213

2

Poles of bandpass filter

So this bandpass filter is a special form of a feedback amplifier.


Complementary Transformation* Can we get a different type of filter from

this bridged-T network? YES.* Need a new injection point.* How do we find a new injection point if

the grounded end of R4 appears to be our only option?

* Have to apply a complementary transformation to redraw the circuit with a new injection point.

* Complementary transformation consists of:1. Nodes of feedback network originally

grounded (c) should be reconnected to op amp output.

2. Nodes of feedback network originally connected to op amp output (b) should be reconnected to ground.

3. Input terminal connections of the op amp should be interchanged.

4. Other input shorted to op amp output instead of being grounded.

=

Note: This is not grounded!

Original ComplementaryTransformed Circuit


Complementary Transformation of Bridged-T Network

* Complementary transformation: Reconnect bottom of R4 to op amp output.

Remove connection of C1 and R3 from op amp output and connect them to ground.

Change connection at op amp input to noninverting (+).

Connect inverting (-) input directly to output.

* Now have two new points (bottom of C1 and R3) that can be used for injection of the input signal to get a new type of active filter.

* Injection at the bottom of C1 gives a high pass filter.

High pass filter


High Pass Filter

V+V0

V-=V0

V1

Ii

I4I2

I3

4321213

2

2

43212

1230

423

0

1320

11

111

423

0

4323

042

432324

0

4

0

4

104

320

3202201

1022

102

323

3

1111

get wegrearrangin and

1111

1

11

11

get we and for ngsubstituti so

11

1

11

so

0since and ideal is amp op since

RRCCCCRss

s

V

VT(s)

RRCCsCCsRV

RsCR

V

sCRsC

VV

sC

IVV

VIZIVV

and

RsCR

V

RRsC

V

R

VIII

RRsC

V

RsC

V

R

V

R

V

R

VVI

RsC

VV

R

V

sCVZIVV

VVsCZ

VVIAlso

IIIR

VI

i

o

oii

iCii

oi

oo

ooC

C

o

* Analysis to do: Find T(s) = Vo(s)/Vi(s)

* Assume op amp is ideal: Input currents are virtually zero. Gain is very large so voltage difference

between inputs is nearly zero. Inverting terminal (-) connected to output

so noninverting (+) is nearly at V0 since op amp gain A is large.

I+ 0

Vi

Vo


High Pass Filter* Circuit analysis gives

* General form of second order filter is

* So a2= 1 and a0 = a1= 0

* Recall form of t(s)

* As expected the numerator of t(s) is the same as the denominator of T(s), so the poles of T(s) are the same as the zeros of t(s)!

4321213

2

2

1111RRCCCCR

ss

s

V

VT(s)

i

o

20

02

12

2

)(

)()(

sQ

s

asasa

sV

sVsT o

i

o

4321413231

2

4321321

2

1111

1111

)(

RRCCRCRCRCss

RRCCRCCss

V

Vst

b

a

43214

321

21

11

RRCCR

RCC

CCQ o

T(dB)

0

0 dBQ(dB)


High Pass Filter Design

00

1

1111

2143213

2

2

sas

jsasjQQ

j

sas

CCRRCCRss

s

V

VT(s) o

o

o

i

o

T(dB)

0

0 dBQ(dB)

ooo

QmRCso

RC

m

RRCCm

R

R

C

C

R

RCC

CCQ

21

2

1

2

1

43214

3

4

321

21

* Given the filter specifications (0 and 0 /Q), we can determine the R’s and C’s.* Two specifications, four parameters so follow same convention and set C1 = C2 and pick a convenient value, say 5 nF.* Define ratio of resistances m = R3/R4 and set R3 = R.


Filter specifications: 0 = 1x103 rad/s, Q = 0.5

High Pass Filter - ExampleT(dB)

0

0 dBQ(dB)

KRm

RRsoQm

RKxx

R

xsradx

QmRC

oo

20015.044

200nF5

sec101

C

sec101

so sec101/101

5.022

34

22

3

33

33

Poles of high pass filter

sradxjsradx

sradxsradx

Q

sradxQ

QQ

jQ

pp

sQ

sCCRRCCR

ss

PP

/1050/105

are sfrequencie pole so

/1025.0

/101 so

/1015.0

4122

,,

01111

33

33

0

30

202121

20

02

2143213

2

NOTE 40 dB/dec

NOTE: since two poles coincide get falloff at 40 dB/dec below ωo.


Other Active Filters Based on Bridged-T Feedback Networks

* Second bridged-T feedback network similar to first, but with the resistors and capacitors interchanged.

* Transfer function t(s) is similar.

* Method of analysis is the same.

* Find T(s) = Vo(s)/Vi(s)

* Get it in standard form

* From specifications (o and Q), determine R’s and C’s.

* NOTE: Need to pick R1 = R2 and define m = C1/C2 to solve in this case.

4321232414

2

4321421

2

1111

1111

)(

CCRRRCRCRCss

CCRRCRRss

V

Vst

b

a

V0

Vi

Rs 20

02

12

2

)(

)()(

sQ

s

asasa

sV

sVsT o

i

o


* Can use the complementary transformation to get other injection points and other filter types.

Other Active Filters Based on Bridged-T Feedback Networks


Butterworth Filters

* Second order filters* Can be low or high pass.* Provide improved

performance: No peak near band edge

that is seen for other filters, i.e. it is maximally flat unlike other second order filters which give the shape shown below

Falloff for Butterworth filter is steeper, i.e. 40 dB/dec rather than 20 db/dec for passive RLC filters.

High Pass Filter

Low Pass Filter


Low Pass Butterworth Filter

222

22

222

222

212

22212

22

2

122

112

11

11

121

21

1)(

21

11

1

1

1

0since

RCCRss

RC

RCssCRV

VsT

so

RCssCRV

sCRsCRVsCRVRIVV

sCRsCVRVCssCVIII

RVCssCRVVsCZ

VVI

sCRVRIVV

III

sCVZ

VI

VVV

i

o

o

ooRi

oooCRR

oooC

oC

oRo

CR

oC

oC

o

20

02

12

2

)(

)()(

sQ

s

asasa

sV

sVsT o

i

o

222

22

1

2

1211

oo

oo

CRa

andQsoRCQRCRC

General form for biquadratic filter This has form for a low pass biquadratic filter

IC2

IC1

IR1IR2

VoVo

V12

Vi


Low Pass Butterworth Filter Design

sas

jsasj

sas

CRRCss

CRV

VT(s) o

i

o

0

2/

01

12

1

222

22

T(dB)

0

0 dBQ(dB)

oo C

RsoRC

11

* Given the filter specification (0), we can determine the R and C.* One specification, two parameters – R and C * Pick a convenient value, say C = 5 nF.* Calculate R from C and ωo.

NOTE 40 dB/dec

16sec)/1026.1)(105(

11

,5

sec/1026.122Given :

79

70

radxFxCR

thennFCChoose

radxMHzExample

o


More Complicated Second Order Active Filters

* Second order filters can employ:

More feedback loops. More op amps

* Can achieve better performance.

* Added complexity In design In implementation

Thomas-Tow Biquad Filter

Can achieve different filter types by the choices of capacitances and resistances.


Second Order Biquadratic Filter

Kerwin-Huelsman-Newcomb (KHN) Biquad Filter

NOTE: Here is a general purpose filter that can be used to achieve any one of five filtering functions depending upon where the output is taken.


Second Order Active Filters

Antoniou Inductor Replacement Biquad Filter


Second Order Active Filters


* Can incorporate other combinations of resistors and capacitors in the feedback network. But may not give a simple filter function.

* Can use op amps with feedback and other circuit configurations.

Other Higher Order Active Filters

From G. Parker, Applied Microwave and Wireless, p.74 Jan 1997.

NOTE: This is a third order filter since it incorporates three capacitors. This will give a 60 dB/dec falloff !

T(dB)

0

0 dBQ(dB)

NOTE 60 dB/dec


Other Higher Order Butterworth Filters

Third Order Filter Fourth Order Filter


Summary of Active Filters

* Can use op amps and simple RC feedback networks to implement a variety of second order filters such as:

Band pass, High pass, Low pass, Notch filter

* As an example, we examined use of the bridged-T network of two resistors and two capacitors.

Illustrated how to analyze such networks to obtain the transfer function T(s) which describes the filter’s characteristics.

Analysis used: Characteristics of ideal op amp Kirchoff’s Laws

* Also analyzed Butterworth filter Similarly used the characteristics of ideal op

amp and Kirchoff’s Laws

* Finally, pointed out third and higher order filters can also be designed using op amps to get superior filter performance.

eces 352 winter 2007 ch. 12 active filters part 2 1 second order active filters based on bridged-t...

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