Cirrus Logic AN048 User Manual

AN48
Application Note
DESIGNNOTESFORA2-POLEFILTERWITH
DIFFERENTIAL INPUT
AIN-
AIN+
R1
R4
R1
R3
C2
C2

Figure 1. 2-Pole Low-Pass Filter with Differential Input

R4
R3
C5
_
+
C5

1. Introduction

Many of today’s Digital-to-Analog Converters (DACs) require a circuit that has a differential input and will remove common-mode errors, reduce the out-of-band noise produced by the delta-sigma modulator and produce a single-ended output. 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.
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(All Rights Reserved)
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 Figure 1 and Figure 3. The differential input function is accom­plished by simply duplicating the component val­ues generated in the filter design
MAR ‘03
AN48REV2
1
AN48
R4
C5
R3R1
C2
Figure 2. Multiple-Feedback Low-Pass
Filter
_
+

2. Design Steps

Step 1: Determine the required pass band gain,
H
. The circuit parameters require that the magni-
o
tude of H also negative due to the inverting op-amp configu­ration.
be greater than or equal to one. Hois
o
R4
R1
R1

Figure 3. Differential Input Circuit

_
+
R4
Step 4: Select convenient values for C5 and C2. Notice in Step 5 that K and H such that is real.
Step 5: Given F
ζ2K 1 Ho–()
,C2,C5, alpha and beta, cal-
c,Ho
must be selected
o
culate R1,R2 and R3 using the following equa- tions.
Step 2: Determine the minimum input impedance.
Step 3: Select the desired filter type, Butterworth,
Bessel, etc. and the corner frequency, F
,forthefi-
c
nal 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.
Table 1: Normalized Pole Locations
FILTER TYPE αβ
Butterworth 0.7071 0.7071
Bessel 1.1030 0.6368
0.01 dB Chebyshev 0.6743 0.7075
0.1 dB Chebyshev 0.6104 0.7106
----------------------=
α
2πFcα2β2+=
C
5
-----=
C
2
R
--------------=
--------------------------------------------------------------------=
ω
oC2
α
2β2
+
4
Ho–()
1
ζζ2K 1 Ho–()±[]
ζ
ω
o
K
R
1
R
3
ζζ2K 1 Ho–()±
R
-------------------------------------------------=
4
ω
oC5
2
AN48
Step 6: Review the resistor and capacitor values.
It is desirable to keep the capacitor values suffi­ciently large to minimize the effects of stray capac­itance. It may be necessary to adjust the capacitor values chosen in Step 4 to achieve this require­ment. Also verify that R1 is larger than the mini­mum required input impedance.
Step 7: The resistor values calculated in Step 5 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. However, it is advisable to calculate the ac­tual filter parameters, using the selected values, using the equations in Step 5.
Step 8: The conversion from the singled-ended cir­cuit to the differential circuit requires duplicating the values of R
1,C2,R3,R4 and C5 in the non-in-
verting input, as shown in Figure 1.

3. Design Example

Step 1: The required pass band gain, Ho,is-1for
this example.
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.) F
=50kHz
c
alpha = .7071 beta = .7071
Step 4: Select convenient values for C
5 and C2.
C5 = 220 pF C
2 = 1000 pF
Step 5: Given F culate R R
1 = 13.77 k
1,R3 and R4.
,Ho,C2,C5, alpha and beta, cal-
c
R3 = 3.343 k R
4 = 13.77 k
Step 6: Review the filter component values and fil­ter parameters. Verify that R
1 is greater than 10 k
and the capacitor values are sufficiently large to negate stray capacitance effects.
Step 7: Select standard resistor values which are nearest the calculated values. R
1 = 13.7 k
R
3 =3.32k
R4 = 13.7 k
Step 2: Minimum input impedance is 10 kΩ.
Step 3: Select a Butterworth response with a cor-
ner frequency of 50kHz. (A two-pole Butterworth
13.7 k
13.7 k
3.32 k
1000 pF

Figure 4. 2-Pole Multiple Feedback Filter Design

Step 8: The conversion from the singled-ended to
the differential circuit requires duplicating the val­ues of R
1,C2,R3,R4 and C5 in the non-inverting
input as shown in Figure 4 and 5.
220 pF
_
+
3
13.7 k
AN48
220 pF
AIN-
13.7 k
13.7 k
1000 pF
3.32 k
3.32 k
_
+
AIN+
1000 pF
13.7 k

Figure 5. 2-Pole Filter with Differential Input

220 pF
.

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
Contacting Cirrus Logic Support
For all product questions and inquiries contact a Cirrus Logic Sales Representative. To find one nearest you go to www.cirrus.com
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• Notes •
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