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ELG4139: Op Amp-based Active Filters
• Advantages:
– Reduced size and weight, and therefore parasitics.
– Increased reliability and improved performance.
– Simpler design than for passive filters and can realize a wider range of
functions as well as providing voltage gain.
– In large quantities, the cost of an IC is less than its passive counterpart.
• Disadvantages:
– Limited bandwidth of active devices limits the highest attainable pole
frequency and therefore applications above 100 kHz (passive RLC
filters can be used up to 500 MHz).
– The achievable quality factor is also limited.
– Require power supplies (unlike passive filters).
– Increased sensitivity to variations in circuit parameters caused by
environmental changes compared to passive filters.
• For applications, particularly in voice and data communications, the
economic and performance advantages of active RC filters far outweigh
their disadvantages.
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First-Order Low-Pass Filter
ffB
Bi
f
ffi
f
i
f
ff
f
f
f
ff
f
f
ff
i
f
i
o
CRf
ffjR
R
CfRjR
R
Z
ZfH
CfRj
RZ
R
CfRj
R
fCj
RZ
Z
Z
V
VfH
2
1
)/(1
1
21
1)(
21
21
2
1
111
)(
A low-pass filter with a dc gain of –Rf /Ri
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dttvRC
tv
t
o in
0
1
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First-Order High-Pass Filter
iiB
B
B
i
f
ii
if
i
i
ii
if
ii
f
i
f
ffi
ii
i
f
i
o
CRf
ffj
ffj
R
R
CRfj
CRfj
R
R
CRfj
CRfj
CfjR
R
Z
ZfH
RZCfj
RZ
Z
Z
v
vfH
2
1
)/(1
)/(
21
2
21
2
2
1)(
2
1
)(
A high-pass filter with a high frequency gain of –Rf /Ri
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Higher Order Filters
n
B
n
i
fn
n
ffjR
R
fHfHfHfH
)/(1
1)1(
)()()()( 21
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Single-Pole Active Filter Designs
High Pass Low Pass
)/1()/1(
1
1
11
1
RCs
s
RCsRC
sRC
sCR
sRC
sCR
v
v
in
out
RCs
RC
v
v
in
out
/1
/1
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7
Two-Pole (Sallen-Key) Filters
-
+
+V
-V
R1
Rf1
Rf2
C1
vin
vout
C2
R2
-
+
+V
-V
R1
Rf1
Rf2
C2
vin
vout
R2
C1
Low Pass Filter High Pass Filter
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8
Three-Pole Low-Pass Filter
-
+
+V
-V
R1
Rf1
Rf2
C1
vin
C2
R2
-
+
+V
-V
R3
Rf3
Rf4
C3
vout
Stage 1 Stage 2
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9
Two-Stage Band-Pass Filter
R2
R1
vin
C1
C2
Rf1
Rf2
C4
C3
R3
R4
+V
-V
vout
Rf3
Rf4
+
-
+
-
+V
-V
Stage 1
Two-pole low-pass
Stage 2
Two-pole high-pass
BW
f1
f2
f
Av
Stage 2
response
Stage 1
response
fo
BW = f2 – f1
Q = f0 / BW
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10
Multiple-Feedback Band-Pass Filter
R1
R2
C1
C2
vin
Rf
+V
-V
-
+v
out
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11
Transfer function H(j)
Transfer
Function
)( jHV
oV
i
)(
)()(
jV
jVjH
i
o
)Im()Re( HjHH
22 )Im()Re( HHH
)Re(
)Im(tan 1
H
HH 0)Re( H
)Re(
)Im(tan180 1
H
HH o
0)Re( H
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12
Frequency Transfer Function of Filters
H(j)
HL
HL
o
o
o
o
ffffjH
fffjH
ffjH
ffjH
ffjH
ffjH
and 0)(
1)(
Filter Pass-Band (III)
1)(
0)(
Filter Pass-High (II)
0)(
1)(
Filter Pass-Low (I)
response phase specific a has
allfor 1)(
Filter shift)-phase(or Pass-All (V)
and 1)(
0)(
Filter (Notch) Stop-Band (IV)
fjH
ffffjH
fffjH
HL
HL
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Bode Plot
To understand Bode plots, you need to use Laplace transforms!
The transfer function of the circuit is:
1
1
/1
/1
)(
)(
sRCsCR
sC
sV
sVA
in
o
v
R
Vin(
s)
b
v
f
fj
RCfjRCjfA
1
1
21
1
1
1)(
where fc is called the break frequency, or corner
frequency, and is given by: RCf
c
2
1
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14
Bode Plot (Single Pole)
o
jCRj
jH
1
1
1
1)(
2
1
1)(
o
jH
2
101011log20)(log20)(
o
dBjHjH
o
dBjH
10log20)(
For >>o
R
C VoV
i
Single pole low-pass filter
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15
dBjH )(
(log)x
x
2x
10
6d
B2
0d
B slope
-6dB/octave
-20dB/decade
o
jH
10log20)(
For octave apart,
1
2
o
dBjH 6)(
For decade apart, 1
10
o
dBjH 20)(
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16
Bode Plot (Two-Pole)
R1
R2
C1
C2
vi v
o
21 oo
2
2
2
1
10111log20)(
oo
jH
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Corner Frequency
• The significance of the break frequency is that it represents the frequency where
Av(f) = 070.7-45
• This is where the output of the transfer function has an amplitude 3-dB below the input amplitude, and the output phase is shifted by -45 relative to the input.
• Therefore, fc is also known as the 3-dB frequency or the corner frequency.
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Bode plots use a logarithmic scale for frequency, where a decade is
defined as a range of frequencies where the highest and lowest
frequencies differ by a factor of 10.
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Magnitude of the Transfer Function in dB
2
/1
1)(
b
v
fffA
b
bb
bdBv
ff
ffff
fffA
/log20
/1log10/1log20
/1log201log20)(
22
2
• See how the above expression changes with frequency:
– at low frequencies f << fb, |Av|dB = 0 dB
• low frequency asymptote
– at high frequencies f >>fb,
|Av(f)|dB = -20 log f/ fb
• high frequency asymptote
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Real Filters
• Butterworth Filters – Flat Pass-band.
– 20n dB per decade roll-off.
• Chebyshev Filters – Pass-band ripple.
– Sharper cut-off than Butterworth.
• Elliptic Filters – Pass-band and stop-band ripple.
– Even sharper cut-off.
• Bessel Filters – Linear phase response – i.e. no signal distortion in pass-band.
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Filter Response Characteristics
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The magnitude response of a Butterworth filter.
Butterworth Filters
Magnitude response for Butterworth filters of various
order with = 1. Note that as the order increases, the
response approaches the ideal brickwall type
transmission.
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Sketches of the transmission characteristics of a representative even- and odd-
order Chebyshev filters.
Chebyshev Filters
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First-Order Filter Functions
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First-Order Filter Functions
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Second-Order Filter Functions
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Second-Order Filter Functions
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Second-Order Filter Functions
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Second-Order LCR Resonator
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The Antoniou inductance-simulation circuit. (b) Analysis of the circuit assuming ideal op
amps. The order of the analysis steps is indicated by the circled numbers.
Second-Order Active Filter: Inductor Replacement
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The Antoniou inductance-simulation circuit. Analysis of the circuit assuming ideal op amps.
The order of the analysis steps is indicated by the circled numbers.
Second-Order Active Filter: Inductor Replacement
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Realizations for the various second-order filter functions using the op amp-RC resonator of
Fig. 11.21 (b). (a) LP; (b) HP; (c) BP, (d) notch at 0;
Second-Order Active Filter: Inductor Replacement
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The Second-Order Active Filter: Inductor Replacement
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Derivation of an alternative two-integrator-loop biquad in which all op amps are used in a
single-ended fashion. The resulting circuit in (b) is known as the Tow-Thomas biquad.
Second-Order Active Filter: Two-Integrator-Loop
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Low-Pass Active Filter Design
Design a fourth-order low-pass Butterworth filter having a frequency cut-off of 100 Hz
k
HzxCfR
B
92.15
)100)(101.0(2
1
2
1
F0.1C Choose
6
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Low-Pass Active Filter Design
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Low-Pass Active Filter Design
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Infinite-Gain Multiple-Feedback (IGMF)
Negative Feedback Active Filter
Vi
Z1
Vx
Z3
Z2
Z4 Z5
Vo
)2( ,at KCLBy
)1( 0 0
4321
x
5
3
3
5
Z
VV
Z
V
Z
V
Z
VVV
VZ
ZVV
Z
ZVvv
oxxxxi
oxxoii
Substitute (1) into (2) gives
V
Z
Z
Z ZV
Z
Z ZV
V
Z
Z
Z ZV
V
Zi
o o
o
o
o
1
3
1 5
3
5 2 5
3
4 5 4
(3)
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rearranging equation (3), it gives,
HV
V
Z Z
Z Z Z Z Z Z Z
o
i
1
1 1 1 1 1 1
1 3
5 1 2 3 4 3 4
H
V
V
YY
Y Y Y Y Y Y Y
o
i
1 3
5 1 2 3 4 3 4
Or in admittance form:
Z1 Z2 Z3 Z4 Z5
LP R1 C2 R3 R4 C5
HP C1 R2 C3 C4 R5
BP R1 R2 C3 C4 R5
Filter Value
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IGMF Band-Pass Filter
H s Ks
s as b( )
2Band-pass:
To obtain the band-pass response, we let
55
44
4
33
32211 11
11
RZsCCj
ZsCCj
ZRZRZ
R1
C3R2
C4 R5H s
sC
R
s C C sC C
R R R R
( )
3
1
2
3 4
3 4
5 5 1 2
1 1 1
This filter prototype has a very low
sensitivity to component tolerance when
compared with other prototypes.
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Simplified Design (IGMF Filter)
R1
C
C R522
551
1
21)(
CsR
Cs
RR
R
sC
sH
Comparing with the band-pass response
22
)(
P
P
P sQ
s
sKsH
It gives,
2
1
5
51
2 2
1
1QH
R
RQ
RRCppp
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Example: IGMF Band Pass Filter
To design a band-pass filter with and 10 Hz512 QfO
)512(21
51
HzRRC
P
2
51 741,662,9 100 RRnFC
102
1
1
5 R
RQ
P
170,62 4.155 51 RR vin v
o
+
-
100nF
4.155
170,62
100nF
With similar analysis, we can choose the following values:
700,621 and 554,1 10 51 RRnFC
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Butterworth Response (Maximally Flat)
n
o
jH2
1
1
where n is the order
Normalize to o = 1rad/s
162.1162.01
124.324.524.524.3
185.1177.0
161.241.361.2
11
122
12
1
22
2345
5
22
234
4
2
23
3
2
2
1
sssss
ssssssB
ssss
sssssB
sss
ssssB
sssB
ssB
Butterworth polynomials
Butterworth polynomials:
)(
1)(ˆ
jBjH
n
njH
21
1)(ˆ
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Second Order Butterworth Response
Started from the low-pass biquadratic function 22
1)(
P
P
P sQ
s
KsH
2
111 QKp
njH
jH
jH
jH
jjH
sssH
222
4
242
222
2
2
1
1
1
1)(
1
1)(
221
1)(
21
1)(
12
1)(
)polynomial butterwothorder (second12
1)(
For
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Second order Butterworth Filter
224321
2
31214
'
11)(
P
P
P sQ
s
K
CCRRsKCsRRRsC
KsH
42
31
31
42
32
41 1
1
CR
CRK
CR
CR
CR
CRQ
P
Setting R1= R2 and C1 = C2
KKKQ
P
3
1
12
1
1111
1
Now K = 1 + RB/ RA
vin
C4
vo
+
-
R1
R2
C3
RB
RA
A
B
A
B
P
R
R
R
RKQ
2
1
13
1
3
1
For Butterworth response:
2
1
PQ
A
B
P
R
RQ
2
1
2
1
We have 414.122 A
B
R
R
We define Damping Factor (DF) as:
414.121
A
B
P R
R
QDF