Question

The decoder used in 32 x 8 ROM is

• A. 8\(\times\)32 decoder
• B. 5\(\times\)32 decoder
• C. 8\(\times\)8 decoder
• D. 5\(\times\)8 decoder

Hint:

A ROM with \(M\) memory locations requires \(\log_2 M\) address lines.
The decoder used takes these \(\log_2 M\) address lines as input and has \(M\) output lines, each selecting one memory location.
So, for \(M\) locations, an \((\log_2 M) \times M\) decoder is used.

Answer 1

The Correct Option is B

A ROM (Read-Only Memory) of size \(M \times N\) has \(M\) memory locations (words), and each location stores an N-bit word. In this case, the ROM is \(32 \times 8\).
Number of memory locations (words) = 32.
Number of bits per word = 8. To select one out of 32 memory locations, we need a certain number of address lines. Let this be 'k'. The relationship is \(2^k = \text{Number of locations}\). So, \(2^k = 32\). Since \(32 = 2^5\), we have \(k=5\). This means there are 5 address lines. A decoder is used to select one specific memory location based on the address input. The decoder will take the 'k' address lines as input and will have \(2^k\) output lines, where each output line enables one memory location. So, for 5 address lines and 32 locations, a \(5 \times 2^5\) decoder is needed, which is a 5-to-32 line decoder. The options are written as \(A \times B\) decoder. This usually means A inputs and B outputs. So, it's a 5-input, 32-output decoder, or \(5 \times 32\) decoder. This matches option (b). \[ \boxed{5 \times 32 \text{ decoder}} \]