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Design >> Mixed-Signal Design >> the relationship between INL and the output residue of each stage
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Message started by Hawick on Mar 24^th, 2011, 7:33pm

Title: the relationship between INL and the output residue of each stage
Post by Hawick on Mar 24^th, 2011, 7:33pm

Hi All,
I am trying to design a 10-bit pipelined ADC with an on-chip sample and hold amplifier that operate at 40MHz with a power supply of 3V.
I don't kown the relationship between INL and the output residue of each stage.
For example, when I simulate the pipelined adc, which is differential input. The positive input is 2V, and the negative one is 1V, while the common mode voltage is 1.5V. The resolution of each stage is 2.5bit.
The requirement of INL is +-0.3LSB,
how does the output residue of each stage meet the requirement of INL?
Please forgive my ignorance.
Thanks.
Best regards.

Title: Re: the relationship between INL and the output residue of each stage
Post by loose-electron on Mar 25^th, 2011, 1:16pm

Suggest - put together a math model of the architecture (Matlab or other tool, your pick) from that you can derive what you want.

Thing is - there are different architectures for pipelines and so a quick formula is not known without the architecture.

Title: Re: the relationship between INL and the output residue of each stage
Post by Hawick on Mar 28^th, 2011, 11:26pm

Hi loose-electron,
Thanks for your reply!
But I still don't understand that.
My pipeline adc consists of front-end SH, four pipeline stages with 2.5bits/stage, 2bits flash adc.

My question is:
1. Assumed INL and DNL have been known.
when I simulate each stage, what requirement does the output residue of each stage meet?

2. The following is my idea.
full scale=2V, resolution=10bit, INL=+-0.3LSB
Typical corner，INL=0.3LSB=0.3*2/(2^10)=0.6mV
definition: the contribution of each stage's error to ADC's INL
sample and hold is I0
from stage1 to stage4 respectively is I1, I2, I3, I4
Then ,
The INL referred to the input of the ADC is found by dividing each INL by the gains preceding the

stage being studied.
I0+I1/4+I2/16+I3/64+I4/256=INL
distribute the contribution of each stage to INL
I0=1/2INL,
I1/4=1/4INL,
I2/16=1/8INL,
I3/64=1/16INL,
I4/256=1/32INL
so, INL=(1/2+1/4+1/8+1/16+1/32)=31/32INL
then,
I0=1/2INL=0.3mV
I1=4*1/4INL=0.6mV
I2=16*1/8INL=1.2mV
I3=64*/16INL=2.4mV
I4=256*1/32INL=4.8mV

the above result is the maximum error of output residue of each stage.
I think my idea is wrong. But I don't kown what is the right one.
Thanks!
Best regards,
Hawick