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Design >> Mixed-Signal Design >> size of quantization levels
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Message started by Croaker on Feb 5^th, 2007, 12:23pm

Title: size of quantization levels
Post by Croaker on Feb 5^th, 2007, 12:23pm

Is there any standard for determining the LSB size? The two definitions I see are:

Vfs/2^N

or

Vfs/(2^N-1)

Is any one better than the other?

Title: Re: size of quantization levels
Post by Vabzter on Feb 27^th, 2007, 3:17am

Hi,
I was going through this paper which highlights major parameters of DAC..Hope this answes your question

http://grouper.ieee.org/groups/1658/NextMeetingArchive/061030DAPONTE/imeko_IWADC_2005.pdf

BR
Vabzter

Title: Re: size of quantization levels
Post by ywguo on Feb 27^th, 2007, 6:32pm

I prefer Vfs/2^N for ADC. :)

I read that paper. What's the definitioin of 1's complement? Why are they named 2's complement and 1's complement respectively?

Thanks
Yawei

Title: Re: size of quantization levels
Post by Croaker on Feb 28^th, 2007, 10:32am

You can find discussions on 1s and 2s complement in any digital electronics book. They are used for representing negative numbers.

http://en.wikipedia.org/wiki/Two's_complement

Title: Re: size of quantization levels
Post by Croaker on Feb 28^th, 2007, 10:34am

Vabzter wrote on Feb 27^th, 2007, 3:17am:

Not really. They are using Vref/2^N. The difference between the two standards that I see are Vref/2^N means your full-scale output is one LSB less than Vref and Vref/(2^N - 1) allows the full-scale output to be equal to Vref. Also, the LSB sizes are slightly different.

Title: Re: size of quantization levels
Post by ywguo on Feb 28^th, 2007, 6:27pm

Hi, Croaker,

Let's discuss it with an N-bit flash ADC. It needs 2^N comparators and references. Normally the references are generated using a resistor ladder, which is tied between the positive reference (ref+) and the negative reference (ref-). If the resitive ladder is composed of 2^N resistors of equal resistance, the most negative reference tied to one of the comparators is V_ref-. The most positive reference tied to one of the comparators is (V_ref+-V_ref-)×(1-1/2^N)+V_ref-. Then one LSB is equal to (V_ref+-V_ref-)×(1/2^N). However, if the resistive ladder is composed of (2^N-1) resistors of equal resistance, the most positive reference tied to one of the comparators is V_ref+. The one LSB is equal to (V_ref+-V_ref-)÷(2^N-1). I think it depends on your choice.

Best regards,
Yawei