Perhaps You could help me confirm my thermal discussion on choosing appropriate LDO?

MIC5225, a 16Vin and 150mA output current quaranteed LDO. So, parameter-wise I would guess that I could use this SOT23-5 package and get it to run from f.i 14Vin to 3.3Vout and 100mA output.

I checked some relevant Q/As about the 'LDO power dissipation' topics:

(can't provide links, since not enough rep)

  • "LDO selection in Thermal point of view"
  • "SOT-223 Thermal Pad and Vias" (many references also)

According to a potentiometer model [http://www.ti.com/lit/an/slva118a/slva118a.pdf], the power dissipated is equivalent to the input and output voltage difference and output current. So

(14Vin - 3.3Vout) * 100mAout = 1.1W. (getting suspicious)

Based on my readings, SOT23-5 (with 235C/W of thermal ambient resistance) can't handle it with such input voltage.

(125C - 25C)/235[C/W] = 0.426 W (So when ambient is 25C, the package could dissipate 426mW) no-airflow

I conclude that it's false to expect the regulator to work up to 16Vin and 3V3, 150mA, even though it's advertised as such (Couldn't find any other package for this either). Why false advertise xD?

I then chose LD1117S33TR with SOT-223 which should manage 1.1W with some heat sinking. Space is also quite scarce. Package thermal resistance is ~110C/W. So perhaps package-only could dissipate:

(125C-25C)/110[C/W] = 0.91W

and with some heatsinking on PCB and airflow it should be fine? I wonder if this train-of-thought is correct?

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    \$\begingroup\$ It is not false advertisement. It can work with 16Vin, it can work with 150mA, just not both at the same time in realistic conditions. I agree that info on datasheets are sometimes unrealistic or a bit inflated, but in this case your wording might be too aggressive. \$\endgroup\$ – Wesley Lee Nov 29 '16 at 15:30
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    \$\begingroup\$ It's not 'false advertising'. They're telling you what each of the individual limiting factors are. They're not telling you that you can combine all of them together at once. If you want 100mA out, then (using your calculations) you can't supply more than 7.5V (0.426/0.1 + 3.3). \$\endgroup\$ – brhans Nov 29 '16 at 15:30
  • \$\begingroup\$ as also for the other comments on 'false adv': Perhaps yes, the wording is a bit strong, but I now understand the 'fine print' on that subject. Thanks! All you gave good input, I guess I chose 'answered' post based on the usefulness to others as well. Thanks! \$\endgroup\$ – crypton Nov 29 '16 at 16:06

Your reasoning is fine, and the calculations are correct. You're perfectly right. And 1.1W can be dissipated by a SOT-223 package, provided that you have sufficient copper area under the package. There are some resources on the internet giving hints on the minimum copper area you need for a given power with this package:

The only thing that needs clarification is that it's not false advertising. MIC5225 can handle 16V input. It can also handle 150mA. The thing is: you can't have both at the same time with this package, indeed. Either you use it with high input voltage but low current (some applications just need a few tens of mA), or, if you need higher current, you need to have lower input voltage.

Note that you can probably have both conditions true for a very short time, however.

A trick that you can use with linear regulators in such a situation is to use a low value, high-power resistor in series with the input. This resistor will dissipate most of the heat when the current raises. You just have to calculate its value so that the voltage drop across this resistor, when the current is at its maximum value, is lower than the difference between your input voltage and the minimum required voltage at the regulator input (use ohm's law). The only drawback would be slightly worse load regulation.

  • \$\begingroup\$ as also for the other comments on 'false adv': Perhaps yes, the wording is a bit strong, but I now understand the 'fine print' on that subject. Thanks! \$\endgroup\$ – crypton Nov 29 '16 at 15:59

That calculation *(14Vin - 3.3Vout) * 100mAout = 1.1W.* is correct, nothing suspisious about it.

That 1.1 W does immediately tells me that you have to do something to get rid of the dissipated heat. For a small LDO in a SOT package, 1.1 W is a bit much. It is OK if you provide adequate cooling (large copper area and/or heatsink, sufficient airflow) but it is pushing the edge.

If you insist on LDOs you could use 2 LDOs in series, on for 14 V to 9 V and a second one for 9V to 3.3 V. That spreads the 1.1 W between 2 devices but the total does remain 1.1 W.

A less wasteful and cooler solution is to go for a switched regulator (a buck converter). Do note that these are a little bit trickier to use than most LDOs, strictly follow the guidelines in the datasheet.

An alternative could be to use a ready-made buck converter module.

  • \$\begingroup\$ Yup, considered a buck as well. Currently the decision for LDO was to minimize component number, since there's not much space ... cost also. Most small form bucks are quite expensive (2-4$ per chip). Seeing that the LDO heat dissipation doesn't really gain me much space either I have to think about it a bit more. Thanks for the informative answer! \$\endgroup\$ – crypton Nov 29 '16 at 16:10
  • \$\begingroup\$ In the case of a space restriction I would definitely go for a switched solution. There are cheap solutions though like this module ebay.nl/itm/… notice how the inductor is the largest component. \$\endgroup\$ – Bimpelrekkie Nov 29 '16 at 16:16

It's quite common to have a part with multiple limits. You have to respect them all.

In the case of a voltage regulator, you have a voltage limit, a current limit, and a temperature limit. All experienced designers know this, so know what to dig into the data sheet for.

The voltage limit and the current limit are simple one-liners, with few if any ifs or buts, and can go right at the top of the data sheet.

The temperature limit (which is always in the order of 150C junction) unfortunately depends on your ambient, and your heatsinking, and your current, and your voltage drop, which takes some work, and cannot be put in a single line. It's not practical to summarise it either. The best summary is what package it's in. A TO-220 will shift more power than a SO-8 before cooking, but only if it's on a heatsink. A 'power pad' SO-8 with a heatsink will outperform a TO-220 without.

Now if you want to point the finger of false advertising, try finding out the capacitance versus voltage bias and temperature for high value high K ceramic caps, there's fun!


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