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schematic

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 ?

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This is correct, or close:

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.

Also note that if the buck regulator is capable of much higher power (say 2-3A) it may not be 85% efficient at 60mA out- maybe only 50%.

In that case, the input current would be 40mA (you'd get 30mA if you used 1.5 * Pin).

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Yes, I think your reasoning is correct.

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