Limiting high current, low voltage - Page 1
Forum section: electronics / beginners

author: derGoldstein
author_rank: Regular Contributor
timestamp: May 17, 2016, 03:33:20 am
content: I've been trying to solve the following problem: How do I create a high constant current source at very low voltages, using a buck converter? When I say low voltage I mean between 0.5V to 2V, and when I say high current I mean the lowest useful output is 20A, and it can get up to 200A.The reason I need this setup (skip this paragraph if you don't care) is to power large, "short" coils to create very powerful magnetic fields. This has to be kept up for at least 5 seconds, but probably no longer than 30 seconds (so a capacitor-discharge setup won't help). It's designed to test the reaction of various chemicals to magnetic fields, as well as separate ferromagnetic materials (like magnetite) suspended in fluids. These coils sometimes have a resistance of less than 0.02 ohms.Right now the power supply that's powering these is a massive, silicon-iron toroidal transformer. It hooks up to the mains and has a massively thick secondary winding (3 or 4 turns of it) that feeds into a brick-sided rectifier. It can deliver a constant current of between 200A and 800A, and looks like something victor frankenstein would use. It's moved around on a cart and weighs maybe 30kg.I realize that there's a practical physical limit to what you can get out of a switchmode converter, due to the thickness of the conductors. However, currents as "low" as 20 amps are useful as well, especially if such a power supply can be made somewhat portable (so they can be moved around the lab more easily).Here's an example of a buck converter that I've been working with:http://www.ti.com/product/TPS40057It can indeed output 20A if it's cooled properly, and deliver voltages as low as around 0.75V, but what I'm missing is the constant-current component. If I feed the output of that circuit through a 0.02-ohm coil, the voltage isn't low enough and the current-limit is tripped, turning off the controller. What I've been doing for testing purposes is the simplest thing possible -- I place a length of nichrome wire in series to limit the current (so much of the power is wasted on heating that wire).There are all sorts of extremely high-current buck converters (like the TPS40090 monster) that can deliver voltages as low as 0.5V, but all of the ones I've seen only have a current-trip limit, not a constant-current function.Is there a way to "add" constant-current functionality to an existing buck converter? (I realize that that's a pretty broad question...)The circuit (device) doesn't have to be too small or light, and its efficiency doesn't have to be extremely high (though ~75% would be nice). Should I pump the output of a non-constant-current buck converter through a very low-voltage constant-current circuit (thereby burning a lot of it as heat)? Is there a better way anyone can think of?Sorry for the extremely long post. Any ideas would be helpful, thanks.

author: danadak
author_rank: Super Contributor
timestamp: May 17, 2016, 10:19:00 am
content: Might be prudent to call FAE at Fairchild Semi, or On Semi, or TI or ST.... with thatquestion.Regards, Dana.

author: sjd.aliyan
author_rank: Contributor
timestamp: May 17, 2016, 10:48:58 am
content: You want a variable voltage constant current?I think LM358 would be good for you but i'm not sure about the voltage range.

author: Monadnock
author_rank: Contributor
timestamp: May 17, 2016, 11:00:42 am
content: Looking at that TPS40057 you can disable its built in current limit and wrap your own current limiting circuit around it. You would need to sense the current and create a circuit to override the voltage feedback loop when in current limit.

author: daddylonglegs
author_rank: Contributor
timestamp: May 17, 2016, 11:32:54 am
content: Yes you can control the switching regulator to a constant current. You put a current sense resistor in series with the output, measure the voltage with a differential amplifier and feed this voltage - with suitable slope and offset adjustments and filtering - into the feedback pin (VFB) of the regulator instead of the normal voltage feedback. Monadnock suggests combining both voltage and current control so that whichever one is at it's limit overrides the control loop.The difficult bit is the frequency response, it is now your job to make the control loop (your filtering + the internal control filtering and control) stable. You might well find yourself slugging the response very heavily.These days there are at least two common uses for constant current supplies in the tens of amperes: battery charging and LED driving, and so you can get controllers that will do constant voltage/ constant current (CV/ CC)  output such as:http://www.linear.com/product/LT3741http://www.linear.com/solutions/3684This is not designed for a very inductive load such as yours so you will probably have to work on the control loop. There is an LT Spice (a simulation program) model for the circuit linked to from the product page so you can simulate your circuit.There are guides to controlling DC-DC converters such as:http://www.linear.com/docs/46311Since the rate of change of current through your magnet will be roughly proportional to the voltage across it, if you keep your maximum output voltage low your circuit won't try to ramp up the current too quickly, which should help.With a high current and an inductive load you will need to protect your circuitry. At a minimum, put a schottky diode in parallel with your output. Also consider overvoltage protection and fusing.I'm sure you know more than I do about the hazards of big magnets and high currents. Stay safe. "constant voltage" "constant current" is a good search term if your looking for more info.

author: derGoldstein
author_rank: Regular Contributor
timestamp: May 18, 2016, 01:52:31 am
content: Thanks for the replies, at least now I know what to look for, battery charging circuits have been a useful reference.A couple of points about the load's inductance -- there isn't much of it. It's not a case where there's a huge coil around a cylinder of iron which would create an initially massive impedance, followed by a huge back emf when the power is switched off. In fact, ferrous metals are removed from the environment prior to tests so that they don't effect the field. The materials effected usually have very low inductance, like 5 grams of Fe3O4 suspended in liquid, or a slightly larger amount of non-ferrous alloys. For example, one of the coils is designed to go around an erlenmeyer flask in order to create a diamagnetic or paramagnetic effect.You can see some examples of similar experiments here:Except in this case there's no iron and no permanent magnets, only an energized coil.What I don't understand is how the TPS40057, which can output voltages down to 0.7, would handle a case where it needs to output less then that. Does it switch off periodically when that happens and lets the inductor's voltage drop further, or does something else happen?I'm attaching a circuit that's close to the one I'm using, it's from the original datasheet. By modifying the value of Rbias in relation to R1 I can get the output voltage down to ~1.35V with no load, and possibly down to 0.75V at maximum load (close to 20A, more around 18.5A -- it's kind of difficult to test this, I have to use resistance wire which means that the resistance increases with temperature and therefor fluctuates). If I were to hook that up to a 0.02 ohm load it would draw over 37A, so it would need to go lower than that. If I can't manually get the voltage down lower by modifying the resistance of Rbias, how would an output of an op-amp do it? Would I have to modify the feedback network?

author: Psi
author_rank: Super Contributor
timestamp: May 18, 2016, 12:23:15 pm
content: Here's what i would do.Look for an old computer motherboard that supports massively power hungry CPUsSome of those old AMD cpus draw 200W at 1.0-1.5VLook up the datasheet for the Vcore switchmode controller and see how the output voltage is set.Figure out how the controller is used on the motherboard. Fets/Diodes/Inductors. It will likely be multiphase.Either re-purpose the motherboard itself or rebuild the circuit on your own PCB. You can unsolder and reuse the existing inductors/diodes/fets since they are perfectly suited for this.Now you have a powersupply that can provide a couple of hundred amps at probably 0.6 - 1.8V (depending on the switchmode chip specs) from 12VDC input (sometimes the circuit uses 5V or 3.3V input but usually its the 12V on the high power motherboards)Note: Multiphase switchmode converters can be a bit temperamental. You need to make sure you design the pcb layout correctly. So its best to tap into the motherboard directly and use it as is. Once you have it working you can look at making your own copy.

author: Monadnock
author_rank: Contributor
timestamp: May 18, 2016, 12:36:36 pm
content: Quote from: Psi on May 18, 2016, 12:23:15 pmThe OP needs a constant current output, those old motherboard controllers will not be set up to do this.

author: Psi
author_rank: Super Contributor
timestamp: May 18, 2016, 12:42:22 pm
content: Adding current regulation should be trivial.Driving large coils of wire doesnt need a fast feedback loop.hell you could do it in software with a 8bit mcu setting the switchmode controller output voltage from a hall effect current sensor on the mcu adc. Just ramp up the voltage until you get the current you want and rampdown if overshoot is detected. The current will change somewhat from temp as the coil heats up but that's still really slow from a feedback perspective and well within what can be done in software.

author: Monadnock
author_rank: Contributor
timestamp: May 18, 2016, 12:46:48 pm
content: Quote from: Psi on May 18, 2016, 12:42:22 pmNothing is trivial at 200A.

author: Psi
author_rank: Super Contributor
timestamp: May 18, 2016, 12:51:47 pm
content: Its just current, dont let it scare youThis would allow a 120khz feedback loophttp://www.digikey.co.nz/product-detail/en/allegro-microsystems-llc/ACS758ECB-200B-PFF-T/620-1323-ND/2042748

author: Marco
author_rank: Super Contributor
timestamp: May 18, 2016, 01:08:16 pm
content: You could add your own high current buck to a low voltage switch mode power supply (5V 400W supplies are relatively cheap). They don't have current limits, wide traces are no limitation ... you can always parallel supplies if need be (can also reduce ripple).That said, how aboutthis monster? It has trim inputs on the outputs, you might be able to abuse that to turn it into a constant current power supply (they tell you to consult an engineer for trimming below 5V, but that's for the 12V module example).

author: Marco
author_rank: Super Contributor
timestamp: May 18, 2016, 04:10:19 pm
content: The5V module datasheetsimply says it can be trimmed down to 10% of 5V, so you should be able to use it for a slightly dodgy 80A current source down to 0.5V with a relatively simple opamp circuit and a shunt. You could combine the outputs too with slightly more complexity.

author: joeqsmith
author_rank: Super Contributor
timestamp: May 18, 2016, 11:08:30 pm
content: What sort $ are you willing to spend? What sort of response time? I am thinking similar to Marco. Driving an off the shelf supply like the Vicor in constant voltage to get your constant current. Post some details about what you are needing (response, accuracy, drift......) Maybe there is something out there you could just buy.

author: derGoldstein
author_rank: Regular Contributor
timestamp: May 19, 2016, 02:30:57 am
content: I've been looking at custom buck converters and I'm beginning to think that would potentially be *less* complicated than modifying the feedback to an existing IC. The reason being that the accuracy of the output is significantly less important than anything that's designed to power sensitive digital electronics. The output amperage can drift by 10% and it wouldn't be that important. I'm also not too concerned about efficiency, and the size of the device can be as large as a shoe box and still be more practical than the beast they're using now (though I'd prefer it to be smaller, obviously).I don't really have too much money to work with, I just saw the device they were using in the lab and thought I could design something much smaller, or at least an additional device they can use. I don't mind spending $100 or so in parts (along with any parts I already have), but this isn't a job, I'm just in it to learn (if I come up with something they can use I'll just give it to them).I think what I'll try to do is construct the "power stage" (I'm probably not using that phrase correctly...) of a non-synchronous buck converter using over-powered parts: a MOSFET driver, one or more very large MOSFETs (I have a few HUF75344G3's laying around), a huge diode (I've used RURG5060 for SMPS stuff in the past), and the highest current-rated inductors I can find. I'll make sure that the traces are huge and that the MOSFETs/diode have large heatsinks attached. Given all of this, and that the power I'll feed into it is current-limited, I think I might have a circuit that (optimistically) "can't" blow up. Worse case is that the MOSFET is locked in the on condition and drives all of the current into the load, but that will be limited by the supply, and wouldn't reach the 285W power dissipation limit of a HUF75344G3.When I've constructed this power-handling part of the circuit I'm left with switching the MOSFET. I'll place a sense-resistor at the output to measure the voltage drop across it, and of course measure the output voltage. Using these two parameters I'll figure out the PWM I need to drive the MOSFET with. I wonder if an arduino would be able to react fast enough or if I'll have to use a fast op-amp or comparator as well (or potentially a faster MCU like a Cortex M3).Do you guys think this is a plausible plan, or would I be in over my head?

author: Marco
author_rank: Super Contributor
timestamp: May 19, 2016, 03:52:50 am
content: Quote from: derGoldstein on May 19, 2016, 02:30:57 amYou will need a low voltage power supply to start from regardless.You could build the circuit for the Vicor on a breadboard (apart from the shunt of course). A high current buck converter not so much.QuotePlenty of people do this with micros for some reason and they do often succeed ... but it seems easier to me to just use buck controllers which can use external switches together with a low offset voltage opamp (even controllers designed for constant current will have too high a feedback voltage at these currents, so you will need some amplification).

author: joeqsmith
author_rank: Super Contributor
timestamp: May 19, 2016, 11:48:25 am
content: So you are not sure on the response time you need or maybe I missed something. Is it really just acting as a resistive load and you want to push a fixed DC current through it for a test, then adjust it to another level for your next test? So the slew rate/ settling time.... really is not important? If this it what you are thinking, then do you only need it to be a current source because of the resistance changes with heat ... and need to keep a constant current through it? In other words, the control loop could be very slow (in 10s of mS)?Sounds like you really want to make your own and just keep the currents to a minimum. I was thinking about about what you could use to replace the old system with something more modern.

author: joeqsmith
author_rank: Super Contributor
timestamp: May 21, 2016, 02:10:24 pm
content: If you are still interested in this subject, I needed a 20+ Amp current source for some testing I plan to run in the near future. It may provide some insight.https://www.eevblog.com/forum/testgear/hear-kitty-kitty-kitty-nope-not-that-kind-of-cat/925/

author: IanB
author_rank: Super Contributor
timestamp: May 21, 2016, 02:56:21 pm
content: Another way to look at this problem is to use a different coil winding for the magnets.For instance, suppose you have 100 turns of a thick wire carrying 100 A. If you have 1000 turns of a thinner wire carrying 10 A occupying the same volume it will produce the same magnetic field intensity. The advantage now is that instead of requiring a power supply to provide 100 A at 1 V, you could use a supply providing 10 A at 10 V. The second power supply is much easier to achieve than the first. For instance, any number of inexpensive bench power supplies can do that.

author: derGoldstein
author_rank: Regular Contributor
timestamp: May 22, 2016, 05:10:19 am
content: joeqsmith: the resistance won't increase too much due to heat, maybe 20%. I'll look through the post you mentioned, I've got a lot of reading piled up to try to understand SMPS principals.IanB: Actually, that's the one thing I *don't* have control over. These aren't coils of wire the way you'd imagine them to be, here are 3 examples:1) Roll coil: imagine a roll of duct tape where the tape is the conductor and the adhesive is the insulator. This isn't a helix -- its cross-section is a spiral. These coils don't usually exceed 40A. Most of them look like a cup with relatively thin walls.2) Bitter-plate coil. These are beautifully constructed, you can see an example of really huge ones here:https://nationalmaglab.org/news-events/feature-stories/making-resistive-magnetsObviously the ones I'd have to power are nowhere near that size, but some of them can get up to 200A. The lowest useful amperage on these is 40A.3) Pipe coils. This is what you'd expect -- copper pipes in a helix that you can pump water through to cool them down. The largest ones they have go up to 3kA (they don't have a supply that can get that high, but 800A is enough for most tests using these coils). I'm not even going to try powering one of these.If there's a way to use a more "typical coil" with more windings then they will, they also have a few lower-amperage power supplies, but the problem I'm trying to help with is the supply used to power the huge, "short" coils mentioned above.From what I've been reading so far, a non-synchronous buck converter will have a problem with very low voltage output -- the voltage drop across the freewheeling diode will turn too much of the power into heat.So I've looked over several introductory documents to synchronous buck converters, but I'm having a noob problem understanding something: Both mosfets in a synchronous buck converter are n-channel. The high-side mosfet works the same as it does in a non-synchronous converter, but the low-side mosfet seems to open a path for the current to flow from the inductor to ground... Isn't this the opposite of what it's supposed to do? Wouldn't it need to allow current to flow from the ground into the inductor, allowing the inductor to act as a current source? I'm sure I'm missing something really trivial here, but all of the synchronous buck converter circuits seem to have the same basic layout, and they all show the low-side mosfet allowing current to flow from the inductor to ground... Why is that?

author: Marco
author_rank: Super Contributor
timestamp: May 22, 2016, 05:54:33 pm
content: When the gate is high on a MOSFET the resistance between drain and source is low impedance, regardless of the direction of the current.So what happens in a synchronous buck converter is that when the upper MOSFET turns off, the body diode of the bottom MOSFET takes over for a moment (this is called the dead time, necessary to prevent shoot through) after which the bottom MOSFET turns on and the inductor current can come from ground with minimum voltage drop.

author: joeqsmith
author_rank: Super Contributor
timestamp: May 23, 2016, 11:41:25 pm
content: I did finally post a video of the programmable voltage power supply being used for a current source. Basically I am driving a small section of wire with it. Don't waste your time with the whole video. Link shows where I explain a little about the controls.https://www.youtube.com/watch?feature=player_detailpage&v=fQowDZstguw#t=464Quote from: derGoldstein on May 22, 2016, 05:10:19 am

author: derGoldstein
author_rank: Regular Contributor
timestamp: May 24, 2016, 02:52:11 am
content: Quote from: Marco on May 22, 2016, 05:54:33 pmThanks, I had no idea of that property of the mosfet, I thought it blocked current both ways. That's a hell of a thing NOT to know after being an EE hobbyist for so many years...Does this mean that the current from the inductor could creep back through the switching mosfet to the power supply? I understand that the voltage would be lower than what the power supply would be providing, but could there be a back-EMF situation?Quote from: joeqsmith on May 23, 2016, 11:41:25 pmThanks, great vids, I'll queue them up and watch.

author: daddylonglegs
author_rank: Contributor
timestamp: May 24, 2016, 09:55:09 am
content: derGoldstein,An inductor stores energy in it's magnetic field that is proportional to the square of the current through it. This means that changing the current through an inductor requires putting energy in or out and so can't happen instantaneously, an inductor 'resists' changes in current.When the switching transistor is on the current through the inductor gradually increases (going from left to right in your diagram), storing energy in the inductors magnetic field. The voltage across the inductor is dropping in the direction of the current flow so power is going from the circuit into the inductor.When the switching transistor is off the current in the inductor will keep on going from left to right (because it can't change instantly) and the voltage across the inductor will change to make this happen. So the voltage will change to an increase in voltage in the direction of the current flow (this is essentially what a back EMF is). The output voltage will not change instantly (because the output capacitors do for voltage what the inductor does for current) and so the voltage at the switching node will drop. When this voltage drops to one diode drop below ground the diode starts conducting and current continues to flow through the inductor from left to right. Because the voltage across the inductor now increases in the direction of the current flow power is now going from the inductor into the circuit and the current through the inductor drops.Switching a transistor in parallel with the diode will save the power dissipation of the current going through the diode drop and the voltage of the switching node will be zero rather than a diode drop below zero.If the circuit is operating in Continuous Conduction Mode (CCM) the current through the inductor never goes to zero and is always flowing from left to right. At low currents the circuit may operate in Discontinuous Current Mode (DCM) and the current through the inductor will drop to zero but should NOT reverse. The current will remain at zero until the switching diode turns on and the next cycle starts.What level of ripple current are you targetting? You will likely find that the ripple current you're getting from your switching converter is quite high. Go for the largest inductor that can handle the current in the switching circuit and consider deliberately adding an inductance in series with your coil (but keep it away from that coil's magnetic field).Hang on to that nichrome wire, having some resistance in series with your coil will be good for damping out resonances between the output capacitors and the inductance of the leads and the coil (and any extra inductance).Using synchronous rectification is good for power efficiency and reducing heat dissipation. I think targetting that efficiency gain would be a false economy in your case so feel free to use diode rectification if your diode can handle the current.

author: derGoldstein
author_rank: Regular Contributor
timestamp: May 25, 2016, 07:56:01 pm
content: daddylonglegs:Thanks for the very detailed explanation, I'm having to relearn analog principals, I've been doing digital for too long.Interesting you brought up ripple current, this is something that I'll have to filter out (specifically the high-frequency range) as much as possible even at the expense of severely reduced efficiency. While gradual variation of current going through the coils during the experiment/process, even as much as 20%, won't effect it that much (provided that this variation is logged), noise is a different matter, because rather than having a subtle change in the intensity of the magnetic field, suddenly you're introducing vibration.This can lead to *very* different results, depending on the process taking place. For example, many of these processes involve introducing ferromagnetic particles into a fluid, stirring it, and then applying a magnetic field in order to force some type of mechanical separation. A slightly stronger or weaker field will change the rate at which this occurs, but noise will cause physical vibration within the effected fluid, potentially leading to something completely different than intended.Actually, this is something that's sometimes done intentionally, usually with far lower currents. You can apply an AC current to one of the coils, causing the target material to vibrate at a certain frequency. If the frequency is low enough you can sometimes see the vibration with the naked eye.So for the purposes of this design, I have to keep the noise to a minimum. I'm going to have to experiment with LC filters.This is really drifting from the original question (not sure about forum etiquette here...), but -- since I have to drive a high-side mosfet, I'm going to either have to use a high-side driver or add a boost converter to the circuit, correct? This isn't a problem, I'm just wondering if there's an alternative.Some of the old integrated buck ICs, like the LM2596, just used darlington transistors instead, but doing that here would be very wasteful I assume. I've seen that some newer integrated circuits that appear to be using mosfets internally, but don't appear to need an external switched capacitor to drive them.

author: Marco
author_rank: Super Contributor
timestamp: May 25, 2016, 09:15:50 pm
content: If you are worried about noise forget about synchronous buck, just use a regular buck. For instance withone of these.If you are designing a high current buck converter some extra supply voltages to efficiently drive the MOSFET shouldn't even be on the radar. just use modules if you don't want to spend the time (bucking from relatively low voltage a bootstrap circuit in a driver or controller might not be enough, if you want 10V gate drive 5V input won't get you there).