LPC1778 EEPROM endurace calculation

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LPC1778 EEPROM endurace calculation

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lpcware
NXP Employee
NXP Employee
Content originally posted in LPCWare by hubert.melchert on Wed Apr 13 01:39:29 MST 2016
I am writing software that store 5 bytes of data in eeprom every minute. EEPROM is used as a cyclic buffer to increase eeprom endurance. After reading reference manual of LPC1778 I am a little confused about eeprom endurance calculation. Every minute,  5 bytes of data are written into EEWDATA (eeprom write data) register and programmed into EEPROM using program command in EECMD register. Everything works fine but I don't understand programming process. Even when I want to program only 5 bytes of one page (64 bytes); Does the whole page has to go trough erase/write cycle ? That would significantly reduce EEPROM endurance in my application.
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lpcware
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NXP Employee
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lpcware
NXP Employee
NXP Employee
Content originally posted in LPCWare by MikeSimmonds on Wed Apr 13 10:41:00 MST 2016
Or go with an external spi (or I2C) eeprom.
For instance the MicroChip 25LC320A (same 4K size -- actually 64 bytes bigger) with 1,000,000
(one million) erase cycles and a 32 byte page size.
33 pence (UK) for 25+ from Farnell.

There are always alternatives. It just depends on cost/time/'stubborn client' factors.

Regards, Mike.
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lpcware
NXP Employee
NXP Employee
Content originally posted in LPCWare by hubert.melchert on Wed Apr 13 05:19:00 MST 2016
12 years assuming that I can use whole eeprom just for storing 5 bytes. It is our client decision if 12 years is enough, not my. At this point 16 pages were dedicated to store this data which gives only 3 years of endurance and that is certainly not enough. I will probably have to reduce frequency of writings to eeprom.
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lpcware
NXP Employee
NXP Employee
Content originally posted in LPCWare by MikeSimmonds on Wed Apr 13 04:22:45 MST 2016
When I read that chapter of the UM, I took it to say that whenever a byte or multiple bytes are programmed the entire 64 byte
page has to be erased.

Thus there are at least (100,000 * 63 pages) cycles guaranteed. [6,300,000] Over 6 million

60  * 24 = 1440.

6300000/1440 = 4375 days = just shy of 12 years (11.986)

I don't think that you have to worry.

Just move your data packet  from page to page.

Cheers, Mike.

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