Holdup Load Current Cross-Scan - Electronic Circuits XIII: Power, Rectification, and Regulation Step by Step

The same stored charge budget supports less time at higher load current. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Holdup time divides stored charge budget by load current

The checked source has four farads, three volts of allowed droop, and two amps of load current. That gives six seconds of holdup.

\Delta t={C\Delta V\over I}=4\ \text{F}\cdot3\ \text{V}/2\ \text{A}=6\ \text{s}

Higher load current spends the same charge budget faster

The capacitance and allowed droop stay fixed. Load current is the denominator, so the supported time steps down.

\begin{array}{c|c|c|c}C&\Delta V&I&\Delta t\\4\ \text{F}&3\ \text{V}&1\ \text{A}&12\ \text{s}\\4\ \text{F}&3\ \text{V}&2\ \text{A}&6\ \text{s}\\4\ \text{F}&3\ \text{V}&3\ \text{A}&4\ \text{s}\\\end{array}