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Background: I'm working on a CANBUS and sensor project for an offroad racing truck. Heat, reliability, and RF interference are all big concerns. I'm designing the PCB's and enclosure, so can be very flexible. The project will be powered off of the truck's battery which typically runs at about 14v. The PCB for the power regulation will be separate from the PCB with all of the logic, primarily for heat isolation. We need about 1A a 5V, and about 600mA at 3.3V.

Question: would a voltage regulator like a LM7805A or a 5v buck converter like a MP2307 be more effective for this particular use?

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    \$\begingroup\$ With a linear regulator, your 5 V load burns 5 W and your regulator burns about 9 W. Your 3.3 V load burns 1.9 W and its regulator burns 6 W. \$\endgroup\$ – The Photon Mar 31 '16 at 5:12
  • \$\begingroup\$ Is your truck powered by the battery, or is it a battery in a truck with an engine that is charging the battery? In the latter case, you have (as Jim Williams put it) the power supply from hell. See ISO7637. \$\endgroup\$ – Peter Smith Mar 31 '16 at 9:06
  • \$\begingroup\$ It is in fact a truck with an engine. \$\endgroup\$ – TheAutomator Mar 31 '16 at 15:28
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When you are dealing with automotive power supply design, there are several factors to keep in mind. 14V truck battery is not just 14V, but it can increase upto 24V if directly connected through ignition. Also the ignition time generates a high current which is first problem to worry. The whole power supply design can be divided into stages.

Stage 1: Protect your PCB from current surge with appropriate PTC fuse.

Stage 2: Use buck converter having appropriate rating (5V/1A in your case, so choose 5V/2A for further current distribution) (Heat sinking/thermal distribution is important point here in PCB design)

Stage 3: Input this 5V/2A to linear regulator to obtain 3.3v/600mA.

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  • \$\begingroup\$ The 5V needs to be 1.6A then. \$\endgroup\$ – Simon Richter Apr 1 '16 at 12:43
  • \$\begingroup\$ Yes Simon, you made a valid point. I modified the original answer accordingly. \$\endgroup\$ – Prasan Dutt Apr 1 '16 at 13:05
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Well, linear regulators will dissipate 15.4 watts of heat (voltage drop times current), so I think it would be advisable to use a switching converter if only to make thermal management simpler. Now, what you could do is build a switcher to step 14 volts down to 5 volts and then use a linear regulator to generate 3.3 volts from the 5 volt rail. The linear regulator in that case would only dissipate about 1 watt.

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