Memory Requirement: How Many 128 × 8 RAM Chips for 2048 Bytes?
A “RAM organization like means the chip has addressable word locations, and each word is bits wide. Therefore, the chip’s total capacity is: Convert to bytes using bits per byte:
Memory capacity target is bytes. So the number of chips needed is:
Answer: chips of are required to provide bytes of storage.
Key ideas: word width (8 bits), depth (128 words), total chip capacity (1024 bits = 128 bytes).
RAM organization (depth × width) basics
Compute required number of 128×8 RAM chips
- 1Step 1
In , the chip has depth and word width bits.
- 2Step 2
Total bits per chip = bits.
- 3Step 3
Total bytes per chip = bytes.
- 4Step 4
Chips needed = .
- 5Step 5
You need 16 chips of to get 2048 bytes.
Common pitfall
Don’t treat as bytes—it’s bits. Always convert using bits byte.
Capacity vs. bus width
This problem only asks for total capacity (bytes). The exact arrangement (e.g., how many chips in parallel to widen to a certain data bus width) depends on the system’s required word size, but for pure capacity, you just divide total bytes.
Capacity per chip vs. total target
Each 128×8 chip stores 128 bytes; 16 chips store 2048 bytes.
From chip spec to required chip count
Read chip organization
Step 1128×8 → 128 words deep, 8 bits per word."
Compute capacity
Step 2128×8 bits = 1024 bits per chip."
Convert to bytes
Step 31024/8 = 128 bytes per chip."
Determine chip count
Step 42048/128 = 16 chips."
Quick checks and related concepts
Memory organization: 128×8
Knowledge Check
A single RAM chip labeled 128×8 can store how many bytes?