Easy way to get 12v DC from 32v DC bus?

I've got a 32 volt DC regulated bus. I also need about 30mA at 12 volts (DC). Is there an easy way to tap some power from the 32v bus?

My first thought was just put in a 12v linear regulator. But those in my parts box have absolute ratings close to (or below) 32v, and I'm not one to push the envelope.

Some divider make up using hairy arsed power resistors might work, but that's not too efficient and generates HEAT.

If the solution gets complicated (e.g. build a buck down converter), I could find a 12v wall wart and tie across the mains supply to my 32v power supply.

I know about voltage multiplier circuits. Is there a neat voltage divider trick somebody can teach a nubie?

• A linear regulator makes no sense with such a large voltage drop. A resistive divider is also a non-starter. You should use a buck converter, but you don't need to design it yourself if you don't want to. You can buy all kinds of all-in-one solutions, even including potted modules complete with inductor designed to plug into a TO-220 footprint. Just go to Digikey or similar and look through their non-isolated DC to DC converters. – user133493 Jan 12 '18 at 2:31
• Finding a 12V solution is not very high tech for an EE forum. Can you explore Amazon or eBay ? Lots of DC-DC solutions there! – Tony Stewart Sunnyskyguy EE75 Jan 12 '18 at 2:41
• @replete That's an answer and should go into the answers box. – pipe Jan 12 '18 at 9:56
• @pipe, indeed, however I expected this to be closed. – user133493 Jan 12 '18 at 11:09
• Turns out there isn't a "magic bullet" for this. All approaches to get tens of milliamps cost a significant amount relative to what you need. A solution which uses just a few components is complicated by most of today's charge pumps and regulators not dealing with more than a decade of volts. The best solution is to deal with the power supply selection so you have the voltages you want and don't have to manufacture them later in the circuit. The Q&D answer is a DC-DC bucker, but price vs quality can also be a problem. TANSTASFL. – HiTechHiTouch Feb 2 '18 at 1:55

this is typically handled with a "switching regulator". A "buck converter" is a type of switching regulator designed to regulate down from one voltage to another. There is any number of prebuilt modules you can find that accept 32V and regulates to 12V. Cheap ones start at a dollar and a month shipping from China, but quality costs more.

The linear approach

A buck converter is the most efficient way to go, and should be easy to find. Other answers cover that. I want to suggest a simpler brute-force method that you may have overlooked from the sound of your question: The good old LM317.

This is a very common linear regulator, but the "trick" compared to its fixed-output 7812 relative is that it's floating. It has a maximum voltage difference of about 40 volts. Your input-to-output difference is 20 volts.

It will dissipate 20 * 0.3 = 6 watts of heat, so it requires a reasonable heat sink, but since you have not given any more details about the project, this may not be a problem. If it is, you need a buck converter.

• 6W of dissipation is a LOT, and results in an efficiency <38%. Using a linear regulator is no different efficiency wise from the OPs "hairy arsed power resistors might work, but that's not too efficient and generates HEAT" approach. – Tom Carpenter Jan 15 '18 at 0:35
• @TomCarpenter Obviously a voltage divider stiff enough to supply 300 mA to a 12 V load will waste massive amounts of power, a lot more than a linear regulator. – pipe Jan 15 '18 at 6:50

Have a look at these buck converters from TI, they're a tested, complete buck converter on a module which meet your input and output voltage and current specifications, with a lot more efficiency than a linear regulator. They're quite expensive, and you could design and implement your own buck converter for less, but they do provide a lot of convenience.

There are other parts which cost less, and are pin compatible with a TO220 linear regulator such as a 7812, but they have more restrictive maximum ratings, so you would need something else as well to lower the input voltage a little. An example of those is this module from TI.

Other manufacturers are available and found by searching for buck converter module.

• I was editing my post as you were commenting I think, @pipe. – Colin Jan 12 '18 at 9:59
• Yep, looks like it! – pipe Jan 12 '18 at 10:00

The Art of Electronics, 3rd ed (Horowitz & Hill) 9.13.2 suggests using a depletion mode FET follower to hold the LDO's Vin a couple of volts above the regulator's required input voltage. Vgs is about -1.5v. Higher values can be obtained by connecting the gate to a resistor divider between the regulator output and FET source terminal.

Here's a portion of the circuit:

simulate this circuit – Schematic created using CircuitLab

Resistors and caps required by the LDO are not shown. I assume, but have not bench checked, the strategy would extend to other regulators.

• Ok for 10mA. Not so great for 300mA. – Tom Carpenter Jan 15 '18 at 0:35