6
\$\begingroup\$

I'm building a series of gadgets for my kids that mostly (microchip, sensors) run off 3.3 but there are some switches and possible servos that run off 12v.

My original idea was just to have two wires, one for 12v and one for 3.3v but I don't like that since most of the items won't need the 12v and it adds clutter in the wiring. The other option is to only have one wire and either step down or up the voltage in the components that need it. The heat from stepping down 12v to 3 worries me and since most will be running of 3v it seems more logical and economical to use step ups when I need the 12v. Is this the best approach or is there another approach that I'm overlooking?

Most of the 12v components are buttons with built in leds so power consumption is pretty low, I guess they are mostly used in auto applications so most of the leds require 9-12v to light up well.

\$\endgroup\$
6
  • 5
    \$\begingroup\$ "The heat from stepping down 12v to 3 worries me" how much power are you going to need on the 3V, hundreds of watts? \$\endgroup\$
    – PlasmaHH
    Feb 15, 2018 at 12:18
  • \$\begingroup\$ Maybe my math is off but I like a few watts would be too much for most cheap regulators when going from 12 to 3. This is enclosed in a box too which doesn't help the heat situation. \$\endgroup\$ Feb 15, 2018 at 12:25
  • 9
    \$\begingroup\$ @RyanDetzel Then don't use cheap linear regulators for any sizeable amount of current. Use buck converters. \$\endgroup\$ Feb 15, 2018 at 12:30
  • 1
    \$\begingroup\$ What is "a few"? You should be able to get rid of 1-2 w of heat, so 5-10W of 3v power is easy \$\endgroup\$
    – PlasmaHH
    Feb 15, 2018 at 12:33
  • 1
    \$\begingroup\$ Olin's answer if great, as usual, but you might also consider researching alternatives to your 12v parts. LEDs don't need that much voltage to light up well, although you may like the convenience of a ready-made push button with built in light. \$\endgroup\$ Feb 15, 2018 at 16:33

3 Answers 3

24
\$\begingroup\$

This is a no-brainer. Bus around 12 V and step that down to lower voltages locally as needed.

The heat shouldn't be a problem. Buck switchers in that voltage range should be over 90% efficient. Even figuring 85% to be pessimistic, there will be very little heat to worry about. Let's say your control circuitry is drawing 100 mA at 3.3 V. That's quite a lot for a modern microcontroller and a little surrounding circuitry. 100 mA times 3.3 V is 330 mW. With a 85% efficient buck regulator, it will draw 388 mW from the 12 V supply, and dissipate 58 mW as heat. That's so little that you'll barely notice it being warm when you put your finger on it.

It seems your high power devices run on 12 V. Due to the higher power, this is where you don't want something between the power source and the device drawing power. Put another way, you get to pick one supply that can be used with 100% efficiency. It should be the one that needs to provide the most power.

It is also useful to bus around a higher rather than lower voltage. At the same power, the current will be lower, meaning you can use smaller wire for the same loss.

Not only will the voltage drop at 12 V be lower, but it will be easier to tolerate. The 12 V can vary a little if you're just running a motor with it, but some digital ICs require a more tightly regulated supply voltage.

Here is a schematic snippet from one of the many projects where I make 3.3 V locally from a higher supply voltage that is bussed around:

I use this basic building block quite a lot. In this case the input was 24 V, but it would work just fine from 12 V and even lower too. With 12 V nominal input, C11 and C12 can be lower voltage. That should allow combining them into a single cap, like 22 µF and 20 V.

To make different output voltages, all you need to do is change R8 and L2. For example, to make 5 V, use 22 µH for L2 and 52.3 kΩ for R8.

I have used this basic circuit in quite a few projects.

\$\endgroup\$
4
  • \$\begingroup\$ Great response, thank you. Does your opinion change if cost is involved? Is it cheaper to step up or down or if I have 100 components and only 20 that need 12 volts it would be cheaper to only outfit those 20 with step ups than the other 80 with step downs, right? Even with that though is it still better (design wise) to down step? \$\endgroup\$ Feb 15, 2018 at 15:21
  • 3
    \$\begingroup\$ @Ryan: Stepping up or down with switchers is the same cost, particularly at these low voltages. Stepping down linearly can be cheaper, but incurs other problems like having to get rid of heat. Unless this is a high volume design, the relative costs are irrelevant. Each component doesn't need its own local regulator. Basically, each location or board does. Generally, stepping down is better, and bussing around higher voltage is better for reasons I mentioned. \$\endgroup\$ Feb 15, 2018 at 15:51
  • \$\begingroup\$ @OlinLathrop: for my own education, can you include how you arrived at 22uH & 52.3k for L2 and R8? \$\endgroup\$ Feb 15, 2018 at 17:30
  • 1
    \$\begingroup\$ @Bry: The inductor and voltage divider ratio is specific to this chip. These values therefore came from the datasheet, or were derived from values given in the datasheet. \$\endgroup\$ Feb 15, 2018 at 18:20
5
\$\begingroup\$

The other option is to only have one wire and either step down or up the voltage in the components that need it.

Sounds simple but a standard step-up (boost) or step-down (buck) don't normally work adequately when Vin = Vout. So, my aims would be: -

  • Input supply voltage transparency - if 3 volts is needed then make a seamless solution that will produce 3 volts if the supply varied from 3 volts to 12 volts.
  • If 12 volts is needed for stuff then it's best not to put a regulator in this line (unless of course it's needed).
  • If batteries are used as the power source make sure you have a solution that can run down to lower than 3 volts at the input.

I'd use a buck-boost chip for the 3.3 volts like this one: -

enter image description here

The picture above shows a 5 volt output but, by altering the ratio of the resistors attached to the FB (feedback) pin, you can get 3 volts or 3.3 volts.

The benefit here is that even if you fed it 3 volts it would output 3 volts so you don't need to worry about switching in or out your buck regulator when you are powering it at 3 volts or 3.3 volts.

Like Olin says, the higher power side will need 12 volts so regulators in this feed are likely to be less efficient than just providing a straight 12 volts.

The other benefit from the LTC3129 is that if you are using batteries, it will still work down to 2.42 volts. The only limitation is that devices fed from this are limited to a maximum current of 200 mA. If this is a problem there are higher output current solutions.

\$\endgroup\$
2
  • \$\begingroup\$ Oh, I like that chip, it's pricey though. Thanks. \$\endgroup\$ Feb 15, 2018 at 15:23
  • \$\begingroup\$ @RyanDetzel there are offerings from TI that can probably do the same for half the price. \$\endgroup\$
    – Andy aka
    Feb 15, 2018 at 15:25
0
\$\begingroup\$

You should also look at the variety of regulator cards that pololu.com offers. They have step-up, step-down, and a combo of the two to stepdown as your source is higher than the desired output and then stepup as your source descreases lower than the desired output. Those switching regulators are great, much better than simple linear regulators.

\$\endgroup\$

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.