Fixed-gain op amps simplify filter design

Dan Christman, Maxim Integrated Products, Sunnyvale, CA -- EDN, 7/6/2000

Simple second-order filters meet many filtering requirements. A low-order lowpass filter, for example, is often adequate for antialiasing in ADC applications or for eliminating high-frequency noise in audio applications. Similarly, a low-order highpass filter can easily remove power-supply noise. When you design such filters with built-in gain, fixed-gain op amps can save space, cost, and time. Figure 1 illustrates the use of fixed-gain op amps in building second-order lowpass and highpass Sallen-Key filters. Filter "cookbooks" are useful in designing these filters, but the cookbook procedures usually break down for a given response, such as Butterworth, if the gain set by R_F and R_G is greater than unity. What's more, the cookbook component-value formulas can yield unrealistic values for the capacitors and resistors.

Butterworth filters, for example, offer the flattest passband. They also provide a fast initial falloff and reasonable overshoot. You can easily design such filters using Table 1 with the following equations: R₂=1/(2?f_CC

and R₁=XR₂. For a gained filter response, the use of a fixed-gain op amp reduces cost and component count. It also reduces sensitivity, because the internal, factory-trimmed, precision gain-setting resistors provide 0.1% gain accuracy. To design a second-order Butterworth lowpass or highpass filter using a fixed-gain op amp, follow these steps:

1. Determine the corner frequency f_C.

2. Select a value for C.

3. For the desired gain value, locate X under the proper column heading in Table 1.

4. Calculate R₁ and R₂ using the equations.

Choosing C and then solving for R₁ and R₂ lets you optimize the filter response by selecting component values as close to the calculated values as possible. C can be lower than 1000 pF for most corner frequencies and gains. Fixed-gain op amps come optimally compensated for each gain version and provide exceptional gain-bandwidth products for systems operating at high frequencies and high gain. Suppose, for example, you must design a lowpass filter with a 24-kHz corner frequency and a gain of 10. Step 1 is complete (f_C=24 kHz). Next, complete Step 2 by selecting a value for C, say 470 pF. In Table 1, note that X=0.076 for a lowpass filter with a gain of 10. Substitute these values in the equations:

R₂=1/(2?f_CC

=1/(2?x24 kHzx470 pFx=51 k?, and R₁=XR₂=0.076x51 k?=3.9 k?. With these component values, the circuit in Figure 1 yields the second-order Butterworth lowpass response in Figure 2. (DI #2551)