EPROM pin count reasoning: A 256 × 4 EPROM provides 256 addresses of 4 data bits each. How many address and data pins does it have?

Introduction / Context:Memory device pin counts follow directly from the required address space and data width. This problem reinforces binary addressing calculations for ROM/EPROM devices.

Given Data / Assumptions:

• Organization: 256 × 4 (256 locations, each 4 bits wide).
• Binary addressing: number of address pins equals log2(number of addresses).
• Data width dictates the number of data pins.

Concept / Approach:Since 256 = 2^8, we need 8 address lines (A0…A7) to uniquely select a word. The data width is 4 bits, so we need 4 data pins (D0…D3) to read out a nibble from the selected location.

Step-by-Step Solution:

Verification / Alternative check:Cross-check with standard memory organizations: 1K × 8 uses 10 address pins; 256 × 4 is smaller, so 8 address pins is consistent.

Why Other Options Are Wrong:

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Final Answer: