topic Re: EEPROM setup MC9S12DP512 in CodeWarrior for MCU

EEPROM setup MC9S12DP512

Leinad — Sun, 03 Apr 2011 12:10:14 GMT

HI, i am still new to freescale product and codewarrior.

I am trying to setup EEPROM, writing 32 bit data to EEPROM.

So far, i know the EEPROM can only consits of 16 bit of data so in order to write

32 biy of data, i need to write modify sector twice. But i do not know how to separate

the bit into two 16 bit data. I'm thinking the function will look like this:

BOOL EEPROM_Write32(UINT32 volatile * const address, const UINT32 data)

*address = data

So, if anyone can tell me how to separate the 32 bit into two 16 bit data, it would be great.

thanks.

Re: EEPROM setup MC9S12DP512

kef — Sun, 03 Apr 2011 14:02:02 GMT

long data;

unsigned short hiword, loword;

long *address;

hiword = data / 0x10000ul;

loword = data;

((short*)address)[0] = hiword;

((short*)address)[1] = loword;

--------

To update whole sector you need not two sector modify commands, but one sector modify and one word program. Sector modify erases whole 2-words sector and updates one word.

Don't forget to word align your data in EEPROM.

Re: EEPROM setup MC9S12DP512

Leinad — Mon, 04 Apr 2011 07:27:28 GMT

i tried it but the compiler gave me a warning

"possible loss of data" in this row:

dataLo = data;

I suspect this is cus data is long and dataLo is short;

How do you do the word alignment?

i've read some guides from here but still don understand

http://www.freescale.com/files/microcontrollers/doc/app_note/AN2400.pdf

Also,

data / 0x10000ul; what does this line do?

Re: EEPROM setup MC9S12DP512

Lundin — Mon, 04 Apr 2011 13:44:46 GMT

The DP12 eeprom has an erase size of 4 bytes, and a write size of 2 bytes. This means that you have to erase a whole 4 bytes before writing, but once that is done you can write to the upper 2 and lower 2 bytes separately. This will not lead to the same sector getting written to twice. It is of course easiest to erase 4 and then write 4 at the same time, however.

"Possible loss of data" in Codewarrior means nothing, that warning is completely broken and irrational, as discussed plenty elsewhere on these forums. I consider it a compiler bug, you should disable it.

Word alignment you do manually. Each sector has to start at an address which is an even multiple of 4.

> data / 0x10000ul; what does this line do?

Apparently it divides data by 0x10000 or 65536 dec. The purpose is to mask out the hiword from the long. It does look odd however... on a dumb compiler, that line will be evaluated to division, making it needlessly ineffective.

An alternative would be:

hiword = (uint16_t) (data >> 16)

loword = (uint16_t) data;

Re: EEPROM setup MC9S12DP512

Jim_P — Sat, 16 Apr 2011 05:44:08 GMT

I tend to use an union for this

union un_long
{
long l;
word w[2];
int i[2];
byte b[4];

};
union un_Long Value;

and reference the value by

Value.L = 232085308;

and thus can do this

T = Value.W[0];

T = value.W[1];

faster code. Simpler to understand and access the variable as needed - - as byte, int or long.

note Word is defined as unsigned Int