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 20 0 5 10 15 20
t (msec)
t (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-frequency component v1(t) andremove the high-frequency component v2(t).
A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequency component v2(t) andremove 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 20 0 5 10 15 20
t (msec)
t (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-frequency component v1(t) andremove the high-frequency component v2(t).
A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequency component v2(t) andremove 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 20 0 5 10 15 20
t (msec)
t (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-frequency component v1(t) andremove the high-frequency component v2(t).
A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequency component v2(t) andremove 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 20 0 5 10 15 20
t (msec)
t (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-frequency component v1(t) andremove the high-frequency component v2(t).
A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequency component v2(t) andremove 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
ω
vo(t)
H(jω)
H(jω)vi(t)
0
LPF
0ωc ωc
ω ω
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
ω
vo(t)
H(jω)
H(jω)vi(t)
0
LPF
0ωc ωc
ω ω
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
ω
vo(t)
H(jω)
H(jω)vi(t)
0
LPF
0ωc ωc
ω ω
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
0
ωc
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
0
ωc
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
0
ωc
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
0
ωc
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 low-pass filter: example
1
0
−1
1.5
0
−1.5
0
1
Filter transfer function Filter output
1
0
−1
0 5 10 15 20
t (msec)
f (kHz) t (msec)
0 5 10 15 20 0 0.5 1 1.5 2 2.5
v
v3
v2
v1
H(jω)
M. B. Patil, IIT Bombay
Ideal high-pass filter: example
1
0
−1
1.5
0
−1.5
Filter transfer function Filter output
1
0
−10
1
0 5 10 15 20
t (msec)
f (kHz) t (msec)
0 5 10 15 20 0 0.5 1 1.5 2 2.5
v
v3
v2
v1
H(jω)
M. B. Patil, IIT Bombay
Ideal band-pass filter: example
1
0
−1
1
0
−1
1.5
0
−1.5
Filter transfer function Filter output
0
1
0 5 10 15 20
t (msec)
f (kHz) t (msec)
0 5 10 15 20 0 0.5 1 1.5 2 2.5
v
v3
v2
v1
H(jω)
M. B. Patil, IIT Bombay
Ideal band-reject filter: example
1.5
0
−1.50
1
1
0
−1
1.5
0
−1.5
Filter transfer function Filter output
0 5 10 15 20
t (msec)
f (kHz) t (msec)
0 5 10 15 20 0 0.5 1 1.5 2 2.5
v
H(jω)
v3
v2
v1
M. B. Patil, IIT Bombay
Practical filter circuits
* In practical filter circuits, the ideal filter response is approximated with a suitable H(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 are listed in filter handbooks. Also, programs for filter designare available on the internet.
M. B. Patil, IIT Bombay
Practical filter circuits
* In practical filter circuits, the ideal filter response is approximated with a suitable H(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 are listed in filter handbooks. Also, programs for filter designare available on the internet.
M. B. Patil, IIT Bombay
Practical filter circuits
* In practical filter circuits, the ideal filter response is approximated with a suitable H(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 are listed in filter handbooks. Also, programs for filter designare available on the internet.
M. B. Patil, IIT Bombay
Practical filters
Low−pass High−pass
0
Practical
1
00 0
00
1
00
IdealPracticalIdeal
Amax|H|
ωc
ω
ωs ωc
ω
Amin
|H|
ω
ωsωc
ω
|H| Amax
ωc
|H|
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, 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 the filter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
Low−pass High−pass
0
Practical
1
00 0
00
1
00
IdealPracticalIdeal
Amax|H|
ωc
ω
ωs ωc
ω
Amin
|H|
ω
ωsωc
ω
|H| Amax
ωc
|H|
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, 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 the filter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
Low−pass High−pass
0
Practical
1
00 0
00
1
00
IdealPracticalIdeal
Amax|H|
ωc
ω
ωs ωc
ω
Amin
|H|
ω
ωsωc
ω
|H| Amax
ωc
|H|
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, 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 the filter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
Low−pass High−pass
0
Practical
1
00 0
00
1
00
IdealPracticalIdeal
Amax|H|
ωc
ω
ωs ωc
ω
Amin
|H|
ω
ωsωc
ω
|H| Amax
ωc
|H|
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, 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 the filter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
Low−pass High−pass
0
Practical
1
00 0
00
1
00
IdealPracticalIdeal
Amax|H|
ωc
ω
ωs ωc
ω
Amin
|H|
ω
ωsωc
ω
|H| Amax
ωc
|H|
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, 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 the filter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
Low−pass High−pass
0
Practical
1
00 0
00
1
00
IdealPracticalIdeal
Amax|H|
ωc
ω
ωs ωc
ω
Amin
|H|
ω
ωsωc
ω
|H| Amax
ωc
|H|
Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, 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 the filter.
* ωc < ω < ωs : transition band.
M. B. Patil, IIT Bombay
Practical filters
For a low-pass filter, H(s) =1
n∑i=0
ai (s/ωc )i.
Coefficients (ai ) for various types of filters are tabulated in handbooks. We now look at |H(jω)| for twocommonly used filters.
Butterworth filters:
|H(jω)| =1√
1 + ε2(ω/ωc )2n.
Chebyshev filters:
|H(jω)| =1√
1 + ε2C2n (ω/ωc )
where
Cn(x) = cos[n cos−1(x)
]for x ≤ 1,
Cn(x) = cosh[n cosh−1(x)
]for x ≥ 1,
H(s) for a high-pass filter can be obtained from H(s) of the corresponding low-pass filter by (s/ωc )→ (ωc/s) .
M. B. Patil, IIT Bombay
Practical filters
For a low-pass filter, H(s) =1
n∑i=0
ai (s/ωc )i.
Coefficients (ai ) for various types of filters are tabulated in handbooks. We now look at |H(jω)| for twocommonly used filters.
Butterworth filters:
|H(jω)| =1√
1 + ε2(ω/ωc )2n.
Chebyshev filters:
|H(jω)| =1√
1 + ε2C2n (ω/ωc )
where
Cn(x) = cos[n cos−1(x)
]for x ≤ 1,
Cn(x) = cosh[n cosh−1(x)
]for x ≥ 1,
H(s) for a high-pass filter can be obtained from H(s) of the corresponding low-pass filter by (s/ωc )→ (ωc/s) .
M. B. Patil, IIT Bombay
Practical filters
For a low-pass filter, H(s) =1
n∑i=0
ai (s/ωc )i.
Coefficients (ai ) for various types of filters are tabulated in handbooks. We now look at |H(jω)| for twocommonly used filters.
Butterworth filters:
|H(jω)| =1√
1 + ε2(ω/ωc )2n.
Chebyshev filters:
|H(jω)| =1√
1 + ε2C2n (ω/ωc )
where
Cn(x) = cos[n cos−1(x)
]for x ≤ 1,
Cn(x) = cosh[n cosh−1(x)
]for x ≥ 1,
H(s) for a high-pass filter can be obtained from H(s) of the corresponding low-pass filter by (s/ωc )→ (ωc/s) .
M. B. Patil, IIT Bombay
Practical filters
For a low-pass filter, H(s) =1
n∑i=0
ai (s/ωc )i.
Coefficients (ai ) for various types of filters are tabulated in handbooks. We now look at |H(jω)| for twocommonly used filters.
Butterworth filters:
|H(jω)| =1√
1 + ε2(ω/ωc )2n.
Chebyshev filters:
|H(jω)| =1√
1 + ε2C2n (ω/ωc )
where
Cn(x) = cos[n cos−1(x)
]for x ≤ 1,
Cn(x) = cosh[n cosh−1(x)
]for x ≥ 1,
H(s) for a high-pass filter can be obtained from H(s) of the corresponding low-pass filter by (s/ωc )→ (ωc/s) .
M. B. Patil, IIT Bombay
Practical filters (low-pass)
Butterworth filters:
Chebyshev filters:
0
1
−100
0
0
1
−100
0
3
4
5
2
n=1
n=1
4
3
2
5
n=1
2
34
5
2
3
n=1
4
5
0 2 3 4 5 1 0.01 0.1 1 10 100
0 2 3 4 5 1 0.01 0.1 1 10 100
|H|(dB
)
ω/ωc
|H|
ω/ωc
ω/ωc
|H|
ω/ωc|H|(dB
)
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
3
4
5
n=1
2
3
4
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
5µFC
100ΩVs Vo
R
(Low−pass filter)
with ω0 = 1/RC → f0 = ω0/2π = 318Hz
H(s) =(1/sC)
R+ (1/sC)=
1
1+ (s/ω0),
f (Hz)
−60
−40
−20
0
20
105104103102101
|H|(dB
)
(SEQUEL file: ee101 rc ac 2.sqproj)
M. B. Patil, IIT Bombay
Passive filter example
5µFC
100ΩVs Vo
R
(Low−pass filter)
with ω0 = 1/RC → f0 = ω0/2π = 318Hz
H(s) =(1/sC)
R+ (1/sC)=
1
1+ (s/ω0),
f (Hz)
−60
−40
−20
0
20
105104103102101
|H|(dB
)
(SEQUEL file: ee101 rc ac 2.sqproj)
M. B. Patil, IIT Bombay
Passive filter example
5µFC
100ΩVs Vo
R
(Low−pass filter)
with ω0 = 1/RC → f0 = ω0/2π = 318Hz
H(s) =(1/sC)
R+ (1/sC)=
1
1+ (s/ω0),
f (Hz)
−60
−40
−20
0
20
105104103102101
|H|(dB
)
(SEQUEL file: ee101 rc ac 2.sqproj)
M. B. Patil, IIT Bombay
Passive filter example
Vs VoR
100Ω
C4µF0.1mF
L
(Band−pass filter)
with ω0 = 1/√LC → f0 = ω0/2π = 7.96 kHz
H(s) =(sL) ‖ (1/sC)
R+ (sL) ‖ (1/sC)=
s(L/R)
1+ s(L/R)+ s2LC
−20
−40
−60
−80
0
f (Hz)
105104103102
|H|(dB
)
(SEQUEL file: ee101 rlc 3.sqproj)
M. B. Patil, IIT Bombay
Passive filter example
Vs VoR
100Ω
C4µF0.1mF
L
(Band−pass filter)
with ω0 = 1/√LC → f0 = ω0/2π = 7.96 kHz
H(s) =(sL) ‖ (1/sC)
R+ (sL) ‖ (1/sC)=
s(L/R)
1+ s(L/R)+ s2LC
−20
−40
−60
−80
0
f (Hz)
105104103102
|H|(dB
)
(SEQUEL file: ee101 rlc 3.sqproj)
M. B. Patil, IIT Bombay
Passive filter example
Vs VoR
100Ω
C4µF0.1mF
L
(Band−pass filter)
with ω0 = 1/√LC → f0 = ω0/2π = 7.96 kHz
H(s) =(sL) ‖ (1/sC)
R+ (sL) ‖ (1/sC)=
s(L/R)
1+ s(L/R)+ s2LC
−20
−40
−60
−80
0
f (Hz)
105104103102
|H|(dB
)
(SEQUEL file: ee101 rlc 3.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters (“Active” filters)
* Op-amp filters can be designed without using inductors. This is a significant advantage since inductors arebulky and expensive. Inductors also exhibit nonlinear behaviour (arising from the core properties) which isundesirable in a filter 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, there are situations in which passive filters are still used.
- high frequencies at which op-amps do not have sufficient gain
- high power which op-amps cannot handle
M. B. Patil, IIT Bombay
Op-amp filters (“Active” filters)
* Op-amp filters can be designed without using inductors. This is a significant advantage since inductors arebulky and expensive. Inductors also exhibit nonlinear behaviour (arising from the core properties) which isundesirable in a filter 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, there are situations in which passive filters are still used.
- high frequencies at which op-amps do not have sufficient gain
- high power which op-amps cannot handle
M. B. Patil, IIT Bombay
Op-amp filters (“Active” filters)
* Op-amp filters can be designed without using inductors. This is a significant advantage since inductors arebulky and expensive. Inductors also exhibit nonlinear behaviour (arising from the core properties) which isundesirable in a filter 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, there are situations in which passive filters are still used.
- high frequencies at which op-amps do not have sufficient gain
- high power which op-amps cannot handle
M. B. Patil, IIT Bombay
Op-amp filters (“Active” filters)
* Op-amp filters can be designed without using inductors. This is a significant advantage since inductors arebulky and expensive. Inductors also exhibit nonlinear behaviour (arising from the core properties) which isundesirable in a filter 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, there are situations in which passive filters are still used.
- high frequencies at which op-amps do not have sufficient gain
- high power which op-amps cannot handle
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo
10 k
10 nF
1 k
f (Hz)
20
0
−20
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 1.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo
10 k
10 nF
1 k
f (Hz)
20
0
−20
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 1.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo
10 k
10 nF
1 k
f (Hz)
20
0
−20
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 1.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo
10 k
10 nF
1 k
f (Hz)
20
0
−20
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 1.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo
10 k
10 nF
1 k
f (Hz)
20
0
−20
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 1.sqproj)M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo1 k 100 nF
10 k
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo1 k 100 nF
10 k
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo1 k 100 nF
10 k
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo1 k 100 nF
10 k
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C
R2
R1
Vo1 k 100 nF
10 k
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 (i.e., f0 = ω0/2π = 1.59 kHz).
(SEQUEL file: ee101 op filter 2.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C1
R2
R1
Vo
100 k
10 k 0.8µF
C2
80 pF
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 .
→ fL = 20 Hz, fH = 20 kHz.
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C1
R2
R1
Vo
100 k
10 k 0.8µF
C2
80 pF
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 .
→ fL = 20 Hz, fH = 20 kHz.
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C1
R2
R1
Vo
100 k
10 k 0.8µF
C2
80 pF
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 .
→ fL = 20 Hz, fH = 20 kHz.
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C1
R2
R1
Vo
100 k
10 k 0.8µF
C2
80 pF
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 .
→ fL = 20 Hz, fH = 20 kHz.
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Op-amp filters: example
Vs
RL
C1
R2
R1
Vo
100 k
10 k 0.8µF
C2
80 pF
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 .
→ fL = 20 Hz, fH = 20 kHz.
(SEQUEL file: ee101 op filter 3.sqproj)
M. B. Patil, IIT Bombay
Graphic equalizer
f (Hz)
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
20
0
−20
R2
C1
a 1−a
R3A R3B
R1A R1B
C2 a=0.90.7
0.5
0.3
0.1
Vs
RL
Vo
105104103102101
R3A = R3B = 100 kΩ
R1A = R1B = 470Ω
R2 = 10 kΩ
C1 = 100 nF
C2 = 10 nF
|H|(dB
)
* Equalizers are implemented as arrays of narrow-band filters, each with an adjustable 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)
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
20
0
−20
R2
C1
a 1−a
R3A R3B
R1A R1B
C2 a=0.90.7
0.5
0.3
0.1
Vs
RL
Vo
105104103102101
R3A = R3B = 100 kΩ
R1A = R1B = 470Ω
R2 = 10 kΩ
C1 = 100 nF
C2 = 10 nF
|H|(dB
)
* Equalizers are implemented as arrays of narrow-band filters, each with an adjustable 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)
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
20
0
−20
R2
C1
a 1−a
R3A R3B
R1A R1B
C2 a=0.90.7
0.5
0.3
0.1
Vs
RL
Vo
105104103102101
R3A = R3B = 100 kΩ
R1A = R1B = 470Ω
R2 = 10 kΩ
C1 = 100 nF
C2 = 10 nF
|H|(dB
)
* Equalizers are implemented as arrays of narrow-band filters, each with an adjustable 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
C1
RB
R1 R2
C2
RA
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
V1
Vs
Vo
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
C1
RB
R1 R2
C2
RA
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
V1
Vs
Vo
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
C1
RB
R1 R2
C2
RA
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
V1
Vs
Vo
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
C1
RB
R1 R2
C2
RA
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
V1
Vs
Vo
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
C1
RB
R1 R2
C2
RA
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL
V1
Vs
Vo
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)
20
0
−20
−40
−60
−80
f (Hz)
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
SEQUEL file: ee101_op_filter_6.sqproj
5.1 n 10 n 62 n
2.2 n 510 p 220 p
2.49 k4.64 k6.49 k8.25 k10.2 k10.7 k
Vs
RL
Vo
105104103102
|H|(dB
)
M. B. Patil, IIT Bombay
Third-order Chebyshev high-pass filter
7.68 k
100 n
(Ref.: S. Franco, "Design with Op Amps and analog ICs")
SEQUEL file: ee101_op_filter_7.sqproj
100 n54.9 k
100 n 15.4 k 154 k
f (Hz)
20
0
−20
−40
−60
−80
Vs
RL
Vo
103102101100
|H|(dB
)M. B. Patil, IIT Bombay
Band-pass filter example
f (Hz)
20
0
−20
−40
40
(Ref.: J. M. Fiore, "Op Amps and linear ICs")
SEQUEL file: ee101_op_filter_8.sqproj
5 k
5 k
5 k
5 k
370 k
5 k
5 k
7.4 n
7.4 n
Vs
Vo
105104103102
|H|(dB
)
M. B. Patil, IIT Bombay
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