A simple regulator claim starts with an exact headroom gate. 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 required input is output plus headroom

The output request is 5 V. The regulator cannot accept the claim until the input valley clears output plus headroom.

Vrequired=Vout+VheadroomV_{\text{required}}=V_{\text{out}}+V_{\text{headroom}}
Regulator headroom setupThe input valley is compared to the required input.Vin 7 VVout 5 VVh 1 VVreq 6 Vgate pass

More headroom can turn the same input into a failure

The input valley stays fixed. The third row asks for too much headroom and is rejected.

VoutVhVrequiredVin,mingate5 V1 V6 V7 Vpass5 V2 V7 V7 Vpass5 V3 V8 V7 Vfail\begin{array}{c|c|c|c|c}V_{\text{out}}&V_h&V_{\text{required}}&V_{\text{in,min}}&\text{gate}\\5\ \text{V}&1\ \text{V}&6\ \text{V}&7\ \text{V}&\text{pass}\\5\ \text{V}&2\ \text{V}&7\ \text{V}&7\ \text{V}&\text{pass}\\5\ \text{V}&3\ \text{V}&8\ \text{V}&7\ \text{V}&\text{fail}\\\end{array}

The regulator needs output plus headroom

The ideal regulator output is accepted only because the input valley clears the exact required input.

5 V+1 V=6 V7 V5\ \text{V}+1\ \text{V}=6\ \text{V}\le7\ \text{V}
Regulator headroom gateThe boundary pass is checked from exact voltages.Vin 7 VVout 5 VVh 1 VVreq 6 Vgate pass