Question

I was solving some old MCQs and found this question:

A RAM chip has 7 address lines, 8 data lines and 2 chip select lines. Then the number of memory locations is _____.

(A) $2^{12}$ (B) $2^{10}$ (C) $2^{19}$ (D) $2^{13}$

Assuming a word consisting of a byte, this should have

2 chip select lines, meaning total $2^2$ chips.

With 7 address lines, we can address $2^7$ memory locations in a chip.

8 data lines should be used to access only the data in the memory location, and not to specify any location.

That'll make for a total of $2^2\times2^7=2^9$ memory locations. But none of the option matches my answer. What should be the correct answer?

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Solution

Here is an actual chip! From the '70s :

Motorola MC6810

https://upload.wikimedia.org/wikipedia/commons/f/f8/Motorola_Microcomputer_Components_1978_pg08.jpg

http://www.icpdf.com/MOTOROLA_datasheet/MCM6810_pdf_874837/#view

128 bytes : 128 * 8bits = 1024 bits = $2^{10}$ "memory cells".

I suppose the author considers each bit in a RAM as a "memory location", the way a memory is addressed, the data bus width, doesn't change its capacity : A 1Mega * 8bits RAM has the same capacity as a 512k * 16bits.

On the MC6810, there are 6 chip select inputs, but they are used to ease the selection of different chips connected in parallel, they don't add to the RAM storage capacity.

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