How many 1024 x 8 RAM chips are needed to provide a memory capacity of 16KB?
a) How many lines of the address bus must be used to access 16KB bytes of memory?
b) How many lines must be used to decode chip select? c) Specify the size of the decoders
The Correct Answer and Explanation is :
Given:
- RAM chip size = 1024 × 8 (1K × 8)
- Required memory capacity = 16 KB
Step 1: Calculate the Number of Required RAM Chips
Each RAM chip has 1K (1024) locations, and each location stores 8 bits (1 byte).
Thus, each RAM chip provides 1 KB (1024 bytes) of storage.
Since 16 KB (16,384 bytes) is required:
[
\frac{16,384 \text{ bytes}}{1024 \text{ bytes per chip}} = 16 \text{ chips}
]
So, 16 chips are needed.
Step 2: Address Bus Lines Required for 16 KB Memory
Each memory address in a byte-addressable system corresponds to one unique location.
Since 16 KB = (2^{14}) addresses, we need 14 address lines to uniquely access each byte.
Thus, 14 address lines are required.
Step 3: Chip Select (CS) Lines Required
Each RAM chip has (2^{10}) = 1024 locations, requiring 10 address lines to select an address within a chip.
Since we have 16 chips, the system must select one out of 16 chips.
This requires (2^4 = 16) combinations, meaning 4 address lines are used for chip selection.
Thus, 4 lines are needed for chip select.
Step 4: Decoder Size for Chip Selection
Since we need to select one out of 16 RAM chips, a 4-to-16 decoder (4 input lines, 16 output lines) is required.
Thus, the required decoder size is 4 × 16.
Final Answers:
a) 14 address lines
b) 4 chip select lines
c) 4 × 16 decoder