CALCULATOR DNL/ INL DAC

Dear mirtaji65,

It appears you did not remove the transient in the beginning of your waveform from "/net055". Hence, I think it will be included in the integration used to determine the threshold option you chose of "auto".  This will not, I believe, provide the threshold you intended. As defined in the manual:

Auto Threshold Value = integral of the waveform divided by the X range.

You have the "delay" set to 0, and hence this transient is considered part of your data. You might either clip the waveform before taking the dnl and or increase the value of delay beyond the transient.

Did either of those thoughts occur to you mirtaji65?

Shawn

Thanks for the reply

lets check together

firs i changed wave form name:

a)     inl(VT("/net055") 62.25u  ?mode "auto"  ?crossType "falling" ?delay 0 ?units "abs" ?nbsamples 16)

The result of the waveform as before  

isn't sampling signal  The amount of output signal increase is actual total for example  62.5u=1v/(16 for 4 bit)

b) i don't know what really it mean and What  value should I put?

Threshold--Threshold value against which the frequency is to be calculated. You need to specify the threshold value only if the mode is auto

c) i choose dealy=.5us half of sample period

Delay--Delay time after which the sampling begins.

inl(VT("/net055") 62.5u  ?mode "user" ?threshold 0  ?crossType "rising" ?delay .5u ?units "abs" ?nbsamples 16)

all of the resualt as the same 

Dear Shawnlogan
Thank you very much for your answer
I want to use INL function in the first post I copied wrong expesion DNL instead of INL function from calculator 
Of course, I also use  DNL function after that.The example you sent for me helped me a lot.
may you check to see i understoor the concept of Sampling signal and threshod 
Sampling signal:It is the sampling time that the converter (óDAC) receives data from theinput signal In this example you sent  for me
sample time converter=10uS 
Threshold: in fact the same amount of LSB value =(full range)/2^n-1      n=resolution
delay:Wait from the beginning of each sampling of the converter to the value of ö delay time, then get the value from plot? or
Ignore a value at the beginning of the waveform Only once at the beginning of the waveform 

I think I could use to get this Another way by calculator
I have a ramp signal at the input of the circuit that produces the output
ramp=>ideal ADC=>DAC
i can use this expression to get INL waveform
abs(VT("/net035")-VT("/net055"))/62.5)
"/net035"=ramp
"/net055"=output
62.5=ideal step size=>1/2^4
And the following waveform is obtained

is it correct?

And the result  of function of dnl is as follows

inl(vtime('tran "/net055") 1u  ?mode "auto"  ?crossType "rising" ?delay 0 ?units "abs" ?nbsamples 16)

inl(vtime('tran "/net055") 1u  ?mode "user" ?threshold 62.5u  ?crossType "rising" ?delay 0 ?units "abs" ?nbsamples 16)

with or without threshold are the same

Please guide me  how can i resolve this problem

Dear mirtaji65,

mirtaji65 said:

Please guide me  how can i resolve this problem

I finally had time to respond. I put together a test bench with a piecewise linear source to model your DAC and created a "noisy" version of your DAC's output voltage versus time. I ran the simulation and created plots of its output voltage, DNL, and INL (the latter two expressed in LSB). I also saved the output expressions in a comma-separated variable file. The input text file used for the simulation and a view of the test bench are included. Download the attached compressed file into a directory and you will find pnd and pdf versions of the output voltage, DNL, and INL. You should be able to use the expressions with a clipped version of your output waveform.

Shawn

m2minl_inverted_noisy_sml_121520.zip

Dear Shawnlogan
Thank you for your help
I implemented the test bench circuit exactly And the file that contained the sawtooth waveform "m2minl_inverted_noisy.txt" load in the vpwlf source

The sawtooth waveform was similar to your waveform

but INL waveform It is completely different from your waveform this expression is

inl(v("/vout" ?result "tran") 1e-06  ?mode "auto"  ?crossType "rising" ?delay 5e-07 ?units "lsb" ?nbsamples nil)

And your waveform is as follows

Why are the  inl not similar?

Thank you for your help my friend

Dear mirtagi65,

mirtaji65 said:

Why are the  inl not similar?

I suspect the number of samples computed for your waveform may not be the same. Did you try setting ?nbsamples to the desired value?

Shawn

Dear Shawnlogan

I think my output waveform of the function is correct
As you can see in the figure below
only at 7us of waveform that The sampling point is below the ideal waveform And elsewhere at the top the ideal waveform

When I read the function guide, I realized that the X-axis is the waveform of the function in terms of time, while our waveform is the X-axis in terms of voltage.

How can  change  x-axis  waveform of the INL function in terms of time instead of voltage?

INL=(V(sample point)-(sample point*LSB))/LSB

Dear mirtaji65,

mirtaji65 said:

How can  change  x-axis  waveform of the INL function in terms of time instead of voltage?

I am not sure what version of the manual you are referring to, but if you examine the latest version of the ViVa function manual (Nov. 2020), it clearly states the x-axis units are the same as the y-axis units of the waveform for which you are computing the INL. In all of your examples, your y-axis units are voltage. Hence, the x-axis of the inl() waveform is voltage - not time. The argument you circled in red refers to whether the inl() output is expressed in "absolute" units or in "LSB". It does not refer to the x-axis of the inl() output function.

The examples in the Nov 2020 manual all show this behavior. If you want to change the x-axis plotted, I believe you will need to use the YvsY() function and create the time axis as a waveform. After creating the time axis as a separate waveform, right-click the X-axis and choose YvsY. The YvsY form appears and you can select the INL waveform as axis the newly created time axis as the second waveform.

Shawn

I made a waveform for testing

The result of the dnl function is as follows
Which is the X-axis and the Y-axis are in terms of voltage

The result of the dnl function in mathlab is as follows
Which is the X-axis in terms of time and the Y-axis are in terms of voltage

As is clear from the signal of the value (0v to 1.8v) after that (1.8v to 0v) after that (0 to 3.2)

And because one axis is in terms of voltage and the other is time, the waveform is not the same.

you can't use the YvsY() function
i use dc signal 0v as the time axis as a waveform (/net011)
when i plot two out put in simulation Draws them in the subwindows form

I can use YvsY for (/net11) and (/net05)  Because both of the in same windows and choose select trace 

i can not plot inl and (/net11) in same window that YvsY

What should I do to solve the problem?

In your opinion, it is not better to write a program VHDL or verilog

Dear mirtaji65,

mirtaji65 said:

What should I do to solve the problem?

I hope the following solves your latest problem. I have included the specific steps I followed and the resulting plots of INL versus voltage and the new plot of INL versus Time.

Shawn

1. Create time values:

value(v("/vout" ?result "tran") 0.5e-06  ?period 1e-06 ?xName "time"  )

Place in Calculator Table (Tools->Table)

2. Create inl values:

value(inl(v("/vout" ?result "tran") 1e-06  ?mode "auto"  ?crossType "rising" ?delay 0.5e-06 ?units "abs" ?nbsamples nil) 0.5e-06 ?period 1e-06 ?xName "time")

Append to table (See below for comma-separated variable file I created)

3. Save table as a comma-separated variable file with <filename> (i.e. nl_xval_inl.csv)

3.Create new waveform:

getAsciiWave("Full path to saved file inl_xval_inl.csv"> 1 4 ?xskip 1 ?yskip 1 ?formatFloat nil ?xName "Time" ?xUnits "sec" ?yName "inl" ?yUnits "LSB" )

4. Plot new waveform

5. Enjoy!

Example "inl_xval_inl.csv"

time (s),xval(value(v("/vout" ?result "tran") 0.5e-06  ?period 1e-06 ?xName "time"  )) (s),V (V),inl
500.00000E-9,500.00000E-9,24.961888E-3,0.0000000
1.5000000E-6,1.5000000E-6,72.934183E-3,-37.478530E-3
2.5000000E-6,2.5000000E-6,123.74225E-3,-18.059857E-3
3.5000000E-6,3.5000000E-6,175.35193E-3,17.442537E-3
4.5000000E-6,4.5000000E-6,223.30014E-3,-20.519130E-3
5.5000000E-6,5.5000000E-6,276.42237E-3,45.331194E-3
6.5000000E-6,6.5000000E-6,323.72824E-3,-5.5186424E-3
7.5000000E-6,7.5000000E-6,374.44416E-3,12.051288E-3
8.5000000E-6,8.5000000E-6,425.94798E-3,45.429609E-3
9.5000000E-6,9.5000000E-6,474.45422E-3,18.664130E-3
10.500000E-6,10.500000E-6,524.10157E-3,14.794140E-3
11.500000E-6,11.500000E-6,576.87473E-3,73.640760E-3
12.500000E-6,12.500000E-6,622.86814E-3,-3.5422922E-3
13.500000E-6,13.500000E-6,675.23440E-3,47.140060E-3
14.500000E-6,14.500000E-6,722.63109E-3,-1.8875072E-3
15.500000E-6,15.500000E-6,772.56539E-3,-555.11151E-18

sml 1/4/2021

Dear Shawnlogan

Your solution for converting X-axis from voltage to time works exactly.

But why do I have to sample the waveform by use <<value function>>  when i want to use INL and DNL function ,why delay of INL function doesn't work itself,

Is this a software bug?

ver virtuoso :6.17.77

This method is not suitable if I want to use the analysis parameter

But your method of  converting X-axis from voltage to time  was great

thanks my friend

Dear Shawnlogan

Your solution for converting X-axis from voltage to time works exactly.

But why do I have to sample the waveform by use <<value function>>  when i want to use INL and DNL function ,why delay of INL function doesn't work itself,

Is this a software bug?

ver virtuoso :6.17.77

This method is not suitable if I want to use the analysis parameter

But your method of  converting X-axis from voltage to time  was great

thanks my friend

Dear mirtaji65,

> Your solution for converting X-axis from voltage to time works exactly.

Great! Thank you for letting us know!

mirtaji65 said:

But why do I have to sample the waveform by use <<value function>>  when i want to use INL and DNL function ,why delay of INL function doesn't work itself,

Is this a software bug?

First, I do not work for Cadence and hence cannot really respond to your comment about the behavior of INL. However, the INL function, as it is documented, does not provide the x-axis as a function of time (which is what I assume you want). Hence, the "workaround" using the value() function.

mirtaji65 said:

This method is not suitable if I want to use the analysis parameter

Sorry, I don't understand exactly what you mean by this comment! You can use this methodology in an Explorer/Assembler expression through the use of an ocean script. This is documented at URL:

https://support.cadence.com/apex/ArticleAttachmentPortal?id=a1Od0000000nXTvEAM&pageName=ArticleContent

and

https://support.cadence.com/apex/ArticleAttachmentPortal?id=a1Od0000000nXS2EAM&pageName=ArticleContent

Hence, you may define the steps in an ocean script and add the ocean script as an Explorer/Assembler output.

I hope this helps mirtaji65,

Shawn