Difficulty: Medium
Correct Answer: the high power dissipation of the resistor
Explanation:
Introduction / Context:Pull-down resistors bias an input to logic LOW when undriven. While useful, they can increase steady-state current draw and power loss, especially when the node is driven HIGH for significant time.
Given Data / Assumptions:
Concept / Approach:When the node is driven HIGH, current flows from VCC through the driver into the resistor to ground, dissipating power P = V^2 / R approximately (neglecting driver drop). Choosing too small a resistor leads to unnecessary power and thermal stress.
Step-by-Step Solution:
1) Consider the worst case: node HIGH continuously with pull-down to ground.2) Compute power in resistor: P ≈ (VCC^2) / R.3) Select R large enough to limit current yet small enough to provide noise immunity when undriven.4) Evaluate logic-family input currents to set the allowable R range.Verification / Alternative check:Thermal check of resistor and supply current measurements validate the choice.
Why Other Options Are Wrong:
Common Pitfalls:Copying pull-up values without considering asymmetry, using very small resistances, and ignoring battery life impacts.
Final Answer:the high power dissipation of the resistor
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