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We are looking to develop a circuit that will reduce 12 V to 5 V @ 5 A using a low drop out (LDO) linear regulator.

The best device that I found is the MIC29751-5.0WWT.

The heat parameters for this device are good, but the total power dissipated is about 36 W.

We don't know which parameter we need to use when calculating the accumulated heat:

  1. θJC = 1.5 °C/W (θ junction-to-case)
  2. θJA = 56 °C/W (θ junction-to-air)

From our calculations: TJ = 2050°C when using θJA or TJ = 128°C when using θJC.

The LDO will be mounted to a metal box directly (without a thermal pad) and the box has a large mass with air blades for transferring heat to the air.

From the literature there is an option to place a high-power resistor in series between input power (12 V) and the input pin of the LDO.

The resistor lowers the voltage at the LDO input at maximum current and still have V-drop-out so the LDO will activate properly.

From my calculations the resistor needs to be 1.2 Ω, 30 W.

Our questions are:

  1. Is there any other device / circuit / module that can convert the power?
  2. What parameter do we need to use for the heat calculations?
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    \$\begingroup\$ Is there some particular need of a linear regulator? Is there a reason why you can't use a buck converter? \$\endgroup\$
    – JRE
    Jan 31, 2023 at 6:45
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    \$\begingroup\$ You should probably look for switching step-down (buck) regulators (up to 97% efficiency). The mentioned part would waste more energy as heat than it provides to your load (always below 41 % efficiency! \$\endgroup\$
    – datenheim
    Jan 31, 2023 at 6:47
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    \$\begingroup\$ I'd recommend to change the heading since you are asking about heat management of a LDO regulator. Also mentioning 12 to 5 V @ 5 A instead of high power would sharpen the focus. \$\endgroup\$
    – datenheim
    Jan 31, 2023 at 7:32
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    \$\begingroup\$ @greybeard Thanks! Not sure what I mistyped on my calculator there, 6*12? Still, using up 35 out of rated absolute max 36 W power dissipation isn’t a good idea for reliability. This begs for a buck converter. \$\endgroup\$
    – winny
    Jan 31, 2023 at 8:42
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    \$\begingroup\$ Please clarify what EMI requirements you need to meet. \$\endgroup\$
    – winny
    Jan 31, 2023 at 8:43

2 Answers 2

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At that current (and power dissipation) level you might consider a regulator (you sure don't need a low dropout regulator) that uses an external pass element.

The chip you would use is classed as a linear regulator controller.

One example would be the LT1575, which would use an N-channel source-follower MOSFET as the pass element, but you'll need to do some research to find the best part for your application.

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  • \$\begingroup\$ If output current has well defined lower limit, say, 3 A, you can even use a power resistor as a "pass element". (At that power level, choice seems even more limited than memory suggested, and price higher. Pity a 12 V incandescent bulb will have a massive "inrush" current.) \$\endgroup\$
    – greybeard
    Jan 31, 2023 at 8:53
  • \$\begingroup\$ @greybeard Yes, that's a good refinement. If a resistor that drops (say) 6V at 5A is used (1.2Ω) then the maximum power dissipation in the regulator pass element would be reduced from 35W to 10.2W (at~2.92A out) with 30W maximum in the resistor (at 5A out) if I did the arithmetic right. Sounds pretty worthwhile for a single 50W chassis-mount < $5 resistor. \$\endgroup\$ Jan 31, 2023 at 9:02
  • \$\begingroup\$ I was arguing a parallel element to lower current rather than a series one to lower voltage and dissipation in the active element - which is discussed in the question. \$\endgroup\$
    – greybeard
    Jan 31, 2023 at 9:32
  • \$\begingroup\$ @greybeard Ah, makes sense. \$\endgroup\$ Jan 31, 2023 at 10:32
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You need to use θJC because you plan to use a heatsink. You also need to do the same calculation for the heatsink - it also must dissipate its heat to the air. A fan can help a lot, even if small.

The tricky part is: they are interconnected, so the air cools the heatsink which cools your regulator. Thus the case temperature of your regulator is higher than the air temperature depending on the size and heat transport capability of the heatsink (given its datasheet).

You need to calculate with worst-case air temperature of course. The application note 9 is for other voltages, but it is a good starting point for you.

The mentioned resistor might need a heatsink too.

Don't overlook the minimum current requirement.

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