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' V4 v* z# _/ }: g7 V
      sync ();
9 F) M6 `+ z) X5 ]- o/ F8 ^       *addr = 0;
) I3 h+ E" ?6 y
$ T1 _( e) G  ?       sync ();% c8 N: Z) S. _$ s1 R% @
      if ((val = *addr) != 0) {" H& \" F5 C& p' J5 W9 Q
            /* Restore the original data before leaving the function.! @/ w: J% d; x/ L! E/ |4 R
               */) e1 h/ \: G) e5 b9 C
            sync ();
. |7 g! l2 {/ s% l* ]8 E+ Y             *addr = save;* B4 n# |4 X+ j8 D. _
            for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {
9 Q0 N& a* S+ T' `  c& r! g& a                      addr  = base + cnt;, l; i9 d3 V- Q; Q0 Z, J1 u
                     sync ();7 S) x' ^! M( F2 _. m- u5 v( D+ ?* @) t
                     *addr = save[--i];
+ K/ ^. x" o8 `- }. ~- S6 o             }
/ j+ [8 a' r* M  E) v2 p6 W( d             return (0);' l$ m. f( C: |* o! n# a
      }
4 ~% R1 B" u6 F% e3 z6 R8 k6 P3 w$ X* }2 ~- {
      for (cnt = 1; cnt < maxsize / sizeof (long); cnt <<= 1) {2 v. V' G  e5 G( D; H' v
            addr = base + cnt;       /* pointer arith! */3 T1 V2 g, g7 ^5 {
            val = *addr;
5 q) n3 j5 i6 o7 t3 Y8 E% Q             *addr = save[--i];0 p, {  N/ @( l: ^
            if (val != ~cnt) {
8 \6 q/ D% ~# }+ X' p% R# w  c1 B                      size = cnt * sizeof (long);
  d. i2 |' O4 x- z8 F$ e                      /* Restore the original data before leaving the function.
% K/ T1 }# k0 ]- j7 ?8 j% K6 S                      */- K/ J1 |6 J% Z6 K0 F. N
                     for (cnt <<= 1; cnt < maxsize / sizeof (long); cnt <<= 1) {$ @- q/ K8 r8 {
                              addr  = base + cnt;
: N; w) K, C& M1 r; l6 g                               *addr = save[--i];
) K0 b, E" \7 A( z8 i* [& {                      }. ^- d6 b. `! H; _
                     return (size);
3 Z$ {7 n" J; l6 U             }
! f3 c# y, j* t( f: M       }
* T+ o$ P- [* i9 G$ o4 ?. X: A/ {% X2 |! }- p3 d% s* \3 G
      return (maxsize);
* E8 E/ S7 V0 D}
. v5 O' r7 V5 q: }4 Iint dram_init(void)
) Q. C2 o6 L3 p. L" |  p; c{
9 q; T; a; N% ]       /* dram_init must store complete ramsize in gd->ram_size */) Q& m2 k+ g1 n- x- P9 B" ]
      gd->ram_size = get_ram_size(2 r! q. P! F0 s
                     (void *)CONFIG_SYS_SDRAM_BASE,
! v& t* z- x( ^( B                      CONFIG_MAX_RAM_BANK_SIZE);* j+ @; c& n& O/ \
      return 0;+ b6 Z* c7 x& M2 G
}
& l" i$ I8 T! z7 a+ Y* T3 O6 |) u; _8 z) j

) u* \' f1 u* O9 n% ~0 L9 g! c$ i! c& g& ]2 x# }7 o

$ T" e. y5 a" D& n$ w" dFLASH是通过检查FLASH内部的ID识别容量，希望对您有帮助！
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9 [( g1 J4 p+ h6 r! y  O$ g, f/ ]4 _, p. ^4 D3 I

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