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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
1
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
ECE 307-10 3
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
)
ECE 307-10 4
2
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
ECE 307-10 6
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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') >>
ECE 307-10 8
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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
ECE 307-10 9
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
ECE 307-10 10
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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 =
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
ECE 307-10 13
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
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
ECE 307-10 17
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
ECE 307-10 18
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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