I would like to control a 2M 5050 12V warm white LED strip from a 3V3 or 5V micro controller, powered by a 240V wall outlet.

To power both the LED strip and MCU from the same switch mode AC adapter "wall wart", I am going to need to use a DC-DC step-down/step-up buck/boost converter for the other device.

I am looking for the most power efficient setup. Candidates:

a) Use a 12V AC adapter and power the LED strip directly, with a buck converter for the 5V MCU.

b) Use a 5V AC adapter and power the 5V MCU directly, with a boost converter for the 12V LED strip.

Common sense tells me option A) would be more efficient as the main load (LED strip) avoids double regulation, stepping all the way down to 5V then stepping back up to 12V.

I would like to both configurations to confirm real world vs common sense. Any advice on how to accurately test each with a multi-meter / temp probe?

  • 1
    \$\begingroup\$ Stepping up is usually less efficient as stepping down. (In a boost converter, you 'waste' more or less current to repeatedly charge the inductor, while in a buck converter, basically every electron pushed into the inductor reaches the load eventually.) \$\endgroup\$
    – JimmyB
    Jul 18, 2016 at 13:10

1 Answer 1


Option B will need a bigger boost converter that can handle 2.5 Amps out (5050 strip current) and a AC-DC adapter of let's say 120% of that in power. So 2.5 A * 12 V = 30 Watts * 120% = 36 Watts / 5 V = 7 + Amps at 5V. Plus the likely minimal current of the microcontroller.

Option A doesn't have those needs. A 12V 3 Amp supply, with a step-down converter of a few hundred milliamps at best for the microcontroller. Let's assume 250mA at 5V and 120% for efficiency. 250 mA * 5V = 1.25 Watts * 120% = 1.5 Watts / 12 Volts = 125 mA at 12V input.

Much smaller regulator is needed, as is the power supply. And that means both will be cheaper and smaller physically, while being easier to find. It's a no brainer. A 7 Amp 5V supply is not a common item, but you could find 2.5 or 3 Amp 12V supplies everywhere.

120% means 80% efficient regulator (20% power loss), and it's more likely to be 90%. My math here is all fuzzy back of the napkin but it's generally correct.

As to your question on how to test, basically just connect everything in ammeter or current sensing mode on a multimeter.

  • \$\begingroup\$ I'm not running the LED strip at full brightness. It's pulling less than 1A at around 10V, ~0.6A at 8V (Using a LDR and MOSFET to govern brightness) \$\endgroup\$ Jul 18, 2016 at 1:28
  • \$\begingroup\$ @mike is the ldr controlling the mosfet directly, or is a microcontroller reading the ldr and providing PWM? \$\endgroup\$
    – Passerby
    Jul 18, 2016 at 3:36
  • \$\begingroup\$ The latter. MCU uses LDR as part of it's decision and provides PWM to the MOSFET. \$\endgroup\$ Jul 18, 2016 at 3:38
  • \$\begingroup\$ Ah good. But keep in mind that it may average out to 8V and 0.6A based on your PWM setting, but the supply is peaking at full voltage and current, so it should still be sized for 100% on. \$\endgroup\$
    – Passerby
    Jul 18, 2016 at 3:40
  • \$\begingroup\$ I thought it was supposed to be 120%, for some extra wiggle room. Either way, I'm well under. \$\endgroup\$ Jul 18, 2016 at 3:42

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