# Is it really the Power required for comparing input and output?

Hi I want to drive a 5W (5V*1A) output... Can it be carried by a 10W input (15V*0.670A) ? Is it just the power needed? Or should I need a larger current than 0.670A ?

Is this applicable for LDO Regulators or/and Switching Regulator?

• Not applicable for any linear regulator, LDO or otherwise. But OK for any decent switcher. – Brian Drummond Jul 12 '18 at 11:52

The power output for any voltage regulator cannot exceed the power input. That's a fundamental rule of nature.

In the real world, the power out is less than the power in. The ratio of power out to power in is the efficiency of the regulator.

A switching regulator can have a higher output current if the output voltage is less than the input voltage, or can have a higher output voltage if the output current is less than the input current. However, a linear voltage regulator must have a higher input voltage than its output voltage and a higher input current than its output current. That means that the efficiency of a linear regulator is usually much lower than a switching regulator.

Power-in to a voltage regulator equals power-out plus reglator losses.

If the regulator is linear then we can also reasonably say that: -

Current-in = current-out

If the input-voltage is 15 volts and the current is 1 amp, the input-power is 15 watts. The output-power is 5 watts and the losses in heat due to regulating the output are 10 watts.

For a switching regulator in the same scenario, the power efficiency is usually much higher and, as a first approximation you can say that power-out equals power-in hence your 15 volt supply would have to supply 333 mA.

As a 2nd (and improved) approximation, you can assume that a switching regulator is 90% efficient hence, power-in will be about 5.6 watts and, the current from the 15 volt input supply will be about 370 mA.