Buck converter or LDO for 9v to 5v for a Teensy codec project - Page 1
Forum section: electronics / beginners

author: jvphotog
author_rank: Contributor
timestamp: August 01, 2025, 09:59:23 pm
content: Hi all. Im making a board with a Teensy 4.1, PCM3168A codec, THAT PGA2500 mic preamp, OPA1656 preamp, and possibly an OLED, encoder, and pot. So far this is what I have budgeted.Power Domains:5v Digital - Teensy, OLED - 300mA5v Analog - THAT PGA2500, PCM3168A, OPA1656 - 300mA-5v Analog - THAT PGA2500( and digital?), OPA1656 - 100mA3v3 Digital - PCM3168A, Encoder, MV Pot - From Teensy(250mA max)The PGA2500 uses a -5v supply for its VD. I plan on using the Teensy's 3.3v converter for my 3.3v needs. I have a 9v power source. Would it be better to use a buck converter to bring that down to around 5.5v and then use LDO's for the three 5v supplies? Or just use LDOs and keep things simpler? Thanks!

author: lakis70
author_rank: Contributor
timestamp: August 02, 2025, 02:02:04 am
content: you need a transformer 2x5V AC 1.2A, rectify and take from the bridge 5V, -5V and regulators for +5V and another for -5V and from 5V a ldo for 3.3V

author: MrAl
author_rank: Super Contributor
timestamp: August 02, 2025, 09:46:00 am
content: Quote from: jvphotog on August 01, 2025, 09:59:23 pmHello,It's a question of efficiency. If you are running from a battery that usually means you want maximum efficiency too.Because the voltage ratio is so high 9/5 which is almost 2, a buck regulator will always provide more efficiency. It's only when you get close to the input voltage (like 8v with a 9v power source) when you might see an LDO or other linear regulator get either nearly the same or better efficiency, since the efficiency of the buck is never 100 percent.So going from 9v to 5.5v would be good with a buck, then you might use a second buck to go from 9v to 3.3v because going from 5.5v to 3.3v is quite a bit of a stretch too.If you don't care about efficiency, anything would work.To find out the difference between a buck and a linear regulator, just calculate the efficiency of the buck and compare that to the efficiency of the linear regulator.An LDO is the same as any linear regulator, it just WORKS at a lower input voltage, it's not more efficient than any other linear regulator.

author: BrianHG
author_rank: Super Contributor
timestamp: August 02, 2025, 09:33:33 pm
content: The other advantage to a buck is if the IC you choose uses a N and P mos push-pull switch architecture, it is possible to generate your negative 5v supply from the same buck's output with an added cap, two second diodes and coil. Though, to keep good regulation, your +5v and -5v need to draw approximately the same current.This means you only need a single positive supply.And, opamp supplies do not necessarily need to be perfectly regulated, they just need to be low noise.Other switching typologies/ICs/wiring exist to generate negative supplies, it depends on how many parts you want to have.

author: MrAl
author_rank: Super Contributor
timestamp: August 03, 2025, 05:44:18 pm
content: Quote from: jvphotog on August 01, 2025, 09:59:23 pmHello again,Here are two text files with information about using buck vs linear.The first includes all combinations of buck and linear comparisons.The second only includes combinations that are better with a buck.Look at the first three columns for your input voltage and output voltage, and output current and buck efficiency rating. Then follow across to find out which is better.If you do not find your exact voltage estimate from the next lower and next higher voltages.For example if you have 12v in and 5.5v out and buck 60 percent efficiency, you'll find these two lines:Vin Vout Iout IinL IinB BEff AEL AEB PIBL12.00 5.00 0.10 0.10 0.04 0.60 0.42 0.60 + 44.012.00 6.00 0.10 0.10 0.05 0.60 0.50 0.60 + 20.0Since the AEL (Actual Efficiency for the Linear) for 5v is 0.42 and for 6v is it 0.50, add the two we get 0.92 and divide by 2 we get 0.46 for the AEL for 5.5 volts out.If that's too hard quote your requirements and I'll post the results here.Funny how often this question comes up.

author: BrianHG
author_rank: Super Contributor
timestamp: August 03, 2025, 10:06:19 pm
content: Just use a multiple output switcher...Example:https://www.ti.com/lit/ds/symlink/tps65135.pdf?ts=1754247675881&ref_url=https%253A%252F%252Fwww.ti.com%252Flit%252Fgpn%252FTPS65135More powerful one, good for a lithium battery, boosts 3.3v through 5v input to +/- 5 through +/-12v out.https://www.ti.com/lit/ds/symlink/tps65131-q1.pdf?ts=1731607105501

author: jvphotog
author_rank: Contributor
timestamp: August 04, 2025, 04:42:13 am
content: Thanks for the responses guys. I don't think that Ti unit will work, unless I step down to 5v first.It's interesting to see how close the LDO can be to the buck in efficiency.I'm powering this off if an ICEamp that has 9v aux power. It's rated for 700mA, so I have plenty of headroom. The ICEamp ALSO has an aux +/- 17v supply, presumably for mic pre amps, but the PGA2500 I want to use needs a -5v supply. So right now, I plan to take the 9v and use an LDO for the digital 5v supply. Then 9v into an inverter to make a -9v. And then from there use 2 LDOs for the bipolar analog supplies. The Teensy will supply the 3.3v for the codec, pot, encoder, etc. But Im having a heck of a time trying to find a part that will work off the EasyEDA site - where Im having my board built and assembled. Im sure you can tell, but im very new to this world, and it seems that component selection takes the majority of the time!

author: PGPG
author_rank: Frequent Contributor
timestamp: August 04, 2025, 08:47:05 am
content: Quote from: jvphotog on August 04, 2025, 04:42:13 amThe efficiency for LDO is simply little less than Vout/Vin. so for 9V->5V it is little less than 55%.For buck I would expect efficiency about 80%.TO220 or DPAK case (without radiator) with 1W dissipated in it got enough hot that you can't touch it for long time.Whenlinearlyregulating 5V/300mA from 9V you have to dissipate 4V*0.3A=1.2W.But to get 5V x 300mA+300mA+250mA (to make then 3V3) you have to dissipate 4V*0.85A=3.4WQuote from: jvphotog on August 04, 2025, 04:42:13 amYou are absolutely right.I remember years ago (when you used paper catalogs and not internet) to spend few weeks searching the regulator to use to get 5V from 12V with maximum efficiency. The main problem was that average load at 5V was 10uA and very short times (like minute per day) 100mA.I found it first day but as it was 5 (or more) times more expensive than 7805 I hoped to find cheaper one. And few weeks later I had to accept that there is no other in comparable price.I used LDO with quiescent current of 5uA so I got efficiency of (5x10)/(12x15)=28% and I think there is no more effective solution.

author: MrAl
author_rank: Super Contributor
timestamp: August 04, 2025, 05:42:06 pm
content: Quote from: BrianHG on August 03, 2025, 10:06:19 pmHi,Just for example, if you input 9v and output 5v an LDO or any linear will be 56 percent efficient, while a buck specified at 80 percent efficient will show 80 percent efficiency.

author: jvphotog
author_rank: Contributor
timestamp: August 06, 2025, 03:48:18 pm
content: After looking for a few days, I think this might be one of my best options:TPS63710https://www.ti.com/lit/ds/symlink/tps63710.pdf?ts=1754456450818&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FTPS63710, and then run that through an LDO. Or something like an LT1931Ahttps://www.analog.com/media/en/technical-documentation/data-sheets/1931fa.pdf, which would need 2 inductors.Or I could do something like an AM3G-0505DLPZhttps://www.digikey.com/en/products/detail/dcomponents/AM3G-0505DLPZ/25596375, which will do -5v, if I can get away with just filers and no LDO. OR a LMS78_05-0.5Rhttps://gaptec-electronic.com/datenblaetter/LMS78_0.5R.pdfis pretty cheap and easy to solder, but I don't know if it will be quiet enough with just a pi filter.

author: jvphotog
author_rank: Contributor
timestamp: August 11, 2025, 02:31:56 am
content: Anyone willing to take a look at my schematic so far? I found an Altium video on YT that made a -5v supply from +9v, so I copied a lot of that section from the video. Thanks!

author: MrAl
author_rank: Super Contributor
timestamp: August 11, 2025, 05:01:56 am
content: Quote from: jvphotog on August 11, 2025, 02:31:56 amWhat are you using that 9vdc at 700ma supply (upper left corner) for?

author: jvphotog
author_rank: Contributor
timestamp: August 11, 2025, 03:05:15 pm
content: I'm powering this off if an ICEamp that has 9v aux power - It's rated for 700mA

author: MrAl
author_rank: Super Contributor
timestamp: August 11, 2025, 05:04:01 pm
content: Quote from: jvphotog on August 11, 2025, 03:05:15 pmOh you need that to be super noise free or something?If it is critical, you may have to go with a differential filter, if that is possible for this project.