Active Filters BW

7/31/2019 Active Filters BW

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Gain (dB)

0BW = f cVo

-20Ideal

0.707 -40

f f 0

BW-

f 10f 100f 1000f f

LPF with different roll-off ratesBasic LPF response


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Gain (dB)

0Vo

-20

0.707 -40Passband

f c f 0

-

0.01f 0.1f f f

Basic HPF response HPF with different roll-off rates


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Vout 21 cco f f f =Centre frequency:1

Quality factor: f Q o=.

Q is an indication of the

f

. Narrow BPF: Q > 10.

Wide-band BPF: < 10.oc1 c2

BW = f c2 - f c1Damping Factor: Q DF

1=


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an op er esponse

Gain (dB) Also known as band 0

-3

re ect, or notc ter.Frequencies within a

certain BW are rejected.

f

Useful for filtering

interferin si nals.oc1 c2

BW


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Characteristics

Av Cheb shev

Bessel

Butterworth

f


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CharacteristicsButterworth: very flat amplitude response in the

phase response however is not linear. (A pole is

Chebyshev: rolloff rate > 20 dB/dec/pole;

response. esse : near p ase response, ere ore no

overshoot on the output with a pulse input; roll o ra e s < 20 ec po e.


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Advantages over passive LC filters :

high Zin and low Zout mean good isolation from source

less bulky and less expensive than inductors when dealin with low fre uenceasy to adjust over a wide frequency range without

altering desired responseDisadvantage: requires dc power supply, and could be


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The damping factor (DF)Frequencyselective +Vin Vout

the response characteristicof the filter. _

R 112

R R

DF =

R 2 Its value depends on the

General dia ram of active filter

filter. For example DF =1.414 for 2 nd order

Butterworth, hence, R 1 =0.586R 2


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R 1+ _

Vin VoutC

2 H RC =

R 1 12

1F A R= +R 2 Roll-off rate for a single-pole

filter is -20 dB/decade.Acl is selectable since DF iso tional for sin le- ole LPF


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The RC network behaves as a voltage divider supplied by v i,and hence the voltage at the non-inverting terminal of the op-amp is given as:

C i

C v v R j X +

=

Where1

2 C X

f C =

The equation for v + then reduces to:

1 2 iv

v j f R C +

= +


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We know that the output of an op-amp non-inverting amplifier

is given by: 12

1o R

v v R

+ = +

u s u ng or v rom e prev ous equa on,

i1

Fo v

fRC2 j1

1

R

R1v

+

+=o F v A=

Where,

i H v +

2 H f

R C =

f H = high cut-off frequency of the filter 1

2

1 p a s s - b a n d g a i n o f t h e f i l t e r F R

A R

= + =

obtained as

v A

( )2

1

o

i H

v f f

=

+

1t a n H

f =

and


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The operation of the low-pass filter can be verified from the

gain magnitude equation:o

F i

v A

v1. At very low frequencies, that is f > f H,

Roll-off Rate :From the gain magnitude equation, we see that, if thefrequency is increased 10 fold (1 decade), the voltage gain is

=.log 10) each time the frequency is increased by 10. Hence theroll-off rate of the first order filter in the stop band is 20

.

At cut-off frequency, f H, the gain falls by 3 dB (= 20 log 0.707).


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Example : Design a first order low-pass filter with a cut-off

frequency at 1 KHz and pass-band gain of 2. Draw thefrequency response of the circuit.

1. From the specified cut-off frequency1

Assume, C = 0.01 F2 H R C

= =

3 6H

1 12 f C 2 ( 1 0 H z ) ( 0 . 0 1 1 0 F ) R x = =

=

2. From the specified pass-band gain R

.

2F R

= + =

This implies, R 1/R 2= 1, or R 1 = R 2

Assume, R 1 = R 2 = 10K


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Frequency Response of the designed low pass filter


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Sallen Key 2nd Order LPF A second order low-pass filter provides a 40 dB/decade roll-off ratein the stop-band. The first order low-pass filter can be convertedn o a secon -or er ype s mp y y us ng an a ona ne wor

The ain of the filter is set bC

A R1 and R 2, while the high cut-off frequency (f H) is set by R A,+ _

Vin VoutA B

C A, B B

1 H f =

R 1

A B A B

The voltage gain magnitude

2

equat on s g ven as:

4

o F v A= H

Where, 12

1 p a s s - b a n d g a i n o f t h e f i l t e r F R

A R

= + =


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Filter Design Proceduresignalinputtheof frequencytheisf

1. To simplify design calculations, set R A=R B=R and C A=C B=C.

2. Choose a value of C 1 F3. Calculate the value of R using the equation

2 H R f C =

. , , 1 =0.586R 2. Choose a value of R 2 100K and calculate the

value of R 1.


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Example : Design a second order low-pass butterworth filter at a

high cut-off frequency of 1KHz. Draw the frequency response of the circuit.

1. Let R A=R B=R and C A=C B=C.

Assume C = 0.0047 F

C A = C B = 0.0047 F

3 6H2 f C 2 ( 1 0 H z ) ( 0 . 0 0 4 7 1 0 F )

R x

= =

R = 33.86K

R A = R B = 33.86K

2. For second order butterworth response we need, R1 = 0.586R 2

=, 2

R1 = 0.586R 2 = 15.82K


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Frequency Response of Designed Low Pass Filter


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CA1

Roll-off rate: -60 dB/dec

+Vin

R A1 R B1 R A2 _

R 1CB1 _ CA2 out

R 23

R 2 oles 1 ole

Third-order (3-pole) configuration


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ng e o e g ass er

C Rolloff

rate,

and

+ _

Vin Vout

R

ormu as or c , an AF are similar to

R 1 t ose

or

.Ideally, a HPF passes R 2 all frequencies above

f c. However,

the

op

amp has an upper frequency limit.


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R AAgain, formulas and

+Vin Vout

BA roll-off rate are similar to those for 2nd-order

_

R 1R B

.To obtain higher roll-

R 2

,can be cascaded.

Basic Sallen-Keysecon -or er HPF


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+Vin

A1

VR A2

_

R 1

R A1 _ CA2

R 23

R Av (dB)

0-3

f LP response

oc1 c2


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Cascading a HPF and a LPF to yield a band pass

c1 c2 sufficiently separated. Hence the resulting

Note that f c1 is the critical frequency for the HPF c2

.Another BPF configuration is the multiple

ee ac w c as a narrower an w and needing fewer components


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C1 Making C 1 = C 2 = C,

_ R 1

2C2

321

31

21

R R R R R

C f o

+=

+

_ n out

R 3Q = f o/BW

;2 21

oooC f

Q R

CA f Q

R ==

2 R

Max. gain:

)2(2 23 oo AQC f Q

R

=

12 Ro

A < 2 2

R 1, C 1 - LP sectionR 2, C 2 - HP section


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Filter

C1The multiple-feedback

R 12

C2

its BP counterpart. For fre uencies between f

+

_ n out

and f c2 the op-amp willtreat V in as a pair of

R 4 common-mode signalsthus rejecting themWhen

C = C =C .1

f o =


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er esponse easuremen s

Discrete Point

Measurement:

Feed

a sine

wave

to t e ter nput w t a vary ng requency ut a constant voltage and measure the output

vo tage at

eac

requency

po nt. A faster way is to use the swept frequency method:

Sweep

Generator Filter

Spectrum

analyzer

The sweep generator outputs a sine wave whose frequency.

Active Filters BW

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