I was looking at TI web bench to find a regulator, my question is my calculation is correct and if it is why we have different value on webbench. The IC is TPS565208

P = (Vout - Vin)*Iout => (12-5)*4 = 28

For Switching Regulator P = (Vout - Vin)*Iout *(1- %Efficiency) => 28(1-%90) = 2.8 W

Thermal Junction = 95.9C/W X 2.8W = 268.52C

enter image description here

But let take look at Webbench which it said 80.42C

Link to webbench (I dont know if it works) enter image description here


No, that's not right. For a switching regulator, for a rough cut number, only the power in multiplied by the efficiency should be used.

\$P_{dissipated} = V_{in}*I_{in}*{efficiency} \$

I think the confusion above is because your using the equation for a regular non-switching regulator. The other problem is the efficiency is for the TPS565208 package and the inductor, so you'll need to find the power dissipated in both.

There should be a number telling you the power dissipation in the part, or I'd just believe the 80.4C given to you by webbench (they don't define the ambient temperature, which is important to know the junction temperature). The problem is TI doesn't define all of their numbers very well so it can be confusing to know how they came up with the numbers.

TI is cheap, and you get what you pay for. The problem that I have with this datasheet is the max power for the package is not specified. This would tell you if the numbers were close.

If we back calculate the junction temperature

80.4C/(60C/W) we get 1.34W for the part from the webbench numbers. It makes me wonder what conditions they used for testing or ambient. The problem is not only do you need to get the heat out of the part, you need to get it out of the board and into air.


Apparently TI thinks that this package can handle 7W, if you take the max operating temp, 125C and divide it by the Junction-to-board characterization parameter (16.4C/W), you get 7W for the part. This seems like way too much power, maybe this part always runs very hot when you run current through it. Be sure your PCB thermal design is very good!

To really verify this, I'd use TI's spice package or simulate it in a spice package of your choice.

  • \$\begingroup\$ Hang on, I need time to answer \$\endgroup\$ – Voltage Spike Aug 2 at 19:16
  • \$\begingroup\$ What is your voltage and current going into the regulator? \$\endgroup\$ – Voltage Spike Aug 2 at 19:17
  • \$\begingroup\$ I have 12v 9Amp ated power supply going into to the regulator which I am looking for 5v 3amp \$\endgroup\$ – Shahreza Aug 2 at 19:19

Your loss estimate \$(V_{in}-V_{out})*I_{out}\$ is for a linear regulator, not for a switcher, and even then it doesn’t account for \$I_{q}\$, the regulator internal current to power its own stuff.

The correct gross loss calculation is \$I_{in} V_{in} - I_{out} V_{out}\$. This is true for either a linear or a switcher, as it will capture \$I_{q}\$ in the gross figure.

Working from there, the losses break down between the chip itself and components external to the chip. A good portion of the loss is IR drop in the inductor and ESR loss in the filter capacitor.

Conveniently, WebBench has tabs with separate power dissipation figures for each major component. They use the IC's figure to calculate \$T_{j}\$, based on their estimate of overall thermal resistance to ambient (\$θ_{ja}\$) based on their reference layout.

Ultimately, the sum of all the power dissipation figures is the gross loss, which should match the figure given by \$I_{in} V_{in} - I_{out} V_{out}\$.

  • \$\begingroup\$ So is my calculation is correct and regulator will get that hot, so how do i calculate for switching regulator \$\endgroup\$ – Shahreza Aug 2 at 19:13
  • \$\begingroup\$ No, it is not correct. Webbench is correctly estimating the losses internal to the chip to calculate Tj. And it’s using the correct method of gross power loss and, net for the chip. \$\endgroup\$ – hacktastical Aug 2 at 19:25

The 95.9 C/W is the junction to ambient value. The majority of the heat will be dissipated through the leads to the copper on the board, so the 16.4 C/W plus whatever the effective thermal resistance of the board is what will determine the temperature rise. You and I don't know what the value is for the layout that TI is using, but it's bound to produce a far lower junction temperature that assuming the dissipation is all directly to ambient.

  • \$\begingroup\$ At 115C thats 7W in a 6 pin SOT, seems a little high to me... \$\endgroup\$ – Voltage Spike Aug 2 at 19:40
  • \$\begingroup\$ @VoltageSpike It does - but even with the huge copper planes that the reference board has, that's not going to give the same rise as being attached to a solid lump of actively cooled copper. \$\endgroup\$ – Phil G Aug 2 at 19:49
  • \$\begingroup\$ Yeah, that's why I'm wondering under what conditions did they measure the 16.4C/W, because a PCB design probably isn't going to machine a big lump of copper and attach it to their board. \$\endgroup\$ – Voltage Spike Aug 2 at 20:02

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