What can i do for using 64M, 128M or even larger ram?

[Henry]

There are 64M ram in my board (arm based), but the system
only use 32M. If I want to use the total size of the ram,
what should i do in my platform builder?

[Yannick Chamming's]

Memory configuration is managed in two step.

First, you need to map your physical memory into virtual space : this is
done in OEMAdressTable which is parsed at the earliest initialization steps
of the kernel. In this array, you will need to map your 64MB instead of
initial 32.

Next, you map your kernel and your RAM (for execution and object store)
into the virtual space in the CONFIG.BIB file. Here, you will dispatch your
64MB between kernel image (if you execute your image in RAM) and "real" RAM
for object store and execution.

In PB Help, you will find all the informations on how to process.

Yannick

--
----------------------------------------------------------------
Yannick Chamming's
ADESET
Windows Embedded Manager
ychammings AT adeset DOT com>
http://www.adeset.com
Tél : +33 (0)4.72.18.57.77
Fax : +33 (0)4.72.18.57.78
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"Henry" <anon...@discussions.microsoft.com> wrote in message
news:071301c3d99c$fc8bc5d0$a401...@phx.gbl...

[anon...@discussions.microsoft.com]

>-----Original Message-----
>Memory configuration is managed in two step.
>
>First, you need to map your physical memory into virtual
space : this is
>done in OEMAdressTable which is parsed at the earliest
initialization steps
>of the kernel. In this array, you will need to map your
64MB instead of
>initial 32.
>

yes.in my OEMAdressTable, 64M sdram has already been
mapped:

OEMAddressTable
;;;---------------------------------------------------
----------
;;; Virt Addr Phys Addr MB
;;;---------------------------------------------------
----------
DCD 0x80000000, 0x02000000, 30 ; 30 MB SROM
(SRAM/ROM) BANK 0
DCD 0x82000000, 0x08000000, 32 ; 32 MB SROM
(SRAM/ROM) BANK 1
DCD 0x84000000, 0x10000000, 32 ; 32 MB SROM
(SRAM/ROM) BANK 2
DCD 0x86000000, 0x18000000, 32 ; 32 MB SROM
(SRAM/ROM) BANK 3
DCD 0x88000000, 0x20000000, 32 ; 32 MB SROM
(SRAM/ROM) BANK 4
DCD 0x8A000000, 0x28000000, 32 ; 32 MB SROM
(SRAM/ROM) BANK 5
DCD 0x8C000000, 0x30000000, 64 ; 64 MB DRAM BANK 0,1
DCD 0x90800000, 0x48000000, 1 ; Memory control
register
.............

**************************
in the config.bib(29M,31M):

NK 8C200000 01D00000 RAMIMAGE
RAM 8E000000 01F00000 RAM

i revised it(29M,63M):

NK 80040000 01D00000 RAMIMAGE
RAM 8c200000 03F00000 RAM

but the result doesn't change. i don't know why?

****************************

[Yannick Chamming's]

> in the config.bib(29M,31M):

> NK 8C200000 01D00000 RAMIMAGE
> RAM 8E000000 01F00000 RAM

This is correct : you use your 64MB for both CE image (NK entry) and RAM for
execution and storage

i revised it(29M,63M):

NK 80040000 01D00000 RAMIMAGE
RAM 8c200000 03F00000 RAM

This is NOT correct :
- NK is mapped in ROM space : imply that you boot your image in ROM
- 8c200000 + 03F00000 = 90100000 instead of 90000000 (i.e 8c000000 + 64MB)

--
----------------------------------------------------------------

Yannick Chamming's
ADESET
Windows Embedded Manager
ychammings AT adeset DOT com>
http://www.adeset.com
Tél : +33 (0)4.72.18.57.77
Fax : +33 (0)4.72.18.57.78
----------------------------------------------------------------

<anon...@discussions.microsoft.com> wrote in message
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