AD7731 filter questions?

Document created by analog-archivist Employee on Feb 23, 2016
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1) On the datasheet I get the impression that the 'CHOP' variable affects the
output rate of the sinc3 filter, and therefore the frequency response. So if SF
= 512 and CHOP = 0, the response is as it is in the datasheet, but then setting
CHOP = 1, will it not stretch the response out such that the first node is at
200Hz. I ask this because the excel model of this filter, which you provide on
your website, the sinc3 filter seems to be unaffected by the value of CHOP.

2) For the calculation of the sinc3 output, the formula that you have used is a
bit different than what I'd expect

3) How would I go about modelling the sinc3 filter in Simulink?

4) Is there any chance you could provide me with the FIR filter coefficients?


1)When chop is off and the FIR filter is not used, the AD7731 does the
inversions needed for chopping. However, it does not do the averaging. You must
take 2 subsequent conversions and average them. The excel model shows the
frequency response before this averaging. However, the averaging that the
customer performs will change the filter response. The averaging will introduce
first order notches at half the output data rate and at odd integer multiples
of half the output data rate. For example, with chop off, FIR filter
not used, the output data rate is 600 Hz i.e. the sinc3 filter is used and is
free running. With chop on and the FIR filter still bypassed, the output data
rate is reduced by a factor of 3. This is because the sinc3 filter has a
settling time of 3 conversions. So, with the analog inputs connected in one
direction, a conversion is generated. The analog inputs are then swapped and
another conversion is generated. These 2 conversions then need to be averaged
by the customer to remove the offset. With chop on, the output data rate is 200
Hz. Therefore, the averaging will introduce notches at 100 Hz, 300 Hz, 500 Hz,

2)The output data rate is listed in the datasheet. The modulator operates at
MCLK/16. So, for a 4.9152MHz clock, the modulator operates at 307.2 kHz. The
output data rate is fmod/SF when chop is off and the sinc3 filter only is used.
So, when SF=512, the output data rate equals 307.2kHz/512 = 600 Hz. When
chopping is enabled, the output data rate is reduced by a factor of 3. So, for
the above conditions, the output data rate is 600/3 = 200 Hz. The excel model
is correct.

3)The sinc3 filter is a standard sinc3 filter.

4)The FIR coefficients are the same for chop off and chop on.