12VAC to 12VDC, small, robust, weatherproof... - Page 1
Forum section: electronics / beginners

author: Purduephotog
author_rank: Regular Contributor
timestamp: May 27, 2020, 05:41:13 pm
content: I have a non unique problem that I was hoping to solve myself, but I've apparently used up all my 'free' time on Circuit Design and I lack the necessary knowledge to continue doing so.A new LED IR illuminator I purchased to light up the front yard is 18W- 12V, 1.5A consumed. It requires a DC power supply; the little housing for the circuit board can accommodate another circuit entirely and easily, once the proper header is added.To reiterate, the voltage coming in is 12 volts, AC. Just because I've been asked that several times.Currently I found an old salvaged bridge rectifier, I added a 25V 400uF cap, and attached that to the input. That gives me about 16V DC out, pretty well filtered, but still not ideal. Adding a power resistor to bleed off the extra 4V is a lot of wasted heat but, in the winter, would certainly keep the box from freezing.I, in all of my free quarantine time, was trying to apply the formulas for a capacitor divider as I wouldn't necessarily have to contend with all of the wasted energy. But here's where I failed- not having a fundamental grasp on these concepts (and my time experimenting and learning circuit lab) I didn't get far enough to do anything worthwhile.So, without going out and buying a buck/boost converter (which I do know are quite cheap) and having to add a bridge rectifier to it anyway to convert to DC, what other options do I have for putting together a 12V converter without wasting some 25% of the energy as heat.Thank you much-Photos of the (input to the LED light)https://imgur.com/a/DwVXWJUThe little IC that is custom made for CCTV lights- pretty cool.edit: Threads I've looked at:https://www.eevblog.com/forum/beginners/ac-voltage-drop-transformer-vs-capacitor-which-is-best/msg1417662/#msg1417662https://www.eevblog.com/forum/beginners/dps3003-power-supply-acdc-converter-output-filter/msg2074579/#msg2074579https://www.eevblog.com/forum/beginners/voltage-regulators-what-am-i-doing-wrong/msg408743/#msg408743https://www.electronics-tutorials.ws/capacitor/capacitive-voltage-divider.htmlhttps://electronics.stackexchange.com/questions/395422/how-can-i-reduce-the-rms-value-for-a-bridge-rectifier

author: themadhippy
author_rank: Super Contributor
timestamp: May 27, 2020, 06:10:48 pm
content: Quotewere did you fall down? its fairly painless,work out the equivalent resistance you need make that value equal 1/2 pi C,do remember capacitor droppers will only work on the AC side ,so wont work after the rectifier,a quick fagcalc recons 470uf will be pretty close to 6 ohms at 50 or 60hz

author: Purduephotog
author_rank: Regular Contributor
timestamp: May 27, 2020, 06:31:43 pm
content: I did do that, but (when I went to simulate it) I failed to get the expected graph (which I was assuming would be a straight-ish line with some ripple left over). Instead I saw my AC line voltage in (not swinging negative though, so I don't know what I did wrong there) and my output with one hump in the beginning, then flat zeros across.But what I couldn't determine was whether or not the amperage would work out. 1.5A isn't piddly, I know, but I'd still rather not chew it up as heat. Probably would melt the hot glue I'll end up having to use (no, I'll stuff it in a pill container and add silicone unless I can get some bad lots of RTV from work).Still the understanding theory is what's really driving all of this. I wish I could remember back to my EE electives in lab when we ddi this but Those memories are just blurred.

author: madires
author_rank: Super Contributor
timestamp: May 27, 2020, 06:44:06 pm
content: Keep the bridge rectifier and filter cap. And add a buck converter with 12V output.

author: Ian.M
author_rank: Super Contributor
timestamp: May 27, 2020, 06:51:09 pm
content: You won't get 16V under load.At 1.5A load, the 400uF capacitor's voltage drops 3.75V per us after it charges to the supply peak, so it wont hold up across the AC zero crossing. A quick sim indicates that if the current was constant the ripple trough would be under 3V. Obviously a nominal 12V DC LED light will stop working if the voltage drops too much, so what's actually going to happen, even if you add a really beefy LDO regulator to limit the max DC voltage to 12V, is the light is going to stobe at 120Hz and averaged over a mains cycle, only deliver a fraction of its normal brightness.To fix this you'll need that buck-boost module you didn't want to use, (it wont have enough headroom for Madires' buck suggestion) and at least 2200uF of capacitance to keep the ripple trough from dropping excessively. Beware of the capacitor ripple current - my sim says a 20W constant power load (i.e. your buck-boost at 90% efficiency feeding the 18W LED load) would cause a 2.7A RMS ripple current in that 2200uF capacitor, so you will almost certainly need more capacitance split between several 105 deg C low ESR capacitors with a good ripple current rating in parallel if you don't want to cook off the capacitor!If you are using LTspice, post your sim .asc file for us to debug. I previously posted myripple.ascsim in another topic - just plug in the capacitor value and load you are using. N.B. for a constant power load replace the sim's constant current load with a BI (behavioural current) source set to P=20 for 20 watts. Seehttp://ltwiki.org/index.php?title=Undocumented_LTspicefor details.

author: madires
author_rank: Super Contributor
timestamp: May 27, 2020, 06:56:43 pm
content: Yep, missed that. I'd suggest 3300µF for the filter cap.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: May 27, 2020, 07:28:00 pm
content: Capacitive droppers are only suitable for low power applications.Efficiency of LED's is is already quite "low", so some 80% of the energy you put into your LED panel is converted into heat anyway. Using a series resistor to drop another 4V in series with the LED's is not such a bad compromise if you want to keep things simple. I.E: With assumed 20% of the LED it's: 0.2*(12/16) = 15% overall efficiency, which is not such a big deal. The difference between a low quality E-core and a high quality Toroidal transformer probably already has more influence.What does the transformer look like? With some transformers it's easy to take of a few secondary windings to lower the output voltage. Adding a few windings and connecting them in "anti series" is also easy, especially for toroidal transformers. Just put the extra winding in series, and then check with a DMM if the extra voltage is added or subtracted.Also keep in mind that Mains voltage is not a constant, and the LED voltage also changes with temperature. If you want to keep the amount of light from the LED's somewhat constant, you want something to linearize it. If you want both constant amount of light and low losses then a SMPS circuit really is the best option. A 1A led driver directly from Mains is probably the best efficiency wise but a Chinese LM2596 or MP1485 SMPS module comes close.

author: Purduephotog
author_rank: Regular Contributor
timestamp: May 28, 2020, 07:10:11 pm
content: Thank you for all of this.I downloaded your sim, changed out the Ideal Current load with an Ideal Power Load (at least some things feel thesame from years of disuse)- the net file shows thisB2 vUnreg 0 P=20which I believe is the correct nomenclature.The IR illuminator is this one-https://www.amazon.com/gp/product/B0771HSGH4/, which comes with a nice power supply but not so-good for 12V AC usage.I have a couple small buck regulators that others have mentioned, and I have several stacks of caps from salvage; I have that cheap LCR/ESR Mega tester so I can probably narrow them down quickly. Going to have a very large bank of caps it seems...So basically my best shot is this, unless I want a big cap and a power resistor to burn some watts as heat (which I can attach to the side of the aluminum box at least). Might give it a nice, long-wave IR look...

author: Ian.M
author_rank: Super Contributor
timestamp: May 28, 2020, 08:20:52 pm
content: Beware of fake Chinese LM2596 modules that cant deliver 1.5A. The fakes run at approx 50KHz vs the 150KHz of the genuine chip. That's three times slower, so would need three times the inductance and capacitance for the same output current. Unfortunately the fakes usually have parts chosen as-if they had a real chip, so need output current derating by at least a factor of 3, in addition to whatever disbelief factor one applies to Chinese Aliexpress/EBAY current or power ratings! You are doing well if you can get 500mA from the fakes.To model bridge rectifier voltage drop in the sim set D1 to a silicon diode of appropriate current rating, and ctrl-right-click it and set Value2 toN=2and visible, for two Vf drops in series. Its still not 100% accurate as it doesn't allow for source impedance or transformer regulation factor, but if you are going to use a buck module rather than a buck-boost, will give you a better idea how close to dropping out of regulation you are. I *DON'T* recommend a plain buck module, but if you must, and you are scrabbling for a smidgin more headroom try the effect of setting D1 to a 3A Schottky diode model in place of an ordinary silicon one, then build your own bridge out of Schottky diodes.

author: Purduephotog
author_rank: Regular Contributor
timestamp: May 29, 2020, 08:34:52 pm
content: Ya know, I am SO sick of supply chain shit.I deal with this at work. Frankly I'm sick of Newark too, jacking up all of my orders.Where would you go to get one that is legit?

author: Purduephotog
author_rank: Regular Contributor
timestamp: May 30, 2020, 12:25:55 am
content: I couldn't believe I didn't have any capacitors in range... until... I came across a box of parts from a salvaged treadmill. Two 3300uF, 50V, capacitors. The little tester indicates 0.13ohm ESR, so I think I lucked out here.

author: Zero999
author_rank: Super Contributor
timestamp: May 30, 2020, 12:29:47 pm
content: What frequency is the 12VAC? Are you sure it's mains frequency? If it powers a halogen lamp it's possible it's an electronic transformer which will work at a much higher frequency than the mains.Stupid question: can't you simply take out the 12VAC power supply and replace it with 12VDC?

author: Purduephotog
author_rank: Regular Contributor
timestamp: May 30, 2020, 01:13:04 pm
content: I have assumed it was a 12VAC at 60hz, since it was an outdoor transformer for Halogen bulbs. I never did measure the frequency though; I think I can do that easily enough.I put together the voltage buck and hooked it up. Measured the voltage coming out at 10.4V DC, going into the light. Now the light itself feeds the LED panel at 9.6V, which I thought was interesting.Attempted to use the pot to turn up the voltage but never got anything higher than 10.4V, so I'm guessing I have a fake unit that can't handle the power draw. I'll have to finangle another one.Then again, working 16 feet in the air is a bit tedious at night.

author: Zero999
author_rank: Super Contributor
timestamp: May 31, 2020, 07:52:15 am
content: Quote from: Purduephotog on May 30, 2020, 01:13:04 pmAn electronic transformer has a high frequency oscillator, enabling a much lighter, cheaper transformer to be used, than would be required for one running at the mains frequency. They're designed for powering halogen lamps, so it's quite likely you have one.https://sound-au.com/lamps/elect-trans.htmlWhy not simply replace the transformer with a 12VDC power supply? The halogen lamps will work of DC, just as well as AC, so it makes more sense to use a single DC power supply for both.

author: Purduephotog
author_rank: Regular Contributor
timestamp: July 09, 2020, 11:22:50 pm
content: Well, it is a 12V AC power supply... a huge one. But I could swap it out with a smaller one, but even then I don't know if it'll supply DC or not.Or do you mean getting an enclosed unit? I guess I could. I have enough PSUs around that in the short term I certainly could, and Meanwells aren't cheap. Wiring up to code, however, might be more expensive (dual isolating boxes).J

author: calzap
author_rank: Frequent Contributor
timestamp: July 10, 2020, 02:28:04 am
content: I faced a similar problem in my firewood shed. Had 12 VAC available, but wanted an LED for lighting. 12 VDC screw-base LED bulbs were readily available. So, I decided to risk it. Has worked perfectly for years. Maybe just lucky. Might not work for you.Mike in California

author: bdunham7
author_rank: Super Contributor
timestamp: July 10, 2020, 03:58:36 am
content: Can the LED driver simply tolerate the unregulated supply? It should be dropping to 13-14 volts at most under load anyway, and the LED driver should be able to manage that.

author: madires
author_rank: Super Contributor
timestamp: July 10, 2020, 09:23:44 am
content: If the OP would post pictures of the PCB we might able to figure out if his LED IR illuminator would also run with AC or with a higher DC voltage.

author: Zero999
author_rank: Super Contributor
timestamp: July 10, 2020, 10:13:35 am
content: Quote from: Purduephotog on July 09, 2020, 11:22:50 pmIf it's a huge one, then it's more likely to be a transformer.Cost is an interesting point. Do you know, if you simply replaced all of the halogen lamps with retrofit LEDs, you'd get the money back within a year, in the form of lower energy costs? The only disadvantage is they will be more prone to icing in the winter.Anyway, that's off-topic. To answer your original question, the easiest way to do this is use a low-dropout regulator, such as the LT1085-12 or LM1085-12, on a good heat sink, with a Schottky diode bridge and a large smoothing capacitor.

author: Purduephotog
author_rank: Regular Contributor
timestamp: January 28, 2023, 04:55:14 pm
content: I will post the shots of the board I took a while back- my apologies for not catching this.I also was going to update that everything is STILL working fine- no issues- no flicker- years since. And we haven't had the best weather.