load-power-sets-average-input-current - Electronic Circuits XIV: Inductors and Switching Energy Step by Step

Ideal converter input current comes from a closed power budget. 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 power ledger starts at the output load

The buck average voltage is already accepted. The load current then sets output power and ideal average input current.

P_{\text{out}}=V_{\text{out}}I_{\text{out}}\quad;\quad I_{\text{in(avg)}}={P_{\text{out}}\over V_{\text{in}}}

More load current asks for more input average current

The voltage conversion stays fixed. Output current changes power, then power sets the ideal input average current.

\begin{array}{c|c|c|c}I_{\text{out}}&V_{\text{out}}&P_{\text{out}}&I_{\text{in(avg)}}\\1\ \text{A}&3\ \text{V}&3\ \text{W}&\tfrac{1}{4}\ \text{A}\\2\ \text{A}&3\ \text{V}&6\ \text{W}&\tfrac{1}{2}\ \text{A}\\3\ \text{A}&3\ \text{V}&9\ \text{W}&\tfrac{3}{4}\ \text{A}\\\end{array}

Load power sets the ideal average input current

Average input current is a power ledger, not a claim about the instantaneous switch waveform.

P_{\text{out}}=V_{\text{out}}I_{\text{out}}=3\ \text{V}\cdot2\ \text{A}=6\ \text{W};\quad I_{\text{in(avg)}}=\tfrac{1}{2}\ \text{A}