Transcript Lecture 5 Active Filter (Part II)

Lecture 5 Active Filter (Part II)
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Biquadratic function filters
Positive feedback active filter: VCVS
Negative feedback filter: IGMF
Butterworth Response
Chebyshev Response
Ref:080225HKN
EE3110 Active Filter (Part 2)
1
Biquadratic function filters
s 
2
Z
s 
2
Z
s  cs  d
QZ
H ( s)  2
K
P
s  as  b
2
2
s 
s  P
QP
2
Realised by:
(I) Positive feedback
Ref:080225HKN
(II) Negative
feedback
EE3110 Active Filter (Part 2)
(III) Band Pass
Biquadratic functions
1
1

K

s 2  as  b
s 2  P s  P2
QP
s2  b
s 2  Z2
H ( s)  K 2
K

s  as  b
s 2  P s  P2
QP
(V) All Pass
(II) High Pass
2
2
s
s
H ( s)  K 2
K

s  as  b
s 2  P s  P2
QP
Ref:080225HKN
s
s

K

s 2  as  b
s 2  P s  P2
QP
(IV) Band Stop
(I) Low Pass
H ( s)  K
H ( s)  K
s2 
Z
s  Z2
s 2  as  b
QZ
H ( s)  K 2
K

s  as  b
s 2  P s  P2
QP
EE3110 Active Filter (Part 2)
Low-Pass Filter
Voltage Gain (dB)
Voltage Gain
Qp = 1.5
1.5
Qp = 1.5
Qp = 1
0
K = 1, p = 1
Qp = 1
Qp =
1
Qp =
1
2
-10
2
Qp =
0.5
-20
Qp
=
Qp =
0.5
0.5
Qp
0
1
0
-30
=0
.1
1
2
Frequency
Ref:080225HKN
3
4
0.1
K = 1, p = 1
0
1
2
Frequency
H s  
1
s
2
s 
1
Qp
EE3110 Active Filter (Part 2)
3
4
High-Pass Filter
Voltage Gain
Voltage Gain (dB)
5
Qp = 1.5
1.5
Qp = 1.5
K = 1, p = 1
0
Qp = 1
1
Qp = 1
1
Qp =
Qp = 0.5
-5
2
Qp =
.5
Q =0
p
2
-10
0.5
Qp =
0
1
2
K = 1, p = 1
3
4
-20
0
Frequency
1
2
Frequency
s2
H s  
s
s2 
1
Qp
Ref:080225HKN
0.1
-15
.1
Qp = 0
0
1
EE3110 Active Filter (Part 2)
3
4
Band-Pass Filter
Voltage Gain (dB)
Voltage Gain
5
1.5
Qp = 1.5
K = 1, p = 1
0
-5
Qp = 1
1
Qp =
1
-10
2
-15
0.5
Qp = 0.1
-25
Qp = 0.1
-30
0
Qp = 0.5
-20
Qp = 0.5
0
Qp = 1.5
Qp = 1
1
Qp =
2
2
4
Frequency
Ref:080225HKN
6
8
H s  
K = 1, p = 1
0
s
s
s2 
1
Qp
EE3110 Active Filter (Part 2)
2
4
Frequency
6
8
Band-Stop Filter
Voltage Gain
Voltage Gain (dB)
5
10
Qp = 1.5
K = 1, p = 1, z = 2
Qp = 1.5
4
0
3
Qp =
Qp =
0.1
0.5
-20
.1
=0
0
-10
2
1
2
Qp
1
1
Qp =
Qp =
Voltage Gain
Qp = 1
2
K = 1, p = 1, z = 2
0
2
4
Frequency
Ref:080225HKN
6
-30
8
0
2
s 2
H s  
s
2
s 
1
Qp
2
2
EE3110 Active Filter (Part 2)
4
Frequency
6
8
Voltage Controlled Votage Source (VCVS)
Positive Feedback Active Filter (Sallen-Key)
Z3
if
ii ia
Z1
Vi
By KCL at Va:
where,
ii  i f  ia  0
if 
ia 
Ref:080225HKN
Z4
Va
V  Va
ii  i
Z1
Vo  Va
Z3
Va
Z2  Z4
K
Z2
Vo
Therefore, we get
Vi  Va Vo  Va
Va


0
Z1
Z3
Z2  Z4
Re-arrange into voltage group gives:
 1
Vi Vo
1
1 

 Va  

 0
Z1 Z3
 Z1 Z3 Z2  Z4 
EE3110 Active Filter (Part 2)
(1)
But,
Vo  Kia Z4  K
Z4Va
Z2  Z4
(2)
Substitute (2) into (1) gives
Vi Vo Vo Z2  Z4   1
1
1 



 
 0
Z1 Z3
KZ4
 Z1 Z3 Z2  Z4 
or
H
Vo
K

ZZ
1
Vi Z1
1  K   1 2  1  Z1  Z2 
Z3
Z3Z 4
Z4
(3)
In admittance form:
H
K
1
1  Y3
Y3Y4


1  Y4    
1 K 
Y1Y2
 Y1 Y2  Y1
(4)
* This configuration is often used as a low-pass filter, so a
specific example will be considered.
Ref:080225HKN
EE3110 Active Filter (Part 2)
1
H (s )  K 2
s  as  b
VCVS Low Pass Filter
In order to obtain the above response, we let:
Z1  R1
Z 2  R2
R
1
Z3 
1
1

jC3 sC3
R
2
Then the transfer function (3) becomes:
H ( s) 
Z4 
1
1

jC4 sC 4
C
3
C
4
K
K'

1  sC 4 R1  R2   sR1C3 1  K   s 2 R1 R2C3C4 s 2  P s   2
P
QP
Ref:080225HKN
EE3110 Active Filter (Part 2)
(5)
we continue from equation (5),
K
H ( s)  2
s R1 R2 C3C4  sC 4 R1  R2   sR1C3 1  K   1
1
K
R1 R2 C 3C 4
H ( s) 
 C 4 R1  R2   R1C 3 1  K  

1
2
s  s



R
R
C
C
R
R
C
C

  1 2 3 4
1
2
3
4
Equating the coefficient from equations (6) and (5), it gives:
P 
1
R1 R2 C3C4

1
1
R1C3
R2 C4
QP 
1
RC
R1C4
R2 C4

 1  K  1 3
R2 C3
R1C3
R2 C4
Now, K=1, equation (5) will then become,
H (s ) 
Ref:080225HKN
1
1  sC 4 R1  R 2   s 2 R1R 2C 3C 4
EE3110 Active Filter (Part 2)
Simplified Design (VCVS filter)
Z1  m R
H ( s) 
Z2  R
Z3 
1
1

j (nC) snC
1
1  sRCm  1  s 2 nm R2C 2
m
R
Comparing with the low-pass response:
1
H ( s)  K

s 2  P s   P2
QP
It gives the following:
p 
1
RC nm
Ref:080225HKN
Z4 
mn
Qp 
m 1
EE3110 Active Filter (Part 2)
1
1

jC sC
R
nC
C
Example (VCVS low pass filter)
1
To design a low-pass filter with f O  512Hz and Q 
2
nC
Let m = 1
mn
1 n
n
1
QP 



m 1 11
2
2
mR
R
+
vin
-
C
 n=2
1
1
P 


 2 (512 Hz )
RC mn RC 1  2 RC 2
vo
1
Choose C  100nF
2k2
200nF
2k2
+
Then
R  2,198 ~ 2.2k
vin
What happen if n = 1?
Ref:080225HKN
EE3110 Active Filter (Part 2)
100nF
-
vo
VCVS High Pass Filter
C1
C2
R3
+
vin
R4
-
vo
KS 2
K' S2
H ( s) 

P
2
2
 1

1
1
1
2
s

s


1  K  
s  s


P
Q
P
 R4C1 R4C2 R3C1
 R3 R4C1C2
Ref:080225HKN
EE3110 Active Filter (Part 2)
VCVS Band Pass Filter
R1
vin
C2
+
R4
C5
K
H ( s) 
R3
-
s
R1C5
s 1
1
1
C 
R1  R3
s   
 1  K   5  
C5  R1 R4 R3
R4C2  R1 R3 R4C2C5
2
Ref:080225HKN
EE3110 Active Filter (Part 2)
vo
K'S

s2 
P
QP
s  P2
Infinite-Gain Multiple-Feedback (IGMF)
Negative Feedback Active Filter
Z
4
Z
1

i
v 0

i
v
V
0 i
By KCL at Vx ,
Z
5
Z
3
V
Z5 Z
Zx
3
Vo   Vx 
V


Vo
2x
Z3
Z5
Vi  Vx Vx Vx Vx  Vo

 
Z1
Z 2 Z3
Z4
(1)
(2)
substitute (1) into (2) gives
Vi
Z3
Z3
Vo
Z3
Vo

V 
V 

V 
Z1 Z1Z5 o
Z 5Z 2 o Z5 Z 4 Z5 o Z 4
Ref:080225HKN
EE3110 Active Filter (Part 2)
V
o
(3)
rearranging equation (3), it gives,
1
V
Z1Z3
H o 
Vi
1  1
1
1
1
1

 
 
Z5  Z1 Z 2 Z3 Z 4  Z 3Z4
Or in admittance form:
H
Vo
Y1Y3

Vi
Y5 Y1  Y2  Y3  Y4   Y3Y4
Filter Value
LP
Z1
Z2
Z3
Z4
Z5
R1
C2
R3
R4
C5
HP
C1
R2
C3
C4
R5
BP
R1
R2
C3
C4
R5
Ref:080225HKN
EE3110 Active Filter (Part 2)
IGMF Band-Pass Filter
s
Band-pass: H ( s )  K 2
s  as  b
To obtain the band-pass response, we let
Z1  R1
Z 2  R2
Z3 
1
1

jC3 sC3
Z4 
1
1

jC4 sC4
sC 3
R1
H (s )  
C  C4 1  1
1
s 2C 3C 4  s 3
   
R5
R 5  R1 R 2 
*This filter prototype has a very low
sensitivity to component tolerance when
compared with other prototypes.
Ref:080225HKN
EE3110 Active Filter (Part 2)
R1
R2
Z 5  R5
C4
C3
R5
Simplified design (IGMF filter)
sC
R1
H (s)  
1
2C
s
 s 2C 2
R1 R5
R5
R1
C
Comparing with the band-pass response
s
H ( s)  K
s2 
P
QP
s  P2
Its gives,
1
p 
C R1 R5
Ref:080225HKN
1
Qp 
2
R5
R1
C
H  p   2Q 2
EE3110 Active Filter (Part 2)
R5
Example (IGMF band pass filter)
To design a band-pass filter with
P 
fO  512Hz and Q  10
1
 2 (512 Hz )
C R1 R5
C  100nF  R1R5  9,662,7412
1 R5
QP 
 10
2 R1
 R1  155.4
100nF
62,170
155.4
R5  62,170
vin
100nF
With similar analysis, we can choose the following values:
C  10 nF R1  1,554 and R5  621,700
Ref:080225HKN
EE3110 Active Filter (Part 2)
+
vo
Butterworth Response (Maximally flat)
H  j  
1

1   
 o 
2n
where n is the order
Normalize to o = 1rad/s
Hˆ ( j ) 
1
1 
2n
Butterworth polynomials:
1
Hˆ ( j ) 
Bn ( j )
Ref:080225HKN
Butterworth polynomials
B1 s   s  1
B2 s   s 2  2 s  1
B3 s   s 3  2s 2  2s  1


 s  1 s 2  s  1
B4 s   s 4  2.61s 3  3.41s 2  2.61s  1



 s 2  0.77s  1 s 2  1.85s  1
B5 s   s 5  3.24s 4  5.24s 3  5.24s 2  3.24s  1



 s  1 s 2  0.62s  1 s 2  1.62s  1
EE3110 Active Filter (Part 2)
Butterworth Response
Ref:080225HKN
EE3110 Active Filter (Part 2)
Second order Butterworth response
Started from the low-pass biquadratic function H ( s )  K
For
p  1
K 1
Q
1
2
1
s2 
P
QP
1
(secondorder butterwothpolynomial
)
2
s  2s  1
1
H ( j ) 
  2  2 j  1
1
H ( j ) 
2
2
1   2  2
H (s) 
 

H ( j ) 
H ( j ) 
H ( j ) 
Ref:080225HKN

1
1  2 2   4  2 2
1
1  4
1
1   
22

1
1   
2n
EE3110 Active Filter (Part 2)
s  P2
Bode plot (n-th order Butterworth)
x

1
1 
2n
In dB form:
Hˆ ( j )  20 log
1 st
-20
1
-60
d
or
d
r
ord
er
er
h
or
-80
de
r
de
-100
or
r
suppose  1
3r
5th
Hˆ ( j )  20 log( 1   2 n )
2nd
-40
or d e
4t
1   2n
voltage gain (dB)
Hˆ ( j ) 
x
Hˆ ( j )  20n log( )
Butterworth response
For decade condition,  10 x
T heButterworth filter would have(-20n)dB/decade
For 2nd order : - 40 dB/decade
For 3rd order : - 60 dB/decade
For n - th order : - 20n dB/.decade
Ref:080225HKN
EE3110 Active Filter (Part 2)
Second order Butterworth filter
K
K'

1  sC 4 R1  R2   sR1C3 1  K   s 2 R1 R2C3C4 s 2  P s   2
P
Q
1
P
R1C4
R2C4
RC

 1  K  1 3
R2C3
R1C3
R2C4
C3
R
R
H ( s) 
QP 
1
2
+
Setting R1= R2 and C1 = C2
1
1
1
QP 


1  1  1  K  1 2  1  K  3  K
C4
vin
-
RB
vo
RA
Now K = 1 + RB/ RA
QP 
1

3 K
1
 R 
3  1  B 
 RA 

1
R
2 B
RA
Therefore, we have 2 
For Butterworth response:
1
1
1
QP 
 QP   R
2 2 B
2
Ref:080225HKN
RA
RB
 2  1.414
RA
We define Damping Factor (DF) as:
DF 
EE3110 Active Filter (Part 2)
R
1
 2  B  1.414
QP
RA
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.
Ref:080225HKN
EE3110 Active Filter (Part 2)
Values for the Butterworth response
B1 s   s  1
B2 s   s 2  2s  1


 2.61s  1  s
B3 s   s 3  2s 2  2s  1  s  1 s 2  s  1
B4 s   s 4  2.61s 3  3.41s 2
B5 s   s 5  3.24s 4  5.24s 3  5.24s 2

 3.24s  1  s  1s
2

 0.77s  1 s 2  1.85s  1
Order
Roll-off
dB/decade
1st stage
poles
DF
1
-20
1
optional
2
-40
2
1.414
3
-60
2
4
-80
5
6
2
2nd stage
poles
DF
1.000
1
1.000
2
1.848
2
0.765
-100
2
1.000
2
-120
2
1.932
2
Ref:080225HKN
EE3110 Active Filter (Part 2)


 0.62s  1 s 2  1.62s  1
3rd stage
poles
DF
1.618
1
0.618
1.414
2
0.518
Forth order Butterworth Filter
R3  R4 , C3  C4
R1  R2 , C1  C2
C1 0.01 F
R1 8.2 k
C3 0.01 F
+15
V
R2 8.2 k
R3 8.2 k
+
R4 8.2 k
-741C
C2
0.01 F
RB 1.5 k
-15 V
RB
 1.848
RA
RB
 0.152
RA
+
- 741C
C4
0.01 F
RA 10 k
DF  2 
+15
V
Vout
RB 27 k
-15 V
RA 22 k
DF  2 
RB
 0.765
RA
RB
 1.235
RA
R B  0.152R A  0.15210k
R B  1.235R A  1.23522k
R B  1.52k
R B  27.17k
Ref:080225HKN
EE3110 Active Filter (Part 2)
Chebyshev Response (Equal-ripple)
H  j  
1
1   2Cn2  
Where  determines the ripple and
Cn2 is the Chebyshev cosine polynomial defined as
Ref:080225HKN
EE3110 Active Filter (Part 2)
Chebyshev Cosine Polynomials
C0    1
C1    
C2    2 2  1
C3    4 3  3
C4    8 4  8 2  1
C5    16 5  20 3  5
Cn    2Cn 1    Cn  2  
Ref:080225HKN
EE3110 Active Filter (Part 2)
Second order Chebychev Response
H  j  
1
1   2Cn2  
Example: 0.969dB ripple gives  = 0.5, C2    2 2  1
H 22   
1
1   2Cn2  
1
2
1  0.52 2  1
1
 4
   2  1.25

H 22 ( )
1 1 5
1
  j
2
2
1
H 2 s H 2  s  
 2 1
 2 1

 s   j  s   j 
2
2



Roots:
s2 
 H 2 s H 2  s 
 s / f

Ref:080225HKN
1
s 4  s 2  1.25
EE3110 Active Filter (Part 2)
Roots
Roots of first bracketed term
Roots of second bracketed term
1

0   4  j 
2

s
2
1
  j
2
1
0   4 
2
s
2
1
  j
2
 1  5  1 
1  5  1 
   
      j 
     
2  4  2 
 2  4  2  

 0.566  j 0.899
H 2 s H 2  s  
or H 2 s  

j

 1  5  1 
1  5  1 
   
      j 
     
2  4  2 
 2  4  2  

 0.566  j 0.899
1
1
．
s  0.556 j 0.899s  0.556 j 0.899 s  0.566 j 0.899s  0.556 j 0.899
1
1
1


s  0.556 j 0.899s  0.556 j 0.899 s  0.5562  0.8992 s 2  0.112s  1.117
Ref:080225HKN
EE3110 Active Filter (Part 2)