The same input minimum can pass, touch, or fail as headroom grows. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Headroom 1 V sets acceptance

Input minimum stays 7 V and output stays 5 V. Headroom 1 V requires 6 V, leaving margin 1 V.

7 V(5 V+1 V)=1 Vpass7\ \text{V}\mathbin{-}\bigl(5\ \text{V}+1\ \text{V}\bigr)=1\ \text{V}\Rightarrow\text{pass}
Regulator headroom boundaryOnly the headroom requirement moves.Vin 7 VVout 5 VVh 1 VVreq 6 Vgate pass

Headroom 2 V sets acceptance

Input minimum stays 7 V and output stays 5 V. Headroom 2 V requires 7 V, leaving margin 0 V.

7 V(5 V+2 V)=0 Vpass7\ \text{V}\mathbin{-}\bigl(5\ \text{V}+2\ \text{V}\bigr)=0\ \text{V}\Rightarrow\text{pass}
Regulator headroom boundaryOnly the headroom requirement moves.Vin 7 VVout 5 VVh 2 VVreq 7 Vgate pass

Headroom 3 V sets acceptance

Input minimum stays 7 V and output stays 5 V. Headroom 3 V requires 8 V, leaving margin negative 1 V.

7 V(5 V+3 V)=1 Vfail7\ \text{V}\mathbin{-}\bigl(5\ \text{V}+3\ \text{V}\bigr)=-1\ \text{V}\Rightarrow\text{fail}
Regulator headroom boundaryOnly the headroom requirement moves.Vin 7 VVout 5 VVh 3 VVreq 8 Vgate fail