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INL & DNL measurement of 10-Bit DAC

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bgpradeep

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inl and dnl measurement

Hi All,

I had doubt in the measurement of INL & DNL.


1)
Is 1LSB = Iref/2^N

Here what should be the Iref value as LSB current
source has 20uA and MSB current source delivers
(20uA * 16). Here LSB = 4bits & MSB = 6bits.

Also should N be different or it should be 10

2)
For DNL measurement , it is 1LSB - [ Iout/2^N]. is this correct?

and how is the INL measurement made because
as per the definition it says draw a line joining
the first and last point.

Thanks
 

inl dnl definition dac

Here the Iref is full scale for 1LSB = Iref/2^N
 

testing of 10bit dac

Hi All,

For checking the INL & DNL, under what consideration the simulation time is set.
Like for checking INL & DNL, i am giving different codes from 00....0 to 11...1 every time. I did not get for how much time the simulation should be run.

.TRAN 0NS 50NS -> my current simulation time.

Clock is 200MHz.

Thanks
 

dnl measurement

the time you need to check all possible bit words is the following. You have 2^10=1024 different values. For a good test to be run you need to run through all levels, each level consists of 10 bits so 10*1024= 10240 bits in total are needed for testing.
Simulating with a clock of 200Mhz gives a sample time of 5ns. This makes the total simulation time 5ns * 10240 = 51.2 µs.
 

dac inl dnl measurement

Thanks for the reply.

I wanted to know whether my measurement of DNL is proper.
This is how i am carrying out the measurement:

1 LSB = Iref/2^10 = 19.5uA/2^10 = 19.043nA

For 0000000001, Iout/2^10 = 19.5uA/2^10 = 19.043nA
Therefore, DNL = 19.043nA - 1LSB = 0LSB

For 0000000010, Iout/2^10 = 39.1uA/2^10 = 38.2nA
Then should this be divided by 2 which becomes,19.1nA

Then, DNL = 19.1nA - 19.043nA = 0.057nA/19.043nA = 0.003LSB

For 0000000011, Iout/2^10 = 59.3uA/2^10 = 57.91nA
Dividing by 3 gives, 19.3nA

Then, DNL = 19.3nA - 19.043nA = 0.257nA/19.043nA = 0.0135LSB

and so on.

Am i going in the right way.

Thanks
 

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