# Current drawn from source by buck reg

simulate this circuit – Schematic created using CircuitLab

The circuit above is that of a buck regulator that has an efficiency of approx 85%. The input and output voltages are 10V and 3.3V resp. The output wattage is 0.2W.

So, the output current is Iout = Power/Voltage = 0.2/3.3 = 60mAmps.

Now, my doubt pertains to the current drawn from the battery or bench supply that gives/supplies the 10V.

Now, since the Pout=0.2W, Pin=0.23W (due to the 85% efficiency value). So, 0.03W is lost or dissipated in heat.

Now, Pin = 0.23W. So, I = 0.23/10 = 0.023Amps = 23mAmps. Is my deduction and understanding correct ?

Efficiency is $\eta = \frac{P_{OUT}}{P_{IN}}$ so $P_{IN} = \frac{P_{OUT}}{\eta}$ = 0.235 for $\eta = 0.85$
The distinction I'm making is that $P_{IN}$ is not 1.15 $\cdot P_{OUT}$- it makes a much bigger difference if the efficiency is << 1.