By the end of this chapter you should be able to:
- Say why an inductorless filter is preferred below about 100 kHz, and what the op-amp supplies in place of the inductor.
- Relate order to roll-off in dB/decade and dB/octave, and locate the cutoff at the −3 dB point of a transfer function.
- Derive the response of first-order active low-pass and high-pass sections and design one to a specified cutoff.
- Derive the Sallen-Key second-order transfer function, and extract \(\omega_0\) and \(Q\) from it.
- Apply the equal-component and unity-gain Sallen-Key design equations, and choose \(K\) to place \(Q\).
- Compare Butterworth, Chebyshev and Bessel responses, and cascade stages of the right \(Q\) to reach order 3 and 4.
- Design a multiple-feedback band-pass, a state-variable filter and a twin-T notch, and verify a finished design numerically.
Chapter 10 filtered a rectifier output with a capacitor and a choke, and Chapter 25 watched coupling and bypass capacitors impose a low-frequency roll-off on an amplifier whether the designer wanted one or not. Both were filters by accident. This chapter builds them on purpose: circuits whose whole job is to pass one band of frequencies and reject another, with the corner placed where the designer chooses and the sharpness chosen independently of it.
A filter can of course be built from resistors, capacitors and inductors alone, and at radio frequencies that is exactly what is done. At audio frequencies it is a poor bargain. A second-order 1 kHz low-pass filter with a 16 kΩ source needs an inductor of the order of a henry; such a component is physically large, expensive, wound with enough wire to have a significant series resistance, magnetically coupled to everything near it, and non-linear if its core saturates. Worse, a passive filter has no gain, its response changes the moment anything is connected to its output, and two of them cascaded do not behave like the product of their individual responses because the second loads the first. An operational amplifier removes every one of those objections at the cost of a few milliamps and a power supply. That is what an active filter is: a network of resistors and capacitors wrapped round an amplifier so that the amplifier synthesises the behaviour the missing inductor would have provided.
1 Why an Active Filter Exists at All
The passive RLC filter is the older technique and it is still the right one above a few hundred kilohertz, where inductors are small, cheap and high-Q, and where op-amps have run out of gain. Below that the arithmetic goes the other way. Take a second-order low-pass section with a corner at 1 kHz built round a 10 kΩ impedance level. The inductor required is \(L = R/(2\pi f_0) = 1.59\) H. An inductor of that value is a heavy iron-cored component with perhaps 200 Ω of winding resistance, which by itself limits the \(Q\) of the section to about 50 and inserts a fixed 0.2 dB of loss; it picks up mains hum from any transformer within a foot of it; and its inductance depends on how much current is flowing through it. To get a corner at 50 Hz instead, that inductor becomes 32 H.
Four separate benefits follow from replacing it with an amplifier.
- No inductors. Only resistors and capacitors are needed, both of which are available as small, cheap, stable, close-tolerance components at any value the design calls for. The op-amp manufactures the phase shift that the inductor would otherwise have supplied.
- Gain is available. A passive filter always attenuates; an active one can have a pass-band gain greater than one, and in the Sallen-Key circuit of Section 4 that gain is not a bolt-on extra but the very quantity that sets the shape of the response.
- Buffering. The op-amp output impedance is milliohms, so the filter’s response does not change when a load is connected, and the input can be made a high-impedance node so the filter does not disturb whatever drives it.
- Cascading works. Because each stage is isolated from the next, the overall response really is the product of the individual responses. A designer can therefore build a sixth-order filter by choosing three second-order sections from a table, which is impossible with passive sections that load one another.
The limits are worth stating in the same breath. An active filter needs a power supply; it adds the op-amp’s noise and offset to the signal; it cannot handle power, so it filters signals and never a loudspeaker feed or a mains supply; and it stops working at the frequency where the op-amp’s own gain-bandwidth product runs out. A good rule is that the loop gain must still be comfortably larger than the section gain at the cutoff frequency, which for a 1 MHz-gain-bandwidth device such as a 741 means keeping \(f_0\) below roughly 10 kHz, and for a modern 10 MHz part below about 100 kHz.
Figure 28.1 sets out the four responses and the vocabulary that goes with them. The low-pass filter passes everything from d.c. up to a cutoff \(f_c\); the high-pass filter does the reverse. A band-pass filter has two cutoffs, and is described either by them or, more usefully when the band is narrow, by its centre frequency \(f_0 = \sqrt{f_1f_2}\), its bandwidth \(\text{BW} = f_2-f_1\) and the ratio of the two, the quality factor \(Q = f_0/\text{BW}\). A band-stop or notch filter removes one narrow band and passes the rest.
The dashed outlines in the figure are what one would like: unity gain in the pass band, zero in the stop band, and a vertical wall between them. No network of finitely many components can do that, because a vertical wall implies a discontinuous transfer function, and a discontinuity in frequency corresponds to a response in time that begins before the input arrives. What is obtainable is a transition band: a finite region between the edge of the pass band and the point where the required stop-band attenuation is finally reached. Every design decision in this chapter is ultimately a decision about how much circuitry to spend narrowing that band and what to accept in exchange.
2 Transfer Function, Cutoff and Roll-Off
A filter is described by its transfer function, the ratio of output to input phasor as a function of complex frequency \(s = j\omega\):
The degree \(n\) of the denominator \(D(s)\) is the order of the filter, and it equals the number of independent energy-storing elements — capacitors, in an active filter — in the signal path.
The cutoff frequency is defined at the point where the magnitude has fallen to \(1/\sqrt{2} = 0.7071\) of its pass-band value. That factor looks arbitrary until one squares it: the output power is halved, so \(f_c\) is the half-power frequency, and in decibels it is
Well beyond the corner every low-pass response settles onto a straight-line asymptote. For an \(n\)th-order function the denominator is dominated by its \(s^n\) term, so \(|H| \propto f^{-n}\) and
A decade is a factor of ten in frequency and an octave a factor of two, so the two numbers describe the same slope. Order 1 gives 20 dB/decade or 6.02 dB/octave; order 2 gives 40 and 12.04; order 4 gives 80 and 24.08.
Figure 28.2 makes the point that matters for design. All three curves are pinned to the same −3 dB point at \(f_c\); ordering does not move the corner, it only steepens the descent beyond it. One decade above cutoff the first-order filter is 20.0 dB down, the second-order 40.0 dB down and the fourth-order 80.0 dB down — and this is precisely how a specification is converted into an order.
An anti-aliasing filter ahead of a sampler must pass 1 kHz with no more than 3 dB of loss and must attenuate everything at and above 10 kHz by at least 45 dB. What order of Butterworth filter is needed?
The stop-band frequency is one decade above the cutoff, so the requirement is \(20n \ge 45\), giving \(n \ge 2.25\) and therefore \(n = 3\). Checking against the exact Butterworth magnitude rather than the asymptote, \(|H| = [1+(f/f_c)^{2n}]^{-1/2}\), a third-order filter at \(f/f_c = 10\) gives \(-10\log_{10}(1+10^{6}) = -60.0\) dB, comfortably inside the requirement; a second-order filter would give only \(-40.0\) dB and would fail by 5 dB.
Now change the specification so that the stop band begins at 2 kHz — one octave up instead of one decade. The requirement becomes \(6.02n \ge 45\), so \(n \ge 7.5\) and an eighth-order filter is needed: four cascaded Sallen-Key sections instead of one and a half. Halving the width of the transition band has cost five extra poles. This is the single most expensive decision in filter design, and it is why a specification should never demand a narrower transition than the application truly needs.
3 First-Order Active Low-Pass and High-Pass
The first-order sections are worth doing carefully, because everything in the rest of the chapter is built from them and because the algebra is short enough to see all of.
Low-pass. Put an RC lag network in front of a non-inverting amplifier of gain \(K = 1 + R_f/R_i\). No current flows into the op-amp input, so the network is unloaded and is a simple potential divider between \(R\) and the capacitor’s impedance \(1/sC\):
The magnitude and phase follow at once:
At \(f = f_c\) the magnitude is \(K/\sqrt2\), which is \(-3.01\) dB relative to the pass-band gain, and the phase is exactly \(-45^\circ\). A decade above, the magnitude is \(-20.04\) dB and the phase \(-84.3^\circ\), on its way to the \(-90^\circ\) that a single pole can never quite reach.
High-pass. Exchange the resistor and the capacitor. The divider now favours the resistor at high frequency:
The same \(f_c = 1/2\pi RC\), the same 20 dB/decade slope but now rising, and a phase that runs from \(+90^\circ\) at d.c. down through \(+45^\circ\) at \(f_c\) towards zero. With \(R = 16\ \text{k}\Omega\) and \(C = 10\) nF the section is 3.01 dB down at 994.7 Hz, 6.95 dB down at 500 Hz and 20.0 dB down at 100 Hz.
Two practical points. First, the gain-setting resistors \(R_f\) and \(R_i\) play no part in the frequency response at all; the corner is set purely by \(R\) and \(C\), and the gain is set purely by the feedback pair, so the two can be adjusted independently. That orthogonality is one of the reasons the active form is pleasant to design with. Second, the resistor \(R\) must have a d.c. path to ground for the op-amp’s input bias current; in the high-pass section that path is \(R\) itself, but in the low-pass section it is \(R\) in series with whatever drives the input, so a capacitively-coupled source needs a bias resistor added.
A first-order section can never be anything but a first-order section, so its roll-off is fixed at 20 dB/decade and its response near the corner is fixed too — there is no free parameter left to shape it with. To do better we need two capacitors and a way of coupling them that does not simply cascade two identical lags. Cascading two identical RC lags gives 40 dB/decade eventually, but the two corners interact and the combined \(-3\) dB point moves down to \(0.644f_c\), with the response already 6 dB down where a single section would be 3 dB down. The Sallen-Key arrangement of the next section fixes this by feeding some of the output back to the first capacitor.
4 The Sallen-Key Second-Order Section
Figure 28.3 shows the circuit. Two RC lags in cascade drive the non-inverting input of an amplifier of gain \(K = 1 + R_f/R_i\), and the top of the first capacitor, instead of going to ground, is returned to the amplifier output. That returned capacitor is the whole trick: near the corner frequency the output is larger than the signal at node A and leads it in phase, so \(C_1\) injects current back into node A and partially cancels the loss through \(R_1\). The network resonates without an inductor.
Write node equations. Let the nodes be A between the resistors, B at the op-amp input, and \(V_o\) at the output, with \(V_B = V_o/K\) because the amplifier is non-inverting. At node B, the current through \(R_2\) equals the current into \(C_2\), and at node A the currents through \(R_1\), \(R_2\) and \(C_1\) sum to zero:
The first equation gives \(V_A = V_B(1+sR_2C_2)\). Substituting that and \(V_B = V_o/K\) into the second, multiplying through by \(R_1\) and collecting terms in \(V_o\) leads, after a page of routine algebra, to
Every second-order low-pass function can be written in the standard form
and comparing coefficients identifies the two quantities that describe the section completely:
\(\omega_0\) is the natural or pole frequency and \(Q\) the quality factor; \(\alpha = 1/Q\) is the damping factor and \(\zeta = \alpha/2\) the damping ratio, all four being names for the same one degree of freedom. Note where \(K\) sits: it appears only in \(Q\), never in \(\omega_0\). Frequency and shape are separately adjustable.
Equal component: put \(R_1 = R_2 = R\) and \(C_1 = C_2 = C\). Then \(\omega_0 = 1/RC\) and \(1/Q = 2 + (1-K) = 3-K\), so \(f_0 = 1/2\pi RC\) and \(Q = 1/(3-K)\). One resistor ratio sets \(Q\); the pass-band gain is forced to equal \(K\).
Unity gain: put \(K = 1\) and \(R_1 = R_2 = R\), leaving the capacitors unequal. Then \(1/Q = 2\sqrt{C_2/C_1}\), so \(Q = \tfrac12\sqrt{C_1/C_2}\) and \(f_0 = 1/(2\pi R\sqrt{C_1C_2})\). The gain is fixed at one, \(Q\) is set by a capacitor ratio, and the amplifier is a plain follower — the lowest-offset, lowest-noise option.
The equal-component form is the one to use when the design must be tuned or when a gain is wanted anyway, and it has an unpleasant sensitivity to watch: as \(K\) approaches 3, \(Q\) approaches infinity, and a small error in \(R_f/R_i\) produces a large error in \(Q\). At \(K = 3\) exactly the denominator loses its damping term altogether and the circuit oscillates. For a Butterworth response \(Q = 0.7071\) requires \(K = 3 - \sqrt2 = 1.5858\), which is a comfortable distance from the cliff; for a high-\(Q\) section such as \(Q = 10\), \(K = 2.9\) and a 1 per cent resistor error shifts \(Q\) by about 30 per cent. High-\(Q\) sections should use the unity-gain form or the state-variable filter of Section 6.
5 Butterworth, Chebyshev, Bessel and Cascading
With \(\omega_0\) and \(Q\) available as independent design variables, the question becomes what values of \(Q\) to choose. There is no single right answer, because three desirable properties — a flat pass band, a fast transition, and a faithful reproduction of the shape of a pulse — conflict, and each classical response family is a different resolution of the conflict.
Butterworth maximises the flatness of the pass band. Its magnitude is
the first \(2n-1\) derivatives of which vanish at d.c., which is what "maximally flat" means. Every pole lies on a circle of radius \(\omega_0\), so all sections share the same pole frequency and differ only in \(Q\), and the response is exactly \(-3.01\) dB at \(f_c\) for every order.
Chebyshev gives up flatness and allows the pass band to ripple by a chosen amount — commonly 0.5 dB or 1 dB — in exchange for a steeper initial descent. Its poles lie on an ellipse, so the sections have different pole frequencies as well as different \(Q\) values, and the \(Q\) values are markedly higher. Bessel abandons selectivity altogether in favour of a maximally flat group delay: all frequency components are delayed by nearly the same time, so a square wave emerges as a square wave with rounded corners rather than with overshoot and ringing. Its \(Q\) values are the lowest of the three.
| Family | Order 2 | Order 3 | Order 4 | Loss at \(2f_c\) | Loss at \(10f_c\) | Group-delay spread |
|---|---|---|---|---|---|---|
| Butterworth | \(Q=0.7071\) \(\alpha=1.4142\) | \(Q=0.5\), \(Q=1.000\) \(\alpha=2.000,\ 1.000\) | \(Q=0.5412,\ 1.3066\) \(\alpha=1.8478,\ 0.7654\) | −12.30 dB | −40.00 dB | 20.7 % over the pass band |
| Chebyshev 0.5 dB | \(Q=0.864\) | \(Q=0.5,\ 1.706\) | \(Q=0.705,\ 2.941\) | −14.23 dB | −42.58 dB | 66.1 % — the worst |
| Bessel | \(Q=0.5774\) | \(Q=0.5,\ 0.691\) | \(Q=0.5219,\ 0.8055\) | −9.82 dB | −35.89 dB | 19.1 % — the best |
Second-order attenuations, all three families scaled so that the −3 dB point falls at the same \(f_c\). At order 4 the same comparison gives −24.10, −34.12 and −13.41 dB at \(2f_c\) — the gap between Chebyshev and Bessel widens rapidly with order.
The 0.5 dB Chebyshev entries need a word of warning. Its natural cutoff is the ripple-band edge, the last frequency at which the response is still within its 0.5 dB band, and the −3 dB point lies further out — at \(1.390f_c\) for order 2 and \(1.093f_c\) for order 4. Tables quote one or the other, and mixing them is a common way of building a filter with the wrong cutoff.
Cascading to higher order. An odd order is built from one first-order section plus \((n-1)/2\) second-order sections; an even order from \(n/2\) second-order sections. The \(Q\) values are not free: they come from the table, and they are not equal. A fourth-order Butterworth is not two Butterworth second-order sections in cascade. Two \(Q = 0.7071\) sections would be 6 dB down at \(f_c\) instead of 3, and the pass band would sag. The correct pair is \(Q = 0.5412\) (a section that is gently over-damped) followed by \(Q = 1.3066\) (a section that peaks by 2.4 dB); the sag of the first exactly cancels the peak of the second and the sum is maximally flat. In the equal-component Sallen-Key form these need \(K = 3 - \alpha\), that is \(K = 1.1522\) and \(K = 2.2346\), so \(R_f/R_i = 0.1522\) and \(1.2346\) respectively.
Put the lowest-\(Q\) section at the input. The high-\(Q\) section peaks near \(f_0\), so if it comes first a large signal near the corner can drive it into clipping even though the overall filter output is well within range. Placing it last means the earlier sections have already attenuated whatever would have caused the trouble. For noise the argument runs the other way, so where dynamic range matters at both ends the answer is a compromise; for undergraduate design work, lowest \(Q\) first is the safe rule.
6 Band-Pass, State-Variable and Notch Filters
A band-pass response can always be made by cascading a high-pass and a low-pass section, and for a wide band — say \(Q\) below about 1 — that is exactly what one does. For a narrow band it is a poor method, because the two corners must be placed close together and the result is sensitive to the difference between two nearly equal numbers. The multiple-feedback (MFB) band-pass filter on the left of Figure 28.4 uses a single op-amp in its inverting configuration and gives a true second-order band-pass response directly.
With both capacitors equal to \(C\), analysis of the inverting node gives
from which, by comparison with the standard band-pass form \(H = -A_0(\omega_0/Q)s/[s^2+(\omega_0/Q)s+\omega_0^2]\),
The bandwidth depends only on \(R_3\) and \(C\), and the centre frequency can then be moved with \(R_2\) without disturbing it — \(R_2\) appears in \(\omega_0\) but in neither \(Q\) nor \(A_0\). That is what makes the circuit convenient to tune.
Inverting the design equations, for a chosen \(f_0\), \(Q\), mid-band gain \(A_0\) and capacitor \(C\):
The expression for \(R_2\) is only positive if \(2Q^2 > |A_0|\), which is the circuit’s one hard constraint: gain and \(Q\) cannot both be made large. Above about \(Q = 10\) the design also becomes very sensitive to component tolerance and the state-variable filter is the better choice.
Design an MFB band-pass with \(f_0 = 1\) kHz, \(Q = 5\) and a mid-band gain of −10, using 10 nF capacitors. Check the constraint first: \(2Q^2 = 50 > 10\), so the design is feasible.
With \(\omega_0 = 2\pi(1000) = 6283.2\) rad/s, \(R_3 = 2(5)/(6283.2\times10^{-8}) = 159.2\ \text{k}\Omega\); \(R_1 = 159\,155/20 = 7958\ \Omega\); and \(R_2 = 5/[6283.2\times10^{-8}(50-10)] = 1989\ \Omega\). The nearest preferred values are \(R_1 = 8.2\ \text{k}\Omega\), \(R_2 = 2.0\ \text{k}\Omega\), \(R_3 = 160\ \text{k}\Omega\).
Recomputing from the standard values gives \(f_0 = 992.3\) Hz, \(Q = 4.99\), \(A_0 = -9.76\) (19.79 dB) and \(\text{BW} = 198.9\) Hz, with half-power points at 897.8 Hz and 1096.7 Hz. Every figure is within one per cent of target, which is a good deal better than the components themselves are specified to and confirms that the design is not sitting near a sensitive point.
The state-variable filter takes three op-amps — a summing amplifier followed by two integrators — and produces high-pass, band-pass and low-pass outputs simultaneously from the same network, because the input of each integrator is the derivative of its output. Its virtues are that \(f_0\) is set by the two integrator time constants, \(f_0 = 1/2\pi RC\), and \(Q\) by an entirely separate damping resistor ratio, typically \(Q = \tfrac13(1+R_2/R_1)\), so the two can be trimmed independently; that \(Q\) values of 50 or 100 are achievable with ordinary components; and that its sensitivity to component tolerance is roughly \(1/Q\) times better than the single-op-amp circuits. A \(Q\) of 5 needs \(R_2/R_1 = 14\). It is the standard building block of graphic equalisers and of laboratory filter instruments, and its cost is three op-amps instead of one.
The twin-T notch on the right of Figure 28.4 is the standard way of removing a single interfering frequency — almost always mains hum at 50 Hz or its harmonics. Two T-networks are placed in parallel: a low-pass T of two resistors \(R\) with \(2C\) to ground, and a high-pass T of two capacitors \(C\) with \(R/2\) to ground. At \(f_0 = 1/2\pi RC\) the two paths deliver signals of equal magnitude and opposite phase, and the output is theoretically zero. With \(R = 33\ \text{k}\Omega\) and \(C = 100\) nF the notch sits at 48.2 Hz; the exact value for 50 Hz would be 31.83 kΩ. The passive twin-T has a fixed and rather poor \(Q\) of 0.25, so it removes a wide swathe of the spectrum around the interference. Bootstrapping the grounded ends from a buffered fraction \(k\) of the output raises this to \(Q = 1/[4(1-k)]\): \(k = 0.9\) gives \(Q = 2.5\) and \(k = 0.95\) gives \(Q = 5\), narrowing the 48.2 Hz notch to a 9.6 Hz bandwidth. The depth of the null depends entirely on how well the two Ts are matched, which is why a twin-T is usually built with one trimmer.
7 A Complete Design, Carried Through and Checked
The pieces are now all in place. This section carries one specification through to a component list and then verifies it, which is the order in which the work should always be done.
Specification. Low-pass, cutoff 1 kHz, maximally flat pass band, at least 20 dB of attenuation at 5 kHz, driven from a low-impedance source, load 10 kΩ.
Step 1 — order and family. Maximally flat means Butterworth. At 5 kHz, that is \(f/f_c = 5\), a second-order Butterworth gives \(-10\log_{10}(1+5^{4}) = -28.06\) dB, which clears the 20 dB requirement with 8 dB to spare. Second order it is, so one Sallen-Key section, and from the table \(Q = 0.7071\), \(\alpha = 1.4142\).
Step 2 — choose the capacitor first. Capacitors come in far fewer values than resistors and their tolerances are worse, so the capacitor is always chosen first and the resistor computed to suit. The working range for audio is 1 nF to 100 nF: too small and stray capacitance (a few picofarads of board and op-amp input capacitance) matters, too large and the resistors become small enough to load the op-amp output. Take \(C_1 = C_2 = C = 10\) nF.
Step 3 — compute R. The equal-component form has \(f_0 = 1/2\pi RC\), so
The nearest E24 value is 16 kΩ, which puts the pole frequency at \(f_0 = 1/[2\pi(16\,000)(10^{-8})] = 994.7\) Hz — an error of 0.53 per cent, far smaller than the ±5 per cent the capacitor is likely to hold.
Step 4 — set Q with the gain. \(Q = 1/(3-K)\) requires \(K = 3 - 1/Q = 3 - \sqrt2 = 1.5858\), so \(R_f/R_i = 0.5858\). Taking \(R_i = 22\ \text{k}\Omega\) gives \(R_f = 12.89\ \text{k}\Omega\); the nearest preferred value is 13 kΩ, so \(K = 1 + 13/22 = 1.5909\) and \(Q = 1/(3-1.5909) = 0.7097\). The \(Q\) error is 0.37 per cent.
Step 5 — verify. With the standard values fitted, the magnitude is
| \(f\) | \(|H|\) | Gain (dB) | Relative to d.c. (dB) | Phase |
|---|---|---|---|---|
| 100 Hz | 1.5909 | 4.03 | 0.00 | \(-8.1^\circ\) |
| 500 Hz | 1.5451 | 3.78 | \(-0.25\) | \(-43.5^\circ\) |
| 994.7 Hz | 1.1291 | 1.05 | \(-2.98\) | \(-90.0^\circ\) |
| 2 kHz | 0.3827 | \(-8.34\) | \(-12.38\) | \(-137.0^\circ\) |
| 5 kHz | 0.0629 | \(-24.02\) | \(-28.06\) | \(-163.7^\circ\) |
| 10 kHz | 0.0157 | \(-36.06\) | \(-40.09\) | \(-171.9^\circ\) |
Reading the check. The response is 2.98 dB down at 994.7 Hz rather than 3.01, because \(Q\) came out marginally high; solving for the true half-power point gives 998.3 Hz, 0.17 per cent below the specified 1 kHz. The phase is exactly \(-90^\circ\) at \(f_0\), as it must be for any second-order section whatever its \(Q\), which makes it the most reliable thing to look for on an oscilloscope when checking a built filter. At 5 kHz the attenuation is 28.06 dB relative to the pass band, meeting the specification. At 10 kHz it is 40.09 dB — one decade above \(f_0\) and 40 dB down, the second-order asymptote arriving exactly on schedule.
The pass-band gain. Note that \(K = 1.5909\) is not a free choice here: it was forced on us by the requirement \(Q = 0.7071\). The filter therefore has 4.03 dB of gain whether or not that is wanted. If unity gain is required, either follow the filter with an attenuator of \(1/1.5909 = 0.629\), or use the unity-gain form instead: \(K = 1\), \(R_1 = R_2 = R\), \(C_1 = 2C_2\). With \(C_2 = 10\) nF and \(C_1 = 20\) nF, \(Q = \tfrac12\sqrt{20/10} = 0.7071\) exactly, and \(R = 1/(2\pi f_0\sqrt{C_1C_2}) = 11.31\ \text{k}\Omega\), for which 11.3 kΩ (E96) gives \(f_0 = 995.9\) Hz. That version gets its \(Q\) from a capacitor ratio, which is easier to hold accurately than an absolute value, and it uses no gain-setting resistors at all.
Two things remain to be checked on the bench before the design is signed off. The first is the op-amp: at \(f_0\) it must still have loop gain to spare, so with a 1 MHz gain-bandwidth product and a closed-loop gain of 1.59 the available loop gain at 1 kHz is about \(10^{6}/(1000\times1.59) = 629\), which is ample. The second is the d.c. path for the input bias current, which here runs through \(R_1\) and \(R_2\) to the source; if the source is capacitively coupled, the 32 kΩ total will float and the output will sit at a supply rail. A filter that works perfectly in simulation and produces a saturated output on the bench has almost always been caught by that.
8 Summary and Key Results
| Quantity | Expression | Value in the worked design |
|---|---|---|
| Cutoff definition | \(|H| = 1/\sqrt2\) of pass band | \(-3.0103\) dB, half output power |
| Roll-off | \(20n\) dB/decade \(= 6.02n\) dB/octave | \(n=2\): 40.0 dB/dec, 12.04 dB/oct |
| First-order section | \(f_c = 1/2\pi RC\) | \(R=16\) kΩ, \(C=10\) nF → 994.7 Hz |
| Sallen-Key pole frequency | \(\omega_0 = 1/\sqrt{R_1R_2C_1C_2}\) | equal-component: \(f_0 = 1/2\pi RC = 994.7\) Hz |
| Sallen-Key Q (equal component) | \(Q = 1/(3-K)\), \(K = 1+R_f/R_i\) | \(R_f=13\) kΩ, \(R_i=22\) kΩ → \(K=1.5909\), \(Q=0.7097\) |
| Sallen-Key Q (unity gain) | \(Q = \tfrac12\sqrt{C_1/C_2}\) | \(C_1=20\) nF, \(C_2=10\) nF → \(Q = 0.7071\), \(R=11.3\) kΩ |
| Butterworth damping, \(n=2\) | \(\alpha = 1.4142\) | \(Q = 0.7071\), \(K = 1.5858\) |
| Butterworth damping, \(n=3\) | first order \(+\) \(\alpha = 1.000\) | \(Q = 1.000\), \(K = 2.000\) |
| Butterworth damping, \(n=4\) | \(\alpha = 1.8478\) then \(0.7654\) | \(Q = 0.5412\) (\(K=1.1522\)), \(Q = 1.3066\) (\(K=2.2346\)) |
| MFB band-pass | \(R_3 = 2Q/\omega_0C\), \(R_1 = R_3/2|A_0|\), \(R_2 = Q/[\omega_0C(2Q^2-|A_0|)]\) | 8.2 kΩ, 2.0 kΩ, 160 kΩ → 992.3 Hz, \(Q=4.99\), \(A_0=-9.76\), BW 198.9 Hz |
| State-variable | \(f_0 = 1/2\pi RC\), \(Q = \tfrac13(1+R_2/R_1)\) | \(Q=5\) needs \(R_2/R_1 = 14\) |
| Twin-T notch | \(f_0 = 1/2\pi RC\), \(Q = 1/[4(1-k)]\) | 33 kΩ, 100 nF → 48.2 Hz; \(k=0.95\) gives \(Q=5\) |
| Verified design | \(|H| = K/\sqrt{[1-(f/f_0)^2]^2+(f/f_0Q)^2}\) | \(-2.98\) dB at 994.7 Hz, \(-28.06\) dB at 5 kHz, \(-40.09\) dB at 10 kHz |
9 Common Mistakes
Two first-order sections each with \(f_c = 1\) kHz do not make a second-order filter with a 1 kHz cutoff. Each is 3.01 dB down at 1 kHz, so the pair is 6.02 dB down there, and the combined half-power point has moved to \(f_c\sqrt{2^{1/n}-1} = 0.644f_c = 644\) Hz. For three identical sections the factor is 0.510 and for four 0.435. The same error appears in a different disguise when a fourth-order Butterworth is built as two \(Q = 0.7071\) Sallen-Key sections: the response sags 6 dB at \(f_c\) instead of 3 and is not maximally flat anywhere. A genuine \(n\)th-order response requires sections of the specific, unequal \(Q\) values in the table — 0.5412 and 1.3066 for fourth-order Butterworth — and the peaking of the high-\(Q\) section is not a defect but the thing that cancels the sag of the low-\(Q\) one.
\(\omega_0\) in the standard second-order form is the pole frequency, and it coincides with the −3 dB point only when \(Q = 0.7071\). For a Bessel section (\(Q = 0.5774\)) the response is already 4.4 dB down at \(f_0\); for a \(Q = 2\) section it is 6.0 dB up. Chebyshev tables compound the confusion by normalising to the ripple-band edge instead, so a 0.5 dB second-order Chebyshev reaches −3 dB only at \(1.390f_0\) and a fourth-order one at \(1.093f_0\). Before using any published table, check which frequency it is normalised to; a filter built from a Chebyshev table read as though it were a Butterworth table has its cutoff 39 per cent too low.
\(Q = 1/(3-K)\) is a hyperbola with a pole at \(K = 3\), so the sensitivity of \(Q\) to \(K\) is \(Q\) itself: a fractional error \(\delta\) in \((3-K)\) produces a fractional error \(\delta\) in \(Q\), but \((3-K)\) is a difference and its fractional error is magnified by \(K/(3-K) = KQ\). At \(Q = 0.71\) a 1 per cent error in \(R_f\) shifts \(Q\) by about 1.1 per cent and nobody notices. At \(Q = 10\), \(K = 2.9\) and the same 1 per cent error shifts \(Q\) by roughly 30 per cent; at \(K = 3\) the section becomes the Wien-bridge oscillator of Chapter 29. Any section needing \(Q\) above about 3 should use the unity-gain form, where \(Q\) depends on a capacitor ratio, or the state-variable filter, where it depends on a resistor ratio that has no pole in it.
10 Chapter Review
1. Design an equal-component Sallen-Key low-pass Butterworth section with \(f_0 = 3.4\) kHz (the top of a telephone channel) using 4.7 nF capacitors, and state the attenuation the finished section gives at 8 kHz.
The equal-component form has \(f_0 = 1/2\pi RC\), so \(R = 1/[2\pi(3400)(4.7\times10^{-9})] = 9959\ \Omega\); the nearest preferred value is 10 kΩ, giving \(f_0 = 1/[2\pi(10^{4})(4.7\times10^{-9})] = 3386\) Hz, 0.41 per cent low. For Butterworth, \(Q = 0.7071\) needs \(K = 3-\sqrt2 = 1.5858\), so \(R_f/R_i = 0.5858\); with \(R_i = 22\ \text{k}\Omega\), \(R_f = 12.89\ \text{k}\Omega\) and the preferred 13 kΩ gives \(K = 1.5909\) and \(Q = 0.7097\). At 8 kHz, \(f/f_0 = 8000/3386 = 2.363\), so \([1-(2.363)^2]^2 = (1-5.583)^2 = 21.00\) and \((2.363/0.7097)^2 = 11.09\), giving \(|H| = 1.5909/\sqrt{32.09} = 0.2809\). Relative to the d.c. gain of 1.5909 that is \(20\log_{10}(0.2809/1.5909) = -15.06\) dB. The section therefore attenuates 8 kHz by just over 15 dB, which for a telephone anti-aliasing job sampling at 8 kHz is not enough on its own — a fourth-order filter would give 30 dB there and is what is actually used.
2. Starting from the Sallen-Key transfer function, show that the equal-component form has \(Q = 1/(3-K)\), and find the value of \(K\) for which the circuit becomes an oscillator. What does \(K < 1\) mean physically?
The general result derived in Section 4 is \(1/Q = \omega_0[C_2(R_1+R_2)+R_1C_1(1-K)]\) with \(\omega_0 = 1/\sqrt{R_1R_2C_1C_2}\). Setting \(R_1 = R_2 = R\) and \(C_1 = C_2 = C\) gives \(\omega_0 = 1/RC\) and \(1/Q = (1/RC)[C(2R) + RC(1-K)] = 2 + 1 - K = 3-K\), so \(Q = 1/(3-K)\). As \(K \to 3\) the coefficient of \(s\) in the denominator vanishes, the poles move onto the imaginary axis at \(\pm j\omega_0\), and the natural response is an undamped sinusoid at \(f_0 = 1/2\pi RC\) — the circuit oscillates. For \(K\) above 3 the poles move into the right half-plane and the oscillation grows until the op-amp clips, which is exactly the start-up condition of Chapter 29. The other end of the range is equally instructive: \(K < 1\) is impossible for a non-inverting amplifier, whose gain is \(1 + R_f/R_i \ge 1\). \(K = 1\) is the minimum and gives \(Q = 0.5\), a critically damped section with two coincident real poles. So the equal-component Sallen-Key section can produce any \(Q\) from 0.5 upwards, but nothing below it; a section needing \(Q < 0.5\), such as the \(Q = 0.5\) first stage of a third-order Butterworth, is realised as two separate real poles instead.
3. An MFB band-pass filter is required with \(f_0 = 400\) Hz, a bandwidth of 50 Hz and a mid-band gain of \(-20\). Test whether the design is possible, and if so compute the components using 22 nF capacitors.
First \(Q = f_0/\text{BW} = 400/50 = 8\). The feasibility constraint is \(2Q^2 > |A_0|\): \(2(64) = 128 > 20\), so the design is possible with a good margin. With \(\omega_0 = 2\pi(400) = 2513.3\) rad/s and \(C = 22\) nF, the product \(\omega_0 C = 5.529\times10^{-5}\) S. Then \(R_3 = 2Q/\omega_0C = 16/(5.529\times10^{-5}) = 289.4\ \text{k}\Omega\); \(R_1 = R_3/(2|A_0|) = 289\,370/40 = 7234\ \Omega\); and \(R_2 = Q/[\omega_0C(2Q^2-|A_0|)] = 8/[5.529\times10^{-5}(128-20)] = 1340\ \Omega\). Preferred values 7.5 kΩ, 1.3 kΩ and 300 kΩ would give \(A_0 = -300\,000/15\,000 = -20.0\) exactly, \(\text{BW} = 1/[\pi(300\,000)(22\times 10^{-9})] = 48.2\) Hz and \(f_0 = (1/C)\sqrt{(R_1+R_2)/(R_1R_2R_3)} = 404.6\) Hz, hence \(Q = 404.6/48.2 = 8.39\). All three are within 5 per cent of specification. Note that \(R_3\) at nearly 300 kΩ is uncomfortably large — it makes the circuit noisy and sensitive to op-amp bias current — and the cure is a larger capacitor: at 100 nF, \(R_3\) falls to 63.7 kΩ.
4. Explain why a Bessel filter is chosen for a pulse channel and a Chebyshev for a channel-separation filter, using the group-delay and attenuation figures of Section 5.
Group delay is \(-\mathrm{d}\phi/\mathrm{d}\omega\), the time by which the filter retards a narrow band of frequencies around \(\omega\). If it is the same at every frequency, every component of a pulse is held up equally and the pulse emerges delayed but undistorted; if it varies, the components arrive at different times and the pulse acquires overshoot and ringing. Over the pass band up to \(f_c\) the second-order figures are 19.1 per cent variation for Bessel, 20.7 per cent for Butterworth and 66.1 per cent for Chebyshev, and the gap widens sharply with order. A digital data channel or an oscilloscope front end is judged entirely by pulse fidelity, so Bessel wins there despite giving only 9.82 dB of attenuation at \(2f_c\) against Butterworth’s 12.30 and Chebyshev’s 14.23. A channel-separation filter is judged by how well it keeps the adjacent channel out, and phase distortion of a steady tone is inaudible and harmless, so Chebyshev wins: at fourth order it delivers 34.12 dB at \(2f_c\) against Butterworth’s 24.10 and Bessel’s 13.41. Put another way, to match a fourth-order Chebyshev’s selectivity with a Bessel response would take roughly ten poles. The 0.5 dB of pass-band ripple that buys this is below the threshold at which a listener notices anything.
5. A 50 Hz twin-T notch built with \(R = 33\ \text{k}\Omega\) and \(C = 100\) nF is found to notch at 48.2 Hz and to attenuate a wanted 100 Hz signal by 2 dB. Diagnose both problems and propose fixes.
The frequency error first. \(f_0 = 1/2\pi RC = 1/[2\pi(33\,000)(10^{-7})] = 48.23\) Hz, 3.5 per cent below 50 Hz, because 33 kΩ is the nearest E24 value to the exact requirement \(R = 1/[2\pi(50)(10^{-7})] = 31\,831\ \Omega\). Using 32 kΩ would give 49.74 Hz and 31.6 kΩ would give 50.37 Hz; in practice the fixed resistors are made 30 kΩ with a 5 kΩ trimmer in series so the notch can be tuned onto the interference while watching it disappear. The matching of the two Ts matters more than their absolute value, since a mismatch limits the depth of the null; the \(R/2\) and \(2C\) elements are the ones to make from paralleled pairs of the main components rather than from separate values. The second problem is not an error at all but the defining property of a passive twin-T: its \(Q\) is fixed at 0.25, so its 3 dB bandwidth is \(f_0/Q = 193\) Hz, and a signal at 100 Hz — only one octave from the notch — is bound to be affected. The fix is to bootstrap the two grounded elements from a buffered fraction \(k\) of the output, which raises \(Q\) to \(1/[4(1-k)]\) without moving \(f_0\). Taking \(k = 0.95\) gives \(Q = 5\) and a bandwidth of \(48.2/5 = 9.6\) Hz, at which the 100 Hz signal is essentially untouched. The price is one extra op-amp and a notch that is now narrow enough to need the trimmer of the first paragraph.