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EE101: Op Amp circuits (Part 3)
M. B. [email protected]
www.ee.iitb.ac.in/~sequel
Department of Electrical EngineeringIndian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Introduction to filters
Consider v(t) = v1(t) + v2(t) = Vm1 sinω1t + Vm2 sinω2t .
−1
0
1
0 5 10 15 20t (msec)
0 5 10 15 20t (msec)
v2
v1v
LPF vo = v1v
HPFv vo = v2
A low-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the low-frequencycomponent v1(t) and remove the high-frequency component v2(t).
A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequencycomponent v2(t) and remove the low-frequency component v1(t).
There are some other types of filters, as we will see.
M. B. Patil, IIT Bombay
Introduction to filters
Consider v(t) = v1(t) + v2(t) = Vm1 sinω1t + Vm2 sinω2t .
−1
0
1
0 5 10 15 20t (msec)
0 5 10 15 20t (msec)
v2
v1v
LPF vo = v1v
HPFv vo = v2
A low-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the low-frequencycomponent v1(t) and remove the high-frequency component v2(t).
A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequencycomponent v2(t) and remove the low-frequency component v1(t).
There are some other types of filters, as we will see.
M. B. Patil, IIT Bombay
Introduction to filters
Consider v(t) = v1(t) + v2(t) = Vm1 sinω1t + Vm2 sinω2t .
−1
0
1
0 5 10 15 20t (msec)
0 5 10 15 20t (msec)
v2
v1v
LPF vo = v1v
HPFv vo = v2
A low-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the low-frequencycomponent v1(t) and remove the high-frequency component v2(t).
A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequencycomponent v2(t) and remove the low-frequency component v1(t).
There are some other types of filters, as we will see.
M. B. Patil, IIT Bombay
Introduction to filters
Consider v(t) = v1(t) + v2(t) = Vm1 sinω1t + Vm2 sinω2t .
−1
0
1
0 5 10 15 20t (msec)
0 5 10 15 20t (msec)
v2
v1v
LPF vo = v1v
HPFv vo = v2
A low-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the low-frequencycomponent v1(t) and remove the high-frequency component v2(t).
A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequencycomponent v2(t) and remove the low-frequency component v1(t).
There are some other types of filters, as we will see.
M. B. Patil, IIT Bombay
Ideal low-pass filter
0
1
0
ωc
vo(t)
ω
H(jω)
H(jω
)
vi(t)
0
LPF
0ωcωc
ωω
Vi(jω
)
Vo(jω
)
Vo(jω) = H(jω) Vi (jω) .
All components with ω < ωc appear at the output without attenuation.
All components with ω > ωc get eliminated.
(Note that the ideal low-pass filter has ∠H(jω) = 1, i.e., H(jω) = 1 + j0 .)
M. B. Patil, IIT Bombay
Ideal low-pass filter
0
1
0
ωc
vo(t)
ω
H(jω)
H(jω
)
vi(t)
0
LPF
0ωcωc
ωω
Vi(jω
)
Vo(jω
)
Vo(jω) = H(jω) Vi (jω) .
All components with ω < ωc appear at the output without attenuation.
All components with ω > ωc get eliminated.
(Note that the ideal low-pass filter has ∠H(jω) = 1, i.e., H(jω) = 1 + j0 .)
M. B. Patil, IIT Bombay
Ideal low-pass filter
0
1
0
ωc
vo(t)
ω
H(jω)
H(jω
)
vi(t)
0
LPF
0ωcωc
ωω
Vi(jω
)
Vo(jω
)
Vo(jω) = H(jω) Vi (jω) .
All components with ω < ωc appear at the output without attenuation.
All components with ω > ωc get eliminated.
(Note that the ideal low-pass filter has ∠H(jω) = 1, i.e., H(jω) = 1 + j0 .)
M. B. Patil, IIT Bombay
Ideal filters
1
0
Low−pass
0ωc
ω
H(jω
)
0
1
0
High−pass
ωc
ω
H(jω
)
0
1
0
Band−pass
ωL ωH
ω
H(jω
)
0
1
0
Band−reject
ωL ωH
ω
H(jω
)
M. B. Patil, IIT Bombay
Ideal filters
1
0
Low−pass
0ωc
ω
H(jω
)
0
1
0
High−pass
ωc
ω
H(jω
)
0
1
0
Band−pass
ωL ωH
ω
H(jω
)
0
1
0
Band−reject
ωL ωH
ω
H(jω
)
M. B. Patil, IIT Bombay
Ideal filters
1
0
Low−pass
0ωc
ω
H(jω
)
0
1
0
High−pass
ωc
ω
H(jω
)0
1
0
Band−pass
ωL ωH
ω
H(jω
)
0
1
0
Band−reject
ωL ωH
ω
H(jω
)
M. B. Patil, IIT Bombay
Ideal filters
1
0
Low−pass
0ωc
ω
H(jω
)
0
1
0
High−pass
ωc
ω
H(jω
)0
1
0
Band−pass
ωL ωH
ω
H(jω
)
0
1
0
Band−reject
ωL ωH
ωH
(jω
)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Ideal filters
1
0
−1
0
1.5
−1.5 0 5 10 15 20
t (msec)
v
v3
v1
v2
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
V1
V2 V3
Let us see the effect ofa few filters on v(t).
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
H(jω)
1
0
−1
0
1
0 0.5 1 1.5 2 2.5
f (kHz)
H(jω)
1.5
0
−1.5 0 5 10 15 20
t (msec)
M. B. Patil, IIT Bombay
Practical filter circuits
* In practical filter circuits, the ideal filter response is approximated with a suitableH(jω) that can be obtained with circuit elements. For example,
H(s) =1
a5s5 + a4s4 + a3s3 + a2s2 + a1s + a0
represents a 5th-order low-pass filter.
* Some commonly used approximations (polynomials) are the Butterworth,Chebyshev, Bessel, and elliptic functions.
* Coefficients for these filters listed in filter handbooks. Also, programs for filterdesign are available on the internet.
M. B. Patil, IIT Bombay
Practical filter circuits
* In practical filter circuits, the ideal filter response is approximated with a suitableH(jω) that can be obtained with circuit elements. For example,
H(s) =1
a5s5 + a4s4 + a3s3 + a2s2 + a1s + a0
represents a 5th-order low-pass filter.
* Some commonly used approximations (polynomials) are the Butterworth,Chebyshev, Bessel, and elliptic functions.
* Coefficients for these filters listed in filter handbooks. Also, programs for filterdesign are available on the internet.
M. B. Patil, IIT Bombay
Practical filter circuits
* In practical filter circuits, the ideal filter response is approximated with a suitableH(jω) that can be obtained with circuit elements. For example,
H(s) =1
a5s5 + a4s4 + a3s3 + a2s2 + a1s + a0
represents a 5th-order low-pass filter.
* Some commonly used approximations (polynomials) are the Butterworth,Chebyshev, Bessel, and elliptic functions.
* Coefficients for these filters listed in filter handbooks. Also, programs for filterdesign are available on the internet.
M. B. Patil, IIT Bombay
Practical filters
0
1
0
IdealIdeal
Practical
Practical
Low−pass High−pass
0
1
00
00
0 ωω ω
ωcωc
ω
ωcωs
Amin
Amax|H||H||H| |H|
ωsωc
Amax
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passbandripple, e.g., Amax = 1 dB.
* Amin is the minimum attenuation to be provided by the filter, e.g.,Amin = 60 dB.
* ωs : edge of the stop band.
* ωs/ωc (for a low-pass filter): selectivity factor, a measure of the sharpness of thefilter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
0
1
0
IdealIdeal
Practical
Practical
Low−pass High−pass
0
1
00
00
0 ωω ω
ωcωc
ω
ωcωs
Amin
Amax|H||H||H| |H|
ωsωc
Amax
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passbandripple, e.g., Amax = 1 dB.
* Amin is the minimum attenuation to be provided by the filter, e.g.,Amin = 60 dB.
* ωs : edge of the stop band.
* ωs/ωc (for a low-pass filter): selectivity factor, a measure of the sharpness of thefilter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
0
1
0
IdealIdeal
Practical
Practical
Low−pass High−pass
0
1
00
00
0 ωω ω
ωcωc
ω
ωcωs
Amin
Amax|H||H||H| |H|
ωsωc
Amax
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passbandripple, e.g., Amax = 1 dB.
* Amin is the minimum attenuation to be provided by the filter, e.g.,Amin = 60 dB.
* ωs : edge of the stop band.
* ωs/ωc (for a low-pass filter): selectivity factor, a measure of the sharpness of thefilter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
0
1
0
IdealIdeal
Practical
Practical
Low−pass High−pass
0
1
00
00
0 ωω ω
ωcωc
ω
ωcωs
Amin
Amax|H||H||H| |H|
ωsωc
Amax
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passbandripple, e.g., Amax = 1 dB.
* Amin is the minimum attenuation to be provided by the filter, e.g.,Amin = 60 dB.
* ωs : edge of the stop band.
* ωs/ωc (for a low-pass filter): selectivity factor, a measure of the sharpness of thefilter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
0
1
0
IdealIdeal
Practical
Practical
Low−pass High−pass
0
1
00
00
0 ωω ω
ωcωc
ω
ωcωs
Amin
Amax|H||H||H| |H|
ωsωc
Amax
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passbandripple, e.g., Amax = 1 dB.
* Amin is the minimum attenuation to be provided by the filter, e.g.,Amin = 60 dB.
* ωs : edge of the stop band.
* ωs/ωc (for a low-pass filter): selectivity factor, a measure of the sharpness of thefilter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
0
1
0
IdealIdeal
Practical
Practical
Low−pass High−pass
0
1
00
00
0 ωω ω
ωcωc
ω
ωcωs
Amin
Amax|H||H||H| |H|
ωsωc
Amax
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passbandripple, e.g., Amax = 1 dB.
* Amin is the minimum attenuation to be provided by the filter, e.g.,Amin = 60 dB.
* ωs : edge of the stop band.
* ωs/ωc (for a low-pass filter): selectivity factor, a measure of the sharpness of thefilter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
For a low-pass filter, H(s) =1
nXi=0
ai (s/ωc )i
.
Coefficients (ai ) for various types of filters are tabulated in handbooks. We now lookat |H(jω)| for two commonly used filters.
Butterworth filters:
|H(jω)| =1p
1 + ε2(ω/ωc )2n.
Chebyshev filters:
|H(jω)| =1p
1 + ε2C2n (ω/ωc )
where
Cn(x) = cosˆn cos−1(x)
˜for x ≤ 1,
Cn(x) = coshˆn cosh−1(x)
˜for x ≥ 1,
H(s) for a high-pass filter can be obtained from H(s) of the corresponding low-passfilter by (s/ωc )→ (ωc/s) .
M. B. Patil, IIT Bombay
Practical filters
For a low-pass filter, H(s) =1
nXi=0
ai (s/ωc )i
.
Coefficients (ai ) for various types of filters are tabulated in handbooks. We now lookat |H(jω)| for two commonly used filters.
Butterworth filters:
|H(jω)| =1p
1 + ε2(ω/ωc )2n.
Chebyshev filters:
|H(jω)| =1p
1 + ε2C2n (ω/ωc )
where
Cn(x) = cosˆn cos−1(x)
˜for x ≤ 1,
Cn(x) = coshˆn cosh−1(x)
˜for x ≥ 1,
H(s) for a high-pass filter can be obtained from H(s) of the corresponding low-passfilter by (s/ωc )→ (ωc/s) .
M. B. Patil, IIT Bombay
Practical filters
For a low-pass filter, H(s) =1
nXi=0
ai (s/ωc )i
.
Coefficients (ai ) for various types of filters are tabulated in handbooks. We now lookat |H(jω)| for two commonly used filters.
Butterworth filters:
|H(jω)| =1p
1 + ε2(ω/ωc )2n.
Chebyshev filters:
|H(jω)| =1p
1 + ε2C2n (ω/ωc )
where
Cn(x) = cosˆn cos−1(x)
˜for x ≤ 1,
Cn(x) = coshˆn cosh−1(x)
˜for x ≥ 1,
H(s) for a high-pass filter can be obtained from H(s) of the corresponding low-passfilter by (s/ωc )→ (ωc/s) .
M. B. Patil, IIT Bombay
Practical filters
For a low-pass filter, H(s) =1
nXi=0
ai (s/ωc )i
.
Coefficients (ai ) for various types of filters are tabulated in handbooks. We now lookat |H(jω)| for two commonly used filters.
Butterworth filters:
|H(jω)| =1p
1 + ε2(ω/ωc )2n.
Chebyshev filters:
|H(jω)| =1p
1 + ε2C2n (ω/ωc )
where
Cn(x) = cosˆn cos−1(x)
˜for x ≤ 1,
Cn(x) = coshˆn cosh−1(x)
˜for x ≥ 1,
H(s) for a high-pass filter can be obtained from H(s) of the corresponding low-passfilter by (s/ωc )→ (ωc/s) .
M. B. Patil, IIT Bombay
Practical filters (low-pass)
−100
0
0
1
−100
0
0
1
Butterworth filters:
Chebyshev filters:
2
3
n=1
45
n=1
2
34
5
n=1
4
3
2
5
3
4
5
2
n=1
0.01 0.1 1 10 100
0 2 3 4 5 1
0.01 0.1 1 10 100
0 2 3 4 5 1
ω/ωc
|H|(
dB)
ω/ωc
|H|
ω/ωc
|H|(
dB)
ω/ωc
|H|
ǫ = 0.5
ǫ = 0.5
M. B. Patil, IIT Bombay
Practical filters (high-pass)
Butterworth filters:
Chebyshev filters:
0
1
−100
0
0
1
−100
0
n=1
2
4
5
n=1
2
3
45
3
n=1
2
345
n=1
2
34
5
0.01 0.1 1 10 100
0.01 0.1 1 10 100
0 1 2 3 4
0 1 2 3 4
ω/ωc
|H|
ω/ωc
|H|(
dB)
ω/ωc
|H|
ω/ωc
|H|(
dB)
ǫ = 0.5
ǫ = 0.5
M. B. Patil, IIT Bombay
Passive filter example
R
C
VoVs100Ω
5µF
(Low−pass filter)
with ω0 = 1/RC .
H(s) =(1/sC)
R + (1/sC)=
1
1 + (s/ω0),
(SEQUEL file: ee101_rc_ac_2.sqproj)
f (Hz)
−60
−40
−20
0
20
105104103102101
|H|(
dB)
M. B. Patil, IIT Bombay
Passive filter example
R
C
VoVs100Ω
5µF
(Low−pass filter)
with ω0 = 1/RC .
H(s) =(1/sC)
R + (1/sC)=
1
1 + (s/ω0),
(SEQUEL file: ee101_rc_ac_2.sqproj)
f (Hz)
−60
−40
−20
0
20
105104103102101
|H|(
dB)
M. B. Patil, IIT Bombay
Passive filter example
R
C
VoVs100Ω
5µF
(Low−pass filter)
with ω0 = 1/RC .
H(s) =(1/sC)
R + (1/sC)=
1
1 + (s/ω0),
(SEQUEL file: ee101_rc_ac_2.sqproj)
f (Hz)
−60
−40
−20
0
20
105104103102101
|H|(
dB)
M. B. Patil, IIT Bombay
Passive filter example
Vs VoR
CL0.1mF
100Ω
4µF
(Band−pass filter)
with ω0 = 1/√
LC .
H(s) =(sL) ‖ (1/sC)
R + (sL) ‖ (1/sC)=
s(L/R)
1 + s(L/R) + s2LC
0
f (Hz)
(SEQUEL file: ee101_lc_1.sqproj)
−20
−40
−60
−80
105104103102
|H|(
dB)
M. B. Patil, IIT Bombay
Passive filter example
Vs VoR
CL0.1mF
100Ω
4µF(Band−pass filter)
with ω0 = 1/√
LC .
H(s) =(sL) ‖ (1/sC)
R + (sL) ‖ (1/sC)=
s(L/R)
1 + s(L/R) + s2LC
0
f (Hz)
(SEQUEL file: ee101_lc_1.sqproj)
−20
−40
−60
−80
105104103102
|H|(
dB)
M. B. Patil, IIT Bombay
Passive filter example
Vs VoR
CL0.1mF
100Ω
4µF(Band−pass filter)
with ω0 = 1/√
LC .
H(s) =(sL) ‖ (1/sC)
R + (sL) ‖ (1/sC)=
s(L/R)
1 + s(L/R) + s2LC
0
f (Hz)
(SEQUEL file: ee101_lc_1.sqproj)
−20
−40
−60
−80
105104103102
|H|(
dB)
M. B. Patil, IIT Bombay
Op Amp filters (“Active” filters)
* Op Amp filters can be designed without using inductors. This is a significantadvantage since inductors are bulky and expensive. Inductors also exhibitnonlinear behaviour (arising from the core properties) which is undesirable in afilter circuit.
* With Op Amps, a filter circuit can be designed with a pass-band gain.
* Op Amp filters can be easily incorporated in an integrated circuit.
* However, at high frequencies (∼ MHz), Op Amps no longer have a high gain→ passive filters.
* Also, if the power requirement is high, Op Amp filters cannot be used→ passive filters.
M. B. Patil, IIT Bombay
Op Amp filters (“Active” filters)
* Op Amp filters can be designed without using inductors. This is a significantadvantage since inductors are bulky and expensive. Inductors also exhibitnonlinear behaviour (arising from the core properties) which is undesirable in afilter circuit.
* With Op Amps, a filter circuit can be designed with a pass-band gain.
* Op Amp filters can be easily incorporated in an integrated circuit.
* However, at high frequencies (∼ MHz), Op Amps no longer have a high gain→ passive filters.
* Also, if the power requirement is high, Op Amp filters cannot be used→ passive filters.
M. B. Patil, IIT Bombay
Op Amp filters (“Active” filters)
* Op Amp filters can be designed without using inductors. This is a significantadvantage since inductors are bulky and expensive. Inductors also exhibitnonlinear behaviour (arising from the core properties) which is undesirable in afilter circuit.
* With Op Amps, a filter circuit can be designed with a pass-band gain.
* Op Amp filters can be easily incorporated in an integrated circuit.
* However, at high frequencies (∼ MHz), Op Amps no longer have a high gain→ passive filters.
* Also, if the power requirement is high, Op Amp filters cannot be used→ passive filters.
M. B. Patil, IIT Bombay
Op Amp filters (“Active” filters)
* Op Amp filters can be designed without using inductors. This is a significantadvantage since inductors are bulky and expensive. Inductors also exhibitnonlinear behaviour (arising from the core properties) which is undesirable in afilter circuit.
* With Op Amps, a filter circuit can be designed with a pass-band gain.
* Op Amp filters can be easily incorporated in an integrated circuit.
* However, at high frequencies (∼ MHz), Op Amps no longer have a high gain→ passive filters.
* Also, if the power requirement is high, Op Amp filters cannot be used→ passive filters.
M. B. Patil, IIT Bombay
Op Amp filters (“Active” filters)
* Op Amp filters can be designed without using inductors. This is a significantadvantage since inductors are bulky and expensive. Inductors also exhibitnonlinear behaviour (arising from the core properties) which is undesirable in afilter circuit.
* With Op Amps, a filter circuit can be designed with a pass-band gain.
* Op Amp filters can be easily incorporated in an integrated circuit.
* However, at high frequencies (∼ MHz), Op Amps no longer have a high gain→ passive filters.
* Also, if the power requirement is high, Op Amp filters cannot be used→ passive filters.
M. B. Patil, IIT Bombay
Op Amp filters: example
1 k
10 k
10 n
Vs
RL
C
R1
R2
Vo
20
0
−20
f (Hz)
105104103102101
|H|(
dB)
Op Amp filters are designed for Op Amp operation in the linear region→ Our analysis of the inverting amplifier applies, and we get,
Vo = −R2 ‖ (1/sC)
R1Vs (Vs and Vo are phasors)
H(s) = −R2
R1
1
1 + sR2C
This is a low-pass filter, with ω0 = 1/R2C .
(SEQUEL file: ee101 op filter 1.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
1 k
10 k
10 n
Vs
RL
C
R1
R2
Vo
20
0
−20
f (Hz)
105104103102101
|H|(
dB)
Op Amp filters are designed for Op Amp operation in the linear region→ Our analysis of the inverting amplifier applies, and we get,
Vo = −R2 ‖ (1/sC)
R1Vs (Vs and Vo are phasors)
H(s) = −R2
R1
1
1 + sR2C
This is a low-pass filter, with ω0 = 1/R2C .
(SEQUEL file: ee101 op filter 1.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
1 k
10 k
10 n
Vs
RL
C
R1
R2
Vo
20
0
−20
f (Hz)
105104103102101
|H|(
dB)
Op Amp filters are designed for Op Amp operation in the linear region→ Our analysis of the inverting amplifier applies, and we get,
Vo = −R2 ‖ (1/sC)
R1Vs (Vs and Vo are phasors)
H(s) = −R2
R1
1
1 + sR2C
This is a low-pass filter, with ω0 = 1/R2C .
(SEQUEL file: ee101 op filter 1.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
1 k
10 k
10 n
Vs
RL
C
R1
R2
Vo
20
0
−20
f (Hz)
105104103102101
|H|(
dB)
Op Amp filters are designed for Op Amp operation in the linear region→ Our analysis of the inverting amplifier applies, and we get,
Vo = −R2 ‖ (1/sC)
R1Vs (Vs and Vo are phasors)
H(s) = −R2
R1
1
1 + sR2C
This is a low-pass filter, with ω0 = 1/R2C .
(SEQUEL file: ee101 op filter 1.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
1 k
10 k
10 n
Vs
RL
C
R1
R2
Vo
20
0
−20
f (Hz)
105104103102101
|H|(
dB)
Op Amp filters are designed for Op Amp operation in the linear region→ Our analysis of the inverting amplifier applies, and we get,
Vo = −R2 ‖ (1/sC)
R1Vs (Vs and Vo are phasors)
H(s) = −R2
R1
1
1 + sR2C
This is a low-pass filter, with ω0 = 1/R2C .
(SEQUEL file: ee101 op filter 1.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
10 k
100 n1 kVs
RL
C R2R1
Vo
f (Hz)
20
0
−20
−40
105104103102101
|H|(
dB)
H(s) = −R2
R1 + (1/sC)=
sR2C
1 + sR1C.
This is a high-pass filter, with ω0 = 1/R1C .
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
10 k
100 n1 kVs
RL
C R2R1
Vo
f (Hz)
20
0
−20
−40
105104103102101
|H|(
dB)
H(s) = −R2
R1 + (1/sC)=
sR2C
1 + sR1C.
This is a high-pass filter, with ω0 = 1/R1C .
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
10 k
100 n1 kVs
RL
C R2R1
Vo
f (Hz)
20
0
−20
−40
105104103102101
|H|(
dB)
H(s) = −R2
R1 + (1/sC)=
sR2C
1 + sR1C.
This is a high-pass filter, with ω0 = 1/R1C .
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
10 k
100 n1 kVs
RL
C R2R1
Vo
f (Hz)
20
0
−20
−40
105104103102101
|H|(
dB)
H(s) = −R2
R1 + (1/sC)=
sR2C
1 + sR1C.
This is a high-pass filter, with ω0 = 1/R1C .
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
10 k
100 n1 kVs
RL
C R2R1
Vo
f (Hz)
20
0
−20
−40
105104103102101
|H|(
dB)
H(s) = −R2
R1 + (1/sC)=
sR2C
1 + sR1C.
This is a high-pass filter, with ω0 = 1/R1C .
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
100 k
10 k80 p
0.8µF
Vs
RL
C2
C1
R2
R1Vo
f (Hz)
20
0
100 102 104 106
|H|(
dB)
H(s) = −R2 ‖ (1/sC2)
R1 + (1/sC1)= −
R2
R1
sR1C1
(1 + sR1C1)(1 + sR2C2).
This is a band-pass filter, with ωL = 1/R1C1 and ωH = 1/R2C2 .
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
100 k
10 k80 p
0.8µF
Vs
RL
C2
C1
R2
R1Vo
f (Hz)
20
0
100 102 104 106
|H|(
dB)
H(s) = −R2 ‖ (1/sC2)
R1 + (1/sC1)= −
R2
R1
sR1C1
(1 + sR1C1)(1 + sR2C2).
This is a band-pass filter, with ωL = 1/R1C1 and ωH = 1/R2C2 .
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
100 k
10 k80 p
0.8µF
Vs
RL
C2
C1
R2
R1Vo
f (Hz)
20
0
100 102 104 106
|H|(
dB)
H(s) = −R2 ‖ (1/sC2)
R1 + (1/sC1)= −
R2
R1
sR1C1
(1 + sR1C1)(1 + sR2C2).
This is a band-pass filter, with ωL = 1/R1C1 and ωH = 1/R2C2 .
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
100 k
10 k80 p
0.8µF
Vs
RL
C2
C1
R2
R1Vo
f (Hz)
20
0
100 102 104 106
|H|(
dB)
H(s) = −R2 ‖ (1/sC2)
R1 + (1/sC1)= −
R2
R1
sR1C1
(1 + sR1C1)(1 + sR2C2).
This is a band-pass filter, with ωL = 1/R1C1 and ωH = 1/R2C2 .
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Op Amp filters: example
100 k
10 k80 p
0.8µF
Vs
RL
C2
C1
R2
R1Vo
f (Hz)
20
0
100 102 104 106
|H|(
dB)
H(s) = −R2 ‖ (1/sC2)
R1 + (1/sC1)= −
R2
R1
sR1C1
(1 + sR1C1)(1 + sR2C2).
This is a band-pass filter, with ωL = 1/R1C1 and ωH = 1/R2C2 .
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Graphic equalizer
f (Hz)
20
0
−20
R2
C1
C2
a 1−a
R3A R3B
R1A R1B
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
a=0.90.7
0.5
0.3
0.1
Vs
RL
Vo
105104103102101
|H|(
dB)
R3A = R3B = 100 kΩ
R1A = R1B = 470Ω
R2 = 10 kΩ
C1 = 100 nF
C2 = 10 nF
* Equalizers are implemented as arrays of narrow-band filters, each with anadjustable gain (attenuation) around a centre frequency.
* The circuit shown above represents one of the equalizer sections.
(SEQUEL file: ee101 op filter 4.sqproj)
M. B. Patil, IIT Bombay
Graphic equalizer
f (Hz)
20
0
−20
R2
C1
C2
a 1−a
R3A R3B
R1A R1B
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
a=0.90.7
0.5
0.3
0.1
Vs
RL
Vo
105104103102101
|H|(
dB)
R3A = R3B = 100 kΩ
R1A = R1B = 470Ω
R2 = 10 kΩ
C1 = 100 nF
C2 = 10 nF
* Equalizers are implemented as arrays of narrow-band filters, each with anadjustable gain (attenuation) around a centre frequency.
* The circuit shown above represents one of the equalizer sections.
(SEQUEL file: ee101 op filter 4.sqproj)
M. B. Patil, IIT Bombay
Graphic equalizer
f (Hz)
20
0
−20
R2
C1
C2
a 1−a
R3A R3B
R1A R1B
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
a=0.90.7
0.5
0.3
0.1
Vs
RL
Vo
105104103102101
|H|(
dB)
R3A = R3B = 100 kΩ
R1A = R1B = 470Ω
R2 = 10 kΩ
C1 = 100 nF
C2 = 10 nF
* Equalizers are implemented as arrays of narrow-band filters, each with anadjustable gain (attenuation) around a centre frequency.
* The circuit shown above represents one of the equalizer sections.
(SEQUEL file: ee101 op filter 4.sqproj)
M. B. Patil, IIT Bombay
Sallen-Key filter example (2nd order, low-pass)
f (Hz)
40
20
0
−20
−40
−60
RARB
R1 R2C1
C2
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
Vo
Vs V1
105104103102101
|H|(
dB)
R1 = R2 = 15.8 kΩ
C1 = C2 = 10 nF
RA = 10 kΩ, RB = 17.8 kΩ
V+ = V− = VoRA
RA + RB
≡ Vo/K .
Also, V+ =(1/sC2)
R2 + (1/sC2)V1 =
1
1 + sR2C2V1 .
KCL at V1 →1
R1(Vs − V1) + sC1(Vo − V1) +
1
R2(V+ − V1) = 0 .
Combining the above equations, H(s) =K
1 + s [(R1 + R2)C2 + (1− K)R1C1] + s2R1C1R2C2.
(SEQUEL file: ee101 op filter 5.sqproj)
M. B. Patil, IIT Bombay
Sallen-Key filter example (2nd order, low-pass)
f (Hz)
40
20
0
−20
−40
−60
RARB
R1 R2C1
C2
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
Vo
Vs V1
105104103102101
|H|(
dB)
R1 = R2 = 15.8 kΩ
C1 = C2 = 10 nF
RA = 10 kΩ, RB = 17.8 kΩ
V+ = V− = VoRA
RA + RB
≡ Vo/K .
Also, V+ =(1/sC2)
R2 + (1/sC2)V1 =
1
1 + sR2C2V1 .
KCL at V1 →1
R1(Vs − V1) + sC1(Vo − V1) +
1
R2(V+ − V1) = 0 .
Combining the above equations, H(s) =K
1 + s [(R1 + R2)C2 + (1− K)R1C1] + s2R1C1R2C2.
(SEQUEL file: ee101 op filter 5.sqproj)
M. B. Patil, IIT Bombay
Sallen-Key filter example (2nd order, low-pass)
f (Hz)
40
20
0
−20
−40
−60
RARB
R1 R2C1
C2
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
Vo
Vs V1
105104103102101
|H|(
dB)
R1 = R2 = 15.8 kΩ
C1 = C2 = 10 nF
RA = 10 kΩ, RB = 17.8 kΩ
V+ = V− = VoRA
RA + RB
≡ Vo/K .
Also, V+ =(1/sC2)
R2 + (1/sC2)V1 =
1
1 + sR2C2V1 .
KCL at V1 →1
R1(Vs − V1) + sC1(Vo − V1) +
1
R2(V+ − V1) = 0 .
Combining the above equations, H(s) =K
1 + s [(R1 + R2)C2 + (1− K)R1C1] + s2R1C1R2C2.
(SEQUEL file: ee101 op filter 5.sqproj)
M. B. Patil, IIT Bombay
Sallen-Key filter example (2nd order, low-pass)
f (Hz)
40
20
0
−20
−40
−60
RARB
R1 R2C1
C2
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
Vo
Vs V1
105104103102101
|H|(
dB)
R1 = R2 = 15.8 kΩ
C1 = C2 = 10 nF
RA = 10 kΩ, RB = 17.8 kΩ
V+ = V− = VoRA
RA + RB
≡ Vo/K .
Also, V+ =(1/sC2)
R2 + (1/sC2)V1 =
1
1 + sR2C2V1 .
KCL at V1 →1
R1(Vs − V1) + sC1(Vo − V1) +
1
R2(V+ − V1) = 0 .
Combining the above equations, H(s) =K
1 + s [(R1 + R2)C2 + (1− K)R1C1] + s2R1C1R2C2.
(SEQUEL file: ee101 op filter 5.sqproj)
M. B. Patil, IIT Bombay
Sallen-Key filter example (2nd order, low-pass)
f (Hz)
40
20
0
−20
−40
−60
RARB
R1 R2C1
C2
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
Vo
Vs V1
105104103102101
|H|(
dB)
R1 = R2 = 15.8 kΩ
C1 = C2 = 10 nF
RA = 10 kΩ, RB = 17.8 kΩ
V+ = V− = VoRA
RA + RB
≡ Vo/K .
Also, V+ =(1/sC2)
R2 + (1/sC2)V1 =
1
1 + sR2C2V1 .
KCL at V1 →1
R1(Vs − V1) + sC1(Vo − V1) +
1
R2(V+ − V1) = 0 .
Combining the above equations, H(s) =K
1 + s [(R1 + R2)C2 + (1− K)R1C1] + s2R1C1R2C2.
(SEQUEL file: ee101 op filter 5.sqproj)
M. B. Patil, IIT Bombay
Sixth-order Chebyshev low-pass filter (cascade design)
10.7 k 10.2 k
5.1 n
2.2 n
8.25 k 6.49 k
10 n
510 p
4.64 k 2.49 k
62 n
220 p
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
20
0
−20
−40
−60
−80
f (Hz)
SEQUEL file: ee101_op_filter_6.sqproj
VsRL
Vo
105104103102
|H|(
dB)
M. B. Patil, IIT Bombay
Third-order Chebyshev high-pass filter
f (Hz)
20
0
−20
−40
SEQUEL file: ee101_op_filter_7.sqproj
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
−60
−80
7.68 k
54.9 k
15.4 k 154 k
100 n 100 n
100 n
VsRL
Vo
103102101100
|H|(
dB)
M. B. Patil, IIT Bombay
Band-pass filter example
f (Hz)
SEQUEL file: ee101_op_filter_8.sqproj
20
0
−20
−40
40
5 k
5 k
5 k
5 k
5 k
370 k5 k
7.4 n
7.4 n
(Ref.: J. M. Fiore, "Op Amps and linear ICs")
Vs
Vo
105104103102
|H|(
dB)
M. B. Patil, IIT Bombay
Notch filter example
1 k
89 k
10 k
10 k10 k
10 k
10 k10 k
10 k
10 k
265 n
265 n
(Ref.: J. M. Fiore, "Op Amps and linear ICs")
f (Hz)
0
−40
−20
SEQUEL file: ee101_op_filter_9.sqproj
Vs
Vo
101 102
|H|(
dB)
M. B. Patil, IIT Bombay