cmlbay54 发表于 2020-3-14 14:46:52

[Question]There must be a way to use Dgus memory ??

an electronic card I am using uses the screen's memory.
I'm sure this.
Because when I connect a new card to the same screen, it shows the old saved value.
However, this value had to be reset when I connected the new card.
This means that the old card has recorded this value in the memory of the screen.
So, this proves that the "value" wasread from dgus memory.

I want to use dgus memory too. because the eeprom memory of the my processor I use is not enough.
But how can it do that?
(my screen DMT80600L104_01WT)

Kingsly0110208 发表于 2020-3-14 16:17:41

You'd better use 90 command "AA 90 ****** CC 33 C3 3C"

cmlbay54 发表于 2020-3-16 14:51:09

thank you so much :D:D

cmlbay54 发表于 2020-3-16 22:12:55

Is there an example to use these commands?

DWIN-ARK 发表于 2020-3-17 09:12:49

The figure below is a detailed introduction of the 0X90 and 0X91 instructions

cmlbay54 发表于 2020-3-17 13:52:33

for ex. i've a variable named uint32_t count = 105776. i want to record on dgus flash.
1*which command should I use??"write in radndom data memory"? or "write in sequential data memory" ?
2*At which address should I write? how will i determine the address?
3*How do I send the value of 'count'?(( 105775 = 0x019D30 hex))
"AA 90 55 AA 5A A5 <addres??> 01 9D 30 CC 33 C3 3C"is this command right?

DWIN-ARK 发表于 2020-3-17 14:27:55

Both Write in Random Data Memory and Write in Sequential Data Memory are possible. The difference between the two is that the first address range and size of the write data memory are inconsistent. You can choose according to the data you want to store. As for the first address, as long as you are writing It can be within the first address range of the data memory, and it is the same as the address read by the 0X91 instruction, such as AA 90 55 AA 5A A5 01 DE 00 00 01 9D 30 CC 33 C3 3C or AA 90 55 AA 5A A5 00 00 00 00 01 9D 30 CC 33 C3 3C

cmlbay54 发表于 2020-3-17 15:39:05

So, Can each address store 1 byte data, can't it?
The value of 01 9D 30 occupies 3 bytes. so while reading this value, i will write '3' instead of <lenght>.
so my reading command is: AA 91 01 DE 00 00 03 CC 33 C3 3C
Am I understand correctly ?

DWIN-ARK 发表于 2020-3-17 16:05:12

Each address can store 2 bytes of data, the unit is also a word when reading, 3 bytes of data can be written as 02, 2 words of length, and 4 bytes of data can be read

cmlbay54 发表于 2020-3-17 16:23:23

Hmm OK. i think i understood. thanks a lot ! :)
I'm trying right now. i hope i can.
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