Analog Circuits and Systems
Prof. K Radhakrishna Rao
Lecture 25: Active Filters
1
Review
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—
—
—
—
—
Inductor Simulation
To convert RLC filters to Active RC filters
Gyrator – Inductor Simulator (L=CR2)
Active and Passive - Parameter Sensitivities in Active RC filters
Effect of finite gain and finite gain bandwidth product on inductor
simulated
f0Q<<Gain Bandwidth Product
2
Review (contd.,)
A0
Q of the inductor simulated =
2ω0 A0 ⎞
⎛
⎜1 − GB ⎟
⎝
⎠
Qa
Q of the filter simulated =
⎛
Qa 2ω0Qa ⎞
−
⎜1 +
⎟
A0
GB ⎠
⎝
3
Q-enhancement- Sallen and Key
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—
—
—
The quality factor Qp of a second order passive RC filter is always
less than 0.5
Qp < 0.5 is unacceptable for a general filter design
Sallen and Key proposed use of negative and positive feedback, and
active devices to enhance Q
Several topologies similar to Sallen and Key filters are possible
4
Second Order Passive filter
—
—
N (s )
Transfer function of second order passive filter =
D (s)
D(s) and N(s) are second order polynomials
with D(s) having Q <0.5
5
Use of feedback to enhance Q
-K ⎡⎣N ( s ) D ( s )⎤⎦
-KN ( s )
Vo
=
=
Vi
1 + K ⎡⎣N ( s ) D ( s )⎤⎦ D ( s ) + KN ( s )
where K is the gain of the active device
6
Use of feedback to enhance Q (contd.,)
For a general second-order passive RC/RL filter
⎡ s2
⎤
s
+ p⎥
⎢m 2 + n
ωp
N(s) ⎢⎣ ωp
⎥⎦
=
D(s)
⎡ s2
⎤
s
+ 1⎥
⎢ 2 +
⎢⎣ ωp ωpQp
⎥⎦
where ωp is the natural frequency of the passive RC filter
Qp is quality factor of the passive second-order RC/RL filter
7
Quality Factor
Qp is always < 0.5
⎡ s2
⎤
s
-K ⎢m 2 + n
+ p⎥
ωp
Vo
⎢⎣ ωp
⎥⎦
=
2
Vi ⎡
⎤
s
s
+ (1 + pK)⎥
⎢(1 + mK) 2 + (1 + nK)
ωp
ωpQp
⎢⎣
⎥⎦
1 + pK
ω0 =
ωp
1 + mK
Qp
Qa =
(1 + pK) (1 + mK)
(1 + nK)
If K is positive, m and p
are positive, and for all
values of n it is a
negative feedback
system
If K is negative m and
p are positive, all
positive values of n it is
a positive feedback
system
8
Enhancement of Qa
—
—
—
Qa can be enhanced by increasing
pK >0 with m = n =0,
mK>0 with p=n=0,
pK>0 and mK>0 with n =0.
These make use of negative feedback.
Qa can also be enhanced by making nK<0 and 0<|nK|<1. This
constitutes using positive feedback.
All types of filters can be designed using any of the Q-enhancement
methods.
9
Second-order low-pass RC filter
N(s)
1
=
2
D(s) (C1R1C2R2 ) s + (C1R1 + C2 (R1 + R2 )) s + 1
ωp =
Qp =
1
R1C1R 2C2
1
⎛
R1 ⎞
⎜1 +
⎟
R2 ⎠
⎝
with R1 = R 2 = R and C1 = C2 = C
C1R1
+
C2R 2
C2R 2
C1R1
1
1
ωp =
and Qp =
RC
3
10
Active Low Pass Filter
m = 0, n = 0 and p = 1
Vo
-K
=
Vi ⎡ s2
⎤
s
+ (1 + K)⎥
⎢ 2 +
R1 = R2 = R and C1 = C2 = C
⎢⎣ ωp ωpQp
⎥⎦
The natural frequency of the active filter is now higher
ω0 = ωp 1 + K ; Qa = Qp 1 + K
Qa can be increased to the required value through
suitable selection of K.
Low-pass passive filter with amplifier gain -K and feedback
11
Structure of Active Low Pass filter
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—
—
Addition required between feedback signal and the input
In order to get 2Vo the gain of VCVS will have to be made 2K
A VCVS with gain -K can be realized by having buffer stage
followed by inverting amplifier
12
Structure of Active Low Pass filter (contd.,)
13
Second order Butterworth low-pass filter
Bandwidth = 40Hz, Qa =
1
2
.
With R1 = R 2 = R and
1
C1 = C2 = C then Qp =
3
Qa = Qp 1 + K =
1+K
1
=
3
2
K = 3.5 and
ω0 = 2π × 40 =
1+K
4.5
=
RC
RC
4.5
RC =
2π × 40
For R = 100kΩ;
4.5
C=
µF = 84nF
2π × 40
14
Frequency Response of the Butterworth LP filter
15
Transient Response of the Butterworth LP filter
16
Frequency response of Low Pass Filter
with Q =5 and f0 = 40 Hz
R=100k; C=0.6mF; K=224
17
Frequency response of Low Pass Filter
with Q =5 and f0 = 400 Hz C = 60 nF Finite GB of the Op.Amp
Finite GB of the Op.Amp
18
Transient response of Low-pass Filter
with Q =5 and f0 = 400 Hz C = 60 nF
Q-Enhancement due to Finite GB
19
Low-pass Filter
with Q =5 and f0 = 600 Hz C = 40 nF
Q-Enhancement due to Finite GB
20
Observations
—
—
—
—
—
Q increases from the specified value
The natural frequency reduces slightly from the specified value
At higher natural frequencies the transient responses are more
oscillatory indicating Q enhancement
Beyond a certain natural frequency the system becomes unstable
and goes into oscillations at the natural frequency
These deviations from the expected behavior are due to finite gain
bandwidth product of the active devices used.
21
Effect of Gain Bandwidth Product of Op Amp
Amplifier using a buffer and an inverting
amplifier of gain K has a transfer function
⎛
2 (1 + K ) s ⎞
Vo
K
=
; K ⎜1 ⎟
⎜
⎟
(1 + 2K)s ⎞ ⎛
s ⎞
Vi ⎛
GB
⎝
⎠
1
+
1
+
⎜
⎟
⎜
⎟
GB
GB ⎠
⎝
⎠⎝
Transfer function of the active low-pass filter
⎛
2 (1 + K ) s ⎞
K ⎜1 ⎟
⎜
⎟
G
B
Vo
⎝
⎠
=
2
Vi
⎛
2 (1 + K ) s ⎞
s
s
+
+ 1 + K ⎜1 ⎟
2
⎜
⎟
GB
ωp ωpQp
⎝
⎠
22
Effect of GB Product of Op Amp (contd.,)
Normalizing
2 (1 + K ) s ⎞
K ⎛
⎜⎜1 ⎟⎟
1+K⎝
GB
⎠
Vo
=
2
Vi
⎛ 2K (1 + K ) s ⎞
s
s
+
-⎜
+1
⎟
2
ω0 ω0Qp 1 + K ⎜⎝ GB(1 + K) ⎟⎠
Qa (due to GB) =
Qp 1 + K
⎛
2K (1 + K ) ω0Qp ⎞
⎜⎜1 ⎟⎟
1 + KGB ⎠
⎝
GB should be large enough
to make
2K (1 + K ) ω0Qp
1 + KGB
2Kω0Qa
=
= 1
GB
23
Examples
Ex:1
Qa (specified) = 5 and f0 = 40Hz and K = 3.5
Qa (due to GB) =
Qp 1 + K
2Kω0Qa ⎞
⎛
⎜1 - GB ⎟
⎝
⎠
= 5.23
Ex:2
Qa (specified)=5 and f0 = 400 Hz and K = 224
Qa (due to GB) =
Qp 1 + K
2Kω0Qa ⎞
⎛
⎜1 - GB ⎟
⎝
⎠
= 48
24
Limitations of GB
2f0Qa
= 1 for the filter to be stable in case inductance simulation
GB
2Kf0Qa
= 1 for the filter to be stable in case of filter using feedback
GB
25
Fourth-order Butterworth Low-pass Filter
1
2
2
⎡
⎤
⎡
s
s
s
s ⎤
+ 2 ⎥ ⎢1 + 1.848
+ 2 ⎥
⎢1 + 0.765
ω0 ω0 ⎥⎦ ⎢⎣
ω0 ω0 ⎥⎦
⎢⎣
26
Effect of finite GB
—
—
—
—
Taking GB into account
with f0 = 3.3 kHz (speech
filter)
Using 741 Op Amp having
a GB of 1 MHz
Q of the 2nd secondorder filter changes by 1%
Q of the first secondorder filter changes by
about 12.4%
27
High Pass Filter
m = 1, n = 0 and p = 0
2
s
-K 2
ωp
Vo
=
2
Vi ⎡
⎤
s
s
+ 1⎥
⎢(1 + K ) 2 +
ωp ωpQp
⎢⎣
⎥⎦
Natural frequency of the
active filter
ωp
ω0 =
(1 + K )
Natural frequency of the
active filter decreases by
a factor 1 + K
Qa = Qp 1 + K
The quality factor Q of the
active filter can be increased
to the required value through
suitable selection of K.
28
Passive HP filter
29
Active HP filter
30
Active HP Filter (contd.,)
2
s
-K 2
Vo
ω0
= 2
where ωp
Vi
s
s
+
+1
2
ω0 ω0Q
( )
Qa =
(C1R1
C2R 2 ) +
(
2
ωp
1
=
; ω0 =
and
C1R1C2R 2
1+K
1+K
(C2R2
)
C1R1 ) (1 + (R1 R 2 ) )
= Qp 1 + K
Required Qa is obtained by selecting K.
ωp is determined for a specified ω0 and K.
31
Topology of active HP filter
32
Example
If the lower cut off frequency is selected as 0.4 Hz.
1
Assuming C1 = C2 = C and R1 = R 2 = R; Qp =
3
1
Qa =
for maximally flat response
2
1+K
1
Qa = Qp 1 + K =
=
;K = 3.5
3
2
1
1
1
ω0 = 2π × 0.4 =
=
;RC =
RC 1 + K RC 4.5
2π × 0.4 4.5
1
For R = 100kΩ;C =
= 1.8 µF
0.8π 4.5
33
Simulation
34
Single Op Amp Topology
—
Buffer amplifier can be removed by suitable adjustment of the
resistances
R1 = R = R3//R4
and C1 = C2 = C
35
Q of active filters
Q of the circuit gets enhanced by a factor of 1 + K
in case of low-pass and high-pass active filters
Natural frequency of the active low-pass filter ω0 =
Natural frequency of the active high-pass filter ω0 =
(
(
)
1 + K ωp
ωp
1+K
)
36
Effect of finite GB
—
—
—
High Pass filter with f0 = 400 Hz and Q = 5
Q = 5 gives K = 224
For f0=400 Hz and R = 100 kW gives C = 265 pF
37
Effect of finite GB (contd.,)
⎡ 2 (1 + K ) s ⎤
K changes because of finite GB to K ⇒ K ⎢1 ⎥
GB
⎢⎣
⎥⎦
⎡ 2 (1 + K ) s ⎤ s2
-K ⎢1 ⎥ 2
GB
ωp
Vo
⎣
⎦
=
2
Vi ⎡⎛
⎤
⎞
2(1
+
K)s
s
s
⎧
⎫
+ 1⎥
⎢⎜1 + K ⎨1 ⎬⎟ 2 +
GB
ω
Q
ω
⎩
⎭
⎢⎝
⎠
⎥⎦
p
p
p
⎣
⎡ 2 (1 + K ) s ⎤ s2
-K ⎢1 ⎥ 2
GB
ωp
Vo
⎣
⎦
=
Vi
⎡
⎤
2Kω0Qa ⎞
s2
s ⎛
1+
+ 1⎥
⎢(1 + K ) 2 +
⎜
⎟
GB ⎠
ωp ωpQp ⎝
⎢⎣
⎥⎦
Q of the high-pass
filter simulated using
741 Op Amp having a
GB of 1 MHz changes
to 2.63 that is by 48%
38
Conclusion
39
Conclusion
40