12V to 24V (up to 28.5V) step-up converter - Page 1
Forum section: electronics / projects

author: brumbarchris
author_rank: Regular Contributor
timestamp: January 31, 2016, 10:28:08 am
content: Hi all,I am looking for a step up converter capable of delivering 10A@28.5V at the output while supplied form a regular car battery; requirement is to work down to an input voltage of 9V.So that is a pretty heavy requirement, and I came across some expensive options:http://www.powerstream.com/dc1-extreme-12-24-15A.htmhttp://www.newmarpower.com/12To24V_Step-Up_Converters/12To24V_Step-Up_Converters.html(both can only go up to approx 27V, but this could be acceptable)I also came across some cheap options:http://www.ebay.com/itm/1200W-DC-DC-Boost-Converter-Power-Supply-8-60V-12V-Step-up-to-12-83V-24V-48V-20A-/171469074723?autorefresh=true(this one a bit limited in terms of input current, but I computed I could still get some 6.3A at the output, which might be marginally acceptable)So the expensive ones are clearly out of my price range (my budget for this could go up to 100EUR). But I am very circumspect of the Chinese eBay product. Their english is so poor, I can hardly understand the item description, so I am wondering how in the world would they have the ability to design a reasonable quality product without enough knowledge of English. There are literally thousands of ways such a product can be designed in a poor way (load step response, output ripple, soft start, inrush current, output short protection etc) without even considering the quality of the manufacturing of the product itself.I am aware you can buy cheap reasonably quality Chinese products off of eBay for a variety of tasks that will perform reasonably, but the point is that this is not a trivial product. It has a lot of unspoken requirements, some of which I mentioned above, which might not be catered for and even testing it is not a 5 minutes job.So I guess I need to ask here if anybody has experience with such a product or bought anything similar from eBay. Alternatively suggesting a different source within my budget could also work.I am also aware there are some Texas Instruments or Linear Technology reference designs which allow getting in excess of 300W at the output without heatsink, but these are mostly relying on a 4 layer PCB, the cost of which alone is probably going to eat out my entire budget for this.Regards,Cristian

author: peter.mitchell
author_rank: Super Contributor
timestamp: January 31, 2016, 11:03:09 am
content: i've done a few designs with the TPS43061 and had no issues - the same kind of power levels you're expecting.To buy an off the shelf induct for that kind of power level you need a fairly high switching frequency, simply because it is easier to build a lower inductance, high current inductor than it is to make a high current high inductance inductor. Choose some FETs that aren't overkill, like 40V, 40A instead of 60v, 200a or something so the TPS43061 doesn't have issues driving them and you'll be off to the races.

author: Fungus
author_rank: Super Contributor
timestamp: January 31, 2016, 01:28:50 pm
content: Quote from: brumbarchris on January 31, 2016, 10:28:08 am??a) They can design it in Chinese, no problem. Chinese has words for electronics.b) The people putting up the eBay auctions probably aren't the circuit designers.PS: When they say "pressure", they mean "voltage" (and you know the page was machine-translated).Quote from: brumbarchris on January 31, 2016, 10:28:08 amThe Chinese products are mostly OK if a) You don't search for the cheapest price possible, and b) You take the power ratings with a pinch of salt.The power ratings you see on eBay aren't usually realistic working loads, they're more like "doesn't catch fire in a well-ventilated environment".eg. If you need to run an appliance at 1200W all day long then don't buy something labelled "1200W", buy "2400W" or more.(nb. This isn't particularly a Chinese thing, everybody does it).Quote from: brumbarchris on January 31, 2016, 10:28:08 amNo heatsink ... but they probably need lots of copper on the PCB and you won't be able to put one in a closed box.Quote from: brumbarchris on January 31, 2016, 10:28:08 amAdding a fuse and extra decoupling capacitors is easy...

author: brumbarchris
author_rank: Regular Contributor
timestamp: January 31, 2016, 03:35:46 pm
content: QuoteUsing a heatsink, I presume, right? I think it is difficult to achieve efficiency above 90% at this kind of power transfer without using a multy-phase topology.But probably this is a path I should consider. At the end of the day, it might be cheaper to go with a smaller PCB and with an off-the shelf heatsink.Regards,Cristian

author: promacjoe
author_rank: Regular Contributor
timestamp: January 31, 2016, 06:47:55 pm
content: something else to consider in your application. Some Boost converters will turn off when it detects an overload or short-circuit. But the Chinese boost converters that I have used will only stop regulating the voltage. They revert to putting out the input voltage at whatever current is available. They will not turn completely off. this is normal for some types of boost converters, And was not mentioned in the listing. in my application this was not acceptable. So like you, I'm still looking for a boost converter at a reasonable price.

author: langwadt
author_rank: Super Contributor
timestamp: January 31, 2016, 08:06:17 pm
content: Quote from: promacjoe on January 31, 2016, 06:47:55 pmlook at the schematic for a boost converter there is no way around it, the is a path from input to output via the inductorand diode

author: promacjoe
author_rank: Regular Contributor
timestamp: January 31, 2016, 08:43:26 pm
content: Quote from: langwadt on January 31, 2016, 08:06:17 pmYes, but the good ones have secondary circuits to turn the system off, when a error occurs. they don't put out a unlimited current at whatever voltage is applied to the input. they turned completely off. I would rather know that it would turn off completely rather than drain your car battery or worse, catch on fire.it is just something to think about. How much safety do you need to be built into the product. And if an error occurs, How much could it costs you.

author: brumbarchris
author_rank: Regular Contributor
timestamp: January 31, 2016, 09:40:54 pm
content: Exactly, that is my point. This turning off of the output in case of a fault is one of the critical requirements I would have from such a device, which is not so obvious when you just say "I want a converter".As designing my own is still an option, I consider at least a fused output as a minimum protection measure, although this is not really the best.Regards,Cristian

author: Fungus
author_rank: Super Contributor
timestamp: January 31, 2016, 10:39:02 pm
content: Quote from: brumbarchris on January 31, 2016, 09:40:54 pmYou can add a fuse to a bought one as well...

author: promacjoe
author_rank: Regular Contributor
timestamp: February 01, 2016, 04:02:43 am
content: Quote from: Fungus on January 31, 2016, 10:39:02 pmunfortunately, I have come across too many circuits, that were protected by fuses, but still managed to burn up the wiring. In some cases catching on fire. in one case, a blower motor locked up. this caused the fan resistor to heat up and catch on fire. the circuit was properly fused, but did not pull enough current to blow the fuse. fuses are a good level of protection. But they cannot protect against everything. In this case, he wants an output of 28.5V at 10A. At 14.5V, this would be about 20A at 100% efficiency. When it comes to energy conversion, nothing is 100% efficient. And when you consider the surge current, you need at least a 30A fuse for this circuit. if an error occurred it would have to draw enough current to blow the 30A fuse. But there are cases that this may not happen. If for some reason it only drew 25A, the fuse would not blow. But power would still be applied to the circuit possibly causing a fire. it is understood that a fuse would be used. That is an absolute must. But turning the power supply off in case of a failure error, would be much better possibly not even blowing the fuse. The fuse would protect the power supply, but the power supply should protect itself.

author: brumbarchris
author_rank: Regular Contributor
timestamp: February 01, 2016, 07:39:03 am
content: Quote...QuoteI am with you on this one. Normally I would have to use a slow blow fuse in order to allow for the large inrush current that is bound to happen, and for slow blow fuses we generally use the assumption that they are only guaranteed to fuse out at a current of 1.4xfuse rating, so 42A in case of a 30A fuse (think worst case and negative temperatures outside, this being simultaneous to a minimum motor winding resistance, hence a large current through the load).Regards,Cristian

author: Zero999
author_rank: Super Contributor
timestamp: February 01, 2016, 07:52:50 pm
content: Fuses are a perfectly good way of protecting against fire. The fuse and cable need to be sized properly. If the cable is not thick enough, then it's possible it could catch fie and the fuse not blow.In this case the boost converter will also have some current limiting. If you're not happy with a fuse, a PTC resistor could be used instead.

author: promacjoe
author_rank: Regular Contributor
timestamp: February 01, 2016, 09:06:40 pm
content: As I said above, A fuse is an absolute must. Anyone who does not consider using an appropriate fuse/circuit breaker, When designing any electronic device is asking for trouble. That is not an issue, not with me anyway. The issue is, in this case, is a fuse enough. If the answer is no, further investigation must be taken to ensure the safety of the device. As noted above, there are certain situations where a properly sized fuse may not be enough to protect the circuit. pretty much all vehicles now have several layers of fuse protection. Starting with a fuse/fuse link at either the starter or distribution point from the battery. from there they are divided into multiple sections using smaller fuses depending on the need of the circuit. the engineers that designed the circuits consider every aspect of the application when assigning the values of these fuses. But in certain cases this is not enough to protect the vehicles. Case in point Ford Motor Company had a defective ignition switch. These ignition switch when turned off with sometimes short out and catch the car on fire even though the circuits are fused. They destroy the cars and sometimes the garages/house that they were parked in. sometimes killing the sleeping occupants. It is not enough just to say a fuse will protect the device, you got to go further. in this case, if the output is drawing too much current, partially turning it off is not enough. you need to turn the output off completely. the fuse would only come into play, if the boost regulator itself was drawing too much current. Having multiple layers of protection is good engineering practice. simply throwing a properly sized fuse in it and assuming that will protect the circuit, and your liability for that circuit is foolish. in this case, you must play the what if game. If this situation occurs, what would be the results. If every scenario you can come up with, blows the fuse, great. If even one scenario fails to blow the fuse, you got problems. and this problem could threaten lives. And in this case, what is the cost of properly turning off the power supply, versus one life.the op has already acknowledged that a fuse is necessary, and will be used. But the question he has, is it enough. In my opinion it is not. In this case there is no device on the input side that can protect 100% of the problems caused by not turning off the output at the boost regulator. having an error occur beyond the regulator but not turning the power off completely is the problem.

author: brumbarchris
author_rank: Regular Contributor
timestamp: February 02, 2016, 07:14:49 am
content: I am considering an off-the shelf part as follows:http://cds.linear.com/docs/en/demo-board-manual/dc1641A.pdfNeeds some modifications, but doable ones.Cannot give me exactly 10A @ 29V, but I suppose it can do 8A @ 29V and that is sufficient (marginally). With a properly appended output protection I think this will fair better than a 20 USD (shipping included) Chinese eBay product. I will just have to clarify if the output can exceed 5A (demo manual and advertising imply it does, but silk screen text on PCB and notes on schematic infer 5A max).Regards,Cristian

author: Fungus
author_rank: Super Contributor
timestamp: February 02, 2016, 09:01:25 am
content: Quote from: brumbarchris on February 02, 2016, 07:14:49 amHow are you intending to use this? ie. How many hours per day?From the datasheet:QuoteKey words: "proper cooling"QuoteSeems like you're complain about 'Chinese' stuff not being reliable but you're proposing something just as bad.

author: brumbarchris
author_rank: Regular Contributor
timestamp: February 02, 2016, 11:07:52 am
content: Yes, but in this case I have an English speaking partner knowledgeable in electronics who can clarify the question marks that I have.And to answer your question, the plan is to use this in a gate automation system where the motor needs 24V in theory (but 29V in practice); the system, however will be supplied by a regular 12V automotive battery. Opening/closing the gate lasts about 15seconds (each way, so 30 seconds in total) and estimation is that an open/close action will be performed 4 times a day in average.Regards,Cristian

author: promacjoe
author_rank: Regular Contributor
timestamp: February 02, 2016, 06:07:27 pm
content: One question, how are you going to charge the batteries. Instead of using the boost converter, would it not be better to use two smaller 12V batteries, and a 24V solar charger. A boost converter is not 100% efficient. You will lose approximately 10% of your power in the conversion. Plus you still have to charge the battery. To me it would make sense to use two, 12V batteries to get 24V. No loss in efficiency. And one, 24V battery charger. Which by itself will give you an equal efficiency to charging one, 12V battery. you don't have to worry about the boost converter or the current loss. using less electronics, with better efficiency. It would also be easier to implement, and maintain.good luck.

author: peter.mitchell
author_rank: Super Contributor
timestamp: February 02, 2016, 06:44:46 pm
content: Quote from: brumbarchris on January 31, 2016, 03:35:46 pmyes, using a heatsink, but not a terribly big one, losses will be somewhere in the 10-35w range - useing 2oz copper and some SMD heatsinks 10W is fairly trivial to dissipate, approaching 35W is much more so - at 35w i would reverse mount to-220s and attach a heatsink to those.

author: brumbarchris
author_rank: Regular Contributor
timestamp: February 03, 2016, 07:06:12 am
content: QuoteWell, that was the previous setup, but it has two major flaws (which are particular to this application):1. A 24V battery bank is not sufficient to actuate the motor under frost conditions, even when fully charged at about 27.6V (this voltage quickly drops to about 25V under load); I was able to get around that by employing an additional lead acid 2V cell in series with these two batteries and then charging the whole bank to 29V; but then I ran into problem number 22. I need to charge a bank of lead acid batteries to 29V from a 21V (18V MPP, 50W MPP in theory, 30W MPP in practice - and even these 30W measured on a sunny day) solar panel; the solar panel is a fixed requirement for this project and cannot be changed. Consequently I need a boost configuration charger. I did implement such a solution using an ST Microelectronics 4 phase charger, not dissipating much power, but the problem is that in order to boost the voltage the charger obviously draws more current from the panel than it delivers to the batteries (normal for a boost converter). And even with this efficient 4-phase boost charger, I found that during a cloudy winter day (of which we have many in a row here) the charge delivered to the batteries is negligible. Hence the whole system ran out of power after a few days.Because of the two problems mentioned above, I decided to go with a 12V battery and a 29V boost converter (external to the main electronics). Because I can then use a buck configuration charger which actually draws less current (not energy) from the solar panel than it delivers to the battery, hence I genuinely believe I would be able to charge a single 12V battery even during a cloudy day.Regards,Cristian

author: promacjoe
author_rank: Regular Contributor
timestamp: February 03, 2016, 06:27:01 pm
content: If the solar charger was originally set up to charge a 24V Battery, it probably will not do any better charging a 12V battery at more than twice the current load. You essentially will have the same wattage available for charging. If you're 24V battery was not retaining enough of the charge, You should look at the type of batteries/size that you are using. Also make sure your charger is adequate to charge the batteries that you are using. Decreasing the battery voltage, and using a boost converter, will give you less available power, not more. Even though the battery may be rated to twice the current capacity. You will still lose approximately 10% or more efficiency through the boost converter. The problem sounds like it is in your solar charger, versus the available current capacity of your batteries.The size of your battery should the able to run the unit for at least 2 days and preferably 4 days. The solar panels should be able to charge the batteries in one day after a 2/4 day rain. as far as the cold weather goes, that is something you'll have to look into. see if you can find batteries that are rated for cold-weather. in the area that I live in, we normally don't get more than 3 inches of snow per year. So cold weather is not a problem. I do know, that the type of batteries that you use is very important to your application. But I cannot advise you on the type of batteries to use.I don't know which type of gate you have, but you should check the mechanics of the gate. Make sure the gate moves freely. Fix any mechanical problems that you find. This will reduce the current load on your batteries. Also check your motor and make sure it runs properly without binding or noise. It should have worked perfectly when it was first installed. Figure out what has changed and fix it.adding a boost converter will only make it harder to fix later on. In this case you need to simplify your circuit, not make them more complex.good luck.

author: brumbarchris
author_rank: Regular Contributor
timestamp: February 03, 2016, 10:28:29 pm
content: I fixed all mechanical problems, but frost is a genuine problem. And we can get a continuous week of cloudy weather, if not longer... Also the electronics of the gate is drawing some continuous 70mA (throughout the night) even in standby and that eats up a lot of battery charge. One solution is to change the electronics and that is the real project. Anyway, aside from making the electronics draw less current in standby, I revised also my charger strategy.For the charge you have Pout=efficiency*Pin orUout*Iout=efficency*Uin*PinIf I need a large Uout (say 29V) and Uin is in a 16...21V range then obviously Iin>Iout. This large Iin is killing the panel, particularly in cloudy weather, the voltage from it plummets and the charger doesnt put much energy in the batteries. I have already tried this. In practice. And it does not work. On the other hand, if I am regulating down the voltage to charge a lower voltage battery then Iin can be smaller, hence getting a better usage of the panel in cloudy weather. Combined with some electronics driving the gate with a real stand-by mode (target is to get below 1mA ins standby mode, cyclicly waking up the micro to sniff for radio signals from the keyfob) I think this will be the winning combination.Regards,Cristian

author: promacjoe
author_rank: Regular Contributor
timestamp: February 03, 2016, 10:57:13 pm
content: Your standby mode is three times more than it should be. Even for 1980s/90s standards. Standby mode should be less than 25 mA.Good luck.

author: brumbarchris
author_rank: Regular Contributor
timestamp: February 04, 2016, 06:35:42 am
content: You are right, it is. But then this system (a TELCOMA gate automation kit) was designed to be supplied from the mains through a transformer. It does not really have a standby mode at all, it just continuously draws its normal operating current, which is at the aforementioned level when the motor is not turning (when it does, current can go to about 5A, but that is not under frost conditions nor motor stall conditions, when it can go considerably higher, to 10A).Anyway, as mentioned before, the plan was (and still is) to redesign the whole electronics, with the intention of placing the battery charger, the boost converter, motor driver and every other needed circuit on the same PCB. In the process of doing so I realized that a 300W output boost converter is not a trivial "just another circuit that goes on the same board" and that it needs special attention. I am therefore in the process of defining the main project to use an external boost converter and then I will design this converter separately (or buy an off-the-shelf part, like the LT evaluation kit mentioned before).Regards,Cristian

author: Zero999
author_rank: Super Contributor
timestamp: February 04, 2016, 08:48:10 pm
content: Quote from: promacjoe on February 01, 2016, 09:06:40 pmYou are correct that a fuse isn't the only thing which protects against fire. If a fault doesn't cause enough current to flow but enough power is dissipated, in a small enough space, near flammable materials, in the presence of air there's a good chance of fire. For example, a poorly designed switch could arc over and cause a fire, even though the current isn't high enough to blow the fuse.However, I don't see how this sort of thing can be prevented when choosing a power supply, other than making sure it's thermally protected.

author: promacjoe
author_rank: Regular Contributor
timestamp: February 05, 2016, 04:43:08 am
content: The issue is with this particular type of power supply, a "Regulated boost power supply", When used with this particular type of application. He is running a motor that apparently has failed to run in adverse conditions. Without the boost power supply this condition would blow a fuse. Adequate protection. When the boost power supply is used, it can be set to shut down during an overload. The fuse does not blow, but the power supply does not completely shut down. It puts out the input voltage instead. considering it is at half voltage, this may or may not blow the fuse. If it does not, it will send out this voltage at whatever current is available. Something is going to be damaged. It may be the motor, It may be the wiring but something will be damaged. At minimum it will kill the battery eventually. Some types of batteries do not like being completely discharged in freezing weather. The batteries tend to freeze, thus will need to be replaced. This can get expensive to say the least.the fix for this can be as simple as a comparator hooked up to a relay. If the output voltage is equal to or less than the input voltage, The relay turns off the power. If the output voltage is higher than the input voltage the relay turns on. Of course there are better ways of doing it but that's the simplest method. Simply making sure if an overload did occur the power turns completely off is the best option. That's what a fuse is design to do. But with a boost converter, it may or may not work.

author: brumbarchris
author_rank: Regular Contributor
timestamp: February 05, 2016, 06:53:10 am
content: I have decided to design a boost converter that will suite my needs. I will initiate a separate thread that will discuss the design of that converter, and I will start with laying out explicit requirements for it (including output undervoltage lockout).Regards,Cristian

author: brumbarchris
author_rank: Regular Contributor
timestamp: February 05, 2016, 07:58:49 am
content: Thread closed. Discussion regarding new boost converter design now in:https://www.eevblog.com/forum/projects/29v-10a-boost-converter-design/Regards,Cristian