# thermometer vs binary DAC DNL expression developing

#### yefj

##### Full Member level 4 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.    Last edited:

#### yefj

##### Full Member level 4 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 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

##### Full Member level 4 #### yefj

##### Full Member level 4 Last edited:

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

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

• yefj

points: 2