74×395A family behavior – role of Output Enable (OE) On the 74395A shift-register device, what function is controlled by the Output Enable pin?

Introduction / Context:
Many 74×× shift registers include tri-state output buffers so their parallel outputs can share a bus. The Output Enable (OE) pin controls whether the device drives the bus or leaves it undriven (Hi-Z).

Given Data / Assumptions:

• Device: 74395A (representative tri-state parallel-out shift register).
• OE is a dedicated control input.
• Mode selection and loading are handled by separate pins.

Concept / Approach:
OE does not change the stored data; it gates the output drivers. When OE is asserted (device-specific polarity), the buffer connects the internal register to the output pins. When deasserted, the outputs go to high-impedance, allowing bus sharing without contention.

Step-by-Step Solution:
OE active → outputs enabled → pins reflect register contents.OE inactive → outputs disabled → pins are Hi-Z regardless of internal data.

Verification / Alternative check:
Datasheets show separate timing for shift/load and OE gating; OE changes the electrical drive state, not the data path inside the register.

Why Other Options Are Wrong:
“Forces HIGH/LOW” would be a direct output override, not a tri-state control.“Determines when data can be loaded” or “selects serial/parallel” are functions of mode or load inputs, not OE.

Common Pitfalls:
Confusing OE with asynchronous clear/load controls. OE only affects whether the device is driving the bus.

Final Answer:
It activates the three-state (tri-state) output buffer