TTL NAND Gate (All Inputs LOW) — Transistor conduction states In a standard TTL NAND gate with a totem-pole output, if all inputs are at LOW level, which ON/OFF combination for the internal transistors (Q1 through Q4) is correct?

Introduction / Context:Understanding the internal operation of a TTL NAND gate clarifies why the output is HIGH when any input is LOW. The gate employs a multi-emitter input transistor (Q1), a phase-splitter (Q2), and a totem-pole pair (Q3 upper pull-up, Q4 lower pull-down).

Given Data / Assumptions:

• All inputs are LOW, so at least one emitter-base junction in Q1 is forward-biased to ground.
• The gate is a conventional TTL NAND with a totem-pole output stage.
• We are identifying conduction states that produce a logic HIGH output.

Concept / Approach:With any input LOW, the internal biasing causes the phase-splitter to turn OFF, which in turn drives the upper transistor ON and the lower transistor OFF in the totem pole. This results in a HIGH logic level at the output.

Step-by-Step Solution:All inputs LOW → Q1 conducts (ON) via the grounded emitter path.Q1 conduction biases Q2 (phase-splitter) OFF.With Q2 OFF → drive to Q3/Q4 results in Q3 (pull-up) ON and Q4 (pull-down) OFF.Output node is driven HIGH by Q3; Q4 is non-conducting.Thus, the correct pattern is Q1-ON, Q2-OFF, Q3-ON, Q4-OFF.

Verification / Alternative check:Logic truth: NAND with any LOW input → HIGH output. The transistor state sequence above is the canonical internal path producing that HIGH.

Why Other Options Are Wrong:

• b: Q2 cannot be ON with all inputs LOW if the output is HIGH.
• c: Both Q3 and Q4 ON would create contention; also contradicts standard operation.
• d: Q4 ON would force a LOW output, contradicting NAND truth for a LOW input.

Common Pitfalls:Confusing the roles of Q3/Q4 or assuming both can be ON simultaneously; proper TTL design avoids extended shoot-through by steering control.

Final Answer:Q1-ON, Q2-OFF, Q3-ON, Q4-OFF