1. 236601 - Coding and Algorithms for Memories Lecture 7 1

2. Constrained Codes for Memories 2

3. Compare cell levels with a threshold (or a sequence of thresholds) 01101010 3 fixed threshol d Read Cycle of Flash Memories

4. 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 4

5. 01101010 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 5 fixed threshol d

6. 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 01101010 6 00100000 fixed fixed threshol d

7. 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 01101010 7 dynamic threshold 00100000 01101010 fixed dynamic fixed threshol d

8. 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 value 3.Problem 1 for two dimensional array 8

9. Memristors 9 L.O. Chua, “Memristor – The Missing Circuit Element,” IEEE Trans., 1971 ResistorCapacitor InductorMemristor

10. Practical Memristors 2008 Hewlett Packard 10 D.B. Strukov et al, “The missing memristor found,” Nature, 2008 R ON R OFF Voltage [V] Current [mA]

11. Crossbar Arrays 11

12. 12 VgVg RLRL VoVo c ij c ij =0  high resistance  low current sensed c ij =1  low resistance  high current sensed

13. 13 VgVg RLRL VoVo 0 c ij =0  high resistance  low current sensed c ij =1  low resistance  high current sensed 1 1 1 1 Desired Path Sneak Path 1

14. An array A has a sneak path of length 2k+1 affecting the (i,j) cell if – a ij =0 a – There exist r 1,…,r k and c 1,…c k such that a ic 1 = a r 1 c 1 = a r 1 c 2 = ⋯ = a r k c k = a r k j = 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 14 11 01 11

15. 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 15 11 01 11 11 101 11 11 001 11

16. 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 16 11 01 11 11 101 11 11 001 11

17. 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 17 111 111 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0

18. Number of Sneak Paths Arrays N(m,n) = number of m ⅹ n isolated 0-rectangle free arrays Lemma 1: Lemma 2: S(k,l) = number of ways to partition k elements into l nonempty subsets aka the Strirling number of second kind 18

19. Number of Sneak Paths Arrays N(m,n) = number of m ⅹ n isolated 0-rectangle free arrays Lemma 1: Lemma 2: N(m,n) ≈ (m+n)log(m+n) 19

20. Motivation: Stuck-at Cells Flash Memories – Electrical charge represents multiple levels in a cell – The level of the cells can only increase (before a block erasure) Phase-Change Memories (PCMs) – Multiple distinct physical states (one amorphous and some partially crystalline) – Cells might be stuck-at crystalline states 20

21. Background Stuck-at cells/memory with defects – Binary cells which can be stuck-at level 0 or 1 – For q-ary cells, every cell can be stuck-at any level s ∊ [q] Partially stuck-at cells – A cell is partially stuck-at level s, if it can store only levels ≥ s Examples: – In flash, cells can represent only high levels if some charge is trapped in the cells – Cells cannot/shouldn’t represent high levels due to degraded reliability – In PCM, a cell cannot reach the crystalline state (can still reach all other states) 21

22. Definitions An (n,M) q code masking u stuck-at cells is a coding scheme with encoder E and decoder D s.t.: – E: input is a message m and locations & values of u stuck-at cells, and output a vector y= E(m,stuck-at cells) which matches the values of the stuck-at cells – D(y) = m 22

23. Codes Masking Stuck-at Cells Theorem (Heegard’85): An [n,k,u+1] code can mask u stuck-at cells Example: [7,4,3] Hamming code – Possible to write 4 bits and masks 2 stuck-at cells – m=(0,1,1,0); cell #2 is stuck-at 1 and cell #6 is stuck-at 0 – w=(0,0,0;0,1,1,0) – Find z ∊ {0,1} 3 s.t. y=w+z  H masks the stuck-at cells; z=(0,1,1) – y = w+(0,1,1)  H =(0,0,0;0,1,1,0)+(0,1,1;1,0,1,0)= (0,1,1;1,1,0,0) 23 1001011 0101101 0010111 H =