Well, let's see what a quick search and a bit of math can do for us...shall we?
10,000 students @ 11 bits each = 88,000 bits.
Dealing with 11 bits isn't worth the pain, so call it 16 bits, in other words, 2 bytes.
10,000 students @ 2 bytes each = 20,000 bytes.
Hmmmm....20,000 bytes....roughly equates to 20KB....Hmmmm....
AHA!!!!
A standard, run of the mill 24LC256, 32K x 8 serial eeprom can easily hold 10,000 students with plenty of room to spare.
Hmm...let's download the datasheet and see what it says...
Datasheet says that it takes about 5ms to write a byte or a page (which according to said datasheet is 64 bytes).
So, I'll just assume we're going to write one byte at a time...
A completely full 24LC256 has 32,768 bytes.
32,768 bytes x 5ms each = 163.84 seconds for a complete write of a 24LC256.
A page write would be about 64 times as fast (because a page is 64 bytes and a byte is one byte).
32,768 / 16 bytes/page x 5ms = 2.56 seconds plus the time it takes to write each byte in a page, maybe another few seconds.
Well now, that wasn't so hard...was it? Didn't even need a calculator.
Notes:
-Reads are much faster than writes.
-An FRAM (www.ramtron.com) would be a much better option here. Write times are much faster, and they don't wear out.
-A good search engine can work wonders
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