|
||||||
May 8, 1997 High-speed active filters made easyMark Sauerwald, Comlinear Products Group,
Passive filters built from resistors, inductors, and capacitors are the dominant filters for high-frequency applications. The performance of these filters is typically limited by the inductors, which are large, expensive, and far from electrically ideal. By using amplifiers and feedback, though, you can achieve the same filter characteristics without using inductors. An active filter's performance is limited by its nonideal amplifiers, so active filters traditionally have been constrained to frequencies below 1 MHz. With very high-speed operational amplifiers now available, it's feasible to build active filters with frequency ranges in the tens of megahertz and even some filters with corner frequencies greater than 100 MHz. The Sallen-Key filter topology employs an op amp as a fixed-gain block, and you can use this filter topology with current- or voltage-feedback amplifiers for lowpass-, highpass-, and bandpass-filter transfer functions. Although many filter responses exist, the three most useful are: c The Butterworth response, which maximizes the flatness in the passband of the filter. The response is flat near dc, begins to slowly roll off so that it is at 3 dB at the cutoff frequency, and approaches an ultimate rolloff rate of 20n dB/decade, where n is the order of the filter. Butterworth filters are a good choice when you must maintain gain flatness, especially at low frequencies. c The Bessel response, which minimizes the phase nonlinearity in the passband. In addition to altering the magnitude of the input signal depending on its frequency, filters introduce a frequency-dependent delay into the signal, which distorts nonsinusoidal signals. This filter minimizes such distortion. c The Chebyshev response, which provides faster rolloff than the Butterworth filter if you are willing to accept some ripple in the passband. Table 1 contains the parameters necessary to design filters as high as order 8 with Butterworth, Bessel, and Chebyshev responses. For the Chebyshev response, Table 1c shows 0.1-dB maximum ripple in the passband, and Table 1d shows 1-dB maximum ripple in the passband. Sallen-Key filter You can characterize a two-pole, Sallen-Key lowpass filter, like any other two-pole filter, by three parameters: K, v0, and Q, where K is the dc gain of the filter, v0 is the corner frequency of the filter, and Q is how close the two poles are in the s plane (Figure 1). The values of K, Q, and v0 are given by: (1a) (1b) (1c) where RF, RG, and m are values from Figure 1. To design a filter, you select the desired values for K, Q, and v0 and then use Equations 1a, 1b, and 1c to determine the component values you need. You obtain Q from Table 1, and your design specification determines v0 and K. In most filters, the actual value of v0 that you design for must be adjusted by the factor found in the filter table to get the proper 3-dB frequency. Because these three parameters are not completely independent, it may be impossible to simultaneously meet all three points. In this case, you should select another value of K and then use an amplifier before or after the filter to make the overall gain meet the system requirements. For example, if you need a Butterworth filter (Q= 0.707) with a dc gain of 4, there is no positive value of m that provides the required filter. Because a negative value of m implies the use of a negative-value resistor, you need to find another solution. If, however, you select K=1.5, by using an m of 0.707, you can obtain the required Q. To obtain the proper overall gain, place an amplifier with a gain of 2.66 in front of the filter. Even when you do all these things, there may be no solution to the desired filter, especially for high-Q sections. In this case, use the design techniques for a filter with unequal capacitors. (See box, "Designing Sallen-Key filters with unequal capacitors.") This box shows a more versatile circuit but one that is somewhat more difficult to design. To select component values, follow these steps:
For example, design a Butterworth filter with a cutoff frequency of 10 MHz and a dc gain of 1.5 using the Comlinear CLC430:
The final filter appears in Figure 2. In some cases, most often when you are trying to design a filter with a high Q, the constraint of equal capacitors can be too restrictive. If this situation occurs, use the information in the box, "Designing Sallen-Key filters with unequal capacitors" to get expressions for Q and v0 in which the capacitors can take on different values. Transform lowpass to highpass You can transform a lowpass filter into a highpass filter by changing capacitors into resistors and resistors into capacitors. Replace each capacitor of value C with a resistor of value 1/v0C, and replace each resistor with a capacitor of value 1/v0R. For example, replace the 100-pF capacitors in Figure 2 with 159V resistors, the 225V resistor with a 71-pF capacitor, and the 112.5V resistor with a 142-pF capacitor. Alternatively, you can calculate component values using a method similar to that shown in the previous seven steps. Figure 3 is a second-order, Sallen-Key highpass filter. The K, Q, and v0 are given by Equation 1a, Cascading reaches higher orders You can realize higher order filters by cascading lower order filters. Cascading n/2 second-order sections gets you a filter of order n. Note that to obtain a fourth-order Butterworth filter, you do not cascade two second-order Butterworth-filter sections. The pole locations for the fourth-order filter are all different, and you need different values of Q and v0 for each section. Cascading active filters of the Sallen-Key topology is easy. Because the output impedances of the filters are low, you can cascade the filters by connecting one section to the next. (Each filter section is designed assuming zero source impedance.) If you need an odd-order filter, add the last pole with a simple passive filter, R3C3, at the output of the active filter (Figure 4). When cascading filters, one question that arises is, "In what order should you place the sections?" The answer to this question is the same as the answer to all filter questions: "It depends." When you cascade filter sections, consider two factors: Are you liable to have intermediate signals that might clip or distort? And how important is noise? In filters with sections where there are large Qs, the maximum amplitude of the signal in between stages may be larger than the input or the output amplitude. This intermediate-stage magnitude can lead to distortion of the signal. To combat this effect, put the stages with the lowest Qs and gains first and cascade the higher gain sections later. Another way to increase the dynamic range is to use an op amp with larger dynamic range. To minimize the broadband noise that is present in an active filter, you should cascade stages with the highest Q, highest gain sections first, followed by lower gain sections. Other low-noise strategies include selecting low-noise op amps and using the lowest possible value resistors to reduce thermal noise. Making the op-amp choice Although many factors go into selecting an op amp, you need to look at just four parameters for most filter applications: bandwidth, recommended gain range, noise, and dynamic range. For bandwidth, you should generally select an op amp that has a bandwidth of at least 10 times and, preferably, 20 times the intended filter frequency. Thus, for a 10-MHz lowpass filter, use an op amp with at least 100 MHz of bandwidth. For a highpass filter, make certain that the bandwidth of the op amp is sufficient to pass all of your signal. Your op amp must have this bandwidth under the same gain, signal-level, and power-supply-voltage conditions that you intend to use it. An op amp that is specified as a 100-MHz device because it has 100-MHz bandwidth with ±15V supplies, 50-mV output level, and a gain of 1 is not likely to be useful in a 10-MHz filter application. If in doubt, configure the amplifier that you are considering at the gain that you are intending to use, and measure its bandwidth. Also, look at compensation for the finite bandwidth amplifier. (See box, "Compensating for finite-bandwidth amplifiers.") For recommended gain range, your filter design dictates the gains at which you will be setting your op amps. If your chosen op amp is a voltage-feedback device, using it at gains outside the recommended gain range is liable to lead to lower than expected bandwidth at best and oscillation at worst. For a current-feedback op amp, operation at gains outside the recommended range will likely force you to use resistor values for RF and RG that are too small or too large to be practical. The input voltage and current noises of the op amp contribute to the noise at the output of the filter. In applications in which noise is a major concern, you need to calculate these contributions (as well as the contributions of the thermal noise of the resistors in the circuit) to determine if you can accept the added noise due to an active filter. In filters that have high-Q sections, there might be intermediate signals that are larger than either the input or the output signal, and you might exceed the dynamic-range specification of the op amp. For the filter to operate properly, all signals must pass without clipping or excessive distortion. As a result, for the filter to operate properly, the dynamic range of the op amps you use must be large enough to pass these amplified signals without distortion. You determine dynamic range by the power supply you use and the headroom required by the op amp you are considering. Some op amps can operate reliably with outputs within a few hundred millivolts of the rails, whereas other op amps require several volts of headroom. In some cases, even though the signals you are filtering are small, you might want to consider wide-dynamic-range op amps for your filter application. Table 2 shows the four primary characteristics for some representative devices. In addition to these specifications, there are several secondary parameters that you probably need to consider. These parameters include phase linearity (which is generally better in current-feedback op amps), packaging (and whether duals or quads are available), power dissipation, price, and availability of technical support.
Mark Sauerwald is a product-marketing manager for the Comlinear Products Group of National Semiconductor Corp (Fort Collins, CO), where he previously was an applications engineer. He has worked on high-speed amplifiers, A/D converters, and serial-digital-communications products. He has a BSEE from the University of California--San Diego. You can build
active filters for use at tens of megahertz with the
newest high-speed op amps, thus avoiding the weaknesses
of passive RLC filters. Starting with the Sallen-Key
filter topology, you can implement a variety of
high-performance filters with a straightforward
procedure.
and
Figure 1 The basic two-pole, lowpass Sallen-Key filter is characterized by dc gain K, corner frequency v0, and spacing of its poles in the s-plane, Q.
The circuit in Figure A is a two-pole, Sallen-Key lowpass filter, which adds the variable n as an additional degree of freedom and so removes the constraint that the two capacitors remain equal. Like the filter in Figure 1, three parameters characterize this filter: K, the dc gain of the filter; Q, how close the two poles are in the s plane; and v0, the corner frequency of the filter. The values of K, Q, and v0 are given by Equation 1a and (2) (3) where Q and K are unitless and v0 is in units of radians per second. (Divide v0 by 2p to get the value in hertz.) To design a filter, you select the desired values for K, Q, and v0 and then to use Equations 1a, 2, and 3 to determine the component values required. You can obtain Q from Table 1. Your design determines v0, the cutoff frequency, as well as K. In most filters, you must adjust the actual value of v0 that you design for by the factor in Table 1 to get the proper 3-dB frequency. To select component values, 1. Select RF and RG to obtain the desired K. If the amplifier that you are using is a current-feedback amplifier, follow the guidelines in the data sheet to select appropriate val- ues of RF and RG. 2. Arbitrarily select a value for n. This value should be small for low-Q filters and should be larger for higher Q filters. Keep in mind when you choose n that the standard val- ues for capacitors are limited. 3. Determine the value of m you need using and the value of Q in Table 1. 4. If you cannot get a positive real solution for m, select another value of K or n and go back to Step 1. 5. Arbitrarily select a value for C. 6. Determine the value of v0 to use by multiplying your desired cutoff frequency by the value of v0 in Table 1. (If your frequency specification is in hertz, remember to multiply the value by 2p to get it into radians per sec- ond.) 7. Determine the value of R you need using 8. If R turns out to be too large or too small for practical pur- poses, select another value for C, then go back to Step 7. (If R is too big, select a larger value for C; if R is too small, select a smaller value of C.) Designing Sallen-Key filters with unequal capacitors
Figure A
and
To design a filter with unequal capacitors, you add variable n as another degree of freedom.
Table 1a--Butterworth lowpass filter Attenuation 2 1 0.707 15 3 1 1 1 21 4 1 0.541 1 1.306 27 5 1 0.618 1 1.62 1 33 6 1 0.518 1 0.707 1 1.932 39 7 1 0.555 1 0.802 1 2.247 1 45 8 1 0.51 1 0.601 1 0.9 1 2.563 51 Table 1b--Bessel lowpass filter n v1 Q1 v2 Q2 v3 Q3 v4 Q4 2 1.274 0.577 3 1.453 0.691 1.327 4 1.419 0.522 1.591 0.806 5 1.561 0.564 1.76 0.917 1.507 6 1.606 0.51 1.691 0.611 1.907 1.023 7 1.719 0.533 1.824 0.661 2.051 1.127 1.685 8 1.784 0.506 1.838 0.56 1.958 0.711 2.196 1.226 Table 1c--0.1-dB-ripple Chebyshev filter Attenuation 2 1.82 0.767 3.31 3 1.3 1.341 0.969 12.24 4 1.153 2.183 0.789 0.619 23.43 5 1.093 3.282 0.797 0.915 0.539 34.85 6 1.063 4.633 0.834 1.332 0.513 0.599 46.29 7 1.045 6.233 0.868 1.847 0.575 0.846 0.377 57.72 8 1.034 8.082 0.894 2.453 0.645 1.183 0.382 0.593 69.16 Table 1d--1.00-dB-ripple Chebyshev filter Attenuation
2 1.05 0.957 11.36 3 0.997 2.018 0.494 22.46 4 0.993 3.559 0.529 0.785 33.87 5 0.994 5.556 0.655 1.399 0.289 45.31 6 0.995 8.004 0.747 2.198 0.353 0.761 56.74 7 0.996 10.899 0.808 3.156 0.48 1.297 0.205 68.18 8 0.997 14.24 0.851 4.266 0.584 1.956 0.265 0.753 79.62 Table 1--Filter-design tables
HIGH-SPEED
ACTIVE FILTERS Figure 2 Figure
3 By going
through the design steps, you get this filter.You can
easily change the Sallen-Key lowpass filter into a
highpass filter. Output CLC412 250 1 to 8 3 12/2 2.9/3.1 CLC416 80 1 to 5 2 2/2 3.5/3.5 CLC418 130 1 to 10 5 13/1.4 3.3/3.3 CLC428 80 1 to 5 2 2/2 3.5/3.5 LM6172 50 1 to 5 12 1/1 8.5/9 LM6182 100 1 to 5 4 16/3 11/11 CLC5602 120 1 to 5 3 8.5/6.9 3.8/3.8 CLC5622 150 1 to 5 2.8 10/7.5 3.8/3.8 CLC432 62 1
to 20 3.3 13/2 6/6 Table 2--Summary of high-speed dual op ampsHIGH-SPEED
ACTIVE FILTERS Figure 4By adding a passive RC network at the output of the Sallen-Key filter, you create an odd-order filter.In all of the analysis so far, the op amps were assumed to have infinite bandwidth. The effect of using a real op amp in place of an ideal one is to slightly lower the corner frequency of the filter. You can compensate for this lowered corner frequency by predistorting the values of the components so that the final filter performance is much closer to what you want. Reference 4 explains the mathematical background for this method. To compensate for the lowered corner frequency, you need to know the op-amp time delay, T, which you can get in a few ways. One way is to look at the frequency-response plots for the op amp and find the frequency at which the phase delay is 1808, multiply this frequency by 2, and take the reciprocal. For example, for the CLC412, the 1808 frequency is approximately 300 MHz. Multiplying by 2 and taking the reciprocal yields a time delay of 1.7 nsec. Another way to get a value for T is to use the vendor-supplied Spice model, set up the amplifier configured as you plan to use it, and measure the delay between the input edge and the output edge of an input-pulse signal. For the filter shown in Figure B, calculate R2 by solving: (4) where (5) (6) (7) Once you have R2, you can get R1 with (8) Compensating for finite-bandwidth amplifiers
Figure
BTo compensate for the finite bandwidth of the
op amps, you must calculate R1
and R2 using Equations 4 through
8.and |
||||||
| EDN Access | Feedback | Table of Contents | |
||||||
| Copyright © 1997 EDN Magazine, EDN Access. EDN is a registered trademark of Reed Properties Inc, used under license. EDN is published by Cahners Publishing Company, a unit of Reed Elsevier Inc. | ||||||
