You are asking two questions: efficiency of regulator into a short circuit and power loss in regulator in short-circuit condition.

The first is easy:

Efficiency is give by formula \$ \eta = \frac {P_{OUT}}{P_{IN}} \$ and power can be calculated from \$ P = VI \$.

For the short-circuit case \$ V_{OUT} = 0 \$ so we have

$$ \eta = \frac {P_{OUT}}{P_{IN}} = \frac {V_{OUT}I_{OUT}}{V_{IN}I_{IN}} = \frac {0 I_{OUT}}{V_{IN}I_{IN}} = 0 $$

The chip power dissipation is not so easy as it may depend on inductor saturation, etc. In any case the regulator will shut down at 150°C and turn back on at 135°C.

You are asking two questions: efficiency of regulator into a short circuit and power loss in regulator in short-circuit condition.

The first is easy:

Efficiency is give by formula \$ \eta = \frac {P_{OUT}}{P_{IN}} \$ and power can be calculated from \$ P = VI \$.

For the short-circuit case \$ V_{OUT} = 0 \$ so we have

$$ \eta = \frac {P_{OUT}}{P_{IN}} = \frac {V_{OUT}I_{OUT}}{V_{IN}I_{IN}} = \frac {0 I_{OUT}}{V_{IN}I_{IN}} = 0 $$

The chip power dissipation is not so easy to calculate as it may depend on inductor saturation, etc., but we don't have to.

Figure 1. Maximum continuous power dissipation is given in the datasheet.

In any case the regulator will shut down at 150°C and turn back on at 135°C.

You are asking two questions: efficiency of regulator into a short circuit and power loss in regulator in short-circuit condition.

The first is easy:

Efficiency is give by formula \$ \eta = \frac {P_{OUT}}{P_{IN}} \$ and power can be calculated from \$ P = VI \$.

For the short-circuit case \$ V_{OUT} = 0 \$ so we have

$$ \eta = \frac {P_{OUT}}{P_{IN}} = \frac {V_{OUT}I_{OUT}}{V_{IN}I_{IN}} = \frac {0 I_{OUT}}{V_{IN}I_{IN}} = 0 $$

You are asking two questions: efficiency of regulator into a short circuit and power loss in regulator in short-circuit condition.

The first is easy:

Efficiency is give by formula \$ \eta = \frac {P_{OUT}}{P_{IN}} \$ and power can be calculated from \$ P = VI \$.

For the short-circuit case \$ V_{OUT} = 0 \$ so we have

$$ \eta = \frac {P_{OUT}}{P_{IN}} = \frac {V_{OUT}I_{OUT}}{V_{IN}I_{IN}} = \frac {0 I_{OUT}}{V_{IN}I_{IN}} = 0 $$

The chip power dissipation is not so easy as it may depend on inductor saturation, etc. In any case the regulator will shut down at 150°C and turn back on at 135°C.