Transforming mains voltage to two rails (48V and 5V), need pointers. - Page 1
Forum section: electronics / beginners

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: March 31, 2015, 09:46:59 pm
content: So, I need to step down (in my case) 230V AC to one 48V DC rail a another rail of 5V DC.I'm currently looking at the option of using one of these5V AC DC convertersand one of these48V AC DC converters.But, this means that I'd spend approx $45per boxjust for the power supply. Seems a bit much to meI have not yet had anything to do with this particular part of a project (all I ever have done is to run micros off of USB chargers...).Could I do this by using a transformer, then rectifying it into a voltage divider resistor, then two separate voltage regulation parts and say "Bob's your uncle!"?Off the cuff (ugly, sorry) schematic:Questions:What would be the cheapest solution??A: Is a 50V 470uF ok here??B: How would I go about regulating 48V? Isthisreally needed??C: Should I wire circuit ground to earth ground? With or without capacitance?What are important figures when looking for a transformer, I have no idea what to look for...Any and all help/pointers much appreciated!

author: TimFox
author_rank: Super Contributor
timestamp: March 31, 2015, 10:05:16 pm
content: To get a useful answer, you first must specify the maximum current required at the +5 V and +48 V rails.

author: mariush
author_rank: Super Contributor
timestamp: April 01, 2015, 12:03:56 am
content: If you want regulated 48v DC, the actual voltage must be a bit higher than that, at least by 1-2 volts. So, after the bridge rectifier and that capacitor (the 470uF in your picture), you need to have at least 50v.The transformer outputs AC, which when rectified using those diodes creates a DC output with a peak voltage equal to about 1.414 x Vac .Also, there will be a bit of voltage loss on the diodes inside the bridge rectifier, usually about 1-1.5v.Also, remember it's only a peak DC voltage of 50v, what comes from the transformer are still some waves of voltage going up and down, and that's why that capacitor is there in the circuit - when the wave goes up towards 50v it fills up and when the wave goes down towards 0v, the capacitor gives the needed energy to maintain the voltage constant. The capacitor should be picked to keep the voltage above the minimum required to keep the 48v regulator happy, which is 50v.So, let's say 50-52v is what you want, in which case you add those 1-1.5v lost in the bridge rectifier and you have a maximum of about 54v. Now if you divide this by 1.414 you know you need a transformer that outputs around 38v AC.In most cases, you'll be fine going with a standard 36v AC transformer or one with two 18v secondary windings that you can put in series.That capacitor should definitely be at least rated for 63v, if you're dabbling with 48v+.--To regulate the 48v, there are plenty of linear regulators which can tolerate a high voltage at the input, for example LM317 withHV(or similar endings in the name) can work with up to 60v. So you just use two resistors to adjust the output of this linear regulator to 48v and you're done.Here's some suggestions, just pick the one that can do your output current :http://uk.farnell.com/webapp/wcs/stores/servlet/Search?catalogId=15001&langId=44&storeId=10151&categoryId=700000004335&sort=P_PRICE&pageSize=25&showResults=true&aa=true&pf=110421493,110421580,111680855,111680864,111680866&min=110421580To get 5v, it's kind of a bad idea to use a linear regulator because lin. regs. dissipate the difference between input and output voltage as heat.. so for example at 100mA, you'd produce (48v-5v)x0.1A= 4.3w of heat.You can buy dc-dc converters from eBay that can work with high voltage (the standard ones usually can only handle up to about 40v), you just have to use the right keywords. Or, you could pick up a switching regulator chip and look in the datasheet and buy the other components needed and make yourself a switching regulator for 5v.Alternatively, if you buy a transformer with two secondary windings, then you could use one and put its own bridge rectifier and capacitor and you'll have the more manageable 25-30v and then you can use really cheap dc-dc switching regulaor chips like MC34063 (dave did videos on how to use one and there's online calculators for it) or you can buy one of those car cigarette usb chargers and pull out the insides and use them right away.No, it's not a smart idea to use resistors as voltage dividers, it can make the resistors very hot (and they're small and not easy to cool) and it generally doesn't work right.

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: April 01, 2015, 12:05:21 pm
content: Wow, thank you both for replyingFor the current needed, it's about 500mA for the 5V rail and not more than 150mA for the 48V rail (it will be used to phantom power 8 microphones;).I might just cut my losses, I'll probably only, make about 20 of these things (remote controlled XLR passive summers) so if I don't feel confident I can make the power supply I'll just buy the converters.I'll investigate further, and if anyone has more pointers I'm all ears!

author: TimFox
author_rank: Super Contributor
timestamp: April 01, 2015, 12:46:57 pm
content: Quote from: alexanderbrevig on April 01, 2015, 12:05:21 pmThe 1k series resistor from the 48V rail to the input of the 7805 is much too high: the 500 mA needs to go through it, and that is a 500 V drop. If you use the 7805 as in the drawing, replace the 1 k resistor with a series string of seven 5.1 V Zeners to drop the input voltage to roughly 13 V, and change the 130 ohm resistor to 10 k. Each zener needs to dissipate 2.6 W at 500 mA, so I would recommend 5 W devices such as the 1N5338 ($0.54 in singles from DigiKey). Also, increase the 0.1 uF capacitor at the 7805 input to 0.47 uF, since the manufacturer requires at least 0.33 uF mounted as close as practicable to the pins. Dropping the voltage to the 7805 is very inefficient, however.

author: AndreasF
author_rank: Frequent Contributor
timestamp: April 01, 2015, 01:04:40 pm
content: I think you might be much better off getting a decently powerful (2A), off-the-shelf 5V DC power supply and then step-up the 5V to 48V with a boost converter.

author: Ian.M
author_rank: Super Contributor
timestamp: April 01, 2015, 01:51:20 pm
content: Since its for microphone system, that may run into noise issues. However the fundamental idea is sound. Start from a convenient low input voltage. Boost it to approx 50V and also buck it to 7V with switching converters inside a screening can. Drop the last two volts in linear regulators to get low noise 5V and 48V rails.

author: mariush
author_rank: Super Contributor
timestamp: April 01, 2015, 02:09:41 pm
content: 50v x 0.150mA = 7.5w ... if you get this from a lower voltage add 1.5w for conversion losses and you're looking at 9w / 5v = 1.8A just for that ghost power for microphones.If you go this route it would make more sense to start with a 7.5v - 9v @ 2A DC adapter ... it means you'd only use up to a more convenient 1-1.2 amps for ghost power, and leave you with 800mA-1000mA available to get 5v. A basic 1117 linear regulator will give you stable 5v with as little as 6v so you'd be fine.Boosting such low voltage to 48-50v isn't quite straight forward, but there are chips or dc-dc converter boards out there.

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: April 01, 2015, 02:49:56 pm
content: Using a step up converter and an off the shelf 5v 2A tranformer and a boost converter seems like a nice option!Would this device do the job?http://www.digikey.com/product-detail/en/LTC3872ETS8%23TRMPBF/LTC3872ETS8%23TRMPBFTR-ND/1620506Reference design for 48V:Also, would it generate a lot of heat, or it most of the current simply used to boost voltage?Since it will be used on a stage I kind of wanted to use the standard C14 as cables for this is plentiful. Though seeing as I'll probably be the only one on the planet using this thing I can live with bringing along a few adapters. Maybe an adapter box (ie 230V in and 5 x 5V out) would be a nice next project?

author: mrkev
author_rank: Regular Contributor
timestamp: April 01, 2015, 03:32:18 pm
content: Quote from: alexanderbrevig on April 01, 2015, 02:49:56 pmI think it would.Quote from: alexanderbrevig on April 01, 2015, 02:49:56 pmNo. Switch mode DC-DC converters are used because they have high efficiency (>85%), which means, that you are usingalmost all the power that goes in for what comes out. Linear regulators (like 7805) regulate voltage by adjusting the voltage drop over themselves, so a lot of power is consumed by the regulator itself (that power is changed to heat).On the other hand, this is something that was usually done by voltage multiplier. So they would use something like 2x8V transformer with 4x Villard cascade multiplier. After all, current needed for the phantom powering of one mic is usually less than 10mA.If you wanna make the more effective and maybe easier way, just use good low pass filter at the output.

author: TimFox
author_rank: Super Contributor
timestamp: April 01, 2015, 03:36:58 pm
content: Assuming the phantom power requirement is very low, the voltage-multiplying rectifier with appropriate capacitors will probably have less noise than the switching power supply. Proper filtration after a SMPS for a noise-sensitive application is an engineering project of its own.

author: mariush
author_rank: Super Contributor
timestamp: April 01, 2015, 03:41:40 pm
content: As my previous post says, remember you can't make energy out of nowhere. If you want 48v at up to 150 mA, you make it with a circuit that has a particular efficiency, which is usually in the 85-90% .. those 10-15% usually show up as heat, waste. So 48 x 0.15A is 7.2 watts and you're probably going to have about 1w lost due to poor efficiency, so overall your input must be able to provide you with 8w just for the 48v part.1w is relatively easy to be dissipated in the pcb and it's spread in the parts around the ltc3872 chip or what you end up using, the diode, the mosfet, the inductor... so you probably won't need any special cooling for 48v.So again... 8w from a 5v power supply means you'll have 8w/5v = 1.6 amps used by the 48v section. If you want 5v @ 500mA for the other stuff, you're exceeding your power supply's capabilities (you say you want 5v @ 2A).Also keep in mind that most cheap 5v @ 2A power supply out there (usb chargers whatever) are not really designed to output 2A continuously and they may even output lower voltage at 2A due to thin cables which results in voltage loss.If you do decide on going straight with a 5v DC power adapter you should really consider choosing one designed for 3-4A.I personally would recommend going with a 7.5v or higher DC adapter.. that ltc3872 for example works with maximum 9.8v so it wouldn't be a problem. But there's other chips that can work with higher input voltages and do what you want.12v @ 1-2A adapters for example are very easy to find and reliable, and that means even with a 1A adapter you'd use about 650mA for the 48v @ 150 mA and you're left with 350mA available for 5v ( 12v * 0.35A / 5v = ~ 0.84a with ideal efficiency)Another advice.. create an account on ti.com and then use their online webbench to design stuff, it can suggest circuits using their chips :http://www.ti.com/lsds/ti/analog/webench/overview.pageHere's one made out for 5v-12v in , 48v @ 150mA out :(attachment)Looks more messy, with more components, but they're cheap resistors and capacitors. The chip itself is probably half the price of the LTC chip, and in an easier to solder package.

author: Ian.M
author_rank: Super Contributor
timestamp: April 01, 2015, 03:54:21 pm
content: It helps considerably if you can afford to drop a couple of volts in a linear regulator as they can be designed to offer good rejection at the switching frequency and its harmonics without bulky high current chokes and large capacitors.I also believe the optimum input voltage for this application if an external mains PSU is used is probably a nominal 12V to feed switching pre-regulators to 50V and 7V followed by linear regulators to the final voltages.In addition to avoiding the high input and switching currents of a 5V in solution that have allready been mentioned, 12V in gives you a much lower risk of damage if the wrong PSU is connected and with an over-voltage shutdown circuit, making it withstand 24V without damage wont be too difficult. It also lets it run off a 12V lead acid battery if mains power is not available. All of the regulatory compliance and most of the safety hassles are dealt with in the off-the-shelf 12V 1A SMPSU so for low volume production an external PSU is likely to be highly desirable.

author: Cliff Matthews
author_rank: Unknown
timestamp: April 01, 2015, 04:18:51 pm
content: Cheapest? - Look at the 48v supply posted here (similarly discussed last week $14) and then use the 7 x 1N5338 zeners into the 7805.https://www.eevblog.com/forum/beginners/how-can-i-double-24vdc/msg638760/#msg638760

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: April 01, 2015, 05:02:31 pm
content: Awesome stuff guys! I think I'll do some thinking and postpone the decision for now.If I had to choose now I think I would go for the 48v external tranformer idea, then filter it for phantom power rail and step it down (with zeners? won't they get very hot?) + regulate it for 5v rail.Thank you all!

author: Ian.M
author_rank: Super Contributor
timestamp: April 01, 2015, 05:17:56 pm
content: Anything except a switching buck converter will get very hot dropping 48V to 5V. That's over 21W dissipation @500mA load current for any linear solution - as much as a low wattage soldering iron!The problem with going from 48V to 5V with a switching buck converter is that it will pollute the 48V rail with its switching noise, making your problem of getting a clean supply for the phantom power much harder. Filtering that out is likely to be bulkier, heavier and more expensive than either starting from 7.5V and using a linear 5V regulator + a boost converter as Mariush suggested or an intermediate voltage and two switcing regulators (one boost, one buck) as I suggested. It would be possible to use a single DC-DC converter circuit with both 5V and 48V (or 7V and 50V if linear post-regulators were needed) but that would need a custom transformer and a lot of design work.

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: April 03, 2015, 10:05:14 pm
content: Soo, I've been thinking and doing some more research. The field of power stages is huge!For now, I think I'll opt for 'outsourcing' this part of the design and go for two separate AC-DC converters.I think I've found the device I'll use:http://www.xppower.com/pdfs/SF_ECE05-10.pdfThey are 5 Watt and do both 5V and 48V. That means 1A@5V and 0.1A@48V right?That should be enough to run an Atmega + ICs at ~5V and eight phantom powered mics (about 10mA each), am I correct?I think I'll do what any self-respecting engineer would do, and order the parts and try to breadboard it and see if it works. I'm too used to planning everything and checking the theory. I'll find my engineering hat and put it on.Thanks for the feedback guys, I'll probably post here again with findings and the like. IF you happen to see something wrong with my thinking please do tell!

author: Cliff Matthews
author_rank: Unknown
timestamp: April 04, 2015, 05:22:26 pm
content: FWIW, stay with a transformer for the mics (cheaper than xppower's 130khz switcher) and small 5v@2amp SMPS are a dime a dozen.Such is the case sometimes, things are cheaper bought (and UL listed too).