To replace my overheating linear regulator DC-DC converter, I've designed this new board. This time I used a couple of switching 12V regulators (LM2575-12V). I'm using two 1 amp regulators because my local supplier doesn't have a 2 amp version.

Here are the schematics.


Here's the board.


There are no pictures of the board because it doesn't exist yet. This time, I did the math first and didn't blindly built the thing.

The red arrows highlight the single-point grounding I tried to make, as recommended in the datasheet. Is that design correct?.

Here is my math for calculating the temperature rise from ambient temperature. According to the LM2575 datasheet, the power dissipation is calculated as follows:

$$PowerDissipated = V_{in} .I_q + (V_o/V_{in}).I_{load}.V_{sat}$$ $$V_o = 12V, V_{in} = 18V, V_{sat} = 1.4V, I_q = 10mA, I_{load} = 1A $$ $$PowerDissipated = 0.18W + 0.93W = 1.11W$$

Again, from the LM2575 datasheet, its termal resistance (junction to ambient, in worst case) is 65°C/W, which would keep the regulators just below 100°C considering a ambient temperature of 25°C. If I use my heatsinks with 20°C/W and the regulator thermal resistance (junction to case) of , that would keep the regulators under 50°C

My questions are:

1. Is my heat dissipation calculations correct?

2. Is my board designed correctly, especially regarding the required single-point grounding for the regulator pins? These points are marked by the red arrows on the board image.

  • 1
    \$\begingroup\$ You are going to overdesign your board (not a bad thing) if you take that 65 C/W number. That is junction to ambient only. Your real world will be a bit better. With heat sinks, you need to use both your heatsink thermal resistance PLUS your junction to case. Not a big difference but real nonetheless. If you put a bit thicker trace around your pins, you'll begin approaching that 45 number. \$\endgroup\$
    – scld
    Commented Dec 9, 2013 at 15:20
  • 1
    \$\begingroup\$ That 65 C/W is for a socket without no pcb heatsinking copper. Look at notes 8 - 11 on page 7 of the pdf. Also, TI has the Switchers made Simple software here: ti.com/ww/en/simple_switcher_dc_dc_converters/… that can help. \$\endgroup\$
    – Passerby
    Commented Dec 9, 2013 at 15:34

1 Answer 1


That 65 C/W is for a SOCKET without no pcb heatsinking copper. If soldered, with appropriate copper layout, it goes down to 45°C/W, or less, Junction to Ambient.

As for your heatsinks, they are ~20°C/W, but you forgot to add the Junction to Case rating of 2°C/W. So 25 Ambient + 20 Heatsink + 2 Junction to Case = 47°C/W * 1.2W = 56.4°C Junction Temperature.

Key points, Look at notes 8 - 11 on page 7 of the pdf, and consider the board layout (you have tons of empty board spacing no need to have everything so close together).

Page 19 also has good information:

In many cases, no heat sink is required to keep the LM2575 junction temperature within the allowed operating range. For each application, to determine whether or not a heat sink will be required, the following must be identified:
1. Maximum ambient temperature (in the application).
2. Maximum regulator power dissipation (in application).
3. Maximum allowed junction temperature (150°C for the LM1575 or 125°C for the LM2575). For a safe, conservative design, a temperature approximately 15°C cooler than the maximum temperature should be selected.
4. LM2575 package thermal resistances θJA and θJC.

But then you realize, The LM2575 is characterized for operation over the virtual junction temperature range of -40°C to 125°C. At 1.2W (I'm rounding up a bit) and worst case 65°C/W Junction to Ambient, that's still only 78°C Junction Temperature. Almost 50°C below it's maximum operating temperature. Worst case, socket, no proper pcb copper sizing, no heatsink, and you're still good to go. ** Rearrange the traces, and throw on your heatsink, and you're golden. You might need to move the L1/L2 inductors or the heatsink won't attach right. Ideally, you would have the Ground Pin 3 connected directly to the large ground plane.**

Just bare in mind, I hope you have selected the right layout for the 2575 you are getting, as it has multiple versions.

Finally, TI has the Switchers made Simple software here: http://www.ti.com/ww/en/simple_switcher_dc_dc_converters/index.html?DCMP=simple_switcher&HQS=switcher that can help (Though the LM2575 is not included). Also, this article http://store.curiousinventor.com/blog/pcb-as-a-heat-sink-calculating-trace-width-for-given-current can help give you some ideas.

  • \$\begingroup\$ Excelent! Thanks for your answer. But what about the single-point grouding? I've updated the image to show where they are (look for the red arrows). \$\endgroup\$
    – Ricardo
    Commented Dec 9, 2013 at 16:15
  • \$\begingroup\$ @Ricardo Updated the answer, and yea, that single ground point is fine, but try adjusting the layout so you can connect Pin 3 to the large ground plane, then worry about the single ground point. You don't have a large/heavily populated pcb with lots of vias, so it's not really that high of a concern. \$\endgroup\$
    – Passerby
    Commented Dec 9, 2013 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.