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AN48
Application Note
Design Notes for a 2-Pole Filter with Differential Input
by Steven Green
R4 R1 AINC2 R1 AIN+ C2 R4 R3 C5 R3
C5
_ +
Figure 1. 2-Pole Low-Pass Filter with Differential Input
Introduction The CS4329 evaluation board required an analog circuit capable of removing common-mode errors and lowering the out-of -band noise produced by the delta-sigma modulator. The circuit in Figure 1 includes a differential input and a two-pole analog filter to achieve these design requirements. This application note outlines the design steps required to select component values. Notice the similarities between Figure 1 and the multiple-feedback low-pass filter shown in Figure 2. The 2-Pole Low-Pass Filter with Differential Input is easily designed using the design equations for the multiple-feedback low-pass filter. Also, notice the similarities between Fig. 1 and Fig. 3. The differential input function is accomplished by simply duplicating the component values generated in the filter design.
MAR '95 AN48REV1 1
Crystal Semiconductor Corporation P.O. Box 17847, Austin, TX 78760 (512) 445-7222 FAX 445-7581
Copyright (c) Crystal Semiconductor Corporation 1996 (All Rights Reserved)
Design Notes for a 2-Pole Filter with Differential Input
R4
R4
C5
R1
_ +
R1
R3
_ +
R1 R4
C2
Figure 2. Multiple-Feedback Low-Pass Filter
Figure 3. Differential Input Circuit
Design Steps Step 1: Determine the required pass band gain, Ho. The circuit parameters require that the magnitude of Ho be greater than or equal to one. Ho is also negative due to the inverting op-amp configuration. Note: For the CS4329, the single-ended output will be 2Vrms for Ho = -1. Step 2: Select the desired filter type, Butterworth, Bessel, etc. and the corner frequency, Fc, for the final design. The filter response and corner frequency determine the pass band phase and amplitude response. The filter type determines the pole-locations and therefore alpha and beta. Table 1 lists the normalized pole locations for several filter types. Step 3: Select convenient values for C5 and C2. Notice in Step 4 that K (C5/C2) and Ho must be selected such that ) is real. 2 - K (1 - Ho Step 4: Given Fc, Ho, C2, C5, alpha and beta, calculate R1, R2 and R3 using the following equations. =
2 + 2
o = 2Fc 2 2 + K= C5 C2 R4 (-Ho) 1 o C2 [ 2 - K(1 - Ho)]
R1 =
FILTER TYPE Butterworth Bessel 0.01 dB Chebyshev 0.1 dB Chebyshev 0.7071 1.1030 0.6743 0.6104 0.7071 0.6368 0.7075 0.7106
R3 =
Table 1. Normalized Pole Locations
2 - K(1 - Ho ) R4 = o C5
2
AN48REV1
Design Notes for a 2-Pole Filter with Differential Input
Step 5: It is recommended that R1 be a minimum of 10 kohm to meet the CS4329 load requirements. Values much larger than 10 kohm can lead to small capacitor values where it is desireable to keep the capacitor values large to minimize the effects of stray capacitance. It may be necessary to adjust the capacitor values chosen in Step 3 to achieve this requirement. Step 6: The resistor values calculated in Step 4 are generally not standard values. Select standard values which are nearest the calculated values. This should not create a large change in the filter characteristics since metal film resistors are available in approximately 2.5% increments which allows for component selection near the calculated values. Step 7: The conversion from the singled-ended circuit to the differential circuit requires duplicating the values of R1, C2, R3, R4 and C5 in the non-inverting input as shown in Figure 1.
Design Example The following example shows the steps required to duplicate the CDB4329 filter. Step 1: The required output level is 2 Vrms, therefore the required pass band gain, Ho, is -1. Step 2: A two-pole Butterworth with a corner frequency of 50 kHz attenuates the signal at 20 kHz by approximately 0.1 dB and has nearly ideal phase linearity within the audio band. A Bessel response would achieve negligible phase improvement at the expense of degraded amplitude response. Fc = 50 kHz alpha = .7071 beta = .7071 Step 3: Select convenient values for C5 and C2. C5 = 220 pF C2 = 1000 pF Step 4: Given Fc, Ho, C2, C5, alpha and beta, calculate R1, R3 and R4. R1 = 13.77 kohm R3 = 3.343 kohm R4 = 13.77 kohm Step 5: Verify that R1 is greater than 10 kohm. R1 = 13.77 kohm Step 6: Select standard values which are nearest the calculated values. R1 = 13.7 kohm R3 = 3.32 kohm R4 = 13.7 kohm Step 7: The conversion from the singled-ended to the differential circuit requires duplicating the values of R1, C2, R3, R4 and C5 in the non-inverting input as shown in Figures 4 and 5.
AN48REV1
3
Design Notes for a 2-Pole Filter with Differential Input
220 pF 13.7 k 13.7 k 3.32 k
_ +
1000 pF
Figure 4. Filter Design
13.7 k
220 pF
13.7 k AIN1000 pF
13.7 k
3.32 k _ +
3.32 k
AIN+ 1000 pF
13.7 k
220 pF
Figure 5. CDB4329 2-Pole Filter with Differential Input
References [1] C. L. Lindquist, "Active Network Design with Signal Filtering Applications", Steward & Sons [2] A. B. Williams, "Electronic Filter Design Handbook", McGraw-Hill [3] "Reference Data for Radio Engineers" (Fourth Edition), International Telephone and Telegraph Corporation
4 AN48REV1

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