Electrical – Use 5v Linear Regulator and LDO 3.3V

3.3vldolinearradiotemperature

I am designing a PCB with a STM32 microcontroller, a GPS, DataLogger and X-Bee module. The power supply is a 12V Battery.
So in my project I need 5V and 3.3V, at the same time.
I don't know if it is a bad ideia to have a 5v linear regulator(input 12V) and using the 5v as input in a 3.3v LDO regulator.
I am aware of the Low Dropout properties of the LDO and the
power dissipation of the voltage regulator.
I came to this idea because the circuit used to shut down due to the temperature. I don't want to use a buck converter because I don't want to put an inductor in the same PCB with the X-bee(Radio Module).
Finally, my project is included in a car, so temperature is a thing(the engine or even sun)

Best Answer

Use a non-linear power supply (i.e. a buck converter (like LM2596) in your case) for conversion from 12V to 5V. Then use a linear regulator (like AMS1117) to convert from 5V to 3.3V.

  • The reason why I have suggested using a non-linear power supply (for 12V to 5V) is because the heat generated by the linear one increases quite a lot with a high voltage difference and the current being consumed. This also means quite a lot of power loss.
  • It is safe to use a linear power supply for 5V to 3.3V as the voltage differnece is less and the consumption of high current is not expected at logic levels. But if the 3.3V devices consume upwards of about 500ma than it is better to use a non-linear one or a linear one with better efficiency.
  • You can design the pcb such that the inductor and xbee radio are far away from each other or you can mount the inductor externally also. I don't think that a inductor of the size found on buck converters will cause a issue.
  • Use the existing schematics of development boards like Arduino, ESP, xbee shields, STM etc. to get a better idea on how to deal with this issue.
Related Topic