failed-headroom-blocks-load-claim - Electronic Circuits XIII & XIV: Power Supplies and Switching Energy Step by Step

A failed-headroom-blocks-load-claim scan separates arithmetic load power from the regulator gate that must accept it. The 5 V output and 2 A load still compute 10 W, but a 5 V input minimum fails the stated 6 V required-headroom boundary. Real regulators have dropout curves, ripple, diode drops, thermal limits, tolerances, current limits, finite precision, and measurement uncertainty.

A computed load power is not accepted as regulated when headroom fails. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Power arithmetic is still gated by headroom

The load arithmetic asks for 10 W, but the input minimum must also clear the required input.

V_{\text{in,min}}\ge V_{\text{out}}+V_h

The same load claim fails until input reaches the boundary

The load power row stays the same; only the input minimum decides whether the regulated claim may proceed.

\begin{array}{c|c|c|c}V_{\text{in,min}}&V_{\text{required}}&P_{\text{load}}&\text{gate}\\5\ \text{V}&6\ \text{V}&10\ \text{W}&\text{fail}\\6\ \text{V}&6\ \text{V}&10\ \text{W}&\text{pass}\\7\ \text{V}&6\ \text{V}&10\ \text{W}&\text{pass}\\\end{array}

Computed load power is blocked by failed headroom

The arithmetic load power still exists, but the regulated supply claim is rejected because headroom fails.

5\ \text{V}\cdot2\ \text{A}=10\ \text{W}\quad\text{while}\quad5\ \text{V}<6\ \text{V}