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I a working on a circuit that has components that need 3.3V and 5V. My power input is 12VDC and can not be changed. The 3.3V components use about 510mA of current and the 5V components use about 155mA. My first attempt at a circuit used two linear voltage regulators in SOT-223 packages. The 5V reg use 12V as it's input voltage and the 3.3V reg used 5V as it's input. Of course the 5V regulator get's extremly hot and then shuts off (luckily it has thermal protection!)

I have read some other posts on this forum about how to dissipate the heat with the PCB traces and also to do it using the case (which is a possibility in my case), however it would be much preferable for this design for the enclosure to not get warm. People using it may complain. So I am considering using a switching-mode voltage regulator like this AD1509. My concern is with the noise. My board has a WiFi module on it and I am wondering if the performance will be affected by the switching noise. The AD1509 switches at 150kHz and I think WiFi is in the 2.5GHz range so I should be okay right?

Also, do you think the switching voltage regulator will also get hot or if I just a big copper pad under it will it be okay?

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A switching power supply will be noisier than a linear regulator, no question. The only way to know if it will disturb the wifi is to power it up and see what happens.

Any switching regulator will dissipate power as a function of conduction losses and switching losses. These parts with an integrated MOSFET do need PCB cooling - usually a multilayer PCB with relatively large 'islands' of copper interconnected with vias. Again, you'll have to do some math to figure out a loss estimate, and gauge the copper size accordingly.

The datasheet says that the 5V part at 2A load and 12V in operates at around 83% efficiency. So, for 10W out you're losing just over 2W in the device (and the external diode that completes the buck converter). This will scale down somewhat with your reduced loading (conduction losses will drop, switching losses probably won't).

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Don't use traces; use a heat sink instead. They are cheap and easy to use.

Don't use the switching regulator directly, use it for 12 to ~6 V then regulate down. That saves power and still gets a clean final output.

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    \$\begingroup\$ Can you please explain how that saves power? Also, is it not possible to get a clean output form a switching regulator? Isn't that what the additional components are for at the output? \$\endgroup\$
    – PICyourBrain
    Commented Mar 10, 2011 at 16:38
  • \$\begingroup\$ 6V outputs are much harder to find. I don't want to add $5 to my board cost... \$\endgroup\$
    – PICyourBrain
    Commented Mar 10, 2011 at 16:42
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    \$\begingroup\$ How will converting to 6V and then using linear regulator below that save power? That's only the case if your switcher is less than 80% efficient, which is probably not the case. \$\endgroup\$
    – Kevin Vermeer
    Commented Mar 10, 2011 at 17:34
  • \$\begingroup\$ As far as sensitive analog parts are concerned you may still not get clean output because of RF coupling of the noise. \$\endgroup\$
    – jpc
    Commented Mar 11, 2011 at 11:00
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If you follow the IC manufacturers tips about component selection and layout you should not have any problems with the WiFi chipset running from a switched power supply.

You may consider going for a switcher at 3.3V while using a linear regulator for 5V.

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