Question

How many address bits are needed to select all memory locations in the 2118 16k 1 ram?

Answer 1

Explanation:

If your RAM is organized as 32 bit words, then you need 4096+16*2^4 or 4352 addresses, if for 64 bit words, 2048+16*2^4 or 2304 addresses. In a system each byte is addressed individually. There are 1024 bytes in one k byte. A 16k ram would therefore need 1024*16 = 16384 addresses.

Answer 2

Address bits are needed to select all memory locations in the 2118 16k 1 ram:

Explanation:

• The size of the memory = N*M

• N = address lines, M = word length, no of registers/memory location required = 2^N
• memory capacity = 16k ⇒2^N=16K

• 1K=1024 memory locations
• 16k=16*1024=16384
• now 2^N=16384
• After factorizing 16384 by 2 we get N as 14.
• 14 is the required address line.