SIGNATURE INPUT PROPERTIES n n n Circuits signature

SIGNATURE INPUT +

PROPERTIES n n n Circuits signature gives us the ability to check Circuits - if they are undamaged. Checking the output of the CUT vs. a known good response is inefficient and not practical. Using Signature Analysis enables us to check CUT efficiently.

The math behind it… + … G (x) + P (x) SIG. Reg. Q (x) Initial State - I (x) = 0 Final State - R (x) It Satisfies this polynomial equation G (x)R (x) : ____ =Q (x) + ____ P (x)

The math behind it(2)… M – Number of bits in stream (input) n. N – Number of bits in Sig. Reg. n. The Num. of streams that produces a specific sig. is 2 M-N (= 2 M / 2 N ) n. The Num. of bad streams that will yield good sig. is 2 M-N - 1 n

The math behind it(3)… n When M>>N the probability for having an unnoticeable mistake is 2 M-N - 1 ≈ 2 -N 2 M - 1 n So, for as the Sig. Reg. is bigger we get a better approximation on the CUT

S-Edit (1_xor_6_SIG)

L-Edit (1_xor_6_SIG)

L-Edit(2) (1_xor_6_SIG)

S – Edit Simulation

S – Edit Simulation

S – Edit Simulation

Example of BIST

Example of BIST n n n In the prev. slide we see a PRBS that produce 3 -bit seq. that are going through 2 CUT and then checked by the Sig. Here P(x) = X 3+X+1 When CUT is fine then the input to Sig is G(x) = X 5+X 4+X final state is - F(x) = X+1 and the output is - Q(x) = X 2+X+1

Example of BIST n n When the circuit inverter is stuck at 1 G(x) = X 5+X 4+X 3+X ; F(x) = 0 ; and Q(x) = X 2+X Both fulfill polynomial eq. as stated above.

Example of BIST n Math - the division is with mod 2! G(x) = X 5+X 4+X, P(x) = X 3+X+1 X 2+X+1 X 5+X 4+X X 3+X+1 X 5+X 3+X 2 X 4+X 3+X 2+X X 4+X 2+X X 3+X+1 n As expected - F(x) = X+1, Q(x) = X 2+X+1