Open-Collector Gates — Purpose of the pull-up resistor Why is an external pull-up resistor required when using an open-collector logic gate output, and what does it provide to the node?

Introduction / Context:
Open-collector outputs act like a switch to ground but cannot drive a logic HIGH level on their own. An external pull-up resistor is therefore mandatory to create a defined HIGH state when the transistor is OFF.

Given Data / Assumptions:

• OC outputs can only sink current (pull LOW) when ON.
• When OFF, the output floats unless externally biased.
• A resistor to a positive rail is used to generate the HIGH logic level.

Concept / Approach:
The pull-up resistor sources current into the node, raising it to the desired HIGH voltage when the OC transistor is OFF. When the transistor turns ON, it sinks current through the resistor to ground, pulling the node LOW.

Step-by-Step Solution:
OFF state → no internal source → external pull-up sets node HIGH.ON state → transistor conducts → node pulled LOW by sinking current.Therefore, the resistor provides the HIGH voltage level.

Verification / Alternative check:
Design equations select Rpull-up to balance rise time and sink current limits: I_sink = (VCC − VOL) / Rpull-up and tr ≈ Rpull-up * Cload.

Why Other Options Are Wrong:

• a, b: VCC and GND pins are supplied directly; the resistor does not power the IC.
• d: The LOW level is provided by the transistor’s sinking action, not the resistor.

Common Pitfalls:
Using an excessively weak pull-up (slow edges) or too strong (over-current when LOW). Also ensure voltage compatibility if level-shifting.

Final Answer:
to provide the HIGH voltage