586 Active Filter Circuits /ASSESSMENT PROBLEM Objective 3—Understand how to use cascaded first- and second-order Butterworth filters 15.4 For the circuit in Fig. 15.25, find values of i? t Answer: R x = 0.707 Q,, R 2 = 1.41 O and R 2 that yield a second-order prototype Butterworth high-pass filter. NOTE: Also try Chapter Problems 15.36,15.38 and 15.39. 15.5 Narrowband Bandpass and Bandreject Filters The cascade and parallel component designs for synthesizing bandpass and bandreject filters from simpler low-pass and high-pass filters have the restriction that only broadband, or low-Q, filters will result. (The Q, of course, stands for quality factor.) This limitation is due principally to the fact that the transfer functions for cascaded bandpass and parallel band- reject filters have discrete real poles. The synthesis techniques work best for cutoff frequencies that are widely separated and therefore yield the lowest quality factors. But the largest quality factor we can achieve with discrete real poles arises when the cutoff frequencies, and thus the pole locations, are the same. Consider the transfer function that results: H(s) s S + (x) c /\S + CO, SOOc s 2 + 2(o c s + at 2 0.5/3.5 s 2 + (3s + ar c ' (15.50) Eq. 15.50 is in the standard form of the transfer function of a bandpass filter, and thus we can determine the bandwidth and center frequency directly: /3 = 2(D C , (15.51) o? 0 = t*l (15-52) From Eqs. 15.51 and 15.52 and the definition of Q, we see that (Jin 00,. 1 Q = 2oO r (15.53) Thus with discrete real poles, the highest quality bandpass filter (or band- reject filter) we can achieve has Q = 1/2. To build active filters with high quality factor values, we need an op amp circuit that can produce a transfer function with complex conjugate poles. Figure 15.26 depicts one such circuit for us to analyze. At the invert- ing input of the op amp, we sum the currents to get K Figure 15.26 • An active high-Q bandpass filter. ^ -Vg X/sC R 3 15.5 Narrowband Bandpass and Bandreject Filters 587 Solving for V. v K = -£-• (15.54) At the node labeled a, we sum the currents to get Solving for V,-, v, v, - R = (1 V V I + 2sR x C - v a v, \/sC 1/sC + RJR 2 )V. A - Ri sR t CV 0 (15.55) Substituting Eq. 15.54 into Eq. 15.55 and then rearranging, we get an expression for the transfer function V 0 / V t : —s where H{S) = - —^ , (15.56) ^ 2 + TTTP + o R 3 C R CH R 3 C 2 R sq = R 1 ]\R 2 = R { + R 2 Since Eq. 15.56 is in the standard form of the transfer function for a bandpass filter, that is, s~ + ps + a>; we can equate terms and solve for the values of the resistors, which will achieve a specified center frequency («„), quality factor (Q), and passband gain (K): fi = ^; (15.57) Kf3 = —; (15.58) ^ = TT7*- (15l59) At this point, it is convenient to define the prototype version of the circuit in Fig. 15.25 as a circuit in which o> 0 = 1 rad/s and C = 1 F. Then the expressions for R^, R 2 , and R 3 can be given in terms of the desired quality factor and passband gain. We leave you to show (in Problem 15.45) that for the prototype circuit, the expressions for Ri, R 2 , and R 3 are Ri = Q/K, Ri = Q/(2Q 2 - K), R 3 = 2Q. Scaling is used to specify practical values for the circuit components. This design process is illustrated in Example 15.12. 588 Active Filter Circuits Designing a High-Q Bandpass Filter Design a bandpass filter, using the circuit in Fig. 15.26, which has a center frequency of 3000 Hz, a quality factor of 10, and a passband gain of 2. Use 0.01 /xF capacitors in your design. Compute the transfer func- tion of your circuit, and sketch a Bode plot of its mag- nitude response. Solution Since Q = 10 and K = 2, the values for R u R 2 , and R 3 in the prototype circuit are Ri = 10/2 = 5, R 2 = 10/(200 - 2) = 10/198, R 3 = 2(10) = 20. The scaling factors are kf = 6OOO77- and k m = lOfyfc/. After scaling, R ] = 26.5 kO, R 2 = 268.0 a, R 3 = 106.1 ka. The circuit is shown in Fig. 15.27. Substituting the values of resistance and capac- itance in Eq. 15.56 gives the transfer function for this circuit: H(s) = -37705 s 2 + 1885.05 + 355 X 10 6 It is easy to see that this transfer function meets the specification of the bandpass filter defined in the example. A Bode plot of its magnitude response is sketched in Fig. 15.28. Figure 15.27 • The high-G bandpass filter designed in ExampLe 15.12. 10 5 0 -3 -10 CQ 5 -15 -20 -25 -30 -35 -40 / / / t / V j • t 1 1 1 Mill 1 ,- 6 dB (gain 0 _ \ \ \ \ \ n f2) ^ 100 500 1000 5000 10,000 50,000100,000 /(Hz) Figure 15.28 • The Bode magnitude plot for the high-Q bandpass filter designed in Example 15.12. The parallel implementation of a bandreject filter that combines low- pass and high-pass filter components with a summing amplifier has the same low-<2 restriction as the cascaded bandpass filter. The circuit in Fig. 15.29 is an active high-Q bandreject filter known as the twin-T notch filter because of the two T-shaped parts of the circuit at the nodes labeled a and b. We begin the analysis of this circuit by summing the currents away from node a: (V n - V,)sC + (K - V 0 )sC + 2(V a -<rV 0 ) R = 0 or V a [2sCR + 2] - VjsCR + 2a] = sCRV;. (15.60) 15.5 Narrowband Bandpass and Bandreject Filters 589 J_ sC 6 R i <vw- sC b R « vv^ ' 1 2sC f o-K, ^—••—• K •+ :(l-«r)J? * <• (TV„\(TR Figure 15.29 • A high-Q active bandreject filter. Summing the currents away from node b yields Vb-V, V b - V a R R + (V b - <rV a )2sC = 0 or V b [2 + 2RCs) -V 0 [l + 2a RCs] = V h (15.61) Summing the currents away from the noninverting input terminal of the top op amp gives V — K (V 0 -V a )sC+ -2-^-t = 0 or sRCV a ~V b + (sRC + l)V 0 = 0. (15.62) From Eqs. 15.60-15.62, we can use Cramer's rule to solve for V 0 : 2(RCs + 1) 0 sCRV; 0 2(RCs + 1) V t -RCs -1 0 K = 2(RCs + 1) 0 -(RCs + 2a) 0 2{RCs + 1) -(2trRCs + 1) -AC* -1 ffCj + 1 (^cv + i)Vi R 2 C 2 s 2 + ARC{\ - a)s + 1' Rearranging Eq. 15.63, we can solve for the transfer function: 1 H(s) = f = S? + /? 2 C 2 2 , 4(1 - o) 1 R 2 C 2 \ (15.63) (15.64) 590 Active Filter Circuits which is in the standard form for the transfer function of a bandreject filter: l i 2 S + 0)() H(s) = -= s 2 + (3s + (4 Equating Eqs. 15.64 and 15.65 gives (15.65) 2,-2 ' R l C (15.66) P = 4(1 ~ cr) RC (15.67) In this circuit, we have three parameters (R, C, and a) and two design constraints (co a and /3). Thus one parameter is chosen arbitrarily; it is usu- ally the capacitor value because this value typically provides the fewest commercially available options. Once C is chosen, R = 1 a> n C (15.68) and A(o (> AQ (15.69) Example 15.13 illustrates the design of a high-Q active bandreject filter. Designing a High-Q Bandreject Filter Design a high-Q active bandreject filter (based on the circuit in Fig. 15.29) with a center frequency of 5000 rad/s and a bandwidth of 1000 rad/s. Use 1 /xF capacitors in your design. Solution In the bandreject prototype filter, co n = 1 rad/s, R = 1 H, and C = 1 F. As just discussed, once <o ( , and Q are given, C can be chosen arbitrarily, and R and cr can be found from Eqs. 15.68 and 15.69. From the specifications, Q = 5. Using Eqs. 15.68 and 15.69, we see that R = 200 il, a = 0.95. Therefore we need resistors with the values 200 ft (/?), 100 ft (i?/2), 190 ft (0-/2),and 10 ft [(1 - a)R]. The final design is depicted in Fig. 15.30, and the Bode magnitude plot is shown in Fig. 15.31. Figure 15.30 A The high-Q active bandreject filter designed in Example 15.13. 15.5 Narrowband Bandpass and Bandreject Filters 591 JLU 5 0 PQ -10 -15 / 1000 5000 10,000 co (rad/s) 50,000 100,000 Figure 15.31 • The Bode magnitude plot for the high-Q active bandreject filter designed in Example 15.13. /ASSESSMENT PROBLEMS Objective 4—Be able to use design equations to calculate component values for prototype narrowband, bandpass, and bandreject filters 15.5 Design an active bandpass filter with Q = 8, K = 5, and <o a = 1000 rad/s. Use 1 /xF capaci- tors, and specify the values of all resistors. Answer: R t = 1.6 kfl, R 2 = 65.04 a, R 3 = 16 kfl. NOTE: Also try Chapter Problem 15.60. 15.6 Design an active unity-gain bandreject filter with (o (} = 1000 rad/s and Q = 4. Use 2 jtF capacitors in your design, and specify the values of R and a. Answer: R - 500 Q, a = 0.9375. Practical Perspective Bass Volume Control We now look at an op amp circuit that can be used to control the amplifica- tion of an audio signal in the bass range. The audio range consists of signals having frequencies from 20 Hz to 20 kHz. The bass range includes frequen- cies up to 300 Hz. The volume control circuit and its frequency response are shown in Fig. 15.32. The particular response curve in the family of response curves is selected by adjusting the potentiometer setting in Fig. 15.32(a). In studying the frequency response curves in Fig. 15.32(b) note the fol- lowing. First, the gain in dB can be either positive or negative. If the gain is positive a signal in the bass range is amplified or boosted. If the gain is negative the signal is attenuated or cut. Second, it is possible to select a 592 Active Filter Circuits Vs* *—• *, (a) (b) Figure 15.32 A (a) Bass volume control circuit; (b) Bass volume control circuit frequency response. •V B Figure 15.33 A The s-domain circuit for the bass volume control. Note that a determines the potentiometer setting, so 0 £ a £ 1. response characteristic that yields unity gain (zero dB) for all frequencies in the bass range. As we shall see, if the potentiometer is set at its midpoint, the circuit will have no effect on signals in the bass range. Finally, as the frequency increases, all the characteristic responses approach zero dB or unity gain. Hence the volume control circuit will have no effect on signals in the upper end or treble range of the audio frequencies. The first step in analyzing the frequency response of the circuit in Fig. 15.32(a) is to calculate the transfer function V 0 /V s . To facilitate this calculation the s-domain equivalent circuit is given in Fig. 15.33. The node voltages V a and V b have been labeled in the circuit to support node volt- age analysis. The position of the potentiometer is determined by the numer- ical value of a, as noted in Fig. 15.33. To find the transfer function we write the three node voltage equations that describe the circuit and then solve the equations for the voltage ratio V () /V s . The node voltage equations are (1 - a)R 2 aR 2 + K -K /?. + (K -V^sC^O; + (V b -V a )sC l + V b ~Vo (1 - a)R 2 + aRo 0; = 0. These three node-voltage equations can be solved to find V 0 as a function of V s and hence the transfer function H(s): ' s It follows directly that -jRi + aR 2 + R&Cis) /?! + (!- a)R 2 + R&Cis' _ -(/?! + gR 2 + jcoR&Ci) W [#! + (1- a)R 2 + jaRiRfit]' Now let's verify that this transfer function will generate the family of fre- quency response curves depicted in Fig. 15.32(b). First note that when a = 0.5 the magnitude of H{j<o) is unity for all frequencies, i.e., |H0»)l \Ri + 0.5R 2 + jaR&Cil | A, + 0.5R 2 + jo}R x R 2 C x \ 1. 15.5 Narrowband Bandpass and Bandreject Filters 593 When w = Owe have \H(jO)\ = /¾ + aR 2 R t + (1 - a)R 2 Observe that \H(jO)\ at a = 1 is the reciprocal of \H(jO)\ at a = 0, that is /¾ + Ri 1 \H{m\a=x = *i \H(Ma=0' With a little thought the reader can see that the reciprocal relationship holds for all frequencies, not just <a = 0. For example a = 0.4 and a = 0.6 are symmetric with a = 0.5 and #0' w )«=o.4 - -(/¾ + 0.4/¾) + fofr/frd (/¾ + 0.6/¾) + j(oRiR 2 C x while Hence tf(;w) a = 0 .6 -(/¾ + 0.6/¾) + j(oRiR 2 Ci (/¾ + 0.4/¾) + jo)RiR 2 C l H(jo>) af= oA = H(j<»)a-0.6 It follows that depending on the value of a the volume control circuit can either amplify or attenuate the incoming signal. The numerical values of R\,R 2 , and C] are based on two design deci- sions. The first design choice is the passband amplification or attenuation in the bass range (as a>-*0). The second design choice is the frequency at which this passband amplification or attenuation is changed by 3 dB. The component values which satisfy the design decisions are calculated with a equal to either 1 or 0. As we have already observed, the maximum gain will be (/¾ + R 2 )/Ri and the maximum attenuation will be R\/(R\ + /¾). If we assume (/¾ + R 2 )/R\ » 1 then the gain (or attenuation) will differ by 3 dB from its maximum value when o> = 1//¾^. This can be seen by noting that H 1 R 2 C X «=i 1/¾ + R 2 + m 1/¾ + iR\\ /¾ + /¾ Ri + /1 and H J R 2 Cy a=0 II + /11 1*1 + /*ll 1/¾ + R 2 + /Xi| II + /11 1 //¾ + /¾ V2\ fll Ri + R 2 Ri V2 + /1 *i *1 + /¾ /V07E; Assess your understanding of this Practical Perspective by trying Chapter Problems 15.61 and 15.62. 594 Active Filter Circuits Summary • Active filters consist of op amps, resistors, and capacitors. They can be configured as low-pass, high-pass, bandpass, and bandreject filters. They overcome many of the disad- vantages associated with passive filters. (See page 560.) • A prototype low-pass filter has component values of Ri = R 2 = 1 H and C = 1 F, and it produces a unity passband gain and a cutoff frequency of 1 rad/s. The prototype high-pass filter has the same component val- ues and also produces a unity passband gain and a cut- off frequency of 1 rad/s. (See pages 561 and 562.) • Magnitude scaling can be used to alter component val- ues without changing the frequency response of a circuit. For a magnitude scale factor of k , the scaled (primed) values of resistance, capacitance, and inductance are R' = k m R, V = k m L, and C = C/k m . (See page 564.) • Frequency scaling can be used to shift the frequency response of a circuit to another frequency region without changing the overall shape of the frequency response. For a frequency scale factor of kf, the scaled (primed) values of resistance, capacitance, and inductance are R' = R, L' = L/kp and C = C/k f . (See page 564.) • Components can be scaled in both magnitude and fre- quency, with the scaled (primed) component values given by R' = k m R, L' = (k m /k f )L, and C = C/{k m k f ). (See page 564.) • The design of active low-pass and high-pass filters can begin with a prototype filter circuit. Scaling can then be applied to shift the frequency response to the desired cutoff frequency, using component values that are com- mercially available. (See page 565.) • An active broadband bandpass filter can be constructed using a cascade of a low-pass filter with the bandpass fil- ter's upper cutoff frequency, a high-pass filter with the bandpass filter's lower cutoff frequency, and (optionally) an inverting amplifier gain stage to achieve nonunity gain in the passband. Bandpass filters implemented in this fashion must be broadband filters (w c . 2 » <a c] ), so that the elements of the cascade can be specified inde- pendently of one another. (See page 568.) • An active broadband bandreject filter can be con- structed using a parallel combination of a low-pass filter with the bandreject filter's lower cutoff frequency and a high-pass filter with the bandreject filter's upper cutoff frequency. The outputs are then fed into a summing amplifier, which can produce nonunity gain in the pass- band. Bandreject filters implemented in this way must be broadband filters {co c2 » o> c .i), so that the low-pass and high-pass filter circuits can be designed independ- ently of one another. (See page 572.) • Higher order active filters have multiple poles in their transfer functions, resulting in a sharper transition from the passband to the stopband and thus a more nearly ideal frequency response. (See page 573.) • The transfer function of an nth-order Butterworth low- pass filter with a cutoff frequency of 1 rad/s can be determined from the equation H(s)H(s) = =- KJK ' i + (-i)V» by • finding the roots of the denominator polynomial • assigning the left-half plane roots to H(s) • writing the denominator of H(s) as a product of first- and second-order factors (See page 578-579.) • The fundamental problem in the design of a Butterworth filter is to determine the order of the filter. The filter specification usually defines the sharpness of the transition band in terms of the quantities A , a p -,A s> and <a r From these quantities, we calculate the smallest integer larger than the solution to either Eq. 15.42 or Eq. 15.46. (See page 583.) • A cascade of second-order low-pass op amp filters (Fig. 15.21) with 1 JQ resistors and capacitor values cho- sen to produce each factor in the Butterworth poly- nomial will produce an even-order Butterworth low-pass filter. Adding a prototype low-pass op amp filter will pro- duce an odd-order Butterworth low-pass filter. (See page 581.) • A cascade of second-order high-pass op amp filters (Fig. 15.25) with 1 F capacitors and resistor values cho- sen to produce each factor in the Butterworth poly- nomial will produce an even-order Butterworth high-pass filter. Adding a prototype high-pass op amp filter will produce an odd-order Butterworth high-pass filter. (See page 585.) • For both high- and low-pass Butterworth filters, fre- quency and magnitude scaling can be used to shift the cutoff frequency from 1 rad/s and to include realistic Problems 595 component values in the design. Cascading an inverting amplifier will produce a nonunity passband gain. (See page 580.) Butterworth low-pass and high-pass filters can be cas- caded to produce Butterworth bandpass filters of any order n. Butterworth low-pass and high-pass filters can be combined in parallel with a summing amplifier to produce a Butterworth bandreject filter of any order n. (See page 585.) If a high-(2, or narrowband, bandpass, or bandreject fil- ter is needed, the cascade or parallel combination will not work. Instead, the circuits shown in Figs. 15.26 and 15.29 are used with the appropriate design equations. Typically, capacitor values are chosen from those com- mercially available, and the design equations are used to specify the resistor values. (See page 586.) Problems Section 15.1 15.1 Find the transfer function V 0 /V; for the circuit shown in Fig. P15.1 if Zf is the equivalent imped- ance of the feedback circuit, Z, is the equivalent impedance of the input circuit, and the operational amplifier is ideal. Figure PI5.1 15.2 a) Use the results of Problem 15.1 to find the trans- fer function of the circuit shown in Fig. PI5.2. b) What is the gain of the circuit asw^ 0? c) What is the gain of the circuit as w —• oo? d) Do your answers to (b) and (c) make sense in terms of known circuit behavior? Figure P15.2 15.3 Repeat Problem 15.2, using the circuit shown in Fig. P15.3. Figure P15.3 DESIGN PROBLEM 15.4 Design an op amp-based low-pass filter with a cut- off frequency of 2500 Hz and a passband gain of 5 using a 10 nF capacitor. a) Draw your circuit, labeling the component val- ues and output voltage. b) If the value of the feedback resistor in the filter is changed but the value of the resistor in the forward path is unchanged, what characteristic of the filter is changed? 15.5 The input to the low-pass filter designed in Problem 15.4 is 2 cos cot V. a) Suppose the power supplies are ±V CC . What is the smallest value of V cc that will still cause the op amp to operate in its linear region? b) Find the output voltage when o> = a) c . c) Find the output voltage when w = 0.2(o c . d) Find the output voltage when oo = 5w t 15.6 a) Using the circuit in Fig. 15.1, design a low-pass filter with a passband gain of 10 dB and a cutoff frequency of 1 kHz. Assume a 750 nF capacitor is available. b) Draw the circuit diagram and label all components.