Module 6: Power Delivery & Grounding
LDO vs switcher, decoupling/PDN, sequencing/inrush, grounding and EMI.
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1conceptAn LDO and a buck switcher both step 12 V down to 3.3 V at 1 A. What is the core trade-off?
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The LDO is the less efficient one: it is the device that burns the voltage drop as heat, not the switcher.
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Correct. A linear regulator burns (12 − 3.3)·1 ≈ 8.7 W as heat with a clean output, while the switcher converts efficiently but its switching action puts ripple/EMI on the rail.
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Their efficiencies differ sharply here: the LDO is ~27% efficient at this drop while the switcher can exceed 90%.
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The switcher does not burn the full (Vin − Vout)·I; it transfers energy via the inductor, which is why it stays efficient.
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2conceptWhere should the local high-frequency ceramic decoupling capacitors be placed, and why?
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Plane inductance between a distant cap and the pin defeats high-frequency decoupling; placement matters.
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Correct. Local ceramics at the pins minimize the loop area and ESL so they can source fast transient current, while a bulk cap handles slower, larger demand. Together they hold the PDN below its target impedance.
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Decoupling only at the regulator leaves the IC's fast transients unsupplied across the board inductance, causing rail collapse.
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Decoupling caps must be close to be effective; distance adds inductance and makes them useless at high frequency.
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3conceptWhy route a signal's return current on the reference plane directly under its trace, in the smallest possible loop?
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Tidiness is not the reason; this is a physics constraint on emissions.
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Correct. A current loop radiates in proportion to its area and the edge rate (di/dt); keeping the return directly under the signal shrinks the loop area, cutting both radiated EMI and noise pickup.
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Loop routing is about magnetic loop area and inductance, not the trace's DC resistance.
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Smaller loop area lowers inductance, and lower inductance is the goal, not higher.
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4calcA 3.3 V rail draws 5 A through a sense path, and you must hold the IR drop in the ground return under 10 mV. What return resistance does that require?
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Correct. R = V / I = 10 mV / 5 A = 2 mΩ, so the return path must be at or below 2 mΩ to keep the drop within budget: a Kelvin-sensed, low-impedance ground.
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20 mΩ at 5 A gives 100 mV of drop, ten times the budget.
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50 mΩ at 5 A gives 250 mV, far over the 10 mV limit.
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200 mΩ at 5 A gives 1 V of drop, wildly over budget.
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5scenarioA board with a large bulk input cap trips the bench supply's current limit at the instant of power-on, before any load runs. Best mitigation?
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Just raising the limit lets a large uncontrolled surge through every time, stressing contacts, the connector, and the cap; it masks the problem rather than fixing it.
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Correct. The surge is inrush charging the bulk cap; an NTC thermistor (or an active soft-start/inrush limiter) ramps the charge current so the cap fills gradually instead of slamming the rail.
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Removing the bulk cap kills the very decoupling you need for transient demand; the fix is to limit inrush, not delete the cap.
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More capacitance increases the charge the rail must deliver at power-on, making the inrush surge worse, not better.
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