236601 - coding and algorithms for memories lecture 9 1
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236601 - Coding and Algorithms for
MemoriesLecture 9
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Constrained Codes for Memories
•Compare cell levels with a threshold (or a sequence of thresholds)
0 1 1 0 1 0 1 0
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fixed threshold
Read Cycle of Flash Memories
•Charge Leakage voltage drift in one direction•Fixed threshold vs dynamic threshold• Dynamic reading thresholds
reduces the BER• A balanced vector satisfies
#0’s = #1’s
Balanced Codes: Motivation
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Balanced Codes: Motivation
0 1 1 0 1 0 1 0
•In writing, half of cells store 0 and the other half store 1•In reading, the n/2 cells with lower voltages are read as 0 The other n/2 cells with higher voltages are read as 1•Relative ranking is most likely preserved
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fixed threshold
•In writing, half of cells store 0 and the other half store 1•In reading, the n/2 cells with lower voltages are read as 0 The other n/2 cells with higher voltages are read as 1•Relative ranking is most likely preserved
Balanced Codes: Motivation
0 1 1 0 1 0 1 0
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0 0 1 0 0 0 0 0fixed
fixed threshold
•In writing, half of cells store 0 and the other half store 1•In reading, the n/2 cells with lower voltages are read as 0 The other n/2 cells with higher voltages are read as 1•Relative ranking is most likely preserved
Balanced Codes: Motivation
0 1 1 0 1 0 1 0
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dynamic threshold
0 0 1 0 0 0 0 00 1 1 0 1 0 1 0
fixeddynamic
fixed threshold
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Balanced Codes: Problems
• Problems:1. How to guarantee that at most half of the
cells have value 1?2. How to guarantee that exactly half of the
cells have value3. Problem 1 for two dimensional array
Memristors
9L.O. Chua, “Memristor – The Missing Circuit Element,” IEEE Trans., 1971
( , )v M x i i
( , )dx
f x idt
Resistor
v R i
Capacitor
q C v
Inductor
L i
Memristor
φ
q
v
i
M q
Practical Memristors
• 2008 Hewlett Packard
10D.B. Strukov et al, “The missing memristor found,” Nature, 2008
2( ) 1 ( )v ON
OFF
RM q R q t
D
RON
ROFF
Voltage [V]
Curr
ent [
mA]
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Crossbar Arrays
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Vg
RL
Vo
cij
cij=0 high resistance low current sensedcij=1 low resistance high current sensed
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Vg
RL
Vo
0cij=0 high resistance low current sensedcij=1 low resistance high current sensed
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1
1 1
Desired PathSneak Path
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Sneak Path• An array A has a sneak path of length 2k+1 affecting the
(i,j) cell if– aij=0
a
– There exist r1,…,rk and c1,…ck such thataic1
= ar1c1 = ar1c2
= ⋯ = arkck = arkj = 1
a
• An array A satisfies the sneak-path constraint if it has no sneak paths and then is called a sneak-path free array
1 1
0 1
1 1
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Characterization of Sneak Paths
• An array A has an isolated zero-rectangle if it contains a rectangle with exactly a single zero
• An array satisfies the isolated zero-rectangle constraint if it has no isolated zero-rectangles and is called an isolated zero-rectangle free array
• Theorem: The sneak path constraint and the isolated zero-rectangle constraint are equivalent
1 1
0 1
1 1
1 1
1 0 1
1 1
1 1
0 0 1
1 1
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Characterization of Sneak Paths• An array A has an isolated zero-rectangle if there is a rectangle with
exactly a single zero• An array satisfies the isolated zero-rectangle constraint if it has no
isolated zero-rectangles and is called an isolated zero-rectangle free array
• Theorem: The sneak path constraint and the isolated zero-rectangle constraint are equivalent
• Lemma: An array is an isolated zero-rectangle free array iff the 1s in every two rows either completely overlap or are disjoint
1 1
0 1
1 1
1 1
1 0 1
1 1
1 1
0 0 1
1 1
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Characterization of Sneak Paths• Theorem: The sneak path constraint and the
isolated zero-rectangle constraint are equivalent• Lemma: An array is an isolated zero-rectangle
free array iff the 1s in every two rows either completely overlap or are disjoint
1 1 1
1 1 1
0 00
0 00
1
1
0 0 0 0 0
0 0 0 0 001 0 0 0 0
00 0 0 0 0
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Number of Sneak Paths Arrays
• N(m,n) = number of mⅹn isolated 0-rectangle free arrays
• Lemma 1:
• Lemma 2:
• S(k,l) = number of ways to partition k elements into l nonempty subsetsaka the Strirling number of second kind
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Number of Sneak Paths Arrays
• N(m,n) = number of mⅹn isolated 0-rectangle free arrays
• Lemma 1:
• Lemma 2:
• N(m,n) ≈ (m+n)log(m+n)