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
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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
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 =
R2 15.9K
=
= 3.18 K Ω
K
5
ECE 307-10 11
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
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
>> 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)')
ECE 307-10 16
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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
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
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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