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I think this is a common situation so hopefully an answer helps more than just me. I have a 3.3v low power MCU which will be controlling a 5v 5A run of LEDs. When the LEDs are off the system will go to low power sleep mode, however the LEDs have a relatively high quiescent current which I’ll need to cut off. I’ve narrowed my power management setup down to two options with different tradeoffs that I’ve list below but I’d love some feedback on which would be better, if I’m mistaken about something or if I’m missing some other solution.

Everything will be powered from a 9.6v-12v supply.

Option 1) 3.3v rail is its own switching regulator and the 5v rail is also its own dedicated switching regulator rated to above the required 5A. Both lines will be powered directly from the 12v supply and the MCU on the 3.3v line will control the enable pin of the 5V regulator giving it the ability to turn of the 5v line and cut the LED quiescent current.

  • Pros - Most efficient on both 5v and 3.3v and I only have to worry about the heat dissipation on the 5v reg.
  • Cons - Most complex option with the most components, which means most likely the most expensive and with two switching regulators should I ever need EMC certs it could be tricky. Also at the low loads of sleep mode on the 3.3v rail my switching reg is not at peak efficiency.

Option 2) 5v line is still a 5A rated switching regulator but now the 3.3v line is just an LDO downstream from the 5v rail. The MCU on the 3.3v rail now would not be able to switch off the enable pin on the 5v rail so I would have to use a power mosfet between the 5v regulator and my 5A led strip to kill the quiescent current.

  • Pros - Simpler design and slightly cheaper replacing the 3.3v switching with the LDO, plus I get back some space on the PCB

  • Cons - I’ve been researching the power mosfets and I’m worried about heat dissipation if I’m actually driving near 5v 5A. I’m also not sure what having the LDO off the 5v switching regulator will do for my overall power efficiency vs. a 5v and 3.3v dedicate switching reg in option 1, could be this option lets me switch off the quiescent current but then I give that power savings right back by sending power to the micro from a 5v reg to a 3.3v LDO.

As I see it I have a tradeoff between complexity and cost vs. simplicity and possible heat issues. On top of that I have to consider PCB space and possible EMC issues with two switching regulators.

Thanks for any input.

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    \$\begingroup\$ The question is in the title but the low power consumption targets hinted at make this question impossible to answer for anyone but a mind reader. \$\endgroup\$ – Andy aka Jul 8 '15 at 12:38
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You will need to compare the current draw of the 5V regulator with a light load vs. the smaller 3.3V switching (or even linear) regulator. Chances are it will draw a fair bit more current. This is your trade-off to make. Remember that when the big print says 86% or whatever efficiency that is the highest efficiency point of operation. The efficiency will go to zero at zero output current and may be worse than even a linear regulator at relatively high output currents (less than ~50mA in some cases for something with several amperes output capability).

You are going to need high current switching regulation anyway so I wouldn't worry too much about any EMC differences. The 5A will be where the issues (if any) will tend to arise.

You will probably require a high-side switch for the 5V/5A if you can't control the regulator directly. That is most simply done with a level shifting transistor (from 3.3V to 5V) plus a power MOSFET. Something like a Diodes DMP2006UFG-7 has maximum 5.5m\$\Omega\$ Rds(on) with 4.5V drive would be okay for a high-side pass transistor. That's less than 150mW dissipation. Of course if you can use the regulator control line directly and meet your specifications it's worth avoiding extra parts. If the regulator will accept a TTL compatible input, the 3.3V may be able to control it directly (read the datasheet carefully on this point).

You could also use a MOSFET at the input of the 5V regulator which would eliminate any shutdown current it might have, but of course the micro would have to have an independent supply.

If your microcontroller draws less than 10mA or so you may find a linear regulator from the supply voltage a good solution- simple and reliable, though some switching regulators use special techniques to get reasonable efficiency even at very low output currents. Again, a trade-off to be made. If your micro draws uA in a sleep mode most of the time, a switching regulator is not going to be of any help, and when the 25W LEDs turn on a linear regulator won't make much difference even if the micro draws tens of mA.

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    \$\begingroup\$ Some regulators, such as the MCP1703A can handle 16v input and 250mA out while only consuming 2uA quiescent. Be mindful of the power dissipation though. I personally like the idea of uC powered from LDO and main battery, and MOSFET high-side switching the 12v input to the 5v regulator. Just pull up the PMOS gate to +12v through a 100k resistor, and pull this to ~0v with an NPN BJT transistor, the base of which is driven by the uC. "High" output = "5v On" logic. \$\endgroup\$ – rdtsc Jul 8 '15 at 14:52
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    \$\begingroup\$ @rdtsc I could imagine a setup with standby power from an linear regulator such as the MCP1703A (for the micro in sleep mode) that would switch over to the 5V source when it became available to reduce power dissipation in the regulator. Possibly a bit complex. \$\endgroup\$ – Spehro Pefhany Jul 8 '15 at 15:19
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Provided I get the correct idea of what you want:

Since I expect the microC te draw only a very small current from it's 3.3 V supply, using an LDO and feed it from the 5 V line can be a good solution.

Then you need a MOSFET to switch off the LEDs. A common mistake is that to think that the MOSFET will dissipate 5 V x 5 A = 25 W, this is UNTRUE ! The 25 W will go to the LEDs ! The voltage across the MOSFET will be less than 100 mV (assuming proper design) so less than 0.5 W power dissipation. And when the MOSFET is OFF, no current so no Power dissipation at all :-)

But what if you supply the microC 3.3 V directly from the 9.6 - 12V, and then switch on/off the 5V regulator with the enable pin as in your option 1. Maybe you will need some levelshifter circuit since the microC is working at 3.3 V and the regulator at 5 V ? This depends on the 5V regulator implementation.

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  • \$\begingroup\$ You have a good point about power dissipation, and much "magic smoke" has undoubtedly been released accidentally. Always remember to check power (Watts = Volts * Amps) for all power-related components. (Even good to do so for small signal lines, as "large" 1/4W passives can be replaced with smaller 1/8th, or 1/16th Watt ones to save PCB area.) LTSpice has a nice feature, alt-click a component during sim to see it's power dissipation. Useful to get a ballpark before testing. \$\endgroup\$ – rdtsc Jul 8 '15 at 18:39

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