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

  1. $2^{12}$
  2. $2^{10}$
  3. $2^{19}$
  4. $2^{13}$
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3 Answers

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With 7 Adress Lines - 2^7 horizontal lines possible

with 2 vertical lines - 2^2 vertical lines possible

Total intersections is 2^7 * 2^2 = 2^9(no of memory location)-----------which gives only data line

To get 8 data lines 2^9 * 2^3 = 2^12(no of memory location). Answer-A

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Here Addressable lines denote the number of a horizontal memory location which is  2^7

2 Chip select means we can select one horizontal line among 2^2

So total lines horizontal*vertical = 2^9

Now data lines are 8 hence total addressable data = 2^9 * 8 ie 2^12
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The question is wrong in the sense that data lines do not specify the number of memory locations, but just the number of bits that can be transferred once a specific address is located. That means $1M\times 8bit$ has same capacity as $512K\times 16bits$ (8Mbits or 1MByte).

So asking number of “memory locations” in this question means asking number of locations possible with 7 bit address with each location of size 8 bits; plus 4 such possible configuration (2 chip select bits $\rightarrow\ 2^2=4$).

Therefore the true answer should be $2^7\times 2^2=2^9$ number of memory addresses.

If however, the question was to give the size of the possible chip configuration, then we have to multiply $2^9$ with 8, which is $2^{12}$ bits or 4Kbits or 512 Bytes.

Answer:

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