256K --> Number of words in each chip
32 --> Size of word in bits (data bus width)
2 --> Number of banks?

So the banks have 256K addresses of 32 bits each (4 bytes) and sums 1Mb of RAM per bank. I don't know what does it means the bank thing... maybe they have 2Mb available per chip at every time, so you'll end up having 16Mb in 8 chips.

Exactly what Zup wrote, these chips as per the datasheet should be 2MB each (2 banks of 1MB).

Ok, so it should be 16mb.

But how does 256K x 2 turn into 2 MegaBytes.

I'm missing something in translation here.

You wrote it yourself correctly initially and Zup also explained it, it's not just 256K x 2 (where are the bits of each word in this?), it's 256K words x 32bit each word x 2 banks.

Let's break it down:
256000 x 32bits = 8192000 bits
8192000 bits / 8 = 1024000 bytes
1024000 bytes x 2 banks = 2MB per chip

Just guessing here, K stands for 1000 and not Kilobytes or anything similar. Like we often write Win2K

konc wrote on 2022-09-07, 12:26:

You wrote it yourself correctly initially and Zup also explained it, it's not just 256K x 2 (where are the bits of each word in […]
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You wrote it yourself correctly initially and Zup also explained it, it's not just 256K x 2 (where are the bits of each word in this?), it's 256K words x 32bit each word x 2 banks.

Let's break it down:
256000 x 32bits = 8192000 bits
8192000 bits / 8 = 1024000 bytes
1024000 bytes x 2 banks = 2MB per chip

Just guessing here, K stands for 1000 and not Kilobytes or anything similar. Like we often write Win2K

Ah, the bits to bytes. That was the part I was missing.