To replace my overheating linear regulator DC-DC converter, I've designed this new board. This time I used a couple of switching 12 V regulators (LM2575-12 V). Here's the data sheet for reference: http://www.kynix.com/uploadfiles/pd...

I'm using two 1 amp regulators because my local supplier doesn't have a 2 amp version.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 details. Is that design correct?.Here is my math for calculating the temperature rise from ambient temperature. According to the LM2575 data sheet, the power dissipation is calculated as follows:

PowerDissipated=Vin.Iq+(Vo/Vin).Iload.VsatPowerDissipated=Vin.Iq+(Vo/Vin).Iload.Vsat

*Vo=12V,Vin=18V,Vsat=1.4V,Iq=10mA,Iload=1AVo=12V,Vin=18V,Vsat=1.4V,Iq=10mA,Iload=1A

PowerDissipated=0.18W+0.93W=1.11W

Again, from the LM2575 data sheet, 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 2°, that would keep the regulators under 50°C

My questions are:

Is my heat dissipation calculations correct?

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

Hi Agjdre, as no one has replied, I suspect for a query such as this, it would be better if placed on a more specialised website / forum.

Google is your friend in such matters. Searching for:

circuit design forum

and

circuit design help forumfinds numerous possibilities.

Good Luck - Keep us posted

message edited by Mike Newcomb

Ask Your Question

Weekly Poll

Do you think Microsoft Office is too confusing to use?

Discuss in The Lounge

Poll History