High-amp buck converter - Page 1
Forum section: electronics / beginners

author: mrmp17
author_rank: Contributor
timestamp: November 25, 2013, 01:51:53 pm
content: Hi!I need to get voltage from 24V, 400W transformer down to about 16V. Output power must be around 300W. I was thinking about switching converter - they have good efficience. Efficience is not number one priority since i've got heatsink with 0.25C/W thermal resistance and 100W to spare, but linear regulator would be to inefficient I think. The circuit shuld be as simple as posible.I'm a bit of beginer in high amperage regulators so I don't know what's the best way to do it. So, if you have any ideas please share it with me.Thanks

author: rbola35618
author_rank: Frequent Contributor
timestamp: November 25, 2013, 02:48:10 pm
content: Is this a homework problem or a home project?RB

author: mrmp17
author_rank: Contributor
timestamp: November 25, 2013, 03:25:15 pm
content: Home project. Power supply for a LiPo Charger

author: fcb
author_rank: Super Contributor
timestamp: November 25, 2013, 04:05:27 pm
content: LTC1775

author: megajocke
author_rank: Regular Contributor
timestamp: November 25, 2013, 08:30:40 pm
content: Is that a 400 VA transformer? 300 W will be difficult to get after rectification unless you use some kind of power factor correction circuit. Typical power factor for a simple transformer-rectifier-capacitor setup will likely be in the range of 0.7-0.8. Throw in a huge inductor and you'll get 0.9 at an output voltage of about 20 V.However, transformers can tolerate gross overloads for quite a long time because they heat up so slowly.

author: mrmp17
author_rank: Contributor
timestamp: November 26, 2013, 11:55:03 am
content: Thank you all!Yes, it's 400VA transformer. If I understand you correctly, power factor is "real" power you get from ractification? What kind of power factor correction circuit are you talking about? Some active or pasive stuff? Also, you said I would get around 20V on the output, but my LiPo charger has a max voltage of 18 volts...You have any idea how to get the voltage down under the max limit on charger?Sory, as I said I've never done such high power regulatoun circuits and I'm a beginer in this stuff

author: megajocke
author_rank: Regular Contributor
timestamp: November 26, 2013, 02:21:19 pm
content: Yes, the rectifier draws nonsinusoidal current so the real power on the input (equal to output power plus diode losses) is lower than the apparent power. Exactly what the power factor will be depends on the source impedance seen by the rectifier but also on how large the smoothing capacitors are, though the effect of the latter usually is negligible if a transformer is involved.Say you use a standard rectifier + a buck converter. If the efficiency of the converter is 90% and the power factor is 70% you'd need about 480 VA. This 20 % overload would increase the transformer losses and thereby temperature rise by about 40 %. Rectified voltage would be around 25 - 40 V depending on load and line voltage.Is continuous operation required in your application? With dissipation just 40 % or so above the continuous rating, intermittent use could be okay depending on how long it needs to operate.Though I think the cheapest (and lightest) option might be to get a 15 V 300 W AC/DC SMPS power supply module from for example Mean Well.

author: mrmp17
author_rank: Contributor
timestamp: November 26, 2013, 08:06:10 pm
content: It would be easiest to buy a SPMPS, but I have that transformer on hand and looking for a simple and cheap circuit to make use of it. Just need to figure out how to lower the voltage a bit...I think it's a bit nonsense to have some high tech switching regulator for lowering voltage for just for just a few VFirstly I wired IRF5305 (31A PFET) to LM2596 through microcontroller with ADC, but things heated up to much with no load (there was correct voltage on output), than I was trying to make LM3485 work with IRF5305 and it didn't really work...Now I'm thinking some kind of linear regulation to get the voltage down a bit.Full load wil not be used used a lot and for 2 hours max (probably less), also I can easily ensure some fan cooling for transformer. Maybe a circuit likehttp://www.reber.si/napajalniki/usmernik_12V_20A/usmernik_12V_20A-sch-300dpi.gifcan solve my problem... what do you think about it?

author: KJDS
author_rank: Super Contributor
timestamp: November 26, 2013, 08:55:34 pm
content: Why not unwind some turns?

author: mrmp17
author_rank: Contributor
timestamp: March 17, 2014, 01:41:50 pm
content: Hi!I decided to buy 15A DC-DC converter and power it from rectified 400VA 24V transformer, still I have some questions... I thought about around 50 000uF of capacitors and a big inductor after the rectifier. Would that work? I also don't know how to calculate power dissipated in rectifier. Is it just voltage drop multiplied by current?Do you think that configuratioon will get me 260W at 17V?

author: mariush
author_rank: Super Contributor
timestamp: March 17, 2014, 03:39:41 pm
content: You have a 24v 400VA transformer ... or 400VA/v = 16.6 A.If you convert this to DC you get a peak voltage (before voltage drop on diodes on rectifier) of 24 x 1.414 = 33.9v and your current drops to about 0.68 x 16.6 = 11.2A.If you convert this to DC and use a large inductor after the bridge rectifer, you get a peak DC voltage of about 0.9 x 24 = ~ 22v but your current remains at around 0.94 x 16.6 = 15.6A . Honestly, I have no idea how large the inductor would have to be for such high current.Going back to the first choice (just bridge rectifier and capacitors) like I said, you'll have a peak voltage of about 34v before the voltage drop on diodes and capacitors.You have two diodes always working in a bridge rectifier, and a bridge rectifier rated for such high currents would have about 0.8-1.1v drop per diode, so you're losing about 2v on the bridge rectifier. Therefore, you'll have a peak voltage of about 32v before the capacitors.Ideally though, you'll put two bridge rectifiers in parallel to reduce the power dissipated in each bridge rectifier, because bridge rectifiers are derated at higher temperatures.For example, look at this 35A bridge rectifier :http://www.digikey.com/product-detail/en/GBJ3510-BP/GBJ3510-BPMS-ND/3191597If you check the datasheet, you can see that it's rated for 35A at up to 100c but it can only do about 7A without a heatsink. It has a voltage drop of about 0.8v at 2A , 1v at 10A per diode..So you know that at 11A it will dissipate 2v x 11 = ~ 22 watts and therefore you need to keep it under 100c with a decent heatsink.If you put two in parallel, they'll each dissipate half that current amount, more or less and you're getting almost to the point where they could work without a heatsink, so you won't need a large heatsink or you won't have to worry much about it.Anyway, going back to the subject... you now have a peak of 32v DC after the diodes and their voltage drop and about 11A. Now you need to figure out how much capacitance to use.The capacitance affects how much the voltage will sag at a particular current used. A "close enough" formula is :C = Current / (2 x FrequencyAC x Vripple) where Frequency AC is the mains frequency (60 Hz in US and some countries, 50 Hz in Europe) and Vripple is how much you're willing to let the voltage go down.For 11A and 2v ripple and EU frequency, we have C = 11 / (2 x 50 x 2) = 11 / 200 = 0.055 Farads or 55000 uF - so with your 50000 uF capacitors, you'll have a voltage hovering around 29.5 and 32v when you use 11A. At lower power consumption, the voltage will be closer to 32v more often.In practice, 55000 uF is A LOT. Normally, it makes little sense to use more than 10000-220000 uF and just accept that the voltage may go down a bit more - a switching regulator will just adjust it's pwm to maintain the output voltage where you want it. It's also usually more expensive to buy such large capacitors and it takes more space, compared to just using a switching regulator or mosfets with better efficiency, or a slightly larger power transformer.So now you have to find a switching regulator that would accept.. let's say 25v - 35v and capable of producing 17v 15-20 amps ... I'm not quite sure you'd be able to do 20 amps, because 17x20 amps is 340 watts and at best you'd have about 30v x 11 = 330 watts and then the switching regulator will be about 90% efficient, so you lose another 20-30 watts, but about 17v @ 17-18a should be doable. 260w like you said in your last post should work.My advice from this point forward would be to go on ti.com and create an account and then use their Webbench software to find something suitable with those parameters (25-35v in, 17v @ 20a out) and see what circuit would work best for you. The software gives you mosfet suggestions (ex pick through hole mosfets instead of surface mount stuff), inductor choices etc etcFor example, you see below two results generated by webbench, that do 28-32v in, 17v @ 18a out, with LM5119 or LM25119, and they both use pretty much the same parts ( four mosfets, some inductors, lots of resistors and capacitors and diodes) :later edit: and if you can go down to about 140w output (17v @ 10A) you can simply the circuit and use just one mosfet and an easier to solder part, LM5088.. see circuit3.png

author: mrmp17
author_rank: Contributor
timestamp: March 17, 2014, 05:32:13 pm
content: Thank you SO much!I was thinking of combination of capacitors and inductor (I think 1000uH would be doable). I allready have rectifier, it's MB3510, 35A rated and 1.1V voltage drop.Speaking of ripple, as I'll have DC-DC converter, the ripple can be big, but it's important that voltage doesn't go below about 18v, so I don't get that ripple on the output of dc-dc converter.I think I will just buy DC-DC module, they are realy chep, like $15 on ebay. I'm fully aware thet it's probably not good quality, but I can simply repair some bad solders, raplace caps, etc... I think it's impossible to make equivalent regulator at home for that cost.Thanks again!

author: mariush
author_rank: Super Contributor
timestamp: March 17, 2014, 05:54:31 pm
content: You'll have to be careful with such dc-dc modules. Yes, they may be capable of 15A or whatever you say they do.. but they don't say in what conditions.. if it's for short periods of time or provided you add heatsink, or for example they may do 12v down to 9v at 15A, but they may not do 32v down to 18v at 15A ... it depends on duty cycle, what inductors they use, how well the main regulator IC is heatsinked ( if you'll check the datasheet for that IC you may notice that for some voltage/current combo a certain amount of pcb heatsink which is undersized in your modules)...It's risky.On the other hand, if 15$ is not much for you by all means try it out. If you do it yourself you need to design a pcb, have it made by some company, then solder the tiny components on pcb and so on. It takes more time and a bit more than $15 to do it.

author: mrmp17
author_rank: Contributor
timestamp: March 17, 2014, 08:48:34 pm
content: I was looking at that one:http://www.ebay.com/itm/300985791638?_trksid=p2055119.m1438.l2649&ssPageName=STRK%3AMEBIDX%3AITHeatsinking should not be a problem, since I will probably do it with with bigger heatsink than included one. If i decide to do it myself, It wil take a lot of time and effort to get it working.So, what do you think about this dc-dc module?