Post on 21-Dec-2015
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