I read in sci.electronics.design that John Larkin wrote (in ) about 'Delta-sigma ADC question', on Tue, 12 Apr 2005:
Look for Audio Engineering Society papers by S Lipshitz and J Vanderkooy. If you are like me, you still won't understand it. (;-)
I read in sci.electronics.design that John Larkin wrote (in ) about 'Delta-sigma ADC question', on Tue, 12 Apr 2005:
Look for Audio Engineering Society papers by S Lipshitz and J Vanderkooy. If you are like me, you still won't understand it. (;-)
Andre wrote in news:d3gkoo$57c$02$ snipped-for-privacy@news.t- online.com:
15do.
This delay is exactly why SAR converters are used in feedback systems.
A very big advantage to sigma delta converters is that the analog antialiasing filters are trivial since the actual sample rate is much high than the "effective" sample rate.
The internal digital filters may have stopband attenuation of 90dB at something near 1/2 the "sample rate" This of course is a major reason for the added delay. The better converters have better filters which have more delay.
You may find that low cost sigma delta converters are still a better choice than a muxed SAR. You can find 4 channel devices that smaple at
100kHz or 200kHz for reasonable cost.
the simple way:
If you model the quantization as as adding white noise with a power proportional to the number of bits, you'll see that for a delta-sigma modulator that noise will show up at the output high-passed, the power is still the same but "pushed" towards higher frequencies. When you then lowpass and decimate you remove the higher frequencies and thus much of the the noise.
quatization noise before lowpass and decimate:
^ | - - | / | / | /
+==------+> f fs/2 quatization noise after lowpass and decimate:^ | | | |
+==+------> f fs/2With a SAR the noise is flat at the output, so lowpass and decimate will remove less of the noise power.
quatization noise before lowpass and decimate:
^ | | | |---------
+---------+>fs/2
quatization noise after lowpass and decimate:
^ | | | |--
+--+-----> f fs/2-Lasse
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