关于MLC的4bit ECC/EDC算法，能否进来指点下

由于硬件不支持MLC的4bit ECC/EDC
所以需要软件来实现

兄弟们觉得这个可行性如何？
软件实现的话，性能要打多大的折扣？

关于MLC用到的4bit ECC/EDC 的算法，兄弟们能给个资料或者什么的吗？
贴个链接也行  

感谢

其实这部分是每一个厂商比较核心的地方。
并且软体去check bit error是非常耗时的，也不现实的。
曾经试验过，速度极其慢。。。。。。
我之前开发的IC也没有支援4bit ecc检验。仅仅有256byte 1bit ecc。
后来使用的IC支援512byte 8bit ECC检验（叫做BCH），MLC的速度依然不是很快。。

不容易，找到一个和我一样写Device驱动的。。交个朋友吧。。呵呵
开发Nand Flash也有2年多了，从底层到管理层。。
做nand driver确实很痛苦。。

qq 112027289

呵呵，谢谢nancyk2h兄指点

看来nancyk2h兄做的比我更底层啊，
我目前的水平也就是在厂家提供的BSP上做做修改而已，惭愧惭愧

我现在连那个SLC上的1bit ECC\EDC都还没咋看明白，这两天困惑呀

SLC的ECC算法就是一个行列的累加计算。这个网络上都有现成的算法分析和代码。
厂家？IC厂家！呵呵我是做IC技术支持的。。

SLC的那算法我是看明白了的
就是下面这段代码里面有部分没看明白

下面是代码，原帖地址在http://blog.2@1ic.com/user1/3633/33493.html#

ECC算法的实现
??static const u_char nand_ecc_precalc_table[] =
??{
????0x00, 0x55, 0x56, 0x03, 0x59, 0x0c, 0x0f, 0x5a, 0x5a, 0x0f, 0x0c, 0x59, 0x03, 0x56, 0x55, 0x00,
????0x65, 0x30, 0x33, 0x66, 0x3c, 0x69, 0x6a, 0x3f, 0x3f, 0x6a, 0x69, 0x3c, 0x66, 0x33, 0x30, 0x65,
????0x66, 0x33, 0x30, 0x65, 0x3f, 0x6a, 0x69, 0x3c, 0x3c, 0x69, 0x6a, 0x3f, 0x65, 0x30, 0x33, 0x66,
????0x03, 0x56, 0x55, 0x00, 0x5a, 0x0f, 0x0c, 0x59, 0x59, 0x0c, 0x0f, 0x5a, 0x00, 0x55, 0x56, 0x03,
????0x69, 0x3c, 0x3f, 0x6a, 0x30, 0x65, 0x66, 0x33, 0x33, 0x66, 0x65, 0x30, 0x6a, 0x3f, 0x3c, 0x69,
????0x0c, 0x59, 0x5a, 0x0f, 0x55, 0x00, 0x03, 0x56, 0x56, 0x03, 0x00, 0x55, 0x0f, 0x5a, 0x59, 0x0c,
????0x0f, 0x5a, 0x59, 0x0c, 0x56, 0x03, 0x00, 0x55, 0x55, 0x00, 0x03, 0x56, 0x0c, 0x59, 0x5a, 0x0f,
????0x6a, 0x3f, 0x3c, 0x69, 0x33, 0x66, 0x65, 0x30, 0x30, 0x65, 0x66, 0x33, 0x69, 0x3c, 0x3f, 0x6a,
????0x6a, 0x3f, 0x3c, 0x69, 0x33, 0x66, 0x65, 0x30, 0x30, 0x65, 0x66, 0x33, 0x69, 0x3c, 0x3f, 0x6a,
????0x0f, 0x5a, 0x59, 0x0c, 0x56, 0x03, 0x00, 0x55, 0x55, 0x00, 0x03, 0x56, 0x0c, 0x59, 0x5a, 0x0f,
????0x0c, 0x59, 0x5a, 0x0f, 0x55, 0x00, 0x03, 0x56, 0x56, 0x03, 0x00, 0x55, 0x0f, 0x5a, 0x59, 0x0c,
????0x69, 0x3c, 0x3f, 0x6a, 0x30, 0x65, 0x66, 0x33, 0x33, 0x66, 0x65, 0x30, 0x6a, 0x3f, 0x3c, 0x69,
????0x03, 0x56, 0x55, 0x00, 0x5a, 0x0f, 0x0c, 0x59, 0x59, 0x0c, 0x0f, 0x5a, 0x00, 0x55, 0x56, 0x03,
????0x66, 0x33, 0x30, 0x65, 0x3f, 0x6a, 0x69, 0x3c, 0x3c, 0x69, 0x6a, 0x3f, 0x65, 0x30, 0x33, 0x66,
????0x65, 0x30, 0x33, 0x66, 0x3c, 0x69, 0x6a, 0x3f, 0x3f, 0x6a, 0x69, 0x3c, 0x66, 0x33, 0x30, 0x65,
????0x00, 0x55, 0x56, 0x03, 0x59, 0x0c, 0x0f, 0x5a, 0x5a, 0x0f, 0x0c, 0x59, 0x03, 0x56, 0x55, 0x00
??};

??//?Creates non-inverted ECC code from line parity
??static void nand_trans_result(u_char reg2, u_char reg3,u_char *ecc_code)
??{
????u_char a, b, i, tmp1, tmp2;

????/* Initialize variables */
????a = b = 0x80;
????tmp1 = tmp2 = 0;

????/* Calculate first ECC byte */
????for (i = 0; i < 4; i++) ????{ ??????if (reg3 & a)??? /* LP15,13,11,9 --> ecc_code[0] */
????????tmp1 |= b;
??????b >>= 1;
??????if (reg2 & a)??? /* LP14,12,10,8 --> ecc_code[0] */
????????tmp1 |= b;
??????b >>= 1;
??????a >>= 1;
????}

????/* Calculate second ECC byte */
????b = 0x80;
????for (i = 0; i < 4; i++) ????{ ??????if (reg3 & a)??? /* LP7,5,3,1 --> ecc_code[1] */
????????tmp2 |= b;
??????b >>= 1;
??????if (reg2 & a)??? /* LP6,4,2,0 --> ecc_code[1] */
????????tmp2 |= b;
??????b >>= 1;
??????a >>= 1;
????}

????/* Store two of the ECC bytes */
????ecc_code[0] = tmp1;
????ecc_code[1] = tmp2;
??}

??//?Calculate 3 byte ECC code for 256 byte block
??void nand_calculate_ecc (const u_char *dat, u_char *ecc_code)
??{
????u_char idx, reg1, reg2, reg3;
????int j;

????/* Initialize variables */
????reg1 = reg2 = reg3 = 0;
????ecc_code[0] = ecc_code[1] = ecc_code[2] = 0;

????/* Build up column parity */
????for(j = 0; j < 256; j++) ????{ ??????/* Get CP0 - CP5 from table */ ??????idx = nand_ecc_precalc_table[dat[j]]; ??????reg1 ^= (idx & 0x3f); ??????/* All bit XOR = 1 ? */ if (idx & 0x40) { // 我就这段没看明白 老兄能指点下不 // 主要是这个reg3 reg2 怎么的就跟j来运算出来就是行累加啊 /* // 下面的注释里面的值是我根据nand_trans_result函数倒推出来的 BIT7 BIT7 BIT7 BIT7 BIT7 BIT7 BIT7 BIT7 reg2 !P64’ !P32’ !P16’ !P8’ !P1024’ !P512’ !P256’ !P128’ reg3 !P64 !P32 !P16 !P8 !P1024 !P512 !P256 !P128 */ ????????reg3 ^= (u_char) j; ????????reg2 ^= ~((u_char) j); ??????} ????} ????/* Create non-inverted ECC code from line parity */ ????nand_trans_result(reg2, reg3, ecc_code); ????/* Calculate final ECC code */ ????ecc_code[0] = ~ecc_code[0]; ????ecc_code[1] = ~ecc_code[1]; ????ecc_code[2] = ((~reg1) << 2) | 0x03; ??} ??//?Detect and correct a 1 bit error for 256 byte block ??int nand_correct_data (u_char *dat, u_char *read_ecc, u_char *calc_ecc) ??{ ????u_char a, b, c, d1, d2, d3, add, bit, i; ????/* Do error detection */ ????d1 = calc_ecc[0] ^ read_ecc[0]; ????d2 = calc_ecc[1] ^ read_ecc[1]; ????d3 = calc_ecc[2] ^ read_ecc[2]; ????if ((d1 | d2 | d3) == 0) ????{ ??????/* No errors */ ??????return 0; ????} ????else ????{ ??????a = (d1 ^ (d1 >> 1)) & 0x55;
??????b = (d2 ^ (d2 >> 1)) & 0x55;
??????c = (d3 ^ (d3 >> 1)) & 0x54;

??????/* Found and will correct single bit error in the data */
??????if ((a == 0x55) && (b == 0x55) && (c == 0x54))
??????{
????????c = 0x80;
????????add = 0;
????????a = 0x80;
????????for (i=0; i<4; i++) ????????{ ??????????if (d1 & c) ????????????add |= a; ??????????c >>= 2;
??????????a >>= 1;
????????}
????????c = 0x80;
????????for (i=0; i<4; i++) ????????{ ??????????if (d2 & c) ????????????add |= a; ??????????c >>= 2;
??????????a >>= 1;
????????}
????????bit = 0;
????????b = 0x04;
????????c = 0x80;
????????for (i=0; i<3; i++) ????????{ ??????????if (d3 & c) ????????????bit |= b; ??????????c >>= 2;
??????????b >>= 1;
????????}
????????b = 0x01;
????????a = dat[add];
????????a ^= (b << bit); ????????dat[add] = a; ????????return 1; ??????} ??????else ??????{ ????????i = 0; ????????while (d1) ????????{ ??????????if (d1 & 0x01) ????????????++i; ??????????d1 >>= 1;
????????}
????????while (d2)
????????{
??????????if (d2 & 0x01)
????????????++i;
??????????d2 >>= 1;
????????}
????????while (d3)
????????{
??????????if (d3 & 0x01)
????????????++i;
??????????d3 >>= 1;
????????}
????????if (i == 1)
????????{
??????????/* ECC Code Error Correction */
??????????read_ecc[0] = calc_ecc[0];
??????????read_ecc[1] = calc_ecc[1];
??????????read_ecc[2] = calc_ecc[2];
??????????return 2;
????????}
????????else
????????{
??????????/* Uncorrectable Error */
??????????return -1;
????????}
??????}
????}

????/* Should never happen */
????return -1;
??}

 

nancyk2h兄
你们做哪方面IC的呀

我做的是消费类电子的IC,主要是手机,mp3,电子字典等.

呵呵 那说不定我们也在用你们的产品呢

哪家的IC？

三星 掌微 凌阳 安凯

晕。。。我是凌阳的工程师。。。

呵呵 北京？深圳？
以前用凌阳的作过一款学习机 不过那时俺才刚毕业 还做应用的

学习机？ 哈哈！用的是16位的IC吧！162001 GPL162002 GPL162003??
很可能你是我的客户哦！我在北京。但是马上要去深圳出差。。

记不得啦 都是一年前做的了 再说那会我还做应用 对硬件不关心呢

请教nancyk2h
nand 的管理驱动里面 实现磨损均衡怎么个实现原理？

写平均是吗?可以实现.创建于一个动态的管理表(好块表),并且做一个空闲链表.
当你写一个blk,就从空闲链表中取出,写入该blk中,并且在c area标记擦写的次数.
并将动态管理表的物理地址和逻辑地址进行更新.标记你刚刚写入的blk和逻辑的对应,将之前对应的物理blk
回收到空闲表中....

绝对的写平均是做不到的,只能是相对的,主要思想是减少擦除次数

我目前用Sniper167贴的算法，但是发现有问题，如下：



(测试所用的数据以及计算得到的ECC校验)

写入的数据(256字节):
79 83 BB 48 24 00 00 00 00 00 00 00 00 A8 C3 C2
D4 AA BE 15 43 DC 74 82 B6 6D 5B BD 8D B9 96 6B
DB 96 6D E3 D8 71 25 37 6F DB 36 6A 9D A7 E5 62
A7 DD DB B6 7D B8 36 7E DA 27 6E 5B A0 05 8F 00
20 00 00 00 00 00 00 50 6D 34 AE 58 BD 5D A8 70
BF B7 58 8B D0 27 0D 3B 24 EF 48 F4 44 70 2D D7
3E 5C FB 70 6D FB B4 AD D3 B6 71 DC B6 DC D3 B6
31 67 E2 37 38 B4 74 8B AA 11 84 10 01 80 00 00
08 E1 35 02 6B A0 EA 96 2E A4 52 85 3F A9 EE A9
C0 CF D5 D8 C5 56 58 C3 EB FE 94 5C DA 6D 4F B6
F6 03 7A A7 73 F7 69 39 87 0B DA F8 A7 1F EE 28
FD D6 5D 43 08 21 02 00 01 00 10 C2 DF A4 3E 34
F5 ED FD E4 3E 91 BF 66 6B DF C9 47 47 22 BF F2
93 30 FB 76 CB 1D 5C EB 70 41 C3 2F 9D 82 B5 12
17 C7 8F D3 C6 6D 83 3F CC 66 B6 2D CC 88 10 63
08 00 44 00 21 84 46 25 AB A5 EB DB F7 62 FB 6D

写入计算得到的3字节ECC校验:65 66 AB

读出的数据(256字节):
69 03 22 08 24 00 00 00 00 00 00 00 00 28 01 00
00 00 2C 14 03 58 74 02 20 6D 5A A8 80 A9 80 63
D3 86 49 82 90 10 01 32 6D 9A 06 00 00 02 40 60
20 1D 58 12 64 10 16 3A 10 02 0A 1A 00 01 03 00
00 00 00 00 00 00 00 10 28 34 22 00 04 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 04 00 00 00 00 00
00 41 25 02 61 20 62 12 2A 20 52 05 3F 28 6C 29
40 43 54 50 41 14 08 80 C0 EC 80 5C DA 6D 45 A4
E2 02 7A A7 73 F7 69 38 87 0B DA F0 A7 1F EE 28
FD C2 4C 00 08 21 02 00 01 00 10 80 90 A4 20 34
C1 CD E8 C4 3C 91 AB 42 40 C2 C1 44 43 22 BA A0
12 10 82 16 89 04 10 60 60 41 41 21 1C 00 31 00
00 80 06 50 46 61 82 03 80 60 96 20 8C 80 10 02
00 00 40 00 21 84 04 21 21 20 48 4A 72 60 8B 09

读入的256字节计算得到的3字节ECC校验:3C F3 0F

更正上面贴的读出的数据为：
读出的数据(256字节):
69 03 22 08 24 00 00 00 00 00 00 00 00 28 01 00  
00 00 2C 14 03 58 74 02 20 6D 5A A8 80 A9 80 63  
D3 86 49 82 90 10 01 32 2D 9A 06 00 00 02 40 60  
20 1D 58 12 64 10 16 3A 10 02 0A 1A 00 01 03 00  
00 00 00 00 00 00 00 10 28 34 22 00 04 00 00 00  
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00  
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00  
00 00 00 00 00 00 00 00 00 00 04 00 00 00 00 00  
00 41 25 02 61 20 62 12 2A 20 52 05 3F 28 6C 29  
40 43 54 50 41 14 08 80 C0 EC 80 5C DA 6D 45 A4  
E2 02 7A A7 73 F7 69 38 87 0B DA F0 A7 1F EE 28  
FD C2 4C 00 08 21 02 00 01 00 10 80 90 A4 20 34  
C1 CD E8 C4 3C 91 AB 42 40 C2 C1 44 43 22 BA A0  
12 10 82 16 89 04 10 60 60 41 41 21 1C 00 31 00  
00 80 06 50 46 61 82 03 80 60 96 20 8C 80 10 02  
00 00 40 00 21 84 04 21 21 20 48 4A 72 60 8B 09