Load Sharing Circuit - Page 1
Forum section: electronics / projects

author: klr5205
author_rank: Regular Contributor
timestamp: February 05, 2014, 05:44:41 pm
content: First post! Found eevblog on youtube about a month back and I've been hooked ever since!Here's the idea, I have in my possession a pair of lsn-5/10-d12h synchronous buck regulators which I would like to run in parallel to create a 5V 20A rated supply for a hobby project I'm working on. Since I don't want the regulators to fight each other, I was trying to come up with an idea for a load sharing circuit. Toward that end, I was thinking of creating a "master/slave" configuration where one regulator output is the reference voltage and I adjust the other regulator's output using a differential amplifier with to monitor the lets say 10-50mV difference in output voltage, and using its output to control the regulator's TRIM pin. This pin can provide up to +/-10% of adjustment.I have a few questions regarding this:Is it wise to actively control the trim pin?With a 1000uF output cap, I've measured ripple to still be about 30mV which is significant... Should I try to filter or possibly rectify this before the diff amp?Should I just run the sense wires and call it a day, leaving the small differences in set points to sort themselves out?Given that wire resistance will be significant at such high currents, should I not even bother and let the voltages balance out over drop in the wires?My load could conceivably draw 14A, but will generally be half that.Thanks, I look forward to hearing your opinions!Edit: I forgot to mention that I am bucking a constant 12V down to 5V, but you'd see that in the datasheet

author: johansen
author_rank: Super Contributor
timestamp: February 05, 2014, 05:58:01 pm
content: it depends on the control loop and topology, they are synchronous buck so what's going to happen is they will always fight each other, according to the rdson of the fets and the inductor.but if they are cycle by cycle current mode controlled.. and if they don't care if the output voltage is high, then they won't backfeed the power supply and they won't fight each other.you'll still have to adjust the trim to get them to share current equally however.

author: klr5205
author_rank: Regular Contributor
timestamp: February 05, 2014, 07:23:53 pm
content: I don't see in the datasheet where it specifically states whether the device is current or voltage controlled, however I do see in the block diagram that the PWM controller is fed a signal from a current sense block, in addition to the error amp. I wonder if that is only for the over current protection or if we can infer current mode control?In any case, given that these regulators are guaranteed regulate within 3% over all line/load/temperature, do you think its overkill to do anything more than trim the outputs to be very closely matched at light loads?

author: kaindub
author_rank: Regular Contributor
timestamp: February 06, 2014, 12:35:22 am
content: One simple way to do what you want, and it,s done in the power supply industry a lot is to just 'or' the outputs using diodes.Connect a diode that can handle the output current of the power supply to each supply output. Connect the cathodes together. That's your output. Adjust each supply separately to get 5v at the cathodes.Using schotky diodes will reduce the losses.This system works quite wellRobert

author: klr5205
author_rank: Regular Contributor
timestamp: February 06, 2014, 12:59:13 am
content: Thanks, you're actually the second person to suggest this to me but I was initially turned off from the idea thinking the forward voltage of the diodes would increase too much as current increased. However after looking over a few datasheets it looks like the difference between 1A and 10A through the diode may only be around 200mV which should be perfectly fine for me, as well as being simple to implement!

author: Richard Head
author_rank: Frequent Contributor
timestamp: February 06, 2014, 06:06:22 am
content: The way it's generally done is by means of a "current bus".The current bus is a third wire that is connected to all PSU's in the system.Each PSU has an additional two op-amps. The one op-amp is wired as a shunt amplifier and accurately measures the load current of its power supply.The other op-amp is wired as a differential amplifier and compares the load current of its PSU to the average load current that is represented by the current bus voltage.The current bus is created by feeding the differential amplifier output via a series resistor.The op-amps feeding the current bus (via a suitable resistor) can both source and sink current. Thus the bus voltage is an analog representation of the average system current.That is, in a nutshell, how it works.I have designed and built systems that use this principle and the individual PSU's share load current to within the tolerance of the components.There are some shortcomings with the principle (such as when a PSU fails) but have been addressed by minor circuit modifications.Dick

author: codeboy2k
author_rank: Super Contributor
timestamp: February 06, 2014, 08:09:38 am
content: Hi Dick Head( please tell me it's not really your name ?? )Thanks for posting that. I was Googling around and I must have lost my Google mojo because I couldn't find anything that described the implementation. Thanks for your post.I'm not experienced in building load sharing supplies, so I have a few questions:1) In the system you describe, you still need the diode-OR, correct? otherwise a shorted or failed output of one PSU will bring down the system.2) is it better to OR at the positive outputs, or at the returns? or does it even matter. I think it doesn't matter, and I've seen designs that do it either way. I just wanted to know if there was any reason to prefer one way over the other way. Perhaps I would use it on the returns if I wanted to make the diode OR an active FET controlled diode.3) why would PSU failure cause problems? if it shorts and disconnects then wouldn't the others take up the load automatically via the current bus voltage? It seems like a rather slick way to share and communicate load currents, actually.Perhaps you have to be careful with the loop bandwidth here, not too fast to make it oscillate but not too slow that it can't react in time.Thanks for your comments!

author: Richard Head
author_rank: Frequent Contributor
timestamp: February 09, 2014, 12:50:19 pm
content: Codeboy2kSorry for the late reply. I normally only come to this site during working hours!I'm not experienced in building load sharing supplies, so I have a few questions:1) In the system you describe, you still need the diode-OR, correct? otherwise a shorted or failed output of one PSU will bring down the system.You are quite correct in that a failed output diode can cause a distributed power system to fail as the faulty module will pull the rest down. However, a failed OR-ing diode AND output diode will also cause a system failure in the same way, so it's debatable whether the OR-ing diode is really required. I've used both approaches before without problems. The OR-ing diode does impact efficiency though. What I doe do however is to use a high current fuse (automotive blade type) between the output diode and the caps. This allows the fuse to open very quickly should a diode short.can also2) is it better to OR at the positive outputs, or at the returns? or does it even matter. I think it doesn't matter, and I've seen designs that do it either way. I just wanted to know if there was any reason to prefer one way over the other way. Perhaps I would use it on the returns if I wanted to make the diode OR an active FET controlled diode.2. I've only ever OR-ed on the positive rail since it allows the op-amps to share a common ground.3) why would PSU failure cause problems? if it shorts and disconnects then wouldn't the others take up the load automatically via the current bus voltage? It seems like a rather slick way to share and communicate load currents, actually.3. If module fails it's op-amp output is pulled to ground and it's output resistor pulls a certain amount of current to ground thereby causing a small current bus error. A MOSFET switch on the output avoids this problem.I normally set the loop bandwidth at one tenth of the voltage loop or slower.Richard

author: codeboy2k
author_rank: Super Contributor
timestamp: February 09, 2014, 01:50:24 pm
content: Thanks for those replies, Richard.Very enlightening.only one other questionYou said you put a fuse between the output diode and the caps. I assume this is a series diode at the output, inside a PSU module, which is why you don't need the OR-ing diodes on the outside then (on the backplane)So if this diode fails short and starts to oversupply the backplane the fuse will blow. Nice!Thanks once again!Cheers!

author: Richard Head
author_rank: Frequent Contributor
timestamp: February 10, 2014, 05:37:53 am
content: Codeboy2kThe fuse arrangement can only really be used if the power system is charging batteries. The batteries can supply the instantaneous fault current required to open the fuse WITHOUT the system voltage dropping too low.Basically the fuse avoids having to use the OR-ing diode thereby improving efficiency.Rich