TL138 uboot是怎么区分配置两款核心板的

Mr.Loser · 发表于 2014-9-11 09:33:39

核心板1：DDR2 128M Byte NAND FLASH 4G bit
核心板2：DDR2 256M Byte NAND FLASH 8G bit
这两个核心板公共一个uboot，DDR2的配置肯定是不同的，uboot怎么检查到是哪个核心板，然后去执行相应的配置呢？

是不是通过读NAND FLASH的ID，两款核心板NAND FLASH不同，ID也不同，这样读到了ID就知道是哪个核心板了？

2532609929 · 发表于 2014-9-11 18:55:41

DDR通过检查最大有效地址来识别容量，具体请看uboot的common/memsize.c文件中的检查代码，也可以看如下：
/*
* Check memory range for valid RAM. A simple memory test determines
* the actually available RAM size between addresses `base' and
* `base + maxsize'.
*/
long get_ram_size(long *base, long maxsize)
{
      volatile long *addr;
      long          save[32];
      long          cnt;
      long          val;
      long          size;
      int          i = 0;

      for (cnt = (maxsize / sizeof (long)) >> 1; cnt > 0; cnt >>= 1) {
            addr = base + cnt;       /* pointer arith! */
            sync ();
            save[i++] = *addr;
            sync ();
            *addr = ~cnt;
      }

      addr = base;
      sync ();
      save = *addr;
) I+ B0 X8 H5 l4 s7 N       sync ();$ U; L4 F- a+ j& b- w2 z
      *addr = 0;
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  Y8 d+ ]- l% c( Q+ i$ b; {8 X       sync ();$ `! L7 J* Q& K" X: A( H6 j
      if ((val = *addr) != 0) {
* P7 P0 b1 P7 T6 k6 h* |             /* Restore the original data before leaving the function.
$ e# N! V: H5 U7 Z0 V* Y* C; \9 q                */) V6 M3 |- |3 D* V6 u; d6 E- S- O
            sync ();1 L" {) d9 d5 Q( ^0 g# d/ E7 z! e1 D
            *addr = save;
; L/ \  \1 D! N, f4 K& m             for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {7 V5 f. T. z) N4 S
                     addr  = base + cnt;. v2 V# ]8 }, _. `: t
                     sync ();6 k' W1 m2 D' f
                     *addr = save[--i];
, F8 P- ]9 k2 o1 `# i             }
# Q- ?% L. ~6 m, Y& P$ a             return (0);
7 |$ T  Z' D2 {       }
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; M  H% d) V7 \       for (cnt = 1; cnt < maxsize / sizeof (long); cnt <<= 1) {3 W. n' t4 ]" o7 m
            addr = base + cnt;       /* pointer arith! */  Q0 ]: P" U2 t
            val = *addr;
, e  K5 @4 O9 X             *addr = save[--i];' A2 h4 \' w% M2 |" t" H
            if (val != ~cnt) {" d% d' C; l$ g0 l7 y
                     size = cnt * sizeof (long);
0 x4 D- B, p$ t5 _; M$ S: k                      /* Restore the original data before leaving the function.2 _& n  s! M9 X* _( Z8 S& U
                     */; Y+ G0 v7 K4 a  ~; j! K" ~- y
                     for (cnt <<= 1; cnt < maxsize / sizeof (long); cnt <<= 1) {: c2 E9 K' w3 S" r$ C% S
                              addr  = base + cnt;
/ @' E, S7 s1 v9 H4 P( X$ _- _                               *addr = save[--i];8 L4 a" ?! B$ \8 |' O
                     }1 h# a  S1 T" O% i) U
                     return (size);
' X# Y9 P2 G% `* G8 h  `" p1 X( J             }) R  E4 D# X) w) J- g0 ~
      }0 s, Z7 b1 Z7 D, G5 {
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      return (maxsize);( H4 ~7 C0 p- m6 T
}5 @/ y& m8 u6 M9 }/ C7 E8 ?' E/ o
int dram_init(void)
3 E. ~6 m' C, H! K+ @{
% e6 q% F) t" ]. c5 P       /* dram_init must store complete ramsize in gd->ram_size */
0 a$ @7 H" Z; X& A4 S       gd->ram_size = get_ram_size(
3 T" T0 O( W- O  j! C* p( x                      (void *)CONFIG_SYS_SDRAM_BASE,5 Y! F# Q1 b. n
                     CONFIG_MAX_RAM_BANK_SIZE);9 B8 H- b/ Y$ |
      return 0;- @' G7 ~  J; S8 i( P) K! B
}
& {8 ]2 A+ M* D1 u, I  e' ~; K( X( n: O4 R; C% N
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5 U  M8 R) c* N2 W' L/ tFLASH是通过检查FLASH内部的ID识别容量，希望对您有帮助！6 o( K( a4 c& w% _- n. d
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