SIGNATURE

+INPUT

PROPERTIES

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)

+

SIG. Reg.

…Q (x)

Initial State - I (x) = 0

Final State - R (x)

G (x)R (x)

____ =Q (x) +____

P (x)P (x)

P (x)

It Satisfies this polynomial equation

:

The math behind it(2)…

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

When M>>N the probability for having an unnoticeable mistake is

2M-N - 1 2-N

2M - 1

The math behind it(3)…

≈

So, for as the Sig. Reg. is bigger weget a better approximation on the CUT

S-Edit (1_xor_6_SIG)

DFF1 XOR DFF6

Clk

Clr

DFF1 DFF2 DFF3

DFF10

DFF4DFF6 DFF5

DFF8 DFF9DFF7

Inpu

t

Out0

Out1

Out2

Out3Out4

Out5Ou

t6

Out7

Out8

Out9Cl

D

Q

Q

ClD

Q

QCl

D

Q

QCl

D

Q

Q

ClD

Q

QCl

D

Q

QCl

D

Q

Q

ClD

Q

QCl

D

Q

QCl

D

Q

Q

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 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) = X3+X+1 When CUT is fine then the input to Sig is

- G(x) = X5+X4+X final state is - F(x) = X+1 and the output is - Q(x) = X2+X+1

Example of BIST

When the circuit inverter is stuck at 1G(x) = X5+X4+X3+X ; F(x) = 0 ; and Q(x) = X2+X

Both fulfill polynomial eq. as stated above.

Example of BIST

Math - the division is with mod 2!

As expected - F(x) = X+1, Q(x) = X2+X+1

G(x) = X5+X4+X, P(x) = X3+X+1

X5+X4+X X3+X+1X2+X+1

X5+X3+X2

X4+X3+X2+XX4+X2+X

X3

X+1X3+X+1