I am looking to use MOSFETs as a switch to run a motor and we ran into a slight problem with the power supply to the circuit board. The main batteries we would like to use output 14.4 V nominally. However I need to supply 3.3 V to the processor. I also have an amplifier that I need to supply a +- voltage to (preferably around +14 and -14). If I had to I could use different batteries to power different components, but that seems silly. The other tricky part is that the solution to this must be rated for operation as 150degC. If anyone has a suggested solution I could look into that would be a great help.

  • \$\begingroup\$ You mean like a voltage regulator? \$\endgroup\$ Jul 27, 2016 at 20:00
  • \$\begingroup\$ It would help if we knew more about your application. 150C is pretty hot for electronics. What's the amplifier for? \$\endgroup\$
    – Adam Haun
    Jul 27, 2016 at 20:39
  • \$\begingroup\$ @AdamHaun we are using a strain gauge with a wheatstone bridge. The amplifier is to amplify the voltage difference so the microcontroller can read it. \$\endgroup\$
    – Carly
    Jul 27, 2016 at 21:57

1 Answer 1


You could use a voltage regulator to create 3.3V from the 14V batteries. Since the device is running from a battery, I would avoid a linear regulator as the power loss would be $$P_{loss} = (14.4V - 3.3V) * I$$ This may be acceptable for your application but I would rather use a switch mode voltage regulator for increased efficiency.

When choosing a suitable regulator, you would need to check the datasheet and make sure it can function at a larger temperature than 150 degrees C. This is especially important when using a linear regulator as the power loss is converted directly to heat.

How much larger the temperature rating should be, we cannot tell without knowing more details of your application.

Without going into much detail about designing systems from a heating perspective (and to learn more about the topic) see this excellent video from Dave Jones at the EEVBlog (https://www.youtube.com/watch?v=8ruFVmxf0zs)

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    \$\begingroup\$ I would add that even if noise would be a problem, it is usually better to handle the large voltage drop with a switching converter and only burn a very small voltage with a linear one. \$\endgroup\$
    – caconyrn
    Jul 27, 2016 at 21:58
  • \$\begingroup\$ @bitshift Would the voltage regulator be able to supply the negative voltage for the amplifier as well? \$\endgroup\$
    – Carly
    Jul 28, 2016 at 16:02
  • \$\begingroup\$ @Carly You would need a so called "inverting regulator" to generate negative voltages such as these from Linear Technology linear.com/products/inverting_regulators \$\endgroup\$
    – bitshift
    Jul 29, 2016 at 7:58

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