Difficulty: Medium
Correct Answer: False
Explanation:
Introduction / Context:Decoupling (bypass) capacitors are critical in digital electronics to provide a low-impedance local energy reservoir and to control high-frequency noise near an integrated circuit. This question checks correct placement of decoupling capacitors relative to each device's power and ground pins.
Given Data / Assumptions:
Concept / Approach:For effective high-frequency decoupling, the loop area formed by the capacitor, the IC's VCC pin, and the IC's ground return must be minimized. The capacitor must be placed physically close to the same device's pins so that current flows in a very short loop, reducing inductive impedance and voltage droop (delta V = L * di/dt).
Step-by-Step Solution:
1) Each IC needs its own local bypass path to ground to handle its switching current spikes.2) Connecting a capacitor from one device's VCC to a different device's ground increases the loop length and inserts shared return paths, which raises inductance and noise coupling.3) The correct practice is to connect the capacitor directly between the same IC's VCC and local ground pins, with the shortest traces possible.4) On multilayer boards, vias to solid planes are fine, but the local loop should remain tight around the IC.Verification / Alternative check:Examine supply ringing on an oscilloscope: with proper local decoupling, high-frequency droop and ground bounce reduce markedly compared to cross-device connections.
Why Other Options Are Wrong:
Common Pitfalls:Placing capacitors far from pins, sharing one capacitor across several ICs, and routing long serpentine traces to the capacitor reduce effectiveness.
Final Answer:False — the decoupling capacitor should be local to the same IC, between its VCC and its ground.
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