| Reply #2
dorus
Spain
7 Posts |
 Posted - 11/06/2019 : 08:30:43
|
Hello, I had already read the publication, but I do not understand the last part of this calculation, referring to the example you published: K9GAG08U0E
This memory is 16GB. In your calculation comes out 2076x128x8628.
What I deduce from the calculations of several del.txt memories would be something like this: (2076x8)x128x8628
Page size is 8K + 436 for the datasheet which is 1024x8 + 436 = 8628 = 21b4h. (Use Windows calculator and selection programmer mode).
PageSize="21b4h".
Block size is based on these two lines
- Page Program : (8K + 436)Byte
- Block Erase : (1M + 54.5K)Byte
1024 (1M) / 8 = 128 = 80h
BlockSize="80h"
Now calculate the size of the flash.(Memory Cell Array : (2,076M x 110.49K) x 8bit)
2076 * 128 * 8628 = 2292701184 = 88A7D800h |
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