Suppose I have a two bit quantizer with the following voltage levels: [0 0.2 0.4 0.6]. Thus 0V corresponds to bits 00 and 0.6 to bits 11. Suppose I use this quantizer in a first order delta sigma loop with a simple integrator y[n] = x[n] + y[n-1], if I use a DC input signal, say 0.25V, I will hit codes 0.2V and 0.4V with certain frequency. Hence after averaging this delta sigma ADC would work better than normal averaging ADC in terms of resolution.
In normal averaging ADC, the quantizer would only hit the code 0.2 V and averaging the values will have no effect on the resolution of the ADC.
Can we know how much resolution improvement is done by the SD ADC in this case? I am a little new to ADC's so I might have missed certain information, but all I really want to know is how much better the SD ADC works as compared to normal averaging ADC for a DC input signal in terms of resolution.
EDIT: Can someone explain how would we calculate the resolution of SD ADC in this case in time domain?
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\$\begingroup\$ I recommend you read this app note from analog devices analog.com/media/en/training-seminars/tutorials/MT-022.pdf \$\endgroup\$– KvegaoroCommented Aug 31, 2015 at 16:42
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\$\begingroup\$ I've got to tell you, DS converters in DC are not well documented. Good luck. \$\endgroup\$– Vladimir CraveroCommented Aug 31, 2015 at 16:51
2 Answers
This picture here taken from analog devices app note MT-022 show why the delta sigma is better than just averaging and its all due to the Delta Sigma modulation that shapes the quantization noise in such fashion that is taken away from the band of interest. If the quantization noise goes down it means you can be more certain that your interpolation can be more granular as shown in figure C. while just filtering and decimation spread. So it all comes down to noise shaping provided by the Delta sigma modulation
The resolution gain comes from the averaging, not from the type of converter. The delta-sigma converter is in a sense self-dithering — to get the same effect in the "normal averaging" converter, you'd explicitly add a ±1 LSB dither signal to the input, after which, the averaging will give you the same resolution as the delta-sigma. The longer you average, the finer the effective resolution gets, in a direct tradeoff against bandwidth.
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\$\begingroup\$ Suppose I average 4 values which come out from the quantizer will this increase my accuracy by 1 bit in SD ADC case or greater? Can we quantify this? \$\endgroup\$– sarthakCommented Aug 31, 2015 at 16:50
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\$\begingroup\$ Each pair of samples you average adds one bit of effective resolution. 4 samples gives 2 bits, 8 samples gives 3 bits, etc. Note that this has no effect whatever on accuracy -- the averaging simply allows you to interpolate between the levels of your quantizer, but they retain whatever accuracy they started with. \$\endgroup\$ Commented Aug 31, 2015 at 16:53
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\$\begingroup\$ But isn't this true for averaging ADC....how is DS ADC better than normal averaging ADC then? And what exactly do you mean by accuracy? \$\endgroup\$– sarthakCommented Aug 31, 2015 at 16:56
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1\$\begingroup\$ The D-S converter is self-dithering, which makes it easier to construct, especially on the analog side. Also, a D-S convert normally uses just a comparator (1-level quantizer), which means that most of the things that adversely affect its linearity simply go away. This is why they're so popular for audio applications. \$\endgroup\$ Commented Aug 31, 2015 at 16:59
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\$\begingroup\$ Can I take this in chat with you? \$\endgroup\$– sarthakCommented Aug 31, 2015 at 17:11