# Sequential ATPG errors that dont go away because of hardware metal faults

1. ## Sequential ATPG errors that dont go away because of hardware metal faults

suppose you had a metal line. Current metal lines are about 22nm. There are about 22000 atoms in 22 nm since an atom is about 1 pm. Now, spectral atpg performs about 10^12 calculate to get 92% coverage with each large circuit with 35000 faults.

Now suppose we look at the microprocessor error that exists in the metal line of the jz instruction, which follows cmp instruction which is used test condtitional of equality. probability that a line is in error = 10^12 /35000 = 1 in 20,000,000 operation.

Now 1/20000000 = ( proability of quantum tunelling) ^ ( 22000 )

Assume the following senario

tunnelling to the first atom + tunnelling to the second atom + tunnelling to the third atom like that for 22000 atoms

p = exp( log(1/(10^12/35000))/(22000))
= 0.99966
Corrected p' = p - 0
= p - limit K-> infinity episoln->0 integral_-K to K episoln dx
= 0.99966 - A
Select A such that p' close to 0 (less than 1/2)

At that rate, error in the metal layer is sufficient to change observed fault coverage on microprocessor based fault simulation during ATPG.  Reply With Quote

•

2. ## Re: Sequential ATPG errors that dont go away because of hardware metal faults Originally Posted by firewireblue suppose you had a metal line. Current metal lines are about 22nm. There are about 22000 atoms in 22 nm since an atom is about 1 pm. Now, spectral atpg performs about 10^12 calculate to get 92% coverage with each large circuit with 35000 faults.

Now suppose we look at the microprocessor error that exists in the metal line of the jz instruction, which follows cmp instruction which is used test condtitional of equality. probability that a line is in error = 10^12 /35000 = 1 in 20,000,000 operation.

Now 1/20000000 = ( proability of quantum tunelling) ^ ( 22000 )

Assume the following senario

tunnelling to the first atom + tunnelling to the second atom + tunnelling to the third atom like that for 22000 atoms

p = exp( log(1/(10^12/35000))/(22000))
= 0.99966
Corrected p' = p - 0
= p - limit K-> infinity episoln->0 integral_-K to K episoln dx
= 0.99966 - A
Select A such that p' close to 0 (less than 1/2)

At that rate, error in the metal layer is sufficient to change observed fault coverage on microprocessor based fault simulation during ATPG.
I would suggest to stop drinking.  Reply With Quote

--[[ ]]--