Fixed-gain operational amplifier simplifies filter design-Fixed-Gain O
Simple second-order filters meet many filtering requirements. A low-order low-pass filter, for example, is often adequate for anTIaliasing in ADC applicaTIons or for eliminaTIng high-frequency noise in audio applicaTIons. Similarly, a low-order high-pass 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 low-pass and high-pass 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 RF and RG is greater than unity. What's more, the cookbook component-value formulas can yield unrealistic values ​​for the capacitors and the resistors.
Figure 1. Sallen-Key filters use fixed-gain op amps to realize a second-order Butterworth response.
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 the table below with the following equations: R2 = 1 / (2πfC √) and R1 = XR2.
Butterworth-Filter-Design Criteria
| Gain | Low-Pass X | High-Pass X |
| 1.25 | * | 1.372 |
| 1.5 | 2 | 1.072 |
| 2 | 0.5 | 0.764 |
| 2.25 | 0.404 | 0.672 |
| 2.5 | 0.343 | 0.602 |
| 3 | 0.268 | 0.5 |
| 3.5 | 0.222 | 0.429 |
| 4 | 0.191 | 0.377 |
| 5 | 0.15 | 0.305 |
| 6 | 0.125 | 0.257 |
| 7 | 0.107 | 0.222 |
| 9 | 0.084 | 0.176 |
| 10 | 0.076 | 0.159 |
| 11 | 0.07 | 0.146 |
| 13.5 | 0.057 | 0.121 |
| 16 | 0.049 | 0.103 |
| twenty one | 0.038 | 0.08 |
| 25 | 0.032 | 0.068 |
| 26 | 0.031 | 0.066 |
| 31 | 0.026 | 0.056 |
| 41 | 0.02 | 0.043 |
| 50 | 0.017 | 0.035 |
| 51 | 0.017 | 0.035 |
| 61 | 0.014 | 0.029 |
| 81 | 0.011 | 0.022 |
| 100 | 0.009 | 0.018 |
| 101 | 0.009 | 0.018 |
For a gained filter response, the use of a fixed-gain op amp reduces cost and component count. It also decreases sensitivity, because the internal, factory-trimmed, precision gain-setting resistors provide 0.1% gain accuracy. To design a second- order Butterworth low-pass or high-pass filter using a fixed-gain op amp, follow these steps: Determine the corner frequency fC. Select a value for C. For the desired gain value, locate X under the proper column heading in the table . 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 1000pF 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 low-pass filter with a 24kHz corner frequenc y and a gain of 10. Step 1 is complete (fC = 24kHz). Next, complete Step 2 by selecting a value for C, say, 470pF. In the table, note that X = 0.076 for a low-pass filter with a gain of 10. Substitute these values ​​in the equations: R2 = 1 / (2π fC √) = 1 / (2π × 24kHz × 470pF × √) = 51kΩ, and R1 = XR2 = 0.076 × 51kΩ = 3.9kΩ.
Figure 2. Using the circuit values ​​in the text, a simulation of the circuit in Figure 1a produces this Butterworth response.
A similar version of this article appeared in the July 6, 2000 issue of EDN.
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