Active Filter Circuits Active Filter Circuits - Cal Poly Pomona

ECE307-10

Active Filter Circuits

Z. Aliyazicioglu Electrical and Computer Engineering Department Cal Poly Pomona

Active Filter Circuits Introduction Filter circuits with RLC are passive filter circuit Use op amp to have active filter circuit Active filter can produce band-pass and band-reject filter without using inductor. Passive filter incapable of amplification. Max gain is 1 Active filter capable of amplification The cutoff frequency and band-pass magnitude of passive filter can change with additional load resistance This is not a case for active filters We look at few active filter with op amps. We look at that basic op amp filter circuits can be combined to active specific frequency response and to attain close to ideal filter response ECE 307-10 2

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Active Filter = Circuits First-Order Low-pass Filters C

Zf

R2

Zi

R1 Vi

OUT

+ Vo

OUT +

+

Transfer function of the circuit

H (s ) =

− Zf Zi

ωc H ( s ) = −K (s + ω c )

−R2 H (s ) = R1(sR2C + 1)

The Gain K=

-

Vi

-

Cutoff frequency

R2 R1

ωc =

H (s ) =

R2 1 − sR SC = 2C + 1 R1 R1

−R2 ||

Transfer function in jω H ( j ω ) = −K

1 R2C

+ Vo

1 (1 + j

ω ) ωc

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Active Filter Circuits Example

• Find R2 and C values in the following active Low-pass filter for gain of 1 and cutoff frequency of 1 rad/s.

C 1F R1 2 R1 Vi

1

From the gain

1 OUT +

K=

+ Vo

R2 =1 R1

R2 = R1 = 1Ω

From the cutoff frequency ωc =

H ( jω ) =

1 (1 + j

ω 1

1 =1 R2C

C=

1 = 1F R2

)

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Active Filter Circuits Example >> w=0.1:.1:10; >> h=20*log10(abs(1./(1+j*w))) ; >> semilogx(w,h) >> grid on >> xlabel('\omega(rad/s)') >> ylabel('|H(j\omega)| dB') >>

ECE 307-10 5

Active Filter Circuits A first order high-pass filter R2 R1

C

-

Vi

OUT +

+ Vo

Transfer function of the circuit H (s ) =

H (s ) =

−R2s s 1 H ( s ) = −K R1(s + ) (s + ω c ) R1C

The Gain K=

R2 R1

− Zf Zi

H (s ) =

−R2 −R2sC = 1 R 1sC + 1 R1 + sC

Transfer function in jω

Cutoff frequency

ωc =

1 R1C

H ( j ω ) = −K

jω ωc (1 + j

ω ) ωc

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Active Filter Circuits Example R1 Vi

R2 C

20 K 0.1 uF

200K OUT +

+ Vo

• Find R2 and R1 values in the above active High-pass filter for gain of 10 and cutoff frequency of 500 rad/s. From the cutoff frequency From the gain

K=

ωc =

1 = 500 R12C

R1 =

1 = 20 K Ω 500C

R2 = 10 R2 = R110 = 200 K Ω R1

Transfer function in jω

H ( jω ) = −10

jω 500

(1 + j

ω

500

) ECE 307-10 7

Active Filter Circuits Example >> w=1:10000; >> h=20*log10(10*(abs((j*w/500 )./(1+j*w/500)))); >> semilogx(w,h) >> grid on >> xlabel('\omega(rad/s)') >> ylabel('|H(j\omega)| dB') >>

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Active Filter Circuits Scaling

• In filter design, we can transform RLC values in to realistic values, this process is called scaling • Two types of scaling, magnitude and frequency scaling • In magnitude scaling, we multiply all L and R by scaling factor km, multiplying all C by 1/km R ' = kmR

L ' = km L

C' =

C km

• km,is positive real number

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Active Filter Circuits Scaling

• frequency scaling, we multiply all L, C by 1/kf where kf is scaling factor. R' = R

L' =

L kf

C' =

C kf

• A circuit can be scaled in both magnitude and frequency in simultanously R ' = kmR

k L' = m L kf

C' =

C kmkf

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Active Filter Circuits Example

• Example 1 , Find R2 and R1 values in the active Low-pass filter for gain of 5 and cutoff frequency of 1Khz and c=0.01 µF

C 1F R1 2 R1 Vi

1

1 OUT +

km =

+ Vo

kf =

ωc ' 2π 1000 = = 6283.185 ωc 1

1 C 1 = = 15915.5 kf C ' 6283.185(10−8 )

R2 ' = kmR2 = 15915.5(1) = 15.9 K Ω

• For gain specification, we need to change R1 R1 =

R2 15.9K = = 3.18 K Ω K 5

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Active Filter Circuits Example

>> f=1:10000; >> w=2*pi*f; >> h=20*log10(5*abs(1./(1 +j*w/(2*pi*1000)))); >> semilogx(f,h) >> grid on >> xlabel(‘f(Hz)') >> ylabel('|H(jf)| dB')

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Active Filter Circuits Op Amp Band-Pass Filters

• • • •

Three components A unity gain low-pass filter, cutoff frequency is ωc2 A unity gain high-pass filter , cutoff frequency ωc1 A gain component to provide the desired level

ωc 2 ≥2 ω c1 Vi

Low-pass filter

High-pass filter

Inverting amp.

Vo

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Active Filter Circuits Op Amp Band-Pass Filters CL RL RH RL

Rf

-

Vi

OUT

RH

CH

-

+

Rf

OUT

OUT

+ +

 −ωc 2   −s   Rf  H (s ) =   −   s + ωc 2   s + ωc1   Ri  H (s ) =

2

−K ω c 2 s

s + (ωc1 + ωc 2 )s + ωc1ωc 2

ωc 2  ωc 1

ωc 2 =

1 RLCL

H (s ) =

H (s ) =

ωc1 =

+ Vo

−K ω c 2 s (s + ωc 2 )(s + ωc1)

βs s + β s + ω02 2

R 1 H ( j ω0 ) = −K = − f Ri max RHCH ECE 307-10 14

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Active Filter Circuits Example:

• Design a band-pass filter for a graphical equalizer that has gain 2 within the frequency between 100 and 10,000 Hz. Use 0.1 µF capacitors • For upper cutoff frequency from LP filter

ωc 2 =

1 RLCL

RL =

1

ωc 2CL

=

1 2π 10000(0.1)10−6

= 80 Ω

• For Lower cutoff frequency from HP filter

ωc1 =

1 RHCH

RH =

1 1 = = 7958 Ω ωc1CL 2π 100(0.1)10−6

• For gain, choose Ri=1KΩ K=

Rf Ri

Rf = Ri K = 1000(2) = 2 K Ω ECE 307-10 15

Active Filter Circuits From transfer function  −2π 1000     2000  − jω H ( jω ) =   −  A = 20log10 | H ( jω ) | ω π ω π + + j 2 1000 j 2 100     1000  dB

>> f=10:80000; >> w=2*pi*f; >> H=((2*pi*10000)./(j*w+2*pi* 10000)).*((j*w)./(j*w+2*pi*100))*( -2); >> A=20*log10(abs(H)); >> semilogx(f,A) >> grid on; >> ylabel ('A_{dB}') >> xlabel ('F (Hz)')

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Active Filter Circuits Op Amp Band-Reject Filters

• • • •

Three components A unity gain low-pass filter, cutoff frequency is ωc1 A unity gain high-pass filter , cutoff frequency ωc2 A gain component to provide the desired level

Low-pass filter

Vi

Vo

Inverting amp. High-pass filter

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Active Filter Circuits Op Amp Band-Reject Filters CL

 −ωc1 −s   Rf  + H (s ) =   − + + s s ω ωc 2   Ri  c1 

RL RL

OUT

Rf

+

Rf

Vi -

RH RH

CH

OUT +

OUT

For ωc 2 >> ωc1

Rf

+

H (s ) =

Rf Ri

 s 2 + 2ω s + ω ω  c1 c1 c 2    (s + ωc1)(s + ωc 2 )   

+ Vo

ωc1 =

1 RLCL

H ( jω )

ωc 2 =

max

1 RHCH

=K =

Rf Ri

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Active Filter Circuits Example:

• Design an active band-reject filter that has gain 5 and the stop frequency between 100 and 2000 Hz. Use 0.5 µF capacitors Fc1 = 100Hz and Fc 2 = 2000Hz

ωc1 =

1 RLCL

ωc 2 =

1 RHCH

RL =

For ωc 2 >> ωc1

1 1 = = 3.18 K Ω ωc1CL 2π 100(0.5)10−6

RH =

1 1 = = 159 Ω ωc 2CH 2π 2000(0.5)10−6

• For gain, choose Ri=1KΩ K=

Rf Ri

Rf = Ri K = 1000(5) = 5 K Ω ECE 307-10 19

Active Filter Circuits  −ωc1 − jω   Rf  H ( jω ) =  +  −  jω + ωc1 jω + ωc 2   Ri 

AdB = 20log10 | H ( jω ) |

>> f=10:80000; >> w=2*pi*f; >> H=(((2*pi*100)./(j*w+2*pi*100)) +((j*w)./(j*w+2*pi*2000)))*(5); >> A=20*log10(abs(H)); >> semilogx(f,A) >> grid on; >> xlabel ('F (Hz)') >> ylabel ('A_{dB}')

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