If I use MP1584 step down IC and short circuit appears how can I calculate the efficiency or the total power dissipation for the IC?


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.

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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.

  • \$\begingroup\$ Thanks. If the input voltage 24V and short circuit apperars on the output how much will the power dissipation if the inductor is 10uH and saturation current is 6A, diode is schottky diode 40V 5A (SK54), Vf: 0.5V (at 5A) \$\endgroup\$ – Bruk Aug 28 '16 at 8:27
  • \$\begingroup\$ You measure it, or you simulate using very detailed models. \$\endgroup\$ – Neil_UK Aug 28 '16 at 8:40

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