Output load current is converted into an input-average current budget check. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

1 A load row

Buck output stays 3 V. Load current 1 A asks for input average 1/4 A against a 1/2 A budget, so margin is 1/4 A.

Iin=14 A,m=14 ApassI_{\text{in}}=\tfrac{1}{4}\ \text{A},\quad m=\tfrac{1}{4}\ \text{A}\Rightarrow\text{pass}
Input-average-current budgetThe current budget is downstream of output power.L 3 HV 9 Vdt 1 sIi 0 AdI 3 AIf 3 AL 3 HV -3 Vdt 3 sIi 3 AdI -3 AIf 0 Asum Vdt 0 V*sgate passVin 12 VD 1/4Vout 3 VVin 12 VVout 3 VIout 1 APout 3 WIin(avg) 1/4 A

2 A load row

Buck output stays 3 V. Load current 2 A asks for input average 1/2 A against a 1/2 A budget, so margin is 0 A.

Iin=12 A,m=0 ApassI_{\text{in}}=\tfrac{1}{2}\ \text{A},\quad m=0\ \text{A}\Rightarrow\text{pass}
Input-average-current budgetThe current budget is downstream of output power.L 3 HV 9 Vdt 1 sIi 0 AdI 3 AIf 3 AL 3 HV -3 Vdt 3 sIi 3 AdI -3 AIf 0 Asum Vdt 0 V*sgate passVin 12 VD 1/4Vout 3 VVin 12 VVout 3 VIout 2 APout 6 WIin(avg) 1/2 A

3 A load row

Buck output stays 3 V. Load current 3 A asks for input average 3/4 A against a 1/2 A budget, so margin is negative 1/4 A.

Iin=34 A,m=14 AfailI_{\text{in}}=\tfrac{3}{4}\ \text{A},\quad m=\tfrac{-1}{4}\ \text{A}\Rightarrow\text{fail}
Input-average-current budgetThe current budget is downstream of output power.L 3 HV 9 Vdt 1 sIi 0 AdI 3 AIf 3 AL 3 HV -3 Vdt 3 sIi 3 AdI -3 AIf 0 Asum Vdt 0 V*sgate passVin 12 VD 1/4Vout 3 VVin 12 VVout 3 VIout 3 APout 9 WIin(avg) 3/4 A