pure linear vs. stepdown + linear - Page 1
Forum section: electronics / beginners

author: ledtester
author_rank: Super Contributor
timestamp: June 28, 2017, 01:51:14 pm
content: I'm curious about things to be aware of when comparing the following methods to obtain 5V from a 12V supply:- using just a linear regulator- using a cheap buck converter module (like you can find on ebay) to first step down the 12V followed by a linear regulatorI'm aware that the pure linear regulator approach will generate more heat, waste more power, etc., but what about the quality of the supply? Are there any other operational things I should be aware of?You can assume I'll need < 500ma from the 5V supply.

author: MosherIV
author_rank: Super Contributor
timestamp: June 28, 2017, 02:27:10 pm
content: HiSorry, you are not being clear on what you are comparing :QuoteDo you mean a straight 5V linear PSU with regulatorQuoteWhat is the 'cheap buck' stepping down from? You cannot simply put mains voltage into a buck regulator (not that I am aware of)It may be that the Chinese vendor has no idea what they are talking about and have labelled a SMPS as a buck.QuoteThe amount of power here is not significant.12V-5V = 7V7V * 0.5A = 3.5W to dissipate by the linear regulator. Perfectly doable'using a buck converter module to first step down the 12V followed by a linear regulator'will not result in a 'clean' PSU like a linear PSU, you still have to heavily filter the buck output. Even then there will still be a small amount of switching noise.It is not economical to make commercial PSUs like this, it may be worth while for a home project there was a project somewhere on the forum the other day.

author: alm
author_rank: Super Contributor
timestamp: June 28, 2017, 03:12:30 pm
content: A basic 7805 with a sufficiently large heat sink should work fine here (Dave did an old EEVBlog video on thermal calculations). The amount of noise should be fine for most circuits, unless you are powering something extremely sensitive like an EEG amplifier (I hope not!). A 12 V to 5 V buck converter will be more efficient and smaller, but as Mosher explains, the output will most likely be noisier, especially the cheap $1 ones from eBay. A buck converter to step it down to something like 7V followed by a linear regulator sounds overcomplicated for this fairly low current. Also you can not necessarily expect a linear regulator to suppress all noise from the buck converter. Ripple rejection is usually excellent at 100/120 Hz but drops off rapidly at higher frequencies. There are linear regulators specifically designed to filter noise from switching power supplies.Any details on what you are powering? How sensitive is it to noise? How clean is the 12 V source? If we are talking automotive 12 V, then you would need more than a simple regulator to survive all the transients. Is available space for a heat sink an issue? Costs?

author: schmitt trigger
author_rank: Super Contributor
timestamp: June 28, 2017, 05:22:05 pm
content: I've seen this technique when stepping down from 24 or 48 volts to 5 volts. In that instance a SMPS is highly advised, if not mandatory.If you simultaneously require a lower noise level, then a downstream linear is used. Preferably a LDO.Important Caveat:Many linear regulators do not have the sufficiently wide bandwidth to attenuate the high frequency noise from a SMPS. You may find that an LC filter may still be required.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: June 28, 2017, 05:38:27 pm
content: Quote from: ledtester on June 28, 2017, 01:51:14 pmQuality will depend on how effective your linear reg (in either case) is when it comes to taming the ripple in the source. 120 Hz ripple (like you get from an AC/DC wall wart) will be cleaned up nicely by most linear regs. 100+ kHz ripple (like you get from a switching regulator) will pass straight through most linear regs, but there are some out there that can handle it well (eg: LT3042). When selecting a linear reg, you should pay close attention to PSRR vs frequency (all linear reg datasheets should have this figure).Essentially you're just trading off efficiency for cost/complexity. Nobody can tell you that either approach is a waste of time, or that the wasted power is insignificant, or if it's not economical, without looking at the entire picture. For a small battery-powered system that needs a clean rail for high precision ADC work, it may make perfect sense to use a switcher+LDO. You're the one that has to make that decision.If you pick the right components you can end up with a clean rail using either approach, but you have to pay attention to specs, especially in the switcher+LDO case.

author: Dave
author_rank: Super Contributor
timestamp: June 28, 2017, 05:44:48 pm
content: Quote from: schmitt trigger on June 28, 2017, 05:22:05 pmJust to confirm, here's the PSRR graph for the good old LM317. At the frequencies where a switcher would operate, you barely have 10 to 20 dB of suppression. Consider the higher harmonics and you will find that there will barely be any attenuation for those at all.To be fair, LM317 is quite old and the specs are not the best, but you get the point.This happens to be my 1000th post. \o/

author: Red Squirrel
author_rank: Super Contributor
timestamp: June 30, 2017, 02:43:23 pm
content: I've used a linear regulator to go from 12v (well around 13.5-14v) to 5v and it barely gets hot. It's only powering a MCU, LCD and a relay though. My heat sink is very crude, I folded it over the protoboard, used a few leads from cut resistors to create a mesh and put a bunch of solder. Stays cool.

author: ledtester
author_rank: Super Contributor
timestamp: June 30, 2017, 06:47:03 pm
content: Thanks everyone for the info - it's all really helpful.The application I had in mind was supplying 5V for a function generator where the +/- 12V comes from a linear supply. Given the concerns about the noise from the buck regulator, I guess I'll just get the 5V from a linear supply (either from the 12V supply or from another transformer.)