# Regulating voltage with low headroom

TLDR: what are the potential issues with boosting a 13V power source using a boost converter (6A) to say 16V, and then reducing it to 12V using a 6A step down regulator?

I am trying to power a 12V circuit off a 12V car/marine battery source. In the circuit is a 12V LCD screen (2.5A) and a 12V data overlay device (0.2A). The manufacturer of the overlay specifically states the device requires tightly regulated 12V.

When I use a typical step down regulator, there is a large voltage drop, which I understand is a function of the regulator (since the battery max voltage is around 13V). At 13V the voltage drops to 11.3, I want to avoid this to protect the overlay. The other issue with this setup is that the voltage swings as the monitor powers up, lights up, goes into standby etc, again which I want to avoid.

Note, would prefer to use off the shelf parts or modules as I am not an engineer.

• If you can tolerate Vbattery when Vbattery <=say 12V then an LDO with a headroom of say 0.1 or 0.2 V is not hard. At 6A to get a 0.1V drop effective resistance = R=V/I = 0.1/6 = 16 milliohms. FETs with Rdson well below 10 milliohms are readily available. LDO circuits based on such are also "readily available. | Eg This MOSFET \$US0.59 Digikey in 1s. P Channel 70A, 30V, 8 milliOhms Rdson. Or this and quite a few more Commented Mar 28, 2020 at 11:54

A buck-boost regulator can work with those voltages. These are available as modules. Sometimes these are called "power stabilizers".

Here's an expensive one that's a 'marine' version: https://www.boatid.com/newmar/6a-12v-isolated-dc-power-stabilizer-mpn-12-12-6i.html

A less expensive one: https://www.ebay.com/c/1683893376

• ok, thanks. And then reducing that again using a step down regulator should be ok? Commented Mar 28, 2020 at 1:51
• You don't need a post-regulator with these. The modules produce 12V, with input voltages above or below 12V. Commented Mar 28, 2020 at 1:52

how about using a 10amp or 20amp stud-mounted rectifier, in series, to provide the high current.

Use an LDO to provide the 0.2 amps.

And, if there is spike/trash on the +13, use 1milliHenry and 1,000uF low-pass filter, the inductor in series with VDD, and the cap shunting to ground. Place a 1 ohm resistor in Parallel with the inductor, to dampen ringing. (use Rdampen = sqrt( L / C) to compute.

• yikes, ok, I should have mentioned in my post, that I am in no way an engineer, so the solution would need to be off the shelf parts/modules. Sorry, hope this answer helps out someone else though. Thanks Commented Mar 28, 2020 at 1:48