Mask ROM programming concept Which statement correctly describes how a bit is stored in a mask ROM cell?

Introduction / Context:Mask ROMs are programmed during fabrication by including or omitting device connections. The readout circuitry interprets the presence of a programmed transistor path as a logical 1 or 0 depending on the architecture. A common MOS mask ROM stores a 1 by providing a transistor gate connection to the word line (address line).

Given Data / Assumptions:

• We are considering a typical MOS mask ROM bit-cell.
• Readout uses a pull-up or pull-down network that senses whether a device is present and enabled by the addressed word line.
• Conventions can vary, but a standard depiction maps the present transistor (gate tied to word line) to a logical 1 at the sense node.

Concept / Approach:

In fabrication, a “mask option” either connects a transistor’s gate to the word line or leaves it unconnected. If connected, when the word line is asserted, the transistor conducts, indicating a stored 1 (with suitable sense circuitry). If left unconnected, the path is absent and the bit reads as 0.

Step-by-Step Solution:

Verification / Alternative check:

Standard cell diagrams in introductory VLSI and memory texts show presence/absence of the gate connection controlling the read path; this is the essence of mask programming.

Why Other Options Are Wrong:

• Options A and B refer to “bipolar base connections”—not typical for MOS mask ROM explanation.
• Option D reverses the usual MOS convention in many pedagogical treatments.

Common Pitfalls:

• Assuming all mask ROMs use identical logic polarity; vendors may swap conventions, but the question targets the common MOS description.

Final Answer:

A 1 is stored by connecting the gate of a MOS cell to the address line.