Linear PSU Design - Page 1
Forum section: electronics / beginners

author: Sythen
author_rank: Contributor
timestamp: November 11, 2013, 09:45:20 pm
content: Evening all!Im in the process of designing a PSU for a project around the RTLSDR.The design specs i need to meet is that its linear based (no switching noise)it can be powered from a pack of AA cells (in this case, 2 packs of 8 AA cells in parallel ), or a 12v lead acid battery.and its output needs to be nice and clean, and it needs to supply at least 1 ampIve put something together, but due to my inexperience in electronics, I'm unsure if I've made any mistakes, or any bad design choices.Any help or pointers would be greatly appreciated.Thanks in advance!

author: fake-name
author_rank: Regular Contributor
timestamp: November 11, 2013, 09:51:17 pm
content: Couple comments.First, the fuse should be on the input side of the regulator, not the output.Also, if you want to drop 12V-5V @ 1A, you're going to be dissipating 7W. You willHAVEto have a heatsink. You should make provisions for heatsinking the regulator.

author: Sythen
author_rank: Contributor
timestamp: November 11, 2013, 09:53:48 pm
content: Hi fake-name, thanks for the input, i intended to add a heatsink to the reg regardless,My intention was to get the design of the circuit correct first, then modify the design to incorporate the heatsink.The project itself wont draw over 500mah, but i also wanted to encorporate a usb charging function, so the PSU can be used to charge USB Devices during travel or a power outage.Thanks much!

author: mariush
author_rank: Super Contributor
timestamp: November 11, 2013, 10:32:06 pm
content: My two cents... the layout is simply bad. For such a simple pcb, there's no reason why this board should be double layer.LT1086 needs about 1 above the 5v output to give you proper 5v, so with 4 alkaline AA batteries, you're gonna have 4 x 1.65 = 6.6v when charged, 4 x 1.5v = 6v - the batteries won't even discharge completely and you're going outside regulation. Forget about rechargeable batteries if you want to stay within 4 batteries.There are much better linear regulators, that only need maybe 100-300mV above the output voltage to give you good quality 5v, so that would allow you to use even rechargeable batteries. The only thing is those that have low vdrop don't like high voltage at the input (most like up to 6-6.5v , maybe 8-10v at best) so you can fix the problem of using 12v as input by putting a cheapo mc34063 to get that 12v+ down to about 6v to the linear regulator.Also, when you have the linear regulator powered from maximum 6-6.5v, at 500mA you'll only waste about 1w which can be safely dissipated into the pcb copper. If you have it double sides, you can have some vias going to the other side and have a double sided heatsink basically.For example, here's some linear regulators that would work great for you :mic37152wr 1.5a max, max 6v input voltage (absolute max 6.5v)... about 250mV drop at 500mAhttp://uk.farnell.com/micrel-semiconductor/mic37152wr/ldo-0-5vdo-1-5a-1-24-5-5v-5spak/dp/2219082mic5219ymm 500mA peak but i wouldn't use it for long periods at that high current as it doesn't heatsink well. on other hand, up to 12v input with no problems.http://uk.farnell.com/micrel-semiconductor/mic5219ymm/ldo-reg-12vin-0-5a-adj-8msop/dp/2100325edit:had this at top, but looking more carefully, this ldo needs a Vbias (powering the internals, using up to 33mA at 1.5A output) that has to be 1.6v above the output voltage so it's not quite suitablemic49150wr adjustable, 1.5a max, max 6.5v input voltage (but absolute max 8v) , about 100-200mV drop at 500mAhttp://uk.farnell.com/micrel-semiconductor/mic49150wr/ic-v-reg-ldo-dual-i-p-1-5a-49150/dp/1100650Anyway, there are also some LDO chips from Microchip that are just great, but can tolerate only a maximum of 6 volts at the input.

author: Sythen
author_rank: Contributor
timestamp: November 11, 2013, 10:41:25 pm
content: Hi mariush, and thanks for your input, 12v input is a requirement I'm afraid. for powering the device from AA cells, i was going to use 2 x 8 cell packs run in parallel, and 1 amp output is required aswell, that's why i specified the LT1086CT-5, as it can provide more than is required for the main project, but can also be used in other applications.The component layout was something i tossed together in about half an hour, its not a final design, it just helps me visualize how the board is connected electrically.Thanks for the input though!

author: AmmoJammo
author_rank: Frequent Contributor
timestamp: November 11, 2013, 10:54:38 pm
content: By powering it from a battery, you effectively have no ripple voltage... The capacitors aren't really doing anything.

author: Sythen
author_rank: Contributor
timestamp: November 11, 2013, 11:03:55 pm
content: I would still need a cap on the output wouldn't i?

author: dr.diesel
author_rank: Super Contributor
timestamp: November 11, 2013, 11:08:32 pm
content: Quote from: Sythen on November 11, 2013, 11:03:55 pmNope, not with a linear reg.

author: Sythen
author_rank: Contributor
timestamp: November 11, 2013, 11:11:14 pm
content: mmmm, ok, ill update the pcb design.

author: AmmoJammo
author_rank: Frequent Contributor
timestamp: November 11, 2013, 11:11:48 pm
content: Quote from: dr.diesel on November 11, 2013, 11:08:32 pmIn other words, you don't need a PCB, you just need 3 bits of wire, some heatshrink, and a heatsink

author: Sythen
author_rank: Contributor
timestamp: November 11, 2013, 11:18:42 pm
content: Yeah, that would work, but id rather have something tidy and on a module that i can drop into an existing box i have.

author: dr.diesel
author_rank: Super Contributor
timestamp: November 11, 2013, 11:24:04 pm
content: You'll need a heck of a heat sink with an input voltage of ~13.6v (normal lead acid battery voltage) regulated down to 5v at 1 amp.Not to mention you'll be wasting a HUGE about of battery on generating heat.

author: Sythen
author_rank: Contributor
timestamp: November 11, 2013, 11:27:58 pm
content: The project this will be powering draws around 450mah, but i want to add USB charging functionality to this PSU, and that is intended for emergencies. I understand that theirs going to be wasted power in forms of heat, but because this is intended for a radio project, i need a clean supply.

author: dr.diesel
author_rank: Super Contributor
timestamp: November 11, 2013, 11:32:36 pm
content: With a linear reg, 13.6 volts in - 5v out = 8.6v * .450ma = 3.87 watts dissipated as waste heat.For USB charging a switcher with a proper inductor would be a better solution.

author: Sythen
author_rank: Contributor
timestamp: November 11, 2013, 11:35:10 pm
content: Indeed it would be, but i require a clean source of power, switch mode is out of the question.

author: AmmoJammo
author_rank: Frequent Contributor
timestamp: November 11, 2013, 11:38:10 pm
content: why are you even asking? do what you like, you seem to have all the answers anyway.

author: dr.diesel
author_rank: Super Contributor
timestamp: November 11, 2013, 11:38:45 pm
content: Dissipating 4 watts via heat your batteries will be toast in no time, depending on what type your using.If you need any sort of life you might consider a switcher down to ~6.5v, then the LDO down to 5v.

author: AG6QR
author_rank: Frequent Contributor
timestamp: November 11, 2013, 11:43:26 pm
content: You don't need a dual layer. You can easily route your orange traces so that they don't have to go over/under the yellow ones. Hint: take the trace from the regulator output down instead of up before going to the right.For mechanical rigidity, your regulator should have some physical support besides the three pins. That's especially true if it's going to have a heat sink on it, which it will need, as has already been pointed out. You need to pick a heat sink that will handle the amount of heat dissipation you'll be doing, and then support that heat sink on the board. Depending on your heat sink choice, you may want to bend the regulator leads and use a screw to go through the regulator, heat sink, and board. Or you might want to leave the regulator vertical, and bolt the regulator to the heat sink, while using an appropriate mechanism to secure the heat sink to the board.The board will work for a while without paying attention to this detail. But over time, mechanical stress on solder joints is not good for reliability.

author: kizzap
author_rank: Unknown
timestamp: November 11, 2013, 11:44:49 pm
content: Quote from: dr.diesel on November 11, 2013, 11:32:36 pmRead the OP, would you seriously suggest a switching power supply to someone with very little experience in electronics? Take a look at the board layout too, again, really?@Slythen:Firstly, I would be leaving the capacitors in there. Primarily so that if they are needed, it is just a simple case of putting them in.Second, if you rotate that voltage regulator 90 degrees you can get a PCB with nice symmetrical components on it.Third, the voltage measurement points aren't really that necessary unless they are going to be permanently wired to something, you can easily measure the voltages at the input connections, either at the solder joints, or even on the voltage regulator itself.-kizzap

author: mariush
author_rank: Super Contributor
timestamp: November 11, 2013, 11:46:31 pm
content: Some linear regulators insist on at least 1uF or 10uF at the output, some are only stable if the ESR of that capacitor is between 0.1 ohm and 1ohm (mic2941 or lm2941 for example). It won't hurt to add a ceramic capacitor at the output , they're cheap.You'll be wasting a lot of energy going down from from 12-14v to 5v, it's just pointlessly wasting the battery. A MC34063 with the inductor and diodes will be less than 1$ and good enough as a sort of pre-regulator, taking down the voltage to 6-7v without losing so much power. From there, the linear regulator can do the job, cleaning the output voltage further.USB charging is stupidly simple, just a few resistors on the tx/rx lines in the usb jack. Seehttp://learn.adafruit.com/minty-boost/ichargingThe USB has a particular tolerance, the voltage can be +/- 5-10% so pretty much anything is supposed to accept 4.7v-5.3v without complaining. There's also the issue of voltage drop on long usb cables so some usb chargers actually output 5.1v-5.2v to counteract that. The point I'm trying to make is that you don't need a linear regulator to provide smooth stable output on a usb port meant for charging stuff. A simple basic switching regulator will do just fine.

author: dr.diesel
author_rank: Super Contributor
timestamp: November 11, 2013, 11:48:18 pm
content: Quote from: kizzap on November 11, 2013, 11:44:49 pmYes, cause the OP mentioned 1 amp min which would generate over 8 watts, not acceptable.

author: Sythen
author_rank: Contributor
timestamp: November 11, 2013, 11:56:01 pm
content: Ok, to keep this simple for me, and to stop my noggin from being overloaded, the USB charging stuffs is out the window.I chose a 12v source because its findable if your in a car, and AA batteries are easy to find, and are cheap aswell. the 1 amp requirement is there so i know I'm going to have enough juice to power the project. the project draws about 450mah, so i dont mind having some losses in heat, the main aspect is to have a super clean 5v supply, with power sources that are common as dog poo, and are cheap.

author: sleemanj
author_rank: Super Contributor
timestamp: November 12, 2013, 01:58:21 am
content: Quote from: Sythen on November 11, 2013, 11:56:01 pmCareful there.A 12v car battery is NOT 12 volts. Normal resting state could be as much as 13.5vOnce the car is running, 14 to 14.5v will be seen because of the alternator running.And that doesn't take into account the various very (VERY) high transients that can and do occur in automotive power systems. Automotive is a hostile environment for electronics.

author: codeboy2k
author_rank: Super Contributor
timestamp: November 13, 2013, 12:25:07 pm
content: Quote from: Sythen on November 11, 2013, 11:27:58 pmThis doesn't make sense. Many radios these days are powered by switch mode regulators. And USB charging doesn't need super clean power either.And the other posters' comments regarding power dissipation going from 13.8V to 5V are quite correct, and not to be ignored. You are dissipating about 8.6 watts max at 1A load, and if you want to design for that, you need to keep the device's junction temperature no more than 120 deg C, and preferably far below that if you can. So lets say you target a junction temperature of 100C ... if ambient is 25C, then you are allowing a temperature rise of 75 deg C. So, to figure out your total thermal resistance allowed, you divide temperature rise/power dissipation, which is 75C / 8.6W = 8.72 degC/Watt. The LT1086 has a Rthjunction-caseof 4 deg C/W and the silpad will add 1 deg C/W. So, 8.72 - 4 - 1 = 3.72 deg C / Watt is the thermal resistance of the heatsink.The common, tiny, heatsink in my picture below won't be good enough...it's about 30 degC/watt, and can maybe handle 2 Watts (3 Watts with a fan)The one below is 3.3 degC per Watt. It's the oneyou need if you want to pull 1A from your power supply:And this is why everyone is suggesting you go with a switch-mode PSU for this purpose.. just look at the size of the heatsink you need. And if you really, really don't care about the wasted power and the excess space taken up by the heatsink and the heat it will give off, well now you know what you will need for a heatsink.

author: dr_p
author_rank: Frequent Contributor
timestamp: November 13, 2013, 01:06:16 pm
content: Quote from: sleemanj on November 12, 2013, 01:58:21 amQuote from: codeboy2k on November 13, 2013, 12:25:07 pmfor the OP: I'd be careful there too. Depending on where it is placed in the car or how sunny it is, ambient can be 40-50C or even higher. I've been burnt once (pun intended) so now for the stuff I build for my car I'm considering battery voltage 7...25V and ambient -20...50C.