Daisy Chain Linear Regulators? - Page 1
Forum section: electronics / beginners

author: squarefrustration
author_rank: Newbie
timestamp: October 22, 2019, 05:32:37 pm
content: Hi there.I'm working on a project where I need both 3.3V and 5V from a 12V supply. The 3.3V and 5V circuits will need between 50-250mA dynamically.I am curious as to if it's more efficient to power the 3.3V regulator from the 5V regulated output, since the VIN will be closer to Vout. However, this would result in more current draw from the 5V regulator which I assume will cancel any benefit from powering the 3.3V regulator with a smaller VIN-VOUT.Can anyone chime in as to if one of these configurations is more efficient than the other (Schematic attached)Thanks!

author: magic
author_rank: Super Contributor
timestamp: October 22, 2019, 05:46:19 pm
content: Efficiency is the same. The sum of currents consumed by both 5V and 3.3V loads will be drawn from the 12V supply and excess voltage turned into heat in one or both regulators.The only difference is that if the 3.3V reg is supplied from 5V, it will run cooler, but the 5V reg will get hotter because of more current. Power dissipation in each regulator is of course (Vin-Vout)·Iout so plug in your numbers and do the math.Sometimes it's beneficial to spread heat between multiple devices, sometimes it's crucial that the higher rail isn't overheated by powering the lower one.

author: squarefrustration
author_rank: Newbie
timestamp: October 22, 2019, 05:59:47 pm
content: Thank you, this is exactly what I was looking for. I may opt for option B just to spread out the load a little more.

author: SeanB
author_rank: Super Contributor
timestamp: October 22, 2019, 06:23:20 pm
content: Option c is to add in a third regulator, probably a small switching converter on a board, pre built, to give you a lower voltage, probably around 8V, so that efficiency is improved considerably. If simply feeding logic from the supplies nothing more filtering wise is needed, otherwise som3e LC filtering of the switching regulator output would be good to reduce the switching noise down to a lower level. Pre made converter modules are cheap, and a simple buck converter will work, as it has adjustable output voltage, making it a drop in part.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 22, 2019, 09:01:45 pm
content: ~3W (or is it 6W, unclear) is rather a lot to put on a linear reg these days. I would at least use a 3-terminal switching reg and a 3.3V LDO, or a dual output reg or two modules.On the other other hand, if it's just a one-off and you don't care about having a heatsink stuck to it, sure, either way. I suppose the independent rather than chained topology will be lighter on power dissipation, since you have two regs to dissipate the drop, rather than putting most of it (12 - 5 = 7V of drop) into one, times whatever the total load is (if the 50-250mA's are independent then that's 500mA peak total).Tim

author: John B
author_rank: Frequent Contributor
timestamp: October 22, 2019, 09:50:29 pm
content: If you don't mind the extra power wasted and don't want to add the complexity of a buck converter, a series resistor before the 3.3V will drop some of the voltage and dissipate some of the heat, rather than the 3.3V regulator dissipating the entirety of the power. Something along the lines of a 22R or 27R resistor. I don't know the usual dropout voltage of those parts, but I assumed around 5V input to the 3.3V reg @ 250mA.