Upload
dinah-young
View
225
Download
2
Embed Size (px)
Citation preview
Active Filter (Part 2) 1
M2-3S Active Filter (Part II)
• Biquadratic function filters
• Positive feedback active filter: VCVS
• Negative feedback filter: IGMF
• Butterworth Response
• Chebyshev Response
Active Filter (Part 2)
22
22
2
2
)(
PP
P
ZZ
Z
sQ
s
sQ
s
Kbass
dcsssH
Realised by:
(I) Positive feedback
(II) Negative feedback
Biquadratic function filters
Active Filter (Part 2)
(I) Low Pass
(II) High Pass
(III) Band Pass
(IV) Band Stop
(V) All Pass
222
11)(
P
P
P sQ
sK
bassKsH
22
2
2
2
)(
P
P
P sQ
s
sK
basss
KsH
222
)(
P
P
P sQ
s
sK
basss
KsH
22
22
2
2
)(
P
P
P
Z
sQ
s
sK
bass
bsKsH
22
22
2
2
)(
P
P
P
Z
Z
Z
sQ
s
sQ
sK
bass
bassKsH
Biquadratic functions
Active Filter (Part 2)
Low-Pass Filter
1
1
2
pQ
ss
sH
0 1 2 3 40
0.5
1
1.5 Qp = 1.5
Qp = 1
Frequency
K = 1, p = 1
Voltage Gain
Qp = 2
1
0 1 2 3 4
-30
-20
-10
0
Frequency
K = 1, p = 1
Qp = 1.5
Qp = 1
Voltage Gain (dB)
Qp = 2
1
Active Filter (Part 2)
High-Pass Filter
12
2
pQ
ss
ssH
0 1 2 3 40
0.5
1
1.5 Qp = 1.5
Qp = 1
Frequency
K = 1, p = 1
Voltage Gain
Qp = 2
1
0 1 2 3 4-20
-15
-10
-5
0
5
Frequency
K = 1, p = 1
Qp = 1.5Qp = 1
Voltage Gain (dB)
Qp = 2
1
Active Filter (Part 2)
Band-Pass Filter
12
pQ
ss
ssH
0 2 4 6 80
0.5
1
1.5 Qp = 1.5
Qp = 0.5
Qp = 0.1
Frequency
K = 1, p = 1
Voltage Gain
Qp = 1
Qp = 2
1
0 2 4 6 8-30
-25
-20
-15
-10
-5
0
5
Frequency
K = 1, p = 1
Qp = 1.5
Qp = 0.5
Qp = 0.1
Voltage Gain (dB)
Qp = 2
1Qp = 1
Active Filter (Part 2)
Band-Stop Filter
1
2
2
22
pQ
ss
ssH
0 2 4 6 80
1
2
3
4
5
Qp = 1.5
Frequency
K = 1, p = 1, z = 2
Vol
tage
Gai
n
Voltage Gain
Qp = 1
Qp = 2
1
0 2 4 6 8-30
-20
-10
0
10
Frequency
K = 1, p = 1, z = 2 Qp = 1.5
Voltage Gain (dB)
Qp = 2
1
Active Filter (Part 2)
Voltage Controlled Votage Source (VCVS) Positive Feedback Active Filter (Sallen-Key)
By KCL at Va: i i ii f a 0
iV VZii a
1
iV
Z Zaa
2 4
where,V VZ
V VZ
VZ Z
i a o a a
1 3 2 4
0
Therefore, we get
VZ
VZ
VZ Z Z Z
i oa
1 3 1 3 2 4
1 1 10
Re-arrange into voltage group gives:
(1)
Z1 Z2
Z3
Z4
K
Vi Vo
Va
if
ii ia
3Z
VVi ao
f
Active Filter (Part 2)
But, V Ki Z KZVZ Zo a
a 44
2 4(2)
Substitute (2) into (1) gives V
ZVZ
V Z Z
KZ Z Z Z Zi o o
1 3
2 4
4 1 3 2 4
1 1 10
or
H
VV
KZZ
KZZZ Z Z
Z Z
o
i
1
3
1 2
3 4 41 21 1
1
In admittance form:
H
K
YY Y
YY
KY YYY
1
1 114
1 2
3
1
3 4
1 2
* This configuration is often used as a low-pass filter, so a specific example will be considered.
(3)
(4)
Active Filter (Part 2)
VCVS Low Pass Filter H s Ks as b
( ) 1
2
In order to obtain the above response, we let:
444
3332211
11
11
sCCjZ
sCCjZRZRZ
R 1 C 3R 2
C 4
Then the transfer function (3) becomes:
224321
2
31214
'
11)(
P
P
P sQ
s
K
CCRRsKCsRRRsC
KsH
(5)
Active Filter (Part 2)
Equating the coefficient from equations (6) and (5), it gives:
42
31
31
42
32
4142314321 1
1
111
CR
CRK
CR
CR
CR
CRQ
CRCRCCRRPP
Now, K=1, equation (5) will then become,
H ssC R R s RR CC
( )
1
1 4 1 22
1 2 3 4
we continue from equation (5),
11)(
312144321
2
KCsRRRsCCCRRsK
sH
43214321
312142
4321
11
1
)(
CCRRCCRR
KCRRRCss
CCRRK
sH
Active Filter (Part 2)
Simplified Design (VCVS filter)
mR nCR
C
22211
1)(
CnmRsmsRCsH
sCCjZ
snCnCjZRZmRZ
11
1
)(
1 4321
Comparing with the low-pass response:
22
1)(
PP
P sQ
sKsH
It gives the following:
1
1
m
mnQ
nmRCpp
Active Filter (Part 2)
Example (VCVS low pass filter)To design a low-pass filter with and
2
1 Hz512 QfO
vin C vo
+
-
mR R
nCLet m = 1
2
1
211
1
1
nn
m
mnQ
P
n = 2
)512(22
1
21
11Hz
RCRCmnRCP
nFC 100Choose
Then kR 2.2~198,2 vin 100nF vo
+
-
2k2 2k2
200nF
What happen if n = 1?
Active Filter (Part 2)
VCVS High Pass Filter
vin
C1
vo
+
-
R3
R4
C2
22
2
2143132414
2
2 '
11
111)(
P
P
P sQ
s
SK
CCRRK
CRCRCRss
KSsH
Active Filter (Part 2)
VCVS Band Pass Filter
vin
R1
vo
+
-
R3
R4
C2
C5
22
52431
31
24
5
3415
2
51'
1111
)(
P
P
P sQ
s
SK
CCRRR
RR
CR
CK
RRRC
ss
CR
sK
sH
Active Filter (Part 2)
Infinite-Gain Multiple-Feedback (IGMF) Negative Feedback Active Filter
V i
Z 1
V x
Z 3
Z 2
Z 4 Z 5
V o
)2( ,at KCLBy
)1( 0 0
4321x
5
3
3
5
Z
VV
Z
V
Z
V
Z
VVV
VZ
ZVV
Z
ZVvv
oxxxxi
oxxoii
substitute (1) into (2) gives VZ
ZZZ
VZZ Z
VVZ
ZZ Z
VVZ
io o
oo
o
1
3
1 5
3
5 2 5
3
4 5 4
(3)
Note: because no current flows into v+, v- terminals of op-amp. Therefore, from KCL at node v- : Vo/Z5+ Vx/Z3 = 0
Active Filter (Part 2)
rearranging equation (3), it gives,
HVV
ZZ
Z Z Z Z Z Z Z
o
i
1
1 1 1 1 1 11 3
5 1 2 3 4 3 4
HVV
YY
Y Y Y Y Y Y Yo
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
Active Filter (Part 2)
IGMF Band-Pass Filter
H s Ks
s as b( )
2Band-pass:
To obtain the band-pass response, we let
5544
433
32211 11
11
RZsCCj
ZsCCj
ZRZRZ
R 1
C 3R 2
C 4 R 5H s
sCR
sCC sC CR R R R
( )
3
1
23 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.
Active Filter (Part 2)
Simplified design (IGMF filter)
R 1
C
C R 5
22
551
1
21)(
CsR
Cs
RR
R
sC
sH
Comparing with the band-pass response
22
)(
P
P
P sQ
s
sKsH
Its gives,
2
11
5
51
2 CR
1- K
2
1
1QjH
R
RQ
RRCppp
Active Filter (Part 2)
Example (IGMF band pass filter)To design a band-pass filter with and 10 Hz512 QfO
)512(21
51
HzRRCP
251 741,662,9 100 RRnFC
102
1
1
5 R
RQ
P
170,62 4.155 51 RR vin vo+
-
100nF
4.155170,62
100nF
With similar analysis, we can choose the following values:
700,621 and 554,1 10 51 RRnFC
Active Filter (Part 2)
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
23455
22
2344
2
233
22
1
sssss
ssssssB
ssss
sssssB
sss
ssssB
sssB
ssB
Butterworth polynomials
Butterworth polynomials:
)(
1)(ˆ
jBjH
n
njH
21
1)(ˆ
Active Filter (Part 2)
Butterworth Response
Active Filter (Part 2)
Second order Butterworth responseStarted from the low-pass biquadratic function
22
1)(
P
P
P sQ
sKsH
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
Active Filter (Part 2)
Bode plot (n-th order Butterworth)
dB/.decade20n - :orderth -nFor
dB/decade 60- :order 3rdFor
dB/decade 40- :order 2ndFor
ecade(-20n)dB/d have ldfilter wouh Butterwort The
10 condition, decadeFor
)log(20 )(ˆ
1 suppose
)1log(20)(ˆ
1
1log20)(ˆ
:form dBIn 1
1)(ˆ
x
2
2
2
njH
jH
jH
jH
n
n
n
Butterworth response
-20
-40
-60
-80
-100
x
x
1st order
2nd order3rd order4th order5th order
volta
ge g
ain
(dB
)
Active Filter (Part 2)
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
vinC4 vo
+
-
R1 R2C3
RB
RA
A
B
A
B
P
R
R
R
RKQ
2
1
13
1
3
1
For Butterworth response:
2
1P
Q
A
B
P
R
RQ
2
1
2
1
Therefore, we have 414.122 A
B
R
R
We define Damping Factor (DF) as:
414.121
A
B
P R
R
QDF
Active Filter (Part 2)
Damping Factor (DF)
• The value of the damping factor required to produce desire response characteristic depends on the order of the filter.
• The DF is determined by the negative feedback network of the filter circuit.
• Because of its maximally flat response, the Butterworth characteristic is the most widely used.
• We will limit our converge to the Butterworth response to illustrate basic filter concepts.
Active Filter (Part 2)
Values for the Butterworth response 162.1162.01124.324.524.524.3
185.1177.0161.241.361.2
11122
12
1
2223455
222344
2233
22
1
sssssssssssB
sssssssssB
sssssssB
sssB
ssB
Roll-off
dB/decade
1st stage 2nd stage 3rd stage
Order poles DF poles DF poles DF
1 -20 1 optional
2 -40 2 1.414
3 -60 2 1.000 1 1.000
4 -80 2 1.848 2 0.765
5 -100 2 1.000 2 1.618 1 0.618
6 -120 2 1.932 2 1.414 2 0.518
Active Filter (Part 2)
Forth order Butterworth Filter2121 , CCRR 4343 , CCRR
kR
kRR
R
R
R
RDF
B
AB
A
B
A
B
52.1
10152.0152.0
152.0
848.12
kR
kRR
R
R
R
RDF
B
AB
A
B
A
B
17.27
22235.1235.1
235.1
765.02
+
-
R2 8.2 k
C1 0.01 F
+
-
Vout
+15 VR1 8.2 k
RB 1.5 k
RA 10 k
R3 8.2 k R4 8.2 k+15 V
-15 V -15 V
C3 0.01 F
C2
0.01 F C4
0.01 FRB 27 k
RA 22 k
741C 741C
Active Filter (Part 2)
Chebyshev Response (Equal-ripple)
221
1
nCjH
Where determines the ripple and is the Chebyshev cosine polynomial defined as 2
nC
Active Filter (Part 2)
Chebyshev Cosine Polynomials
21
355
244
33
22
1
0
2
52016
188
34
12
1
nnn CCC
C
C
C
C
C
C
Active Filter (Part 2)
Second order Chebychev Response
Example: 0.969dB ripple gives = 0.5, 12 22 C
221
1
nCjH
25.1
1
125.01
1
1
1
24
22
2222
nCH
js
2
1
2
5112Roots:
jsjs
sHsH
21
21
1
2222
25.1
1
)(
24
22/
22
ss
sHsHHfs
969.0)1
1(log20
210
Note:
Active Filter (Part 2)
Roots
Roots of first bracketed term
899.0566.0
2
1
4
5
2
1
2
1
4
5
2
1
2
1
2
21
40
j
j
j
j
s
Roots of second bracketed term
899.0566.0
2
1
4
5
2
1
2
1
4
5
2
1
2
1
2
21
40
j
j
j
j
s
899.0556.0899.0566.0
1
899.0556.0899.0556.0
122 jsjs
.jsjs
sHsH
or 117.1112.0
1
899.0556.0
1
899.0556.0899.0556.0
12222
sssjsjs
sH
))2
sin()2
(cos(2)/(tan22 1 jCeCCeebabja jjabj
Note: