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So my previous question here confirmed what I thought, to calculate the power dissipated in a component, take the voltage across the device and multiply by the current.

But why do some devices have a current limit and some a power limit?

For example, Low Drop Out Regulators such as this one: LDO40L, do I have to be concerned about power, or as long as I am within the voltage and current specs will I be ok? For example, could I run this at 37V at 380mA with an output voltage of 3.3V?

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  • \$\begingroup\$ That depends what you plan to have as the output voltage. You can't calculate power dissipation in the regulator with 37V in at 380mA only, so the question is unanswerable. Current limit and power limit are completely separate properties of any chip. Wires or transistors can pass only the planned amount of current without damage, and it heats based on the power dissipation and has a maximum operating temperature limit. \$\endgroup\$ – Justme Apr 11 at 16:05
  • \$\begingroup\$ The output voltage is 3.3V \$\endgroup\$ – B.Baker Apr 11 at 16:07
  • \$\begingroup\$ Voltage regulators tend to have both power and current limits; you must stay within both limits. \$\endgroup\$ – user_1818839 Apr 11 at 16:59
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How much current it can pass is one property, and how much it can heat up by the power dissipated is another property.

You can input 5V and make 380mA 3.3V, and the regulator would heat up only at 0.65W, which is certainly doable.

If you input 37V, and output output 3.3V at 380mA, the poor chip has to dissipate 12.8W, which is completely absurd amount of power wasted, heating up the chip.

The chip has a Tja thermal resistance of 42 degrees C per watt, so dissipating 12.8W would make the chip heat up by 500 degrees C from the ambient. Which of course is impossible, as it would shut down at 150 degrees C, and if it didn't, it would desolder itself long before reaching 500 degrees C.

So, no, linear regulators are completely out of question in the above case, even with a big hefty heat sink.

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  • \$\begingroup\$ You got there a minute before I posted. The wierd thing is my proposed answer was practically word for word yours, which very rarely happens. So have +1 for faster typing. \$\endgroup\$ – Neil_UK Apr 11 at 16:18
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In general, you can run this at 37V input and 380mA. This would still comply with the absolute maximum ratings where neither a maximum current nor maximum power dissipation are specified directly.

But: as you might have seen, this device has some protection features, like short circuit and overtemperature protection. Those implicitly limit both the current (>1.5A are considered a short) and power dissipation (device shuts down when temperature rises above 175°C).

So, wether or not this device will operate in a specific case all depends on thermal management. In your case, the device will dissipate around 10W at least (380mA@12V), which is far more than what can dissipate from an ordinary pcb. You'd need to have a large and thick metal core pcb, heatsink and maybe a fan to keep this at an acceptable temperature.

Beside the thermal problem, this would still dissipate 2/3 of the input power. A better solution would be to use a buck type converter and add a linear regulator/LDO after that if required.

Edit: if 380mA are only the peak current and the average is far below that, things might look a bit different. However, the challenge here is thermal management.

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  • \$\begingroup\$ Yes, it can be run at 37V input and it can pass 380mA, but for a 100 degree C temperature rise, the output voltage must roughly be about 31 volts, so not very useful if the intention is to make 3.3V \$\endgroup\$ – Justme Apr 11 at 16:35
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But why do some devices have a current limit and some a power limit?

  • The current limit will be determined by the maximum allowable current density in the bonding wires and internal components of the chip. As current density increases so will heating and there has to be a physical limit to that.
  • The power limit will also be due to thermal considerations but this time it will be for the whole chip. The semiconductor material has a maximum operating temperature.

For example, Low Drop Out Regulators such as this one: LDO40L, do I have to be concerned about power, or as long as I am within the voltage and current specs will I be ok?

Yes, you have to remain within all limits.

For example, could I run this at 37V at 380mA with an output voltage of 3.3V?

That would be about 34 V across the device so P = 34 × 0.38 = 13 W. Section 6.4 of the datasheet shows how to calculate the maximum power dissipation allowable - but it won't be anywhere near 13 W.

For 37 to 3.3 V conversion a linear regulator is the wrong part for the task. A switching regulator will be much more efficient and will avoid all the thermal problems you would have with the linear regulator. LDO regulators are designed to solve problems where the output voltage is very close to the input voltage. This is not your use case.

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Limits for absolute maximum Voltage Current and Temperature are “absolutes” and all these and others must be avoided. However for reliability, you will learn how to derate these limits or stick with suggested operation levels.

If you think about it, it’s a little ironic to have a 34V drop for a “low dropout regulator” so if that really was a requirement, a wise designer would use a SMPS regulation mode.

The power limit and Rja thermal resistance and Cu layout are major design issues for absolute maximum temperature with 50% margin at room temp.

Applying 37V x 0.38 A means supplying 12.8W of power and only 3.3/37 x100% is used by the load. So using the thermal resistance from junction to ambient and recommended Cu layout for heatsink, you can expect thermal shutdown to occur.

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