Using 64 × 8 ROM chips with an enable input, construct a 512 × 8 ROM with eight chips and a decoder.
The correct answer and explanation is :
Constructing a 512 × 8 ROM Using 64 × 8 ROM Chips and a Decoder
Solution
To construct a 512 × 8 ROM using 64 × 8 ROM chips, we need to meet the requirement of storing 512 words of 8-bit data.
Each 64 × 8 ROM chip can store 64 words (or addresses), each being 8 bits wide. To reach 512 words, we divide:
[
512 \div 64 = 8
]
Thus, eight (8) ROM chips are required.
To select one of the eight chips at a time, we need a 3-to-8 decoder since:
[
2^3 = 8
]
The three most significant address lines (A6, A7, A8) from the 9-bit address bus (since ( 2^9 = 512 )) are used as inputs to this 3-to-8 decoder. The decoder activates exactly one of the eight ROM chips at a time through its chip enable (CE) input.
Each ROM chip receives the lower six address lines (A0 – A5) as input to access one of its 64 memory locations. The output data buses of all ROM chips are connected together in a shared tri-state configuration, meaning only the selected ROM will place data on the bus.
Connections:
- Address Bus (9-bit)
- A0 – A5 go to all ROM chips.
- A6 – A8 go to the 3-to-8 decoder.
- Decoder
- Takes A6, A7, A8 as inputs.
- Outputs are connected to the chip enable (CE) of each ROM chip.
- Data Bus (8-bit)
- Outputs of all ROM chips are tied together in tri-state mode.
- Control Lines
- Common read and write control lines.
This setup ensures that at any moment, only one ROM chip is active, enabling access to the full 512 × 8 memory.
Now, I will generate an image illustrating this configuration.
Here is the digital circuit diagram illustrating the construction of a 512 × 8 ROM using eight 64 × 8 ROM chips and a 3-to-8 decoder.