I am using a 5V regulator (LP2950) from ON Semiconductor. I am using this for USB power and I'm feeding in 9V from an adapter. USB requires maximum of 500mA right? So the maximum power dissipation in the regulator is (9V - 5V) * 500mA = 2W. My Question is whether this regulator can take that much power with 10x10mm copper as heatsink on the PCB?

Also I was looking at another regulator (LM2940S-5.0/NOPB) from TI. Which one would you thing would be a better regulator to use with the same 10x10mm copper area as heatsink. Remember I am using this to power up a USB port (not on the computer side because that provides 5V, 500mA)

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    \$\begingroup\$ Regardless of heatsink, the LP2950 is only rated for 100ma. It will not be able to handle 500ma! \$\endgroup\$ – DoxyLover Aug 27 '13 at 22:01
  • \$\begingroup\$ awesome. i know that too but my mate here did not believe me and said with bigger heatsink you can do anything (yeah right).. thanks \$\endgroup\$ – David Norman Aug 27 '13 at 22:05
  • \$\begingroup\$ @DavidNorman You might want to pull your friend's leg a bit about liquid nitrogen cooling of the linear regulator. \$\endgroup\$ – Anindo Ghosh Aug 27 '13 at 22:17
  • \$\begingroup\$ @AnindoGhosh, lol i wish. I'm new to this job and he wouldn't like it if I pull his leg. surprisingly he has been doing this for past 10 years \$\endgroup\$ – David Norman Aug 27 '13 at 22:25
  • \$\begingroup\$ If the adapter is putting out half-wave rectified, unregulated DC, then the 9V nominal value is misleading and your calculation is wrong. The actual input voltage to the regulator will be whatever the reservoir capacitor charges to (or the average of its peak to peak swing). \$\endgroup\$ – Kaz Aug 27 '13 at 23:02

few points:

the datasheet says part is internally current limited and thermally current limited, but no information is given regarding the limits, generally they will give Thermal trip point and Internal current limit at particular condition.

the maximum current given by LDO will depend upon the series elements current handling capacity also. basically we need to do thermal analysis.

Worst case Power dissipated across = VinminIoutmax-(VoutmaxIoutmax)-PQ(Quiscent Power). So, here if Vinmin to Voutmax difference reaching to min dropout voltage, we can have maximum current which will not exceed the Tj of the device package.

from datasheet, in section Electrical characterstics, Ilimit(Vout shrted to GND) is 300mA.

from Ruff calc, by neglecting the quiscent current (~4mA from graph in datasheet ==> 4*9 = 36mW) considering DPAK with thejaJ-A as 92 enter image description here

Vinmin =9V, Voutmax = 5.5V(USB), TA =25 (assumed), Tj=150(from datasheet) maximum current can give = (150-25)/(92)*(9-5.5) = 0.388A. AS we don't much info regarding current limit, and thermal shutdown point, also block diagram doesn't given any circuitry regarding current Limit, thermal shutdown, You can confirm these things from vendor.

There are very good application notes from different vendors regarding thermal analysis. http://www.ti.com/lit/an/slva118/slva118.pdf

I have seen sometimes people will take more current from LDO than rated, unless it will not exceed the thermal trip point, but most of the times it mostly transient current.


there is no chance of getting 500 mA from this regulator for your input(9V), Output(5V) combination.


The quick answer is: You're probably over the temp/power limit for that part, on that PCB.

The not-so-quick-but-more-accurate answer is:

Using the numbers from the datasheet, combined with some thermal conductivity numbers for your PCB, factoring in the ambient air temp and ventilation you should be able to calculate the junction temp of the regulator (and this, if you are exceeding the max temp).

That sounds easy, but it is not for most engineers. The problem is that you don't really know your airflow, or the thermal conductivity of the copper on your PCB. You could make some guesses and get somewhat close, but in the end you'd have to just try it and see.

Or, you can use experience and a rough guide. This isn't super accurate either, but it gets you in the right ballpark. My experience says that 1 cm^2 is not enough to dissipate 2 watts. I would go with at least 4 cm^2 and 6 cm^2 would be much better. But even if you get 6 cm^2, that doesn't mean you're safe. If you are running with high ambient temperatures or don't have much airflow, or there are other hot components nearby then you might still have issues.

Most professional EE's in this situation would do as much math as possible, make an educated guess, then make the copper as large as possible. Then, after the prototypes are built, verify the design by running it at max current, at max ambient temperature. Only then would an EE be satisfied that all the requirements have been met.

If this were my design, I would use a switching regulator instead of a linear regulator. It will be smaller, and waste less power. But it is more expensive and more complicated.

  • \$\begingroup\$ I got another question, this regulator (LM2940S-5.0/NOPB) is rated at 1A so if you dissipate say 900mA with minimum dropout, that would generate so much heat that means your PCB will be 'huge' as you would require about 10cm^2 of copper to dissipate that much heat right? \$\endgroup\$ – David Norman Aug 27 '13 at 22:23
  • \$\begingroup\$ @DavidNorman It depends on your definition of "minimum dropout". \$\endgroup\$ – user3624 Aug 27 '13 at 22:40
  • \$\begingroup\$ by minimum dropout I mean 7V for this regulator \$\endgroup\$ – David Norman Aug 28 '13 at 0:02
  • \$\begingroup\$ @DavidNorman I think your idea of "minimum dropout" is different than mine. When I think of dropout voltage, I think of the spec at the bottom of page 5 of this datasheet: ti.com/lit/ds/symlink/lm2940-n.pdf As you see, it is nowhere near 7v. A Linear Reg with 7v minimum dropout would be laughed out of the market. \$\endgroup\$ – user3624 Aug 28 '13 at 4:16
  • \$\begingroup\$ I would assume the poster means that 7V is the minimum nominal input to guarantee a 5V nominal output, so the drop-out is 2V. That's typical of "old school" and high-current regulators. 2V * 500 mA is 1.0W. Rather than using a surface mount device, this is what TO-220 is for -- it can do 1W in plain air. It's easy to add heat sinks, too! However, if you have the budget, use a Murata OKI-78SR. It's a 1.5A switching regulator that's a drop-in replacement for the 7805, with > 90% efficiency! Under five bucks in singles, too! \$\endgroup\$ – Jon Watte Aug 28 '13 at 5:16

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