三星2440 wince5.0 nandflash驱动分析
creator
sz111@126.com
若干天的实验和测试，终于对三星2440 wince5.0的nandflash有了一个初步的认识。
首先解决了一个疑问：
1.我的开发板仅仅有nandflash驱动，把stepldr烧录到0块，eboot烧录到第二块，看
程序是从nandflash的第二块开始读的啊。怎么就不行了呢？最后看到长高网站介绍要放到
第16块。
仔细研究发现：
#define NAND_BLOCK_SIZE_BYTES 0x00020000
#define NAND_PAGE_SIZE_BYTES 0x00000200
#define NAND_PAGES_PER_BLOCK (NAND_BLOCK_SIZE_BYTES / NAND_PAGE_SIZE_BYTES)
// NOTE: we assume that this Steppingstone loader occupies *part* the first (good) NAND flash block. More
// specifically, the loader takes up 4096 bytes (or 8 NAND pages) of the first block. We'll start our image
// copy on the very next page.
#define NAND_COPY_PAGE_OFFSET 2*NAND_PAGES_PER_BLOCK
NAND_BLOCK_SIZE_BYTES为0x00020000，而我的64M的nandflash本来是1(block)=1(page)x32=512x32=0x4000
为何这里却是0x00020000呢？
最后发现cfnand.h中：
static S2440Spec astNandSpec[] = {
/*************************************************************************/
/* nMID, nDID, */
/* nNumOfBlks */
/* nPgsPerBlk */
/* nSctsPerPg */
/* nNumOfPlanes */
/* nBlksInRsv */
/* nBadPos */
/* nLsnPos */
/* nECCPos */
/* nBWidth */
/* nTrTime */
/* nTwTime */
/* nTeTime */
/* nTfTime*/
/*************************************************************************/
/* 8Gbit DDP NAND Flash */
{ 0xEC, 0xD3, 8192, 64, 4, 2,160, 0, 2, 8, BW_X08, 50, 350, 2000, 50},
/* 4Gbit DDP NAND Flash */
{ 0xEC, 0xAC, 4096, 64, 4, 2, 80, 0, 2, 8, BW_X08, 50, 350, 2000, 50},
{ 0xEC, 0xDC, 4096, 64, 4, 2, 80, 0, 2, 8, BW_X08, 50, 350, 2000, 50},
//{ 0xEC, 0xBC, 4096, 64, 4, 2, 80, 0, 2, 8, BW_X16, 50, 350, 2000, 50},
//{ 0xEC, 0xCC, 4096, 64, 4, 2, 80, 0, 2, 8, BW_X16, 50, 350, 2000, 50},
/* 2Gbit NAND Flash */
{ 0xEC, 0xAA, 2048, 64, 4, 1, 40, 0, 2, 8, BW_X08, 50, 350, 2000, 50},
{ 0xEC, 0xDA, 2048, 64, 4, 1, 40, 0, 2, 8, BW_X08, 50, 350, 2000, 50},
//{ 0xEC, 0xBA, 2048, 64, 4, 1, 40, 0, 2, 8, BW_X16, 50, 350, 2000, 50},
//{ 0xEC, 0xCA, 2048, 64, 4, 1, 40, 0, 2, 8, BW_X16, 50, 350, 2000, 50},
/* 2Gbit DDP NAND Flash */
{ 0xEC, 0xDA, 2048, 64, 4, 2, 40, 0, 2, 8, BW_X08, 50, 350, 2000, 50},
{ 0xEC, 0xAA, 2048, 64, 4, 2, 40, 0, 2, 8, BW_X08, 50, 350, 2000, 50},
//{ 0xEC, 0xBA, 2048, 64, 4, 2, 40, 0, 2, 8, BW_X16, 50, 350, 2000, 50},
//{ 0xEC, 0xCA, 2048, 64, 4, 2, 40, 0, 2, 8, BW_X16, 50, 350, 2000, 50},
/*1Gbit NAND Flash */
{ 0xEC, 0xA1, 1024, 64, 4, 1, 20, 0, 2, 8, BW_X08, 50, 350, 2000, 50},
{ 0xEC, 0xF1, 1024, 64, 4, 1, 20, 0, 2, 8, BW_X08, 50, 350, 2000, 50},
//{ 0xEC, 0xB1, 1024, 64, 4, 1, 20, 0, 2, 8, BW_X16, 50, 350, 2000, 50},
//{ 0xEC, 0xC1, 1024, 64, 4, 1, 20, 0, 2, 8, BW_X16, 50, 350, 2000, 50},
/* 1Gbit NAND Flash */
{ 0xEC, 0x79, 8192, 32, 1, 4,120, 5, 0, 6, BW_X08, 50, 350, 2000, 50},
{ 0xEC, 0x78, 8192, 32, 1, 4,120, 5, 0, 6, BW_X08, 50, 350, 2000, 50},
//{ 0xEC, 0x74, 8192, 32, 1, 4,120,11, 0, 6, BW_X16, 50, 350, 2000, 50},
//{ 0xEC, 0x72, 8192, 32, 1, 4,120,11, 0, 6, BW_X16, 50, 350, 2000, 50},
/* 512Mbit NAND Flash */
{ 0xEC, 0x76, 4096, 32, 1, 4, 70, 5, 0, 6, BW_X08, 50, 350, 2000, 50},
{ 0xEC, 0x36, 4096, 32, 1, 4, 70, 5, 0, 6, BW_X08, 50, 350, 2000, 50},
//{ 0xEC, 0x56, 4096, 32, 1, 4, 70,11, 0, 6, BW_X16, 50, 350, 2000, 50},
//{ 0xEC, 0x46, 4096, 32, 1, 4, 70,11, 0, 6, BW_X16, 50, 350, 2000, 50},
/* 256Mbit NAND Flash */
{ 0xEC, 0x75, 2048, 32, 1, 1, 35, 5, 0, 6, BW_X08, 50, 350, 2000, 50},
{ 0xEC, 0x35, 2048, 32, 1, 1, 35, 5, 0, 6, BW_X08, 50, 350, 2000, 50},
//{ 0xEC, 0x55, 2048, 32, 1, 1, 35,11, 0, 6, BW_X16, 50, 350, 2000, 50},
//{ 0xEC, 0x45, 2048, 32, 1, 1, 35,11, 0, 6, BW_X16, 50, 350, 2000, 50},
/* 128Mbit NAND Flash */
{ 0xEC, 0x73, 1024, 32, 1, 1, 20, 5, 0, 6, BW_X08, 50, 350, 2000, 50},
{ 0xEC, 0x33, 1024, 32, 1, 1, 20, 5, 0, 6, BW_X08, 50, 350, 2000, 50},
//{ 0xEC, 0x53, 1024, 32, 1, 1, 20,11, 0, 6, BW_X16, 50, 350, 2000, 50},
//{ 0xEC, 0x43, 1024, 32, 1, 1, 20,11, 0, 6, BW_X16, 50, 350, 2000, 50},
{ 0x00, 0x00, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
};
说明三星已经对多个flash（包括大页，小页）进行了支持。为了统一起见，1(BLOCK)=0x00020000
这样就可以保证在文件系统中大页的flash和小页的flash都可以采用统一的block的大小。而在进行一个
逻辑转换就行。这样，对文件系统来说，1block=128k，对应大页的nand是一对一，对应小页的nand是一对八。
