嵌入式开发者社区
标题:
TL138 uboot是怎么区分配置两款核心板的
[打印本页]
作者:
Mr.Loser
时间:
2014-9-11 09:33
标题:
TL138 uboot是怎么区分配置两款核心板的
核心板1:DDR2 128M Byte NAND FLASH 4G bit
6 @2 ~( ^+ U2 e% B
核心板2:DDR2 256M Byte NAND FLASH 8G bit
! _. ~2 G; x1 y
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?
* o# M' P& u! m1 e1 z; A6 @
# T& v3 d2 u, [6 b4 L6 P0 B) h r
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?
9 G2 E P+ e6 G a. y: u% C
* I) ?) m/ ^( T. ` T/ I4 O5 x
作者:
2532609929
时间:
2014-9-11 18:55
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:
5 K1 I$ X; H: o6 [2 H, f2 N5 D
/*
; x. k# P) k8 x9 {
* Check memory range for valid RAM. A simple memory test determines
h6 d X( n7 Z5 l4 F$ u$ ^+ D
* the actually available RAM size between addresses `base' and
4 p, f1 `9 A3 c) B: W- `
* `base + maxsize'.
$ N6 s* C3 P; r0 _% u2 @; [1 w6 V
*/
3 R. c4 F5 o7 v$ k' ?
long get_ram_size(long *base, long maxsize)
4 y' K9 C: G8 K5 r
{
z7 C$ Z# D0 _5 c& z: Q
volatile long *addr;
1 g% e7 D- F: q* L+ D% i# `; {+ E
long save[32];
Q& B- B$ |* f3 }0 t# ]; k
long cnt;
" Z) V. l) {5 v, {2 u$ S4 i
long val;
; E6 q5 ]& x" I) \2 P1 R
long size;
: S* n$ j# Q4 l
int i = 0;
5 N+ L4 _7 j8 I4 ~
7 x$ u8 T( A+ }: `! I
for (cnt = (maxsize / sizeof (long)) >> 1; cnt > 0; cnt >>= 1) {
2 ?3 L+ l2 }8 k% s+ O8 q- K
addr = base + cnt; /* pointer arith! */
4 ?, }# R' m: C5 M
sync ();
0 @7 v9 G" x* j2 k/ W1 v
save[i++] = *addr;
, i( _( _& e$ @! U
sync ();
/ b( L+ [/ R: |- b
*addr = ~cnt;
2 _5 V# l5 B5 J
}
0 g7 v+ \/ u) v
9 H8 n; }8 ?( P. b, k) X
addr = base;
, m9 B3 r8 z: V, G7 V
sync ();
/ t. }0 E, U; ~& C% p
save
= *addr;
- d8 G. X q/ Y: b3 {8 y2 b
sync ();
0 T! [6 c: @/ a6 }4 e2 k
*addr = 0;
& y0 I& Z! |+ P* F1 x
# ]1 V. ^, b( e% j6 Q c
sync ();
. c* e" t3 Y& p- K! _. b# U
if ((val = *addr) != 0) {
/ B- u; U! f# f$ f* m2 A' t4 B
/* Restore the original data before leaving the function.
) m8 V# r+ _2 O0 B6 `
*/
: U. t0 {# _5 d% ?; x6 N4 p5 N
sync ();
; c1 B0 y8 U" Y& n1 U7 a; w [
*addr = save
;
& ~5 M X& H% T
for (cnt = 1; cnt < maxsize / sizeof(long); cnt <<= 1) {
3 W; I% k+ Z) e) i/ A
addr = base + cnt;
S" e* s* z+ l) i: f
sync ();
, f7 A4 N/ c: @: v5 m- h
*addr = save[--i];
4 R5 N( q2 T# e7 Z
}
5 | A. l u4 f$ G0 c; F
return (0);
" Q. }( G' J$ i" t5 R/ ~7 E/ i
}
% Z3 z$ j/ X, ^& G
' k" P" D, c' L9 f# P9 i, s" Q
for (cnt = 1; cnt < maxsize / sizeof (long); cnt <<= 1) {
6 Y3 ?) p8 e0 p' E2 W
addr = base + cnt; /* pointer arith! */
- }; I, A, a9 d
val = *addr;
, F6 g, x8 R& T+ [
*addr = save[--i];
* Z0 E$ p4 D9 x
if (val != ~cnt) {
( G, t* P8 i, K/ m7 l2 B2 y' u
size = cnt * sizeof (long);
8 I J) T1 x9 P1 u' Q
/* Restore the original data before leaving the function.
3 t) A/ J/ ]$ i
*/
, {& A. B7 n- y/ f& c! I
for (cnt <<= 1; cnt < maxsize / sizeof (long); cnt <<= 1) {
0 f4 v: u2 @+ m" k, h# A
addr = base + cnt;
7 U: N9 A5 j9 o4 S/ {5 T
*addr = save[--i];
9 F& |( Z+ w9 K1 ~, G" X
}
" ^+ R- h8 g- p5 k( a% {, X
return (size);
* q6 Q" `3 Q7 p
}
8 b9 G9 Q2 ]. N( S! H7 u# }
}
( Z, h) u# h, T& C; f
6 @; G" Y: D7 ^# l. r9 U
return (maxsize);
$ Z7 U5 h# H2 F; j% Y& U
}
& W8 n& [: c8 B' X9 e) V
int dram_init(void)
7 Y$ K" ?5 V0 X8 T# ]
{
# ?) w: T+ T+ h$ _. R
/* dram_init must store complete ramsize in gd->ram_size */
6 h4 A# t( F% e L
gd->ram_size = get_ram_size(
# [0 V9 h6 v' F1 j6 r+ O3 i
(void *)CONFIG_SYS_SDRAM_BASE,
' m7 H2 r3 ]( y! @: }; d1 ~/ M3 C' A2 b
CONFIG_MAX_RAM_BANK_SIZE);
- S* k* X8 W, @1 D4 A" a7 P/ Q; e
return 0;
! a9 o- A9 b' w& @) D
}
# @4 {) f) G& X+ [) h& W
9 B, b, [' u+ i% l+ c2 g; k* |' j
2 J; t. T# z7 c4 {0 g. L
" v0 F1 D7 }- G" G$ B8 S4 D( q
% ]. c, U2 d, T2 U: t8 X$ X
FLASH是通过检查FLASH内部的ID识别容量,希望对您有帮助!
+ K( J. T4 ~1 ~9 L
0 K* w& C* k+ }7 h5 @! A
* B F) J6 [7 w. j) G3 W
* l! E5 R( e: H* b! @9 l i
欢迎光临 嵌入式开发者社区 (https://www.51ele.net/)
Powered by Discuz! X3.4