0
\$\begingroup\$

So, I have a wall-wart power supply giving regulated 12v 500mA to a 2.1mm Jack.

I'm using this to power a 12v 160mA fan, and an Arduino Pro Mini (at 5V). I understand that the Mini's regulators to drop from 12 to 5 will produce a lot of waste heat (which I'd rather avoid, as am keen to minimise the system's power consumption).

What I was planning was for something like:

Wall Wart (12v, 500mA) ------------------------> Relay -> Fan
                \ (1) -> Arduino Pro Mini @5V -> /

Would it be more power efficient to put a switching regulator at point (1) in the diagram (with its surrounding components, or as an integrated circuit) and supply power to the mini's VCC input?

Also, I'm not sure how to search for an integrated circuit switching regulator (if such exist) as this might be a neater solution than self soldering a whole support circuit?

\$\endgroup\$
6
  • 1
    \$\begingroup\$ What current is taken thru the proposed regulator at (1) i.e. what load current? If it's a trivial amount then don't bother with a switcher use a 78L05. \$\endgroup\$
    – Andy aka
    Feb 27, 2015 at 13:42
  • \$\begingroup\$ So other than the baseline draw of the Arduino Pro Mini (which I can't seem to find anywhere reliably here suggests 16 mA before sleep/modification) That said, however, the baseline draw needs to include the onboard LED and here suggests a baseline chip draw of 50mA. I will additionally require about 20mA from one of the pins to drive the relay when HIGH. \$\endgroup\$
    – jvc26
    Feb 27, 2015 at 13:58
  • \$\begingroup\$ What @Andy says. Does the linear regulator on the ProMini get hot? If not, compared to the fan, there's no significant saving. Sleep helps a lot, look for a library called Narcoleptic Sleep. And remove the LED. \$\endgroup\$
    – tomnexus
    Feb 27, 2015 at 13:58
  • \$\begingroup\$ Cool, thanks for the above - currently experimenting, if it doesn't heat that much I'll just leave it, if we're getting roasting will try the above - thanks @Andyaka \$\endgroup\$
    – jvc26
    Feb 27, 2015 at 13:59
  • \$\begingroup\$ Assume 100mA for a standard Arduino at full load. \$\endgroup\$
    – Passerby
    Feb 27, 2015 at 14:34

1 Answer 1

Reset to default
2
\$\begingroup\$

Yes, a switching regulator, buck converter specifically in this case, would be efficient in providing 5 V to the arduino from your 12 V input.

Be careful in how you feed this 5 V to the arduino. First, check its schematic to see if it is using a linear or switching regulator. If the latter, they you can feed it 12 V directly without a lot of loss. If it has a linear regulator, then you may have to remove it so that you can drive its output with nothing on its input.

However, if you're so keen to minimize power consumption, then use a transistor instead of a relay to control the fan, and use your own microcontroller circuit instead of something off the shelf and with extra things you don't need, like the arduino.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ Info only: The Pro Mini is rather minimalist. 15 components total (excl PCB & connectors). 2 LEDS, 3R, 6C, Xtal, LDO, uP, reset button. At as little as $1.85 assembled with bootloader loaded you could well find a repurposed use for them. You can plug an TTDI or similar USB/serial bridge into a provided connector if desired, but bit bashed USB on 2 I/O is also available for bootloader-ing. The only thing that makes it an Arduino is the standardised connector layout and the C compiler variant come GUI that is usually used with it. eg GCC or assembler can be used. \$\endgroup\$
    – Russell McMahon ♦
    Feb 27, 2015 at 13:58
  • \$\begingroup\$ Schematic - I'm using these (and the Nano with added USB-serial bridge) for grab-one-and-do-something-minorish-(almost)-instantly tasks. I pay $US3 each in 10's for the USB ready version (Nano). I cannot build anything equivalent for the cost. So far I'm just using their IDE and compiler - access to essentially everything except assembler, and that can be added when needed. || I'm actually FAR more at home in assembler than in C but the convenience does grow on you just a little. \$\endgroup\$
    – Russell McMahon ♦
    Feb 27, 2015 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.