minimum-valley-must-feed-regulator - Electronic Circuits XIII: Power, Rectification, and Regulation Step by Step

A regulator claim needs the reservoir valley to clear the required input. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

The valley is peak minus ripple

The checked peak is 10 V and the checked ripple is 3 V, so the valley is fixed before any minimum is tested.

V_{\text{valley}}=V_{\text{peak}}-\Delta V

The same valley can pass or fail different minimums

The inclusive boundary row passes exactly at the valley; a higher required minimum fails.

\begin{array}{c|c|c|c|c}V_{\text{peak}}&\Delta V&V_{\text{valley}}&V_{\min}&\text{gate}\\10\ \text{V}&3\ \text{V}&7\ \text{V}&6\ \text{V}&\text{pass}\\10\ \text{V}&3\ \text{V}&7\ \text{V}&7\ \text{V}&\text{pass}\\10\ \text{V}&3\ \text{V}&7\ \text{V}&8\ \text{V}&\text{fail}\\\end{array}

The valley must stay above the regulator minimum

The same valley can pass one declared minimum and fail a higher one. The gate is inclusive and exact.

10\ \text{V}\mathbin{-}3\ \text{V}=7\ \text{V}\ge6\ \text{V}\quad\text{but}\quad7\ \text{V}<8\ \text{V}