30A OR-ing Circuit? - Page 1
Forum section: electronics / beginners

author: PeteD
author_rank: Newbie
timestamp: June 26, 2015, 03:47:10 pm
content: I'm looking for information on OR-ing circuits.I have a 3D printer that works well for the most part, but it is underpowered when it comes to heating the bed. I'm usually waiting about 45 minutes to an hour for the bed to heat up to 100C, and the wait is irritating me. Others with my type of printer recommend using a 24V 30A PSU and a 100A SSR to get the bed heat up time to less than 10 minutes. Unfortunately, I'm on a budget and don't want to spend the $200+ that that solution would cost.I have a couple of cheap Chinese 24V 15A switching power supplies, but linking power supplies is a little more complicated than putting batteries in parallel. My google-fu has found some information on low voltage, low amp OR-ing circuits, but it has failed me in finding anything for higher currents and voltages.Could someone who knows more about this subject than I (that wouldn't be hard, I've pretty much said it all in this post) point me towards more information on OR-ing design? Also, are there any cheap solutions already out there?Thanks,Pete

author: krish2487
author_rank: Frequent Contributor
timestamp: June 26, 2015, 04:25:22 pm
content: Yes.Just use any standard diode bridge (medium to high A rating) and you are good to goPSU1 ---|>---\_________ TO LOAD/PSU2 ---|>---As simple as that. For your particular application it is not critical about load sharing and such. Just make sure that you put a heatsinkon the bridge. Or if you can also go with a high A rating Schottky diode bridges like MBR20100 and short out the anodes too to get a 20A rated Schottky diode. Use several such to parallel your Power suppliesLike this.|-----|>-----\|-----|>------\------------------- TO LOAD|-----|>------/|-----|>-----/

author: macboy
author_rank: Super Contributor
timestamp: June 26, 2015, 04:29:29 pm
content: You basically do just need to connect them in parallel.To ensure current sharing between them, use ballast resistors. The simplest ballast resistor is simply the length of wire between the supply and the load. So, instead of connecting the two power supplies in parallel near the supplies then using a single length of wire to the load, use one length of wire to the load from each supply, making the parallel connection only at the load. The voltage drop across each length of wire helps ensure current sharing. The other important thing is to adjust the two supplies' output voltage to match as close as possible.If you want true "OR"-ing of the outputs, you put a series diode on each output. Then you could technically use only one without having the disabled supply see the voltage from the enabled supply. You need really beefy diodes here. I have salvaged several suitable diodes from dead computer power supplies. They are in TO-247 (TO-3P) packages like transistors, but are actually high current Schokttky diodes. One that comes to mind isMBR3060CTMUR3060PT, a 30 A, 60 V (reverse) diode.

author: PeteD
author_rank: Newbie
timestamp: June 26, 2015, 05:06:59 pm
content: Thanks, Krish, Macboy!It's good to know I was over complicating this. About how much heat should I expect from a couple of parallel MBR20100 diodes in normal operation? Would attaching them to the sheet steel legs of my printer's stand be enough to dissipate the heat, or should I get some heatsinks as well?

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 26, 2015, 05:21:05 pm
content: QuoteDefinenormalAt 15A you get a voltage drop of about 900mV (depending of the tempreatur), therefore 13,5W power dissipation. And this is a lot if you have no heatsink. In fact the diode would go up in smoke pretty soon.QuoteHard to say. It would probably do the job.You should try it first with the method macboy suggested. Additionally you should trim (almost every SMPS has a little trimmer near the output terminal) the output voltage of the supplies to match each other. Ideally you also measure the current of both supplies simultaneously while they are at full load, to make sure, they spread the load current evenly.However a diode bridge (also heatsinked of course), like KBPC5010 for instance, worth also a try.

author: krish2487
author_rank: Frequent Contributor
timestamp: June 26, 2015, 05:26:07 pm
content: Actually diodes decouple the PSUs from interfering with each others control loop.If there is even a slight mismatch in the output voltage setting or length of the wires then the power supplies willconstantly "fight" with each other.The diodes, on the other hand, ensure that each power supplies senses and regulates its own output.13.5 Watts is not really a lot of power to dissipate. It can be managed with a decent sized heatsink.PS: Scratch my suggestion with the MBR20100. The MUR3060 is more likely a better diode to do the job.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 26, 2015, 05:52:16 pm
content: QuoteAs I mentioned

author: tszaboo
author_rank: Super Contributor
timestamp: June 26, 2015, 06:42:09 pm
content: Any reason not to connect them in series?

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 26, 2015, 07:16:11 pm
content: QuoteThat gives a nominal resistance of 0.8Ohm.At 48V you would get a current of 60A and a power of 2880W.These supplies are not designt to limit the output current, they just switch of, if there nominal current (times a faktor of 1.1 ... 1.5 or so) is exceeded.In general connecting SMPS in series is not more or less critical than connecting them in parallel if they are not designed accordingly.

author: PeteD
author_rank: Newbie
timestamp: June 26, 2015, 08:43:24 pm
content: OK, I'll stop trying to be too cheap and geek for some heat sinks.Just curious, why do you think the MUR3060 will perform better than the MBR20100? Better Peak Voltage? It's unlikely these diodes will see 100V, let alone 600V, isn't it?

author: Ian.M
author_rank: Super Contributor
timestamp: June 26, 2015, 09:23:58 pm
content: As power dissipation is a major issue here, you want to look for diodes with the lowest possible Vf at a rated current greater than the max PSU output current. I would recommend at least a 20A If rating, preferably 25A, as if one PSU trips out, the other one's current is likely to surge very briefly over its normal max rating before it also trips. The worst case scenario voltage wise is one PSU running and the other tripped with 24V across the tripped PSU's diode, and any diodes with a voltage rating greater than 35V should handle that OK. Avoid diodes with a grossly excessive voltage rating as they usually have a higher Vf as well.Due to the fact that the PSUs cant operate in a current limited mode, its essential to have ballast resistors as macboy described above. The output wires for each supply should be as far as possible absolutely identical, especially the length and both the +ve and -ve wires should be separate all the way to the diodes and build platform for optimum current sharing.

author: madires
author_rank: Super Contributor
timestamp: June 26, 2015, 09:40:59 pm
content: There are dedicated ORing controllers which drive MOSFETs, like the LTC4357 or LTC4359 for example.

author: dom0
author_rank: Super Contributor
timestamp: June 26, 2015, 09:44:16 pm
content: What moron had the idea to heat these things with low-voltage DC?To contribute something meaningful also: get a fat transformer in the right range and do full-wave control with a SSR (on the primary!). Should give you >98 % efficiency, especially with a toroidal (xformer switching relays required then).

author: krish2487
author_rank: Frequent Contributor
timestamp: June 27, 2015, 12:59:28 am
content: Not the voltage. The current. The Die itself is rated for 75A for the MUR3060.Quote from: PeteD onTodayat 06:43:24 AM

author: edpalmer42
author_rank: Super Contributor
timestamp: June 27, 2015, 02:16:16 am
content: Instead of buying diodes and a heatsink and making your own, check at your local automotive supply store. They probably have a battery isolator which allows you to have two batteries that charge together but are discharged seperately. It's typically used when you have auxiliary loads like an inverter. You power the inverter off one battery and start the vehicle with the other one. The auxiliary loads can't drain the starting battery. Electrically, it's nothing more than two diodes and a heatsink!

author: rx8pilot
author_rank: Super Contributor
timestamp: June 27, 2015, 04:15:36 am
content: Build an 'ideal diode' with a controller and a MOSFET. Wont need a heatsink and can do it with about $15 or so - probably less than the value of the heat sink needed for a normal diode.I have been using LM5050 but there are tons of options out there. Find a low Rds(on) mosfet and a few other parts. It

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 27, 2015, 07:14:12 am
content: QuoteWhat do you get, if you switch the MOSFET on? Basically a piece of wire. Meaning you are shorting out both supplies, which is the exact opposite of what we want here.The topic in this case is chosen rather confusing. He actually wants an ADDing circuit since both supplies are intended to deliver power simultaneously

author: madires
author_rank: Super Contributor
timestamp: June 27, 2015, 11:07:51 am
content: Quote from: PSR B1257 on June 27, 2015, 07:14:12 amHave you ever read any datasheets and application notes of ORing controllers? They got a reverse current protection, i.e. if current flows the wrong direction, the MOSFET is switched off. Most controllers use the R_DS_on as current sense resistor.

author: rx8pilot
author_rank: Super Contributor
timestamp: June 27, 2015, 04:29:03 pm
content: Quote from: PSR B1257 on June 27, 2015, 07:14:12 amSort of....you are not shorting the supplies - you are putting them in parallel. This will only happen if current is flowing out of each supply. If one supply dies or is a lower voltage the high side supply will not backfeed into it. The only purpose of the diodes (or ideal diodes) is to prevent reverse current that will damage the supply.Quote from: madires on June 27, 2015, 11:07:51 amIt does nothing to help balance the load, which is an additional effort if the load needs to be evenly split. First things first - don't allow one SMPS to feed into another and ideal diodes are designed for that with very little heat dissipation or voltage drop. The title is asking about an OR-ing circuit which is exactly what a diode (OR'ing controller) does. Perfectly matching the supplies is a longer conversation. If it were me, I would use two isolated heaters that each SMPS can handle on it's own.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 27, 2015, 06:17:16 pm
content: QuoteOf course not shorted in terms of shorted from V+ to V- but shorted in terms of a quasi zero ohm connection from V+ to V+ and V- to V-

author: madires
author_rank: Super Contributor
timestamp: June 27, 2015, 07:28:46 pm
content: Quote from: rx8pilot on June 27, 2015, 04:29:03 pmSharing the load is quite simpleAssuming that both PSUs got different output voltages, the PSU with the higher voltage wins. But if the load is too large for one PSU the PSU's output voltage drops until it's not far off from the output voltage of the second PSU. The second PSU starts providing current. Get a dual-channel lab PSU, add two diodes and a shared load. Then set the output voltages to be 1V apart. Slowly decrease the higher voltage and watch the currents. Yes, it's simple like this.

author: rx8pilot
author_rank: Super Contributor
timestamp: June 27, 2015, 07:35:08 pm
content: I understand how the highest voltage wins. The problem is that each supply is only capable of carrying a partial load. Because of that, you cant have one supply win over the other because it will ve overloaded and possibly shutdown. When it does shutdown, the other supply will also be overloaded and fail.

author: dom0
author_rank: Super Contributor
timestamp: June 27, 2015, 07:42:36 pm
content: This can work with OVPd supplies, too, but only if the output resistance of the supplies is high enough to ensure balanced currents before OVP kicks in (or the supply fails). Now, with a good supply output resistance is very small and consists mostly of irreproducible contact resistances...In redundant server supplies they do it with a current-sharing line, which acts as a control line for all supplies. Basically each supply puts a voltage proportional to it's load current via a resistor on that line and an error amp ensures that every supplies load current is about current-sharling-line / n.

author: LukeW
author_rank: Frequent Contributor
timestamp: June 28, 2015, 12:51:20 pm
content: Do you really need 720 watts (30A at 24V) of power to get a 3D printer's heated platform up to an acceptable temperature quickly?I seriously doubt it, unless it's one of those "print something the size of a house" things.What kind of bed heater are you using exactly?What voltage is the heating element designed to run on?What is its resistance?

author: PeteD
author_rank: Newbie
timestamp: June 29, 2015, 03:09:06 pm
content: Quote from: LukeW on June 28, 2015, 12:51:20 pmThe heated bed is a 12" diameter plate with the heating element in 1/8 inch thick fiberglass. On top of the heating plate there is a 1/8" thick aluminum disc, a 1/8" thick borosilicate glass disc, and a 1/32" thick disc of PEI. The 100K thermistor is attached to the heating plate.The Onyx heating plate was designed to run with 12V and 24V PSUs. I am currently using a 12V 450W power supply.I haven't measured the resistance of my heating plate, but it is advertised to have a 1.1 Ohm resistance at room temperature. Some users have seen a higher RT resistance on their plates.

author: Magicmushroom666
author_rank: Contributor
timestamp: July 02, 2015, 05:01:24 pm
content: Hi, I'm running an onyx heated plate for my printer too. I'm using 18V at whever current that takes, not actually measured it recently, but 18V was enough to get good heat up times without needing silly sized cables etc.You could save on the SSR and just use a mechanical relay to save costs, as long as the controller is just off-on. I've used a relay salvaged from a car which happily manages the current.

author: PeteD
author_rank: Newbie
timestamp: July 06, 2015, 05:46:41 pm
content: Quote from: Magicmushroom666 on July 02, 2015, 05:01:24 pmSomeone please correct me if I'm wrong, but I believe wire gauge is determined by amperage not voltage.I did look into using an EMR instead of an SSR, but too may people who have used EMRs in their printers have had issues with false signals messing up the print when the EMR switches. Avoiding that problem is worth the cost of an SSR to me.

author: ruffy91
author_rank: Regular Contributor
timestamp: July 06, 2015, 06:57:12 pm
content: Quote from: PeteD on July 06, 2015, 05:46:41 pmThe heated bed is fixed resistance so more volts means more ampere.

author: PeteD
author_rank: Newbie
timestamp: July 06, 2015, 07:32:52 pm
content: Quote from: ruffy91 on July 06, 2015, 06:57:12 pmThe heated bed is a fixed resistance, so more volts means morepower dissipated, which is the purpose of this exercise.However, the amperage flowing through the wires to the heated bed should max out at 30A, and the wire gauge won't change between 12V or 18V or 24V.

author: ruffy91
author_rank: Regular Contributor
timestamp: July 06, 2015, 09:40:02 pm
content: No. If the bed is 1.1 Ohm you will get a max. current of 12V/1.1Ohm=11A for 12V or 22A for 24V

author: Riotpack
author_rank: Regular Contributor
timestamp: July 06, 2015, 10:42:01 pm
content: big rectangle power resistor bolted to the plate and connected to the mains through a thermostat

author: LukeW
author_rank: Frequent Contributor
timestamp: July 07, 2015, 03:20:08 am
content: Quote from: PeteD on July 06, 2015, 07:32:52 pmHigher voltage means less current for the same power transfer, meaning thinner wires and less supply current needed, less voltage drop in the wires and less heating,ifthe heating element is designed for the appropriate voltage.Let's suppose that your particular 3D printer design needs 100W of bed heater power. This depends on the area, the materials, airflow, thermodynamics, desired temperature at the surface etc.But I think 100W is plausible - not 720 watts.Suppose you've got a 12V power supply available.So you use a 1.44 ohm heating element, and this gives you 100W of heat power, ignoring loom resistance. And your power supply needs to deliver 8.33A.But you need to watch the loom resistance - suppose your total loom resistance is 0.5 ohms.Now your power delivered to the heating element is only 55 watts, so it won't get hot quickly enough. And you're dissipating about 20 watts in the wires.So suppose you keep the same heating element, and crank the supply voltage up to 24 volts.Ignoring loom resistance, you're now dissipating 400W in the heated bed, so it will get way too hot. The PSU also needs to be able to deliver 4x the power, twice the current it did before - 16.7 amps at 24V.If there's 0.5 ohms of loom resistance then the heater power has dropped to 220 watts - still probably too large.And you're dissipating 77 watts in the wiring loom.One solution is to use thicker cable, but a pound of copper costs a bit.A better solution is to use a higher supply voltage if you can, the highest voltage that is practical, andchoose the bed heater resistance appropriately. And use relatively thick connecting wires.Suppose the supply voltage is 24V and the heating element resistance is 4.7 ohms and the loom resistance is 0.5 ohms. The element resistance should be high relative to the cables.Then you'll have 100W at the heating element, a manageable 10W in the wiring loom, and a supply current requirement of 4.6A which is pretty manageable.