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thermometer vs binary DAC DNL expression developing

yefj

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Hello,I am trying to develop the expression for the DNL of DAC
the worst case DNL is the transtion between 011..11 and 1..000 (assume between the MSB is N+1 )
for 01..111
For binary weighted:
\[ I_{dac}=Vref*2^{n+1}*N(G,\delta G) \\ 2^0+..+2^n=\frac{2^0(2^{n+1}-1)}{2-1}=2^{n+1} \]
so for 1000..0
\[ I_{dac}=Vref*2^{n}*N(G,\delta G) =Vref*N(G,\sqrt{2^{n}}\delta G) \\ \]
so if we substract the two current for DNL the variance of the DNL is:
\[ \sqrt{2^{n+1}}-\sqrt{2^{n}}=\sqrt{2^{n}}(\sqrt{2}-1) \]
which is different form the formal expression shown bellow.Where did i go wrong?
Thanks.

1600278595257.png
1600278638265.png1600274434613.png
1600274516125.png
 
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sutapanaki

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For an Nbit binary DAC, the DNL and its variance are

1600287012621.png
 

yefj

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I am talking about worst case DNL(the major code transition)
the transistion from 011..111 and 100..00
on one member we only have 1one bit of 1 the rest are zero,we do not have two sums
only one?
why i am wrong in my calculation?
 

sutapanaki

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I am talking about worst case DNL(the major code transition)
the transistion from 011..111 and 100..00
on one member we only have 1one bit of 1 the rest are zero,we do not have two sums
only one?
why i am wrong in my calculation?


Yes, what I've written is for the worst case DNL.
 

yefj

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from the from formula show belloe 01111 is \[ 2^{B-1} -1 \]
\[ 2^0...2^{n-2} \]
100...00 is
\[ 2^{n-1} \]
how did they get this variance expression shown bellow?
1600325654772.png
1600324915030.png


1600324375828.png
 
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yefj

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Hello,the subtraction of two randon numbers is not subtraction of the variances.
Where am i wrong?

UPdate: those are corelated so we need to subtract.
1600350397794.png
 
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sutapanaki

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Subtraction of two random variables leads to the sum of their variances.

To your 1st question

2^(B-1)+2^(B-1)-1=2*2^(B-1) - 1 = 2^B - 1
 
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    yefj

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