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Section 7 Active FIlters.pdf Electrical fundamentals

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madventre16

This document discusses active filters, which use operational amplifiers to build filters with advantages over passive filters such as adjustable Q-factors and gain. It introduces the fundamental building blocks of op-amp integrators and differentiators, analyzing them in the time, Laplace, and frequency domains. Practical implementations are discussed to limit problems with infinite gain. First-order active low-pass and high-pass filters are presented using these building blocks. Higher-order filters can be made by cascading first-order stages or using second-order filter topologies like the Sallen-Key filter.

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ENGR 203 – Electrical Fundamentals III SECTION 7: ACTIVE FILTERS
K. Webb ENGR 203 2 Introduction  In ENGR 202 we studied different types of first- and second-order passive filters  Passive, because they contain only passive components:  Resistors, capacitors, and inductors  Can also construct filters using opamps  Active filters
K. Webb ENGR 203 3 Introduction  Active filters have advantages over passive filters:  Can build high-Q filters without inductors  Low output impedance  Easily adjustable: 𝑓𝑓𝑐𝑐, 𝑄𝑄  Can provide gain ( > 0 dB )  Before getting into the design of active filters, we will look at two fundamental filter building blocks:  Opamp integrators  Opamp differentiators
K. Webb ENGR 203 Opamp Integrators 4
K. Webb ENGR 203 5 Integrators and Differentiators  Opamp circuits can perform many different mathematical operations  Operational amplifiers  Multiplication  Inverting and non-inverting amplifiers  Addition and subtraction  Summing and difference amplifiers  Can also perform integration and differentiation  Feedback controllers  Building block of active filters
K. Webb ENGR 203 6 Opamp Integrator – Time Domain 𝑖𝑖 𝑡𝑡 = 𝑣𝑣𝑖𝑖 𝑡𝑡 𝑅𝑅  Capacitor integrates input current to give output voltage 𝑣𝑣𝑜𝑜 𝑡𝑡 = − 1 𝐶𝐶 � 0 𝑡𝑡 𝑖𝑖 𝜏𝜏 𝑑𝑑𝑑𝑑 𝑣𝑣𝑜𝑜 𝑡𝑡 = − 1 𝑅𝑅𝑅𝑅 � 0 𝑡𝑡 𝑣𝑣𝑖𝑖 𝜏𝜏 𝑑𝑑𝑑𝑑  Output is the (scaled and inverted) integral of the input  Analyze the opamp integrator in the time domain  Virtual ground at inverting input, so
K. Webb ENGR 203 7 Opamp Integrator – Laplace Domain 𝐼𝐼 𝑠𝑠 = 𝑉𝑉𝑖𝑖 𝑠𝑠 𝑅𝑅  Output voltage: 𝑉𝑉 𝑜𝑜 𝑠𝑠 = −𝐼𝐼 𝑠𝑠 1 𝐶𝐶𝐶𝐶 = − 𝑉𝑉𝑖𝑖 𝑠𝑠 𝑅𝑅𝑅𝑅𝑅𝑅 = − 1 𝑠𝑠 ⋅ 𝑉𝑉𝑖𝑖 𝑠𝑠 𝑅𝑅𝑅𝑅  Recall that multiplication by 1/𝑠𝑠 in the Laplace domain corresponds to integration in the time domain  Analyze the opamp integrator in the Laplace domain  Again, a virtual ground at inverting input, so
K. Webb ENGR 203 8 Opamp Integrator – Frequency Response  Transfer function: 𝐺𝐺 𝑠𝑠 = − 1 𝑅𝑅𝑅𝑅𝑅𝑅  Single pole at 𝑠𝑠 = 0  Gain: constant slope of -20 dB/dec  Infinite DC gain  Phase: -90° from integrator pole + 180° from inversion yields constant +90°
K. Webb ENGR 203 9 Ideal Integrator - Problem  Laplace domain step response of the ideal integrator 𝑉𝑉 𝑜𝑜 𝑠𝑠 = 1 𝑠𝑠 ⋅ 𝐺𝐺 𝑠𝑠 = − 1 𝑅𝑅𝑅𝑅 ⋅ 1 𝑠𝑠2  Inverse transforming to the time domain 𝑣𝑣𝑜𝑜 𝑡𝑡 = − 1 𝑅𝑅𝑅𝑅 ⋅ 𝑡𝑡  Output increases linearly with time  Opamp will quickly saturate in response any DC input component  Infinite DC gain  Not a practical circuit  Inputs will always have some non-zero offset  Real (non-ideal) opamps have non-zero offset voltages and input bias currents
K. Webb ENGR 203 10 Practical Opamp Integrator  Now there is a feedback path for DC signals  DC gain limited to 𝑅𝑅𝑓𝑓/𝑅𝑅  Behaves as an inverting opamp at low frequencies  Still behaves as an integrator at high frequencies  A practical or lossy integrator circuit  Problem with ideal integrator is infinite DC gain  No DC feedback  Open-loop at DC  Add a feedback resistor in parallel with the capacitor
K. Webb ENGR 203 11 Opamp Integrator – Frequency Response  Transfer function: 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝐶𝐶𝐶𝐶 𝑅𝑅𝑓𝑓 + 1 𝐶𝐶𝐶𝐶 𝑅𝑅 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 1 𝑅𝑅𝑓𝑓𝐶𝐶𝐶𝐶 + 1  Pole (corner frequency) set by the feedback network: 𝜔𝜔𝑐𝑐 = 1 𝑅𝑅𝑓𝑓𝐶𝐶  For 𝜔𝜔 ≫ 𝜔𝜔𝑐𝑐, still behaves like an integrator  Gain: rolls off at -20 dB/dec  Phase: ~90° Integrator Amplifier
K. Webb ENGR 203 Opamp Differentiators 12
K. Webb ENGR 203 13 Opamp Differentiator – Time Domain 𝑖𝑖 𝑡𝑡 = 𝐶𝐶 𝑑𝑑𝑣𝑣𝑖𝑖 𝑑𝑑𝑑𝑑  Ohm’s law gives the output voltage 𝑣𝑣𝑜𝑜 𝑡𝑡 = −𝑅𝑅𝑅𝑅 𝑡𝑡 = −𝑅𝑅𝑅𝑅 𝑑𝑑𝑣𝑣𝑖𝑖 𝑑𝑑𝑑𝑑  Output is the (scaled and inverted) derivative of the input  Analyze the opamp differentiator in the time domain  Virtual ground at inverting input, so
K. Webb ENGR 203 14 Opamp Differentiator – Laplace Domain 𝐼𝐼 𝑠𝑠 = 𝐶𝐶𝐶𝐶 ⋅ 𝑉𝑉𝑖𝑖 𝑠𝑠  Output voltage: 𝑉𝑉 𝑜𝑜 𝑠𝑠 = −𝑅𝑅𝑅𝑅 𝑠𝑠 = −𝑅𝑅𝑅𝑅𝑅𝑅𝑉𝑉𝑖𝑖 𝑠𝑠 = −𝑠𝑠 ⋅ 𝑅𝑅𝑅𝑅𝑉𝑉𝑖𝑖 𝑠𝑠  Recall that multiplication by 𝑠𝑠 in the Laplace domain corresponds to differentiation in the time domain  Analyze the differentiator in the Laplace domain  Again, a virtual ground at inverting input, so
K. Webb ENGR 203 15 Opamp Differentiator – Frequency Response  Transfer function: 𝐺𝐺 𝑠𝑠 = −𝑅𝑅𝑅𝑅𝑅𝑅  Single zero at 𝑠𝑠 = 0  Gain: constant slope of +20 dB/dec  Very large high-frequency gain  Phase: +90° from zero at the origin + 180° from inversion yields constant +270° = -90°
K. Webb ENGR 203 16 Ideal Differentiator - Problem  Gain continues to increase with frequency  High-frequency gain is very large  Any input signal will include some noise  Better to limit the gain above some upper frequency
K. Webb ENGR 203 17 Practical Opamp Differentiator  High-frequency gain limited to 𝑅𝑅𝑓𝑓/𝑅𝑅  Still behaves as a differentiator at low frequencies  Behaves as an inverting opamp at high frequencies  A practical or lossy differentiator circuit  Problem with ideal differentiator:  Low input impedance at high frequency  Excessive high-frequency input current  Add a resistor in series with the input capacitor
K. Webb ENGR 203 18 Practical Opamp Differentiator  Transfer function: 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 + 1 𝐶𝐶𝐶𝐶 = − 𝑅𝑅𝑓𝑓𝐶𝐶𝐶𝐶 𝑅𝑅𝑅𝑅𝑅𝑅 + 1  Pole (corner frequency) set by the input network: 𝜔𝜔𝑐𝑐 = 1 𝑅𝑅𝑅𝑅  For 𝜔𝜔 ≪ 𝜔𝜔𝑐𝑐, still behaves like a differentiator  Gain: increases at +20 dB/dec  Phase: ~-90° Differentiator Amplifier
K. Webb ENGR 203 First-Order Opamp Active Filters 19
K. Webb ENGR 203 20 First-Order Active Filters  Practical integrator and differentiator circuits  Additional resistors fix problems with ideal circuits  First-order low pass and high pass filters
K. Webb ENGR 203 21 First-Order Low Pass Filter  Transfer function 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 1 𝑅𝑅𝑓𝑓𝐶𝐶𝐶𝐶 + 1 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 1 𝑅𝑅𝑓𝑓𝐶𝐶 𝑠𝑠 + 1 𝑅𝑅𝑓𝑓𝐶𝐶  Corner frequency 𝑓𝑓𝑐𝑐 = 1 2𝜋𝜋𝑅𝑅𝑓𝑓𝐶𝐶  Pass-band gain 𝐴𝐴𝑣𝑣 = − 𝑅𝑅𝑓𝑓 𝑅𝑅
K. Webb ENGR 203 22 First-Order High Pass Filter  Transfer function 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓𝐶𝐶𝐶𝐶 𝑅𝑅𝑅𝑅𝑅𝑅 + 1 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 𝑠𝑠 𝑠𝑠 + 1 𝑅𝑅𝐶𝐶  Corner frequency 𝑓𝑓𝑐𝑐 = 1 2𝜋𝜋𝜋𝜋𝜋𝜋  Pass-band gain 𝐴𝐴𝑣𝑣 = − 𝑅𝑅𝑓𝑓 𝑅𝑅
K. Webb ENGR 203 Higher Order Active Filters 23
K. Webb ENGR 203 24 Higher-Order Active Filters  Higher order active filters can be constructed by:  Cascading first-order active filters  Using second-order active filter stages  Cascading second- and first-order stages  Create higher order band pass/stop filters similarly:  Cascade first-order high/low pass filters  Use and/or cascade second-order band pass/stop stages  Many different second-order active filter topologies  We’ll look at the Sallen-Key circuit
K. Webb ENGR 203 25 Sallen-Key Filter – Generalized Form  Sallen-Key filter topology  Low pass and high pass filters  Band-pass, and notch filters with slight modifications  We’ll look first at the filter in its most generalized form, then consider the specific low pass and high pass filter forms  Type of filter depends on the location of components – resistors and capacitors
K. Webb ENGR 203 26 Sallen-Key Filter – Generalized Form  Transfer function  Nodal analysis  KCL at V+ and Vf  Virtual short at opamp inputs  After a lot of ugly algebra: 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽 𝑍𝑍1𝑍𝑍2 𝑍𝑍3𝑍𝑍4 + 𝛽𝛽 𝑍𝑍2 𝑍𝑍4 + 𝛽𝛽 𝑍𝑍1 𝑍𝑍4 + 𝛽𝛽 − 1 𝑍𝑍1 𝑍𝑍3 + 𝛽𝛽 where 𝛽𝛽 is the feedback path gain 𝛽𝛽 = 𝑅𝑅𝑓𝑓𝑓 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓𝑓
K. Webb ENGR 203 Sallen-Key Low Pass Filter 27
K. Webb ENGR 203 28 Sallen-Key Second-Order Low Pass Filter  Transfer function 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2𝑠𝑠2 + 𝛽𝛽𝑅𝑅2𝐶𝐶2𝑠𝑠 + 𝛽𝛽𝑅𝑅1𝐶𝐶2𝑠𝑠 + 𝛽𝛽 − 1 𝑅𝑅1𝐶𝐶1𝑠𝑠 + 𝛽𝛽 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2 𝑠𝑠2 + 1 𝑅𝑅1𝐶𝐶1 + 1 𝑅𝑅2𝐶𝐶1 + 𝛽𝛽 − 1 𝛽𝛽𝑅𝑅2𝐶𝐶2 𝑠𝑠 + 1 𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2  Z1 and Z2 are resistors  Z3 and Z4 are capacitors
K. Webb ENGR 203 29 Sallen-Key Low Pass Filter  Generalized second-order low pass transfer function: 𝐺𝐺 𝑠𝑠 = 𝐾𝐾 ⋅ 𝜔𝜔0 2 𝑠𝑠2 + 𝜔𝜔0 𝑄𝑄 𝑠𝑠 + 𝜔𝜔0 2  Equating coefficients with the Sallen-Key transfer function gives  Resonant frequency: 𝜔𝜔0 = 1 𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2  Quality factor: 𝑄𝑄 = 𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2 𝑅𝑅2𝐶𝐶2 + 𝑅𝑅1𝐶𝐶1 + 𝛽𝛽 − 1 𝛽𝛽 𝑅𝑅1𝐶𝐶1  DC gain: 𝐾𝐾 = 1 𝛽𝛽 = 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓𝑓 𝑅𝑅𝑓𝑓𝑓
K. Webb ENGR 203 30 Sallen-Key Low Pass Filter  𝜔𝜔0, Q, and gain all set by appropriate component selection, but  There are more degrees of freedom than we need  Transfer function is a bit more complicated than we’d like  Simplify by setting component values equal  Transfer function becomes 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽 𝑅𝑅𝑅𝑅 2 𝑠𝑠2 + 3 − 1 𝛽𝛽 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  Where now 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 and 𝑄𝑄 = 1 3− 1 𝛽𝛽
K. Webb ENGR 203 31 Sallen-Key Low Pass Filter 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽 𝑅𝑅𝑅𝑅 2 𝑠𝑠2 + 3 − 1 𝛽𝛽 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  We can also write the transfer function in terms of DC gain, 𝐾𝐾 𝐺𝐺 𝑠𝑠 = 𝐾𝐾 𝑅𝑅𝑅𝑅 2 𝑠𝑠2 + 3 − 𝐾𝐾 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  𝜔𝜔0 and 𝑄𝑄 in terms of 𝐾𝐾: 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 and 𝑄𝑄 = 1 3−𝐾𝐾  The filter’s DC gain is dependent on the filter’s 𝑸𝑸 and vice versa  For independent control of DC gain, cascade an additional gain stage
K. Webb ENGR 203 32 Sallen-Key Low Pass Filter  Note dependence of 𝑄𝑄 and 𝐾𝐾  Both set by feedback path gain  𝑄𝑄 and gain are independent of 𝜔𝜔0  𝜔𝜔0 set by capacitors and resistors at the input  Second-order  Gain roll-off: -40 dB/dec
K. Webb ENGR 203 Sallen-Key High Pass Filter 33
K. Webb ENGR 203 34 Sallen-Key High Pass Filter  Here, we will jump straight to the simplified circuit with equal-valued components  Location of resistors and capacitors swapped relative to low pass filter  High pass transfer function 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽 𝑠𝑠2 𝑠𝑠2 + 3 − 1 𝛽𝛽 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  Again, 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 and 𝑄𝑄 = 1 3− 1 𝛽𝛽
K. Webb ENGR 203 35 Sallen-Key High Pass Filter  As with the low pass filter, we can write the transfer function in terms of gain, 𝐾𝐾  𝐾𝐾 still represents passband gain, but now it is the high-frequency gain, not the DC gain 𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝑠𝑠2 𝑠𝑠2 + 3 − 𝐾𝐾 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  𝜔𝜔0 and 𝑄𝑄 are the same as for the low pass filter: 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 and 𝑄𝑄 = 1 3−𝐾𝐾  Same dependence between passband gain, resonant frequency, and Q
K. Webb ENGR 203 36 Sallen-Key High Pass Filter  Note dependence of 𝑄𝑄 and 𝐾𝐾  Both set by feedback path gain  𝑄𝑄 and gain are independent of 𝜔𝜔0  𝜔𝜔0 set by capacitors and resistors at the input  Second-order  Gain roll-off: -40 dB/dec
K. Webb ENGR 203 37 Sallen-Key Filter – Stability  Sallen-Key filter has two feedback paths:  Negative feedback  Generally stabilizing  Positive feedback  Generally destabilizing Positive feedback Negative feedback  Relative amount of negative and positive feedback determines stability  Net negative feedback: circuit is stable  Behaves as a linear filter/amplifier  Net positive feedback: circuit is unstable  Will oscillate or saturate
K. Webb ENGR 203 38 Sallen-Key Filter – Stability  Overall net feedback must remain negative  But, we can vary just how negative by varying 𝛽𝛽  Varying 𝛽𝛽 allows us to vary 𝑄𝑄: 𝑄𝑄 = 1 3 − 1 𝛽𝛽 = 1 3 − 𝐾𝐾 Positive feedback Negative feedback  As 𝛽𝛽 increases:  Negative feedback increases  Overall feedback becomes more negative  Quality factor, 𝑄𝑄, decreases  Damping ratio, 𝜁𝜁, increases  Pass band gain, 𝐾𝐾, decreases
K. Webb ENGR 203 39 Sallen-Key Filter – Stability Positive feedback Negative feedback  As 𝛽𝛽 decreases:  Negative feedback decreases  Overall feedback becomes less negative  Quality factor, 𝑄𝑄, increases  Damping ratio, 𝜁𝜁, decreases  Pass band gain, 𝐾𝐾, increases  There is an upper limit on 𝐾𝐾:  For 𝐾𝐾 = 3, 𝑄𝑄 = ∞ and 𝜁𝜁 = 0  An un-damped circuit  Negative and positive feedback cancel  The border between stability and instability  For stability: 𝐾𝐾 ≤ 3
K. Webb ENGR 203 Filter Families 40
K. Webb ENGR 203 41 Filter Families  Higher-order filters of all types can be designed with transfer functions that fit into one of several families of filters  Butterworth  Chebyshev  Elliptic  Bessel  Each filter family defined by the nature of its characteristic polynomial  Equivalently, each defined by pole locations, e.g.,  Butterworth poles lie evenly spaced on a circle in the left half of the complex plane
K. Webb ENGR 203 42 Filter Families – Frequency Response  Butterworth  Maximally-flat pass band  Slow roll off  Chebyshev  Steeper roll off  Pass band ripple  Elliptic  Very steep roll off  Pass band ripple  Stop band ripple  As always, all about trade offs
K. Webb ENGR 203 43 Filter Families – System Poles  Butterworth  Poles lie on a semi- circle in the LHP  Equally spaced  Equal magnitude, 𝜔𝜔0  Chebyshev/elliptic  Poles lie on semi- ellipses in the LHP  Varying magnitudes
K. Webb ENGR 203 44 Butterworth Poles  Butterworth poles:  Magnitude: 𝜔𝜔0  Order: 𝑁𝑁  Separation angles: 180°/𝑁𝑁  Poles for 𝑘𝑘 = 1 … 𝑁𝑁 𝑠𝑠𝑘𝑘 = 𝜔𝜔0 − sin 𝜋𝜋 2𝑘𝑘 − 1 2𝑁𝑁 + 𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗 𝜋𝜋 2𝑘𝑘 − 1 2𝑁𝑁  Each complex conjugate pair are the poles of a single second-order Sallen-Key stage  All with equal 𝜔𝜔0  Each with different 𝜁𝜁
K. Webb ENGR 203 Filter Synthesis 45
K. Webb ENGR 203 46 Filter Synthesis Procedure 1. Determine filter order, 𝑁𝑁, and cutoff frequency, 𝜔𝜔𝑐𝑐 2. Determine 𝜔𝜔0 and 𝑄𝑄 or 𝜁𝜁 for each stage by utilizing either a) Design tables, or b) MATLAB 3. For each stage, select R and C to yield the required 𝜔𝜔0 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 4. For each stage, select 𝑅𝑅𝑓𝑓𝑓 and 𝑅𝑅𝑓𝑓𝑓 to set gain, 𝐾𝐾, to provide the required 𝑄𝑄 𝐾𝐾 = 3 − 1 𝑄𝑄
K. Webb ENGR 203 47 Filter Design Tables  Design tables exist for different filters of different orders from different filter families  Pole locations, 𝜔𝜔0, and 𝑄𝑄 given for each second- and first-order stage for a given filter order, 𝑁𝑁  Only second-order stages for even 𝑁𝑁  Second-order plus one first-order stage for odd 𝑁𝑁  Frequencies are normalized  Multiply 𝜔𝜔0 by the cutoff frequency, 𝜔𝜔𝑐𝑐  Multiply 𝜎𝜎 and 𝜔𝜔𝑑𝑑 by 𝜔𝜔𝑐𝑐
K. Webb ENGR 203 48 Butterworth Design Table Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝑲𝑲 𝝈𝝈 𝝎𝝎𝒅𝒅 2 1 0.7071 0.7071 1.00 0.7071 1.5858 3 1 0.5000 0.8660 1.00 1.0000 1.0000 2 1.0000 - 1.00 - - 4 1 0.9239 0.3827 1.00 0.5412 1.1522 2 0.3827 0.9239 1.00 1.3065 2.2346 5 1 0.8090 0.5878 1.00 0.6180 1.382 2 0.3090 0.9511 1.00 1.6182 2.382 3 1.0000 - 1.00 - - 6 1 0.9659 0.2588 1.00 0.5176 1.0681 2 0.7071 0.7071 1.00 0.7071 1.5858 3 0.2588 0.9659 1.00 1.9319 2.4824
K. Webb ENGR 203 49 Chebyshev Design Table – 0.5 dB ripple Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝑲𝑲 𝝈𝝈 𝝎𝝎𝒅𝒅 2 1 0.71281 1.004 1.2313 0.8638 1.8422 3 1 0.3123 1.0219 1.0689 1.7062 2.4139 2 0.6265 - 0.6265 - - 4 1 0.4233 0.4210 0.5970 0.7051 1.5818 2 0.1754 1.0163 1.0313 2.9406 2.6599 5 1 0.2931 0.6252 0.6905 1.1778 2.1510 2 0.1120 1.0116 1.0177 4.5450 2.7800 3 0.3623 - 0.3623 - 6 1 0.2898 0.2702 0.3962 0.6836 1.5372 2 0.2121 0.7382 0.7681 1.8104 2.4476 3 0.0777 1.0085 1.0114 6.5128 2.8465
K. Webb ENGR 203 50 Chebyshev Design Table – 1.0 dB ripple Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝑲𝑲 𝝈𝝈 𝝎𝝎𝒅𝒅 2 1 0.5489 0.8951 1.0500 0.9565 1.9545 3 1 0.2471 0.9660 0.9771 2.0177 2.5044 2 0.4942 - 0.4942 - - 4 1 0.3369 0.4073 0.5286 0.7846 1.7254 2 0.1395 0.9834 0.9932 3.5590 2.7190 5 1 0.2342 0.6119 0.6552 1.3988 2.2851 2 0.0895 0.9901 0.9941 5.5564 2.8200 3 0.2895 - 0.2895 - - 6 1 0.2321 0.2662 0.3531 0.7609 1.6857 2 0.1699 0.7272 0.7468 2.1980 2.5450 3 0.0622 0.9934 0.9954 8.0037 2.8751
K. Webb ENGR 203 51 Chebyshev Design Table – 3.0 dB ripple Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝑲𝑲 𝝈𝝈 𝝎𝝎𝒅𝒅 2 1 0.3225 0.7772 0.8414 1.3047 202335 3 1 0.1493 0.9038 0.9161 3.0677 2.6740 2 0.2986 - 0.2986 - - 4 1 0.2056 0.3921 0.4427 1.0765 2.0711 2 0.0852 0.9465 0.9503 5.5789 2.8208 5 1 0.1436 0.5970 0.6140 2.1375 2.5322 2 0.0549 0.9659 0.9675 8.8178 2.8866 3 0.1775 - 0.1775 - - 6 1 0.1427 0.2616 0.2980 1.0443 2.0425 2 0.1044 0.7148 0.7224 3.4581 2.7108 3 0.0382 0.9764 0.9772 12.7800 2.9218
K. Webb ENGR 203 52 Filter synthesis in MATLAB  MATLAB has built-in filter design functions, e.g.,  butter.m  cheby1.m  ellip.m  Design procedure: 1. Use functions to get transfer function coefficients for given filter specifications 2. Create MATLAB transfer function object 3. Determine filter poles, 𝜔𝜔0, and 𝑄𝑄 from transfer function – place low-Q stages first 4. Determine component values from 𝜔𝜔0 and 𝑄𝑄
K. Webb ENGR 203 53 Butterworth Filter – butter(…) [b,a] = butter(N,wn,ftype,’s’)  Inputs:  N: filter order  wn: cutoff frequency [rad/sec]  ftype: filter type: ‘low’, ‘bandpass’, ‘high’, ‘stop’ – optional – default: ‘low’  ‘s’: specifies analog filter  Outputs:  b: coefficients of the transfer function’s numerator polynomial  a: coefficients of the transfer function’s denominator polynomial
K. Webb ENGR 203 54 Chebyshev Filter – cheby1(…) [b,a] = cheby1(N,R,wn,ftype,’s’)  Inputs:  N: filter order  R: pass band ripple [dB]  wn: cutoff frequency [rad/sec]  ftype: filter type: ‘low’, ‘bandpass’, ‘high’, ‘stop’ – optional – default: ‘low’  ‘s’: specifies analog filter  Outputs:  b: coefficients of the transfer function’s numerator polynomial  a: coefficients of the transfer function’s denominator polynomial
K. Webb ENGR 203 55 Elliptic Filter – ellip(…) [b,a] = cheby1(N,Rp,Rs,wn,ftype,’s’)  Inputs:  N: filter order  Rp: pass band ripple [dB]  Rs: stop band attenuation [dB]  wn: cutoff frequency [rad/sec]  ftype: filter type: ‘low’, ‘bandpass’, ‘high’, ‘stop’ – optional – default: ‘low’  ‘s’: specifies analog filter  Outputs:  b: coefficients of the transfer function’s numerator polynomial  a: coefficients of the transfer function’s denominator polynomial
K. Webb ENGR 203 56 Transfer Function Model – tf(…) sys = tf(b,a)  b: vector of numerator polynomial coefficients  a: vector of denominator polynomial coefficients  sys: transfer function model object  Transfer function is assumed to be of the form 𝐺𝐺 𝑠𝑠 = 𝑏𝑏1𝑠𝑠𝑟𝑟 + 𝑏𝑏2𝑠𝑠𝑟𝑟−1 + ⋯ + 𝑏𝑏𝑟𝑟𝑠𝑠 + 𝑏𝑏𝑟𝑟+1 𝑎𝑎1𝑠𝑠𝑛𝑛 + 𝑎𝑎2𝑠𝑠𝑛𝑛−1 + ⋯ + 𝑎𝑎𝑛𝑛𝑠𝑠 + 𝑎𝑎𝑛𝑛+1  Inputs to tf(…) are  Num = [b1,b2,…,br+1];  Den = [a1,a2,…,an+1];
K. Webb ENGR 203 57 Getting 𝜔𝜔0 and 𝑄𝑄 – damp(…) [wn,zeta,p] = damp(sys)  sys: transfer function system model object  wn: vector of natural frequencies (magnitudes) of poles  zeta: vector of damping ratios, 𝜁𝜁, of poles  p: vector of poles  Use wn values for 𝜔𝜔0 of each filter stage  Calculate 𝑄𝑄 of each stage from 𝜁𝜁 values 𝑄𝑄 = 1 2𝜁𝜁
K. Webb ENGR 203 Filter Design Example 58
K. Webb ENGR 203 59 Sallen-Key Filter – Example  Design a Butterworth (maximally-flat) low pass active filter to satisfy the following specifications:  Corner frequency: fc = 1MHz  Frequency response roll off beyond fc: 80dB/dec  Pass band (DC) gain: 12dB (4)  Roll off spec of 80 dB/dec tells us we need a fourth- order filter – cascade two Sallen-Key stages  Add a constant gain stage if necessary to meet gain specification
K. Webb ENGR 203 60 Sallen-Key Filter – Example  Fourth-order filter  Cascade two second-order Sallen-Key stages  Additional gain stage necessary to meet gain specification  Non-inverting opamp amplifier  Note that the circuit in this example has been simplified by setting 𝑅𝑅𝑓𝑓𝑓 equal in each stage  Not necessarily the right choice
K. Webb ENGR 203 61 Sallen-Key Filter – Example  Butterworth filter, so, for both stages, 𝜔𝜔0 = 𝜔𝜔𝑐𝑐 = 2𝜋𝜋 ⋅ 𝑓𝑓𝑐𝑐 = 2𝜋𝜋 ⋅ 1 𝑀𝑀𝑀𝑀𝑀𝑀  Determine 𝑅𝑅 and 𝐶𝐶 for desired 𝜔𝜔𝑐𝑐  Arbitrarily choose 𝐶𝐶 = 1 𝑛𝑛𝑛𝑛 𝑅𝑅 = 1 2𝜋𝜋𝑓𝑓𝑐𝑐𝐶𝐶 = 1 2𝜋𝜋 ⋅ 1 𝑀𝑀𝑀𝑀𝑀𝑀 ⋅ 1 𝑛𝑛𝑛𝑛 = 159 Ω  If using ±1% resistors, 158 Ω is a standard value 𝑅𝑅 = 158 Ω and 𝐶𝐶 = 1 𝑛𝑛𝑛𝑛
K. Webb ENGR 203 62 Sallen-Key Filter – Example  To determine gain of each stage, consult the Butterworth design table Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝝈𝝈 𝝎𝝎𝒅𝒅 4 1 0.9239 0.3827 1.00 0.5412 2 0.3827 0.9239 1.00 1.3065  Calculate 𝐾𝐾 for each stage from its 𝑄𝑄 𝐾𝐾1 = 3 − 1 𝑄𝑄1 = 3 − 1 0.5412 = 1.152 𝐾𝐾2 = 3 − 1 𝑄𝑄2 = 3 − 1 1.3065 = 2.235
K. Webb ENGR 203 63 Sallen-Key Filter – Example  Alternatively, use MATLAB to determine 𝜔𝜔0 and 𝐾𝐾 values for each stage  Note that we would put the low-Q stage first
K. Webb ENGR 203 64 Sallen-Key Filter – Example  Arbitrarily choose 𝑅𝑅𝑓𝑓𝑓 = 5.11 𝑘𝑘Ω  Calculate 𝑅𝑅𝑓𝑓𝑓 and 𝑅𝑅𝑓𝑓𝑓 to give the required 𝐾𝐾1 and 𝐾𝐾2 𝐾𝐾1 = 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓𝑓 𝑅𝑅𝑓𝑓𝑓 → 𝑅𝑅𝑓𝑓𝑓 = 𝑅𝑅𝑓𝑓𝑓 𝐾𝐾1 − 1 = 5.11 𝑘𝑘Ω ⋅ 0.152 = 778 Ω 𝐾𝐾2 = 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓𝑓 𝑅𝑅𝑓𝑓𝑓 → 𝑅𝑅𝑓𝑓𝑓 = 𝑅𝑅𝑓𝑓𝑓 𝐾𝐾2 − 1 = 5.11 𝑘𝑘Ω ⋅ 1.235 = 6.31 𝑘𝑘Ω  Again, assuming ±1% resistors, we choose the closest standard values: 𝑅𝑅𝑓𝑓𝑓 = 787 Ω and 𝑅𝑅𝑓𝑓𝑓 = 6.34 𝑘𝑘Ω
K. Webb ENGR 203 65 Sallen-Key Filter – Example  Finally, set the gain of the third stage to satisfy the gain requirement  Overall gain given by 𝐾𝐾 = 𝐾𝐾1𝐾𝐾2𝐾𝐾3 = 4 → 𝐾𝐾3 = 4 𝐾𝐾1𝐾𝐾2 = 4 1.152 ⋅ 2.235 = 1.554  Calculate 𝑅𝑅𝑓𝑓𝑓 to give the required 𝐾𝐾3 𝐾𝐾3 = 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓4 𝑅𝑅𝑓𝑓𝑓 → 𝑅𝑅𝑓𝑓𝑓 = 𝑅𝑅𝑓𝑓𝑓 𝐾𝐾3 − 1 = 5.11 𝑘𝑘Ω ⋅ 0.554 = 2.83 𝑘𝑘Ω  Again, assuming ±1% resistors, we choose the closest standard value: 𝑅𝑅𝑓𝑓𝑓 = 2.8 𝑘𝑘Ω
K. Webb ENGR 203 66 Sallen-Key Filter – Example The complete 4th-order Sallen-Key Butterworth low pass filter:
K. Webb ENGR 203 67 Sallen-Key Filter – Example  DC gain: ~12 dB  𝑓𝑓𝑐𝑐 ≈ 1 𝑀𝑀𝑀𝑀𝑀𝑀  Gain rolloff: -80 dB/dec  Stage 1:  Low Q  Low gain  Stage 2:  Higher Q  Higher gain  Stage 3:  Constant gain

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Section 7 Active FIlters.pdf Electrical fundamentals

  • 1. ENGR 203 – Electrical Fundamentals III SECTION 7: ACTIVE FILTERS
  • 2. K. Webb ENGR 203 2 Introduction  In ENGR 202 we studied different types of first- and second-order passive filters  Passive, because they contain only passive components:  Resistors, capacitors, and inductors  Can also construct filters using opamps  Active filters
  • 3. K. Webb ENGR 203 3 Introduction  Active filters have advantages over passive filters:  Can build high-Q filters without inductors  Low output impedance  Easily adjustable: 𝑓𝑓𝑐𝑐, 𝑄𝑄  Can provide gain ( > 0 dB )  Before getting into the design of active filters, we will look at two fundamental filter building blocks:  Opamp integrators  Opamp differentiators
  • 4. K. Webb ENGR 203 Opamp Integrators 4
  • 5. K. Webb ENGR 203 5 Integrators and Differentiators  Opamp circuits can perform many different mathematical operations  Operational amplifiers  Multiplication  Inverting and non-inverting amplifiers  Addition and subtraction  Summing and difference amplifiers  Can also perform integration and differentiation  Feedback controllers  Building block of active filters
  • 6. K. Webb ENGR 203 6 Opamp Integrator – Time Domain 𝑖𝑖 𝑡𝑡 = 𝑣𝑣𝑖𝑖 𝑡𝑡 𝑅𝑅  Capacitor integrates input current to give output voltage 𝑣𝑣𝑜𝑜 𝑡𝑡 = − 1 𝐶𝐶 � 0 𝑡𝑡 𝑖𝑖 𝜏𝜏 𝑑𝑑𝑑𝑑 𝑣𝑣𝑜𝑜 𝑡𝑡 = − 1 𝑅𝑅𝑅𝑅 � 0 𝑡𝑡 𝑣𝑣𝑖𝑖 𝜏𝜏 𝑑𝑑𝑑𝑑  Output is the (scaled and inverted) integral of the input  Analyze the opamp integrator in the time domain  Virtual ground at inverting input, so
  • 7. K. Webb ENGR 203 7 Opamp Integrator – Laplace Domain 𝐼𝐼 𝑠𝑠 = 𝑉𝑉𝑖𝑖 𝑠𝑠 𝑅𝑅  Output voltage: 𝑉𝑉 𝑜𝑜 𝑠𝑠 = −𝐼𝐼 𝑠𝑠 1 𝐶𝐶𝐶𝐶 = − 𝑉𝑉𝑖𝑖 𝑠𝑠 𝑅𝑅𝑅𝑅𝑅𝑅 = − 1 𝑠𝑠 ⋅ 𝑉𝑉𝑖𝑖 𝑠𝑠 𝑅𝑅𝑅𝑅  Recall that multiplication by 1/𝑠𝑠 in the Laplace domain corresponds to integration in the time domain  Analyze the opamp integrator in the Laplace domain  Again, a virtual ground at inverting input, so
  • 8. K. Webb ENGR 203 8 Opamp Integrator – Frequency Response  Transfer function: 𝐺𝐺 𝑠𝑠 = − 1 𝑅𝑅𝑅𝑅𝑅𝑅  Single pole at 𝑠𝑠 = 0  Gain: constant slope of -20 dB/dec  Infinite DC gain  Phase: -90° from integrator pole + 180° from inversion yields constant +90°
  • 9. K. Webb ENGR 203 9 Ideal Integrator - Problem  Laplace domain step response of the ideal integrator 𝑉𝑉 𝑜𝑜 𝑠𝑠 = 1 𝑠𝑠 ⋅ 𝐺𝐺 𝑠𝑠 = − 1 𝑅𝑅𝑅𝑅 ⋅ 1 𝑠𝑠2  Inverse transforming to the time domain 𝑣𝑣𝑜𝑜 𝑡𝑡 = − 1 𝑅𝑅𝑅𝑅 ⋅ 𝑡𝑡  Output increases linearly with time  Opamp will quickly saturate in response any DC input component  Infinite DC gain  Not a practical circuit  Inputs will always have some non-zero offset  Real (non-ideal) opamps have non-zero offset voltages and input bias currents
  • 10. K. Webb ENGR 203 10 Practical Opamp Integrator  Now there is a feedback path for DC signals  DC gain limited to 𝑅𝑅𝑓𝑓/𝑅𝑅  Behaves as an inverting opamp at low frequencies  Still behaves as an integrator at high frequencies  A practical or lossy integrator circuit  Problem with ideal integrator is infinite DC gain  No DC feedback  Open-loop at DC  Add a feedback resistor in parallel with the capacitor
  • 11. K. Webb ENGR 203 11 Opamp Integrator – Frequency Response  Transfer function: 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝐶𝐶𝐶𝐶 𝑅𝑅𝑓𝑓 + 1 𝐶𝐶𝐶𝐶 𝑅𝑅 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 1 𝑅𝑅𝑓𝑓𝐶𝐶𝐶𝐶 + 1  Pole (corner frequency) set by the feedback network: 𝜔𝜔𝑐𝑐 = 1 𝑅𝑅𝑓𝑓𝐶𝐶  For 𝜔𝜔 ≫ 𝜔𝜔𝑐𝑐, still behaves like an integrator  Gain: rolls off at -20 dB/dec  Phase: ~90° Integrator Amplifier
  • 12. K. Webb ENGR 203 Opamp Differentiators 12
  • 13. K. Webb ENGR 203 13 Opamp Differentiator – Time Domain 𝑖𝑖 𝑡𝑡 = 𝐶𝐶 𝑑𝑑𝑣𝑣𝑖𝑖 𝑑𝑑𝑑𝑑  Ohm’s law gives the output voltage 𝑣𝑣𝑜𝑜 𝑡𝑡 = −𝑅𝑅𝑅𝑅 𝑡𝑡 = −𝑅𝑅𝑅𝑅 𝑑𝑑𝑣𝑣𝑖𝑖 𝑑𝑑𝑑𝑑  Output is the (scaled and inverted) derivative of the input  Analyze the opamp differentiator in the time domain  Virtual ground at inverting input, so
  • 14. K. Webb ENGR 203 14 Opamp Differentiator – Laplace Domain 𝐼𝐼 𝑠𝑠 = 𝐶𝐶𝐶𝐶 ⋅ 𝑉𝑉𝑖𝑖 𝑠𝑠  Output voltage: 𝑉𝑉 𝑜𝑜 𝑠𝑠 = −𝑅𝑅𝑅𝑅 𝑠𝑠 = −𝑅𝑅𝑅𝑅𝑅𝑅𝑉𝑉𝑖𝑖 𝑠𝑠 = −𝑠𝑠 ⋅ 𝑅𝑅𝑅𝑅𝑉𝑉𝑖𝑖 𝑠𝑠  Recall that multiplication by 𝑠𝑠 in the Laplace domain corresponds to differentiation in the time domain  Analyze the differentiator in the Laplace domain  Again, a virtual ground at inverting input, so
  • 15. K. Webb ENGR 203 15 Opamp Differentiator – Frequency Response  Transfer function: 𝐺𝐺 𝑠𝑠 = −𝑅𝑅𝑅𝑅𝑅𝑅  Single zero at 𝑠𝑠 = 0  Gain: constant slope of +20 dB/dec  Very large high-frequency gain  Phase: +90° from zero at the origin + 180° from inversion yields constant +270° = -90°
  • 16. K. Webb ENGR 203 16 Ideal Differentiator - Problem  Gain continues to increase with frequency  High-frequency gain is very large  Any input signal will include some noise  Better to limit the gain above some upper frequency
  • 17. K. Webb ENGR 203 17 Practical Opamp Differentiator  High-frequency gain limited to 𝑅𝑅𝑓𝑓/𝑅𝑅  Still behaves as a differentiator at low frequencies  Behaves as an inverting opamp at high frequencies  A practical or lossy differentiator circuit  Problem with ideal differentiator:  Low input impedance at high frequency  Excessive high-frequency input current  Add a resistor in series with the input capacitor
  • 18. K. Webb ENGR 203 18 Practical Opamp Differentiator  Transfer function: 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 + 1 𝐶𝐶𝐶𝐶 = − 𝑅𝑅𝑓𝑓𝐶𝐶𝐶𝐶 𝑅𝑅𝑅𝑅𝑅𝑅 + 1  Pole (corner frequency) set by the input network: 𝜔𝜔𝑐𝑐 = 1 𝑅𝑅𝑅𝑅  For 𝜔𝜔 ≪ 𝜔𝜔𝑐𝑐, still behaves like a differentiator  Gain: increases at +20 dB/dec  Phase: ~-90° Differentiator Amplifier
  • 19. K. Webb ENGR 203 First-Order Opamp Active Filters 19
  • 20. K. Webb ENGR 203 20 First-Order Active Filters  Practical integrator and differentiator circuits  Additional resistors fix problems with ideal circuits  First-order low pass and high pass filters
  • 21. K. Webb ENGR 203 21 First-Order Low Pass Filter  Transfer function 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 1 𝑅𝑅𝑓𝑓𝐶𝐶𝐶𝐶 + 1 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 1 𝑅𝑅𝑓𝑓𝐶𝐶 𝑠𝑠 + 1 𝑅𝑅𝑓𝑓𝐶𝐶  Corner frequency 𝑓𝑓𝑐𝑐 = 1 2𝜋𝜋𝑅𝑅𝑓𝑓𝐶𝐶  Pass-band gain 𝐴𝐴𝑣𝑣 = − 𝑅𝑅𝑓𝑓 𝑅𝑅
  • 22. K. Webb ENGR 203 22 First-Order High Pass Filter  Transfer function 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓𝐶𝐶𝐶𝐶 𝑅𝑅𝑅𝑅𝑅𝑅 + 1 𝐺𝐺 𝑠𝑠 = − 𝑅𝑅𝑓𝑓 𝑅𝑅 𝑠𝑠 𝑠𝑠 + 1 𝑅𝑅𝐶𝐶  Corner frequency 𝑓𝑓𝑐𝑐 = 1 2𝜋𝜋𝜋𝜋𝜋𝜋  Pass-band gain 𝐴𝐴𝑣𝑣 = − 𝑅𝑅𝑓𝑓 𝑅𝑅
  • 23. K. Webb ENGR 203 Higher Order Active Filters 23
  • 24. K. Webb ENGR 203 24 Higher-Order Active Filters  Higher order active filters can be constructed by:  Cascading first-order active filters  Using second-order active filter stages  Cascading second- and first-order stages  Create higher order band pass/stop filters similarly:  Cascade first-order high/low pass filters  Use and/or cascade second-order band pass/stop stages  Many different second-order active filter topologies  We’ll look at the Sallen-Key circuit
  • 25. K. Webb ENGR 203 25 Sallen-Key Filter – Generalized Form  Sallen-Key filter topology  Low pass and high pass filters  Band-pass, and notch filters with slight modifications  We’ll look first at the filter in its most generalized form, then consider the specific low pass and high pass filter forms  Type of filter depends on the location of components – resistors and capacitors
  • 26. K. Webb ENGR 203 26 Sallen-Key Filter – Generalized Form  Transfer function  Nodal analysis  KCL at V+ and Vf  Virtual short at opamp inputs  After a lot of ugly algebra: 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽 𝑍𝑍1𝑍𝑍2 𝑍𝑍3𝑍𝑍4 + 𝛽𝛽 𝑍𝑍2 𝑍𝑍4 + 𝛽𝛽 𝑍𝑍1 𝑍𝑍4 + 𝛽𝛽 − 1 𝑍𝑍1 𝑍𝑍3 + 𝛽𝛽 where 𝛽𝛽 is the feedback path gain 𝛽𝛽 = 𝑅𝑅𝑓𝑓𝑓 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓𝑓
  • 27. K. Webb ENGR 203 Sallen-Key Low Pass Filter 27
  • 28. K. Webb ENGR 203 28 Sallen-Key Second-Order Low Pass Filter  Transfer function 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2𝑠𝑠2 + 𝛽𝛽𝑅𝑅2𝐶𝐶2𝑠𝑠 + 𝛽𝛽𝑅𝑅1𝐶𝐶2𝑠𝑠 + 𝛽𝛽 − 1 𝑅𝑅1𝐶𝐶1𝑠𝑠 + 𝛽𝛽 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2 𝑠𝑠2 + 1 𝑅𝑅1𝐶𝐶1 + 1 𝑅𝑅2𝐶𝐶1 + 𝛽𝛽 − 1 𝛽𝛽𝑅𝑅2𝐶𝐶2 𝑠𝑠 + 1 𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2  Z1 and Z2 are resistors  Z3 and Z4 are capacitors
  • 29. K. Webb ENGR 203 29 Sallen-Key Low Pass Filter  Generalized second-order low pass transfer function: 𝐺𝐺 𝑠𝑠 = 𝐾𝐾 ⋅ 𝜔𝜔0 2 𝑠𝑠2 + 𝜔𝜔0 𝑄𝑄 𝑠𝑠 + 𝜔𝜔0 2  Equating coefficients with the Sallen-Key transfer function gives  Resonant frequency: 𝜔𝜔0 = 1 𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2  Quality factor: 𝑄𝑄 = 𝑅𝑅1𝑅𝑅2𝐶𝐶1𝐶𝐶2 𝑅𝑅2𝐶𝐶2 + 𝑅𝑅1𝐶𝐶1 + 𝛽𝛽 − 1 𝛽𝛽 𝑅𝑅1𝐶𝐶1  DC gain: 𝐾𝐾 = 1 𝛽𝛽 = 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓𝑓 𝑅𝑅𝑓𝑓𝑓
  • 30. K. Webb ENGR 203 30 Sallen-Key Low Pass Filter  𝜔𝜔0, Q, and gain all set by appropriate component selection, but  There are more degrees of freedom than we need  Transfer function is a bit more complicated than we’d like  Simplify by setting component values equal  Transfer function becomes 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽 𝑅𝑅𝑅𝑅 2 𝑠𝑠2 + 3 − 1 𝛽𝛽 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  Where now 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 and 𝑄𝑄 = 1 3− 1 𝛽𝛽
  • 31. K. Webb ENGR 203 31 Sallen-Key Low Pass Filter 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽 𝑅𝑅𝑅𝑅 2 𝑠𝑠2 + 3 − 1 𝛽𝛽 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  We can also write the transfer function in terms of DC gain, 𝐾𝐾 𝐺𝐺 𝑠𝑠 = 𝐾𝐾 𝑅𝑅𝑅𝑅 2 𝑠𝑠2 + 3 − 𝐾𝐾 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  𝜔𝜔0 and 𝑄𝑄 in terms of 𝐾𝐾: 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 and 𝑄𝑄 = 1 3−𝐾𝐾  The filter’s DC gain is dependent on the filter’s 𝑸𝑸 and vice versa  For independent control of DC gain, cascade an additional gain stage
  • 32. K. Webb ENGR 203 32 Sallen-Key Low Pass Filter  Note dependence of 𝑄𝑄 and 𝐾𝐾  Both set by feedback path gain  𝑄𝑄 and gain are independent of 𝜔𝜔0  𝜔𝜔0 set by capacitors and resistors at the input  Second-order  Gain roll-off: -40 dB/dec
  • 33. K. Webb ENGR 203 Sallen-Key High Pass Filter 33
  • 34. K. Webb ENGR 203 34 Sallen-Key High Pass Filter  Here, we will jump straight to the simplified circuit with equal-valued components  Location of resistors and capacitors swapped relative to low pass filter  High pass transfer function 𝐺𝐺 𝑠𝑠 = 1 𝛽𝛽 𝑠𝑠2 𝑠𝑠2 + 3 − 1 𝛽𝛽 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  Again, 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 and 𝑄𝑄 = 1 3− 1 𝛽𝛽
  • 35. K. Webb ENGR 203 35 Sallen-Key High Pass Filter  As with the low pass filter, we can write the transfer function in terms of gain, 𝐾𝐾  𝐾𝐾 still represents passband gain, but now it is the high-frequency gain, not the DC gain 𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝑠𝑠2 𝑠𝑠2 + 3 − 𝐾𝐾 𝑅𝑅𝑅𝑅 𝑠𝑠 + 1 𝑅𝑅𝑅𝑅 2  𝜔𝜔0 and 𝑄𝑄 are the same as for the low pass filter: 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 and 𝑄𝑄 = 1 3−𝐾𝐾  Same dependence between passband gain, resonant frequency, and Q
  • 36. K. Webb ENGR 203 36 Sallen-Key High Pass Filter  Note dependence of 𝑄𝑄 and 𝐾𝐾  Both set by feedback path gain  𝑄𝑄 and gain are independent of 𝜔𝜔0  𝜔𝜔0 set by capacitors and resistors at the input  Second-order  Gain roll-off: -40 dB/dec
  • 37. K. Webb ENGR 203 37 Sallen-Key Filter – Stability  Sallen-Key filter has two feedback paths:  Negative feedback  Generally stabilizing  Positive feedback  Generally destabilizing Positive feedback Negative feedback  Relative amount of negative and positive feedback determines stability  Net negative feedback: circuit is stable  Behaves as a linear filter/amplifier  Net positive feedback: circuit is unstable  Will oscillate or saturate
  • 38. K. Webb ENGR 203 38 Sallen-Key Filter – Stability  Overall net feedback must remain negative  But, we can vary just how negative by varying 𝛽𝛽  Varying 𝛽𝛽 allows us to vary 𝑄𝑄: 𝑄𝑄 = 1 3 − 1 𝛽𝛽 = 1 3 − 𝐾𝐾 Positive feedback Negative feedback  As 𝛽𝛽 increases:  Negative feedback increases  Overall feedback becomes more negative  Quality factor, 𝑄𝑄, decreases  Damping ratio, 𝜁𝜁, increases  Pass band gain, 𝐾𝐾, decreases
  • 39. K. Webb ENGR 203 39 Sallen-Key Filter – Stability Positive feedback Negative feedback  As 𝛽𝛽 decreases:  Negative feedback decreases  Overall feedback becomes less negative  Quality factor, 𝑄𝑄, increases  Damping ratio, 𝜁𝜁, decreases  Pass band gain, 𝐾𝐾, increases  There is an upper limit on 𝐾𝐾:  For 𝐾𝐾 = 3, 𝑄𝑄 = ∞ and 𝜁𝜁 = 0  An un-damped circuit  Negative and positive feedback cancel  The border between stability and instability  For stability: 𝐾𝐾 ≤ 3
  • 40. K. Webb ENGR 203 Filter Families 40
  • 41. K. Webb ENGR 203 41 Filter Families  Higher-order filters of all types can be designed with transfer functions that fit into one of several families of filters  Butterworth  Chebyshev  Elliptic  Bessel  Each filter family defined by the nature of its characteristic polynomial  Equivalently, each defined by pole locations, e.g.,  Butterworth poles lie evenly spaced on a circle in the left half of the complex plane
  • 42. K. Webb ENGR 203 42 Filter Families – Frequency Response  Butterworth  Maximally-flat pass band  Slow roll off  Chebyshev  Steeper roll off  Pass band ripple  Elliptic  Very steep roll off  Pass band ripple  Stop band ripple  As always, all about trade offs
  • 43. K. Webb ENGR 203 43 Filter Families – System Poles  Butterworth  Poles lie on a semi- circle in the LHP  Equally spaced  Equal magnitude, 𝜔𝜔0  Chebyshev/elliptic  Poles lie on semi- ellipses in the LHP  Varying magnitudes
  • 44. K. Webb ENGR 203 44 Butterworth Poles  Butterworth poles:  Magnitude: 𝜔𝜔0  Order: 𝑁𝑁  Separation angles: 180°/𝑁𝑁  Poles for 𝑘𝑘 = 1 … 𝑁𝑁 𝑠𝑠𝑘𝑘 = 𝜔𝜔0 − sin 𝜋𝜋 2𝑘𝑘 − 1 2𝑁𝑁 + 𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗 𝜋𝜋 2𝑘𝑘 − 1 2𝑁𝑁  Each complex conjugate pair are the poles of a single second-order Sallen-Key stage  All with equal 𝜔𝜔0  Each with different 𝜁𝜁
  • 45. K. Webb ENGR 203 Filter Synthesis 45
  • 46. K. Webb ENGR 203 46 Filter Synthesis Procedure 1. Determine filter order, 𝑁𝑁, and cutoff frequency, 𝜔𝜔𝑐𝑐 2. Determine 𝜔𝜔0 and 𝑄𝑄 or 𝜁𝜁 for each stage by utilizing either a) Design tables, or b) MATLAB 3. For each stage, select R and C to yield the required 𝜔𝜔0 𝜔𝜔0 = 1 𝑅𝑅𝑅𝑅 4. For each stage, select 𝑅𝑅𝑓𝑓𝑓 and 𝑅𝑅𝑓𝑓𝑓 to set gain, 𝐾𝐾, to provide the required 𝑄𝑄 𝐾𝐾 = 3 − 1 𝑄𝑄
  • 47. K. Webb ENGR 203 47 Filter Design Tables  Design tables exist for different filters of different orders from different filter families  Pole locations, 𝜔𝜔0, and 𝑄𝑄 given for each second- and first-order stage for a given filter order, 𝑁𝑁  Only second-order stages for even 𝑁𝑁  Second-order plus one first-order stage for odd 𝑁𝑁  Frequencies are normalized  Multiply 𝜔𝜔0 by the cutoff frequency, 𝜔𝜔𝑐𝑐  Multiply 𝜎𝜎 and 𝜔𝜔𝑑𝑑 by 𝜔𝜔𝑐𝑐
  • 48. K. Webb ENGR 203 48 Butterworth Design Table Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝑲𝑲 𝝈𝝈 𝝎𝝎𝒅𝒅 2 1 0.7071 0.7071 1.00 0.7071 1.5858 3 1 0.5000 0.8660 1.00 1.0000 1.0000 2 1.0000 - 1.00 - - 4 1 0.9239 0.3827 1.00 0.5412 1.1522 2 0.3827 0.9239 1.00 1.3065 2.2346 5 1 0.8090 0.5878 1.00 0.6180 1.382 2 0.3090 0.9511 1.00 1.6182 2.382 3 1.0000 - 1.00 - - 6 1 0.9659 0.2588 1.00 0.5176 1.0681 2 0.7071 0.7071 1.00 0.7071 1.5858 3 0.2588 0.9659 1.00 1.9319 2.4824
  • 49. K. Webb ENGR 203 49 Chebyshev Design Table – 0.5 dB ripple Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝑲𝑲 𝝈𝝈 𝝎𝝎𝒅𝒅 2 1 0.71281 1.004 1.2313 0.8638 1.8422 3 1 0.3123 1.0219 1.0689 1.7062 2.4139 2 0.6265 - 0.6265 - - 4 1 0.4233 0.4210 0.5970 0.7051 1.5818 2 0.1754 1.0163 1.0313 2.9406 2.6599 5 1 0.2931 0.6252 0.6905 1.1778 2.1510 2 0.1120 1.0116 1.0177 4.5450 2.7800 3 0.3623 - 0.3623 - 6 1 0.2898 0.2702 0.3962 0.6836 1.5372 2 0.2121 0.7382 0.7681 1.8104 2.4476 3 0.0777 1.0085 1.0114 6.5128 2.8465
  • 50. K. Webb ENGR 203 50 Chebyshev Design Table – 1.0 dB ripple Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝑲𝑲 𝝈𝝈 𝝎𝝎𝒅𝒅 2 1 0.5489 0.8951 1.0500 0.9565 1.9545 3 1 0.2471 0.9660 0.9771 2.0177 2.5044 2 0.4942 - 0.4942 - - 4 1 0.3369 0.4073 0.5286 0.7846 1.7254 2 0.1395 0.9834 0.9932 3.5590 2.7190 5 1 0.2342 0.6119 0.6552 1.3988 2.2851 2 0.0895 0.9901 0.9941 5.5564 2.8200 3 0.2895 - 0.2895 - - 6 1 0.2321 0.2662 0.3531 0.7609 1.6857 2 0.1699 0.7272 0.7468 2.1980 2.5450 3 0.0622 0.9934 0.9954 8.0037 2.8751
  • 51. K. Webb ENGR 203 51 Chebyshev Design Table – 3.0 dB ripple Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝑲𝑲 𝝈𝝈 𝝎𝝎𝒅𝒅 2 1 0.3225 0.7772 0.8414 1.3047 202335 3 1 0.1493 0.9038 0.9161 3.0677 2.6740 2 0.2986 - 0.2986 - - 4 1 0.2056 0.3921 0.4427 1.0765 2.0711 2 0.0852 0.9465 0.9503 5.5789 2.8208 5 1 0.1436 0.5970 0.6140 2.1375 2.5322 2 0.0549 0.9659 0.9675 8.8178 2.8866 3 0.1775 - 0.1775 - - 6 1 0.1427 0.2616 0.2980 1.0443 2.0425 2 0.1044 0.7148 0.7224 3.4581 2.7108 3 0.0382 0.9764 0.9772 12.7800 2.9218
  • 52. K. Webb ENGR 203 52 Filter synthesis in MATLAB  MATLAB has built-in filter design functions, e.g.,  butter.m  cheby1.m  ellip.m  Design procedure: 1. Use functions to get transfer function coefficients for given filter specifications 2. Create MATLAB transfer function object 3. Determine filter poles, 𝜔𝜔0, and 𝑄𝑄 from transfer function – place low-Q stages first 4. Determine component values from 𝜔𝜔0 and 𝑄𝑄
  • 53. K. Webb ENGR 203 53 Butterworth Filter – butter(…) [b,a] = butter(N,wn,ftype,’s’)  Inputs:  N: filter order  wn: cutoff frequency [rad/sec]  ftype: filter type: ‘low’, ‘bandpass’, ‘high’, ‘stop’ – optional – default: ‘low’  ‘s’: specifies analog filter  Outputs:  b: coefficients of the transfer function’s numerator polynomial  a: coefficients of the transfer function’s denominator polynomial
  • 54. K. Webb ENGR 203 54 Chebyshev Filter – cheby1(…) [b,a] = cheby1(N,R,wn,ftype,’s’)  Inputs:  N: filter order  R: pass band ripple [dB]  wn: cutoff frequency [rad/sec]  ftype: filter type: ‘low’, ‘bandpass’, ‘high’, ‘stop’ – optional – default: ‘low’  ‘s’: specifies analog filter  Outputs:  b: coefficients of the transfer function’s numerator polynomial  a: coefficients of the transfer function’s denominator polynomial
  • 55. K. Webb ENGR 203 55 Elliptic Filter – ellip(…) [b,a] = cheby1(N,Rp,Rs,wn,ftype,’s’)  Inputs:  N: filter order  Rp: pass band ripple [dB]  Rs: stop band attenuation [dB]  wn: cutoff frequency [rad/sec]  ftype: filter type: ‘low’, ‘bandpass’, ‘high’, ‘stop’ – optional – default: ‘low’  ‘s’: specifies analog filter  Outputs:  b: coefficients of the transfer function’s numerator polynomial  a: coefficients of the transfer function’s denominator polynomial
  • 56. K. Webb ENGR 203 56 Transfer Function Model – tf(…) sys = tf(b,a)  b: vector of numerator polynomial coefficients  a: vector of denominator polynomial coefficients  sys: transfer function model object  Transfer function is assumed to be of the form 𝐺𝐺 𝑠𝑠 = 𝑏𝑏1𝑠𝑠𝑟𝑟 + 𝑏𝑏2𝑠𝑠𝑟𝑟−1 + ⋯ + 𝑏𝑏𝑟𝑟𝑠𝑠 + 𝑏𝑏𝑟𝑟+1 𝑎𝑎1𝑠𝑠𝑛𝑛 + 𝑎𝑎2𝑠𝑠𝑛𝑛−1 + ⋯ + 𝑎𝑎𝑛𝑛𝑠𝑠 + 𝑎𝑎𝑛𝑛+1  Inputs to tf(…) are  Num = [b1,b2,…,br+1];  Den = [a1,a2,…,an+1];
  • 57. K. Webb ENGR 203 57 Getting 𝜔𝜔0 and 𝑄𝑄 – damp(…) [wn,zeta,p] = damp(sys)  sys: transfer function system model object  wn: vector of natural frequencies (magnitudes) of poles  zeta: vector of damping ratios, 𝜁𝜁, of poles  p: vector of poles  Use wn values for 𝜔𝜔0 of each filter stage  Calculate 𝑄𝑄 of each stage from 𝜁𝜁 values 𝑄𝑄 = 1 2𝜁𝜁
  • 58. K. Webb ENGR 203 Filter Design Example 58
  • 59. K. Webb ENGR 203 59 Sallen-Key Filter – Example  Design a Butterworth (maximally-flat) low pass active filter to satisfy the following specifications:  Corner frequency: fc = 1MHz  Frequency response roll off beyond fc: 80dB/dec  Pass band (DC) gain: 12dB (4)  Roll off spec of 80 dB/dec tells us we need a fourth- order filter – cascade two Sallen-Key stages  Add a constant gain stage if necessary to meet gain specification
  • 60. K. Webb ENGR 203 60 Sallen-Key Filter – Example  Fourth-order filter  Cascade two second-order Sallen-Key stages  Additional gain stage necessary to meet gain specification  Non-inverting opamp amplifier  Note that the circuit in this example has been simplified by setting 𝑅𝑅𝑓𝑓𝑓 equal in each stage  Not necessarily the right choice
  • 61. K. Webb ENGR 203 61 Sallen-Key Filter – Example  Butterworth filter, so, for both stages, 𝜔𝜔0 = 𝜔𝜔𝑐𝑐 = 2𝜋𝜋 ⋅ 𝑓𝑓𝑐𝑐 = 2𝜋𝜋 ⋅ 1 𝑀𝑀𝑀𝑀𝑀𝑀  Determine 𝑅𝑅 and 𝐶𝐶 for desired 𝜔𝜔𝑐𝑐  Arbitrarily choose 𝐶𝐶 = 1 𝑛𝑛𝑛𝑛 𝑅𝑅 = 1 2𝜋𝜋𝑓𝑓𝑐𝑐𝐶𝐶 = 1 2𝜋𝜋 ⋅ 1 𝑀𝑀𝑀𝑀𝑀𝑀 ⋅ 1 𝑛𝑛𝑛𝑛 = 159 Ω  If using ±1% resistors, 158 Ω is a standard value 𝑅𝑅 = 158 Ω and 𝐶𝐶 = 1 𝑛𝑛𝑛𝑛
  • 62. K. Webb ENGR 203 62 Sallen-Key Filter – Example  To determine gain of each stage, consult the Butterworth design table Order, N Section Poles 𝝎𝝎𝟎𝟎 𝑸𝑸 𝝈𝝈 𝝎𝝎𝒅𝒅 4 1 0.9239 0.3827 1.00 0.5412 2 0.3827 0.9239 1.00 1.3065  Calculate 𝐾𝐾 for each stage from its 𝑄𝑄 𝐾𝐾1 = 3 − 1 𝑄𝑄1 = 3 − 1 0.5412 = 1.152 𝐾𝐾2 = 3 − 1 𝑄𝑄2 = 3 − 1 1.3065 = 2.235
  • 63. K. Webb ENGR 203 63 Sallen-Key Filter – Example  Alternatively, use MATLAB to determine 𝜔𝜔0 and 𝐾𝐾 values for each stage  Note that we would put the low-Q stage first
  • 64. K. Webb ENGR 203 64 Sallen-Key Filter – Example  Arbitrarily choose 𝑅𝑅𝑓𝑓𝑓 = 5.11 𝑘𝑘Ω  Calculate 𝑅𝑅𝑓𝑓𝑓 and 𝑅𝑅𝑓𝑓𝑓 to give the required 𝐾𝐾1 and 𝐾𝐾2 𝐾𝐾1 = 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓𝑓 𝑅𝑅𝑓𝑓𝑓 → 𝑅𝑅𝑓𝑓𝑓 = 𝑅𝑅𝑓𝑓𝑓 𝐾𝐾1 − 1 = 5.11 𝑘𝑘Ω ⋅ 0.152 = 778 Ω 𝐾𝐾2 = 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓𝑓 𝑅𝑅𝑓𝑓𝑓 → 𝑅𝑅𝑓𝑓𝑓 = 𝑅𝑅𝑓𝑓𝑓 𝐾𝐾2 − 1 = 5.11 𝑘𝑘Ω ⋅ 1.235 = 6.31 𝑘𝑘Ω  Again, assuming ±1% resistors, we choose the closest standard values: 𝑅𝑅𝑓𝑓𝑓 = 787 Ω and 𝑅𝑅𝑓𝑓𝑓 = 6.34 𝑘𝑘Ω
  • 65. K. Webb ENGR 203 65 Sallen-Key Filter – Example  Finally, set the gain of the third stage to satisfy the gain requirement  Overall gain given by 𝐾𝐾 = 𝐾𝐾1𝐾𝐾2𝐾𝐾3 = 4 → 𝐾𝐾3 = 4 𝐾𝐾1𝐾𝐾2 = 4 1.152 ⋅ 2.235 = 1.554  Calculate 𝑅𝑅𝑓𝑓𝑓 to give the required 𝐾𝐾3 𝐾𝐾3 = 𝑅𝑅𝑓𝑓𝑓 + 𝑅𝑅𝑓𝑓4 𝑅𝑅𝑓𝑓𝑓 → 𝑅𝑅𝑓𝑓𝑓 = 𝑅𝑅𝑓𝑓𝑓 𝐾𝐾3 − 1 = 5.11 𝑘𝑘Ω ⋅ 0.554 = 2.83 𝑘𝑘Ω  Again, assuming ±1% resistors, we choose the closest standard value: 𝑅𝑅𝑓𝑓𝑓 = 2.8 𝑘𝑘Ω
  • 66. K. Webb ENGR 203 66 Sallen-Key Filter – Example The complete 4th-order Sallen-Key Butterworth low pass filter:
  • 67. K. Webb ENGR 203 67 Sallen-Key Filter – Example  DC gain: ~12 dB  𝑓𝑓𝑐𝑐 ≈ 1 𝑀𝑀𝑀𝑀𝑀𝑀  Gain rolloff: -80 dB/dec  Stage 1:  Low Q  Low gain  Stage 2:  Higher Q  Higher gain  Stage 3:  Constant gain