yadunandankasu yadunandankasu

Reg: HCS12 Memory Re-Map

Discussion created by yadunandankasu yadunandankasu on May 26, 2008
Latest reply on May 29, 2008 by yadunandankasu yadunandankasu
Dear Everyone,
 
I am working with MC9S12DJ 64 MCU
 
By default, when I create a new project using Codewarrior Ver5.7.0. the .prm file assigns 3Kbytes to RAM i.e from 0x0400 TO 0x0FFF.
 
I want to remap this to have 4Kbytes.
 
And I came to know the default Start12.c code has his remap technique if we enable Serial Monitor according to the following code snippset.
 
Code:
#if defined(_HCS12_SERIALMON)   /* for Monitor based software remap the RAM & EEPROM to adhere      to EB386. Edit RAM and EEPROM sections in PRM file to match these. */   ___INITRG = 0x00;  /* lock registers block to 0x0000 */   ___INITRM = 0x39;  /* lock Ram to end at 0x3FFF */   ___INITEE = 0x09;  /* lock EEPROM block to end at 0x0fff */#endif

But, i have the follwoing two questions on this,

1. Can anyone please tell me how to enable this Serial Monitor code to re-map the memory in the MCU.

2. And imagine if the Serial Monitor is enabled, and the initialization of registers (INITRG = 0x00, INITRM = 0x39, and INITEE = 0x09) has become effective, what will be the starting address for the RAM or what are the boundaries of RAM in the whole memory map.

 According to the above snippset, it says the ending address of the RAM is 0x3FFF, and the total RAM size of this MCU is 4Kbytes, so the starting address will be 0x3000. (i.e. form 0x3000 to 0x3FFF = 4Kbytes)

Here, I am confused, why the INITRM was assigned 0x39 and why not 0x31...?? and what is the starting address of RAM if INITRM = 0x39; (Why the RAM11 bit of INITRM register has been assigned as 1).

Actually, I am pretty much confused with RAM11 bit, I referred to AN2881.pdf, it says...for example

if INITRM = 0x20; then RAM Boundary will be 0x2000 ~ 0x2FFF

if INITRM = 0x30; then RAM Boundary will be 0x3000 ~ 0x3FFF

so what will be the starting ddress format if we use RAM11 bit ...???

As I am not very experieced in this filed, all the help from you will be greatly helpful to me...

Thanks to you all in advance

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