Logic gate power dissipation (PD) Define the power dissipation of a logic gate and select the correct expression: PD equals the product of the dc supply voltage (Vcc) and which current?

Introduction / Context:
Power dissipation of digital ICs is a key parameter for thermal design, reliability, and power budgeting. In logic families such as TTL and CMOS, the static and dynamic currents drawn from the dc supply determine the average power consumed by the device during typical operation.

Given Data / Assumptions:

• Single-supply logic gate with dc supply voltage Vcc.
• Current drawn varies with switching activity, loading, and temperature.
• We need an average measure representative of normal operation.

Concept / Approach:
Average power for a dc-supplied device is PD = Vcc * Icc(avg). The average supply current includes both static leakage/bias currents and dynamic switching currents averaged over time. Peak current is not representative of continuous dissipation, and ac supply descriptors are not applicable to standard logic Vcc rails.

Step-by-Step Solution:

Verification / Alternative check:
Thermal calculations use PD to estimate junction temperature: Tj ≈ Ta + θJA * PD. Since heating depends on average power, the average current is the relevant quantity, not instantaneous peak values.

Why Other Options Are Wrong:

• Peak current: Overstates instantaneous draw; not representative of average dissipation.
• ac supply options: Logic gates use dc supplies; ac product is inapplicable for PD specification.

Common Pitfalls:
Confusing short transient peaks with continuous thermal load; ignoring that dynamic power in CMOS also depends on switching frequency and load capacitance (Pdynamic ≈ Cload * V^2 * f), but datasheet PD is still framed against Vcc and average current.

Final Answer:
dc supply voltage and the average supply current