How does current limiting work in a PSU? - Page 1
Forum section: electronics / beginners

author: Red Squirrel
author_rank: Super Contributor
timestamp: May 04, 2016, 07:12:24 am
content: Let's say I want to dynamically limit current and voltage in a PSU, let's just say a buck converter. For voltage, I just adjust the duty cycle of the high frequency wave and smooth it out, and get my voltage, then I monitor that voltage for any changes and adjust the duty cycle accordingly.But let's say I want to output a certain voltage but also limit the current, how is that done? For example battery chargers (smart ones) how do they do it? I know a lot of them will just reduce the voltage but smarter ones will actually control the current. How do they do that?I've also heard references to controlling the input or output impedance, for example a MPPT solar controller will actually change the input impedance to match the output impedance of the solar panel, what exactly does that mean?

author: Zero999
author_rank: Super Contributor
timestamp: May 04, 2016, 07:59:01 am
content: Current limiting is done by monitoring the output current and adjusting the duty cycle of the PWM to keep it within the desired limit.

author: tron9000
author_rank: Frequent Contributor
timestamp: May 04, 2016, 09:14:56 am
content: short version is how Hero describes.Long version is that there is usually a shunt resistor in series with the output, usually a very low value, high power resistor (order of 1R to 0.01R depending on how much current).Across the shunt resistor is an amplifier: commonly a differential amplifier or one specifically designed for current monitoring application based on differential configuration.When current flow through the shunt resistor, it gives rise to voltage (ohms law), this voltage is then amplified by the amplifier and then passed back to the power control cct to limit the current and or voltage (CV or CC modes).Its essentially creating a feedback signal. so this can be used to control PWM of Buck/Boost converter, or linear reg, up to you. But that's how its commonly done.

author: Red Squirrel
author_rank: Super Contributor
timestamp: May 04, 2016, 10:11:35 am
content: Yeah I knew about using a shunt for current measurement, but if you are controlling the PWM isin't that going to lower the voltage and not the current? I guess it would indirectly lower the current too as it would reduce how much power the load is taking. Would you then have a boost converter that brings the voltage back up to the selected level? Ex: if you have a 12v PSU that delivers 1amp vs one that delivers 12 amps, what does the 1 amp one do differently?

author: jitter
author_rank: Frequent Contributor
timestamp: May 04, 2016, 10:25:32 am
content: Do you realize that V = I * R means that if you keep R constant and vary I that V has no choice but to change as well?

author: wraper
author_rank: Unknown
timestamp: May 04, 2016, 10:27:34 am
content: Quote from: Red Squirrel on May 04, 2016, 10:11:35 amQuoteDidn't know you can control the current without changing the voltage.

author: tron9000
author_rank: Frequent Contributor
timestamp: May 04, 2016, 10:31:03 am
content: If you connect a load to the output of your constant current supply, that load is going to give rise to a voltage (again ohms law), you want that voltage to be 12V, you monitor the voltage instead of the current, and use that signal to adjust the current from whatever topology power supply your using: Like a simple buck converter.

author: Psi
author_rank: Super Contributor
timestamp: May 04, 2016, 10:33:07 am
content: The real question to think about is...What happens if you power a 1A constant current source from a 950mA current limited powersupply.

author: CJay
author_rank: Super Contributor
timestamp: May 04, 2016, 10:39:14 am
content: Quote from: Red Squirrel on May 04, 2016, 10:11:35 amDifferent questions.Current limit and maximum available current are two different things.Maximum current is limited by what the components are rated for, a 1 amp rectifier diode for instance won't give an awful lot more than 1 amp before it overheats and fails, similarly a transformer rated for 1 amp will not be able to deliver much more than that before it overheats or you run into limits imposed by the resistance of the windings, magnetic effects with the transformer core saturating etc.Exceeding maximum current ratings is likely to shorten the life of the supply, possibly to an extreme of a few seconds or less.Current limit is a designed in feature that prevents you drawing more than a supply is designed to deliver to prevent damage (to the supply and/or load) and it can do this in a number of ways but they all AFAIK give the same results in a DC supply, a reduction in output voltage (possibly to zero).

author: Red Squirrel
author_rank: Super Contributor
timestamp: May 04, 2016, 10:44:28 am
content: Quote from: wraper on May 04, 2016, 10:27:34 amIs that not what current limiting does? Limit the current at a specified voltage? At least that's what I thought, or is it really as simple as lowering the voltage? Assuming we're not talking about fuses, then it just kills the flow completely.

author: tron9000
author_rank: Frequent Contributor
timestamp: May 04, 2016, 10:59:58 am
content: current limiting does what it says on the tin! it limits the current! by whatever means.you try and pull 1A out of a 500mA supply, you'll see the voltage will fall. If monitored from 500mA to 1A : you can actually characterise the impedance of the supplies output, which you touched on in your OP.

author: Red Squirrel
author_rank: Super Contributor
timestamp: May 04, 2016, 11:46:05 am
content: Hmmm ok that makes it much easier to understand then, that it actually lowers the voltage. I knew lowering the voltage would in effect lower how much current the device is pulling, such as a light bulb would go dimmer so not drawing as much current, but I always figured a current limiting feature in a PSU was more "fancy" than that, and that it managed to limit current while keeping voltage at a set point but just not delivering as much current to the load.Not too sure what low/high in/out put impedance really means though. Is it just a fancy term for load? So a PSU that is limiting the current would be high impedance as it's acting kinda like a high value resistor vs a PSU that is not limiting current will let any amount of current through so kinda like a low ohm resistor? I've read a lot on impedance and I can just never grasp the concept but hear it thrown around a lot in various tutorials and such that I should probably really understand it.

author: Zbig
author_rank: Frequent Contributor
timestamp: May 04, 2016, 12:00:16 pm
content: Quote from: Red Squirrel on May 04, 2016, 10:44:28 amNo, assuming constant equivalent resistance at the load, there is no magical way to alter the current without altering the voltage. There's no "simple" and "sophisticated" way of doing Constant Current; you can't get around the Ohm's Law or physics in general. By putting the power supply in CC mode, you simply state that current is the value you care about at the moment and the control circuit will lower the voltage as needed in order not to exceed the current you have set. Like others mentioned before, the current capacity stated on the PSU brick is just that - a capacity to provide stated current without being damaged. It doesn't mean that you'll damage your laptop by replacing original 19V 3A PSU with a 19V 4A PSU - the laptop will take only the current that it "needs" and the PSU will just work cooler as it won't need to pump all the 4A it's capable to.

author: jeroen79
author_rank: Frequent Contributor
timestamp: May 04, 2016, 01:47:10 pm
content: Quote from: Psi on May 04, 2016, 10:33:07 amIn trying to draw 1A from the .95A supply the load will reduce it's impedance until it is a short circuit and then the power supply will source 950mA through that short.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 04, 2016, 03:06:11 pm
content: If the architecture is voltage mode, your question is an excellent one.If the architecture is current mode, it simply can't deliver any more current than it's designed to go up to. There's no independent "PWM" variable, it only switches to maintain whatever current is required.You can see which is the superior method!Tim

author: jitter
author_rank: Frequent Contributor
timestamp: May 04, 2016, 04:15:28 pm
content: Quote from: Red Squirrel on May 04, 2016, 10:44:28 amYou really must try and understand Ohm's Law V = I * R, especially the implication that V and I cannot change independently. The change in one always results in a proportional change in the other. Even if you consider ideal voltage and current sources, there is no way around Ohm's Law.Perhaps an example will help:If you have a 10 ohm resistor and connect it to a 12 V supply capable of 100 A, it will not draw 100 A. Ohm's Law (rearranged as I = V / R) calculates that the current that will flow through the resistor will be 12 / 10 = 1.2 A. The PSU is comfortably capable of supplying it with lots of capacity to spare. It is operating in constant voltage (CV) mode.If you lower the resistance to 5 ohms, the current will rise to 2.4 A, but the voltage will remain at 12 V.Now assume you can limit the current to 1 A, but keep the voltage set to 12 V.Before you connect the resistor, the PSU will output 12 V just fine. As soon as the 10 ohm is connected, it would like to draw 1.2 A from the PSU. This doesn't happen, as soon as the 1 A current limit is hit, it will stop rising. So, V = 1 * 10 = 10 V. The PSU is now operating in constant current (CC) mode, and the voltage that results depends on the load connected.If you lower the resistance to 5 ohms, the voltage across it will drop to 5 V, but the current through it will remain 1 A.As you can see, a PSU operates either in CV or in CC mode. The changeover happens at the point where the current limit is hit.

author: Red Squirrel
author_rank: Super Contributor
timestamp: May 05, 2016, 02:22:06 am
content: Quote from: jitter on May 04, 2016, 04:15:28 pmYeah I think I get it now, that makes sense. So current limiting is nothing more than simply monitoring the current and then reducing voltage if the load tries to pull more than the limit, and by doing that the load will then draw less current. In the case of a constant current supply it would basically be doing the same except it's trying to keep the current the same and does not care about voltage at all so it will change the voltage till the current is at the set point. So a constant current PSU set to 100ma will raise voltage as high as it needs so it's delivering 100ma or as low as it needs. I guess if it's open that would be an exception though as you can't deliver any amps to nothing. I guess a constant current psu would also have an upper voltage limit so it won't necessarily always deliver the max current but will do so if the load is decent enough. Am I on the right track?Quote from: jeroen79 on May 04, 2016, 01:47:10 pmThis part I still don't quite understand, I know that the PSU may not be able to deliver what the load is trying to take, so it will either 1: fail (fuse, component etc) or 2: limit current by reducing the voltage (now I understand that part). But what exactly does it mean when you say it reduces the impedance?I have a lot to learn obviously but I do want to look at making my own PSU(s) mostly for fun, and one feature I do want to add is current limiting so that if I short it out or put a huge load it won't just blow up. So now I know I can simply do that by monitoring current and lowering voltage. So the programming logic would be to first lower/raise pwm till desired voltage is achieved as this will always be changing based on load, then check current and lower pwm if current is higher than desired. I imagine there are pure electronic ways of doing it as well such as using a separate shunt that if the voltage hits a certain point it makes a transistor take the load or something. Basically as a fail safe if MCU fails or for instant load spikes etc. But I can read up on those methods more now that I understand the main concept and that it simply entails lowering the voltage.

author: jitter
author_rank: Frequent Contributor
timestamp: May 05, 2016, 07:20:21 am
content: Quote from: Red Squirrel on May 05, 2016, 02:22:06 amYes, indeed, you clearly get it.Actually the working principles of voltage and current sources are the same, the only difference is what's compared to a reference and then adjusted if needed.Simply said, in a constant voltage source, a voltage reference is set and compared to the actual output voltage. If they differ, an amplifier tries to correct that. The result is a stable output despite varying loads.In a current source, a reference is set for the current (which is also a voltage, BTW) and then compared to the voltage across the shunt resistor in the output. The shunt resistor is a known value, so a voltage across it represents a current through it. Again these two values are compared and the output is corrected if necessary. The result is a constant current, despite varying loads.QuoteYes, everything must be within the limits of a CV or CC supply.In the example in my previous post with the 100 A PSU set to 12 V and a current limit of 1 A, you will have noticed that it operates EITHER in CV mode OR in CC mode, depending on the load. With too low a load, it cannot work in CC mode. With too high a load, it cannot work in CV mode.A fixed constant current source (e.g. a 350 mA LED driver) will have a range specified in which it can do this (e.g. 12-50 V).A fixed voltage source (e.g. a 12 V wall wart) will have a max current specified it can deliver (e.g. 1 A).Please note that fixed sources may not fully work as Ohm's Law predicts because there will be limiting circuits at work that may abruptly kick in to prevent damage.The above example can be tested with a lab supply with adjustable limits for both current and voltage.

author: dentaku
author_rank: Frequent Contributor
timestamp: May 05, 2016, 03:40:06 pm
content: This has been very educational. I was wondering about this but wasn't sure how to ask the question, especially without getting people jumping on me and saying "well, don't you now bla bla bla...). This is a beginners forum for a reason.I just found this page on the B&K website which explains this too.http://www.bkprecision.com/support/downloads/power-supply-guide.htmlpart way down where it says Power Supply Specifications Constant Current and Constant Voltage Mode

author: Buriedcode
author_rank: Super Contributor
timestamp: May 05, 2016, 04:52:05 pm
content: Would also like to add that switch-mode power supplies store 'packets' of energy in an inductor, which dumps these into the output capacitor. The inductor is essentially a current source, which means the voltage at the output can rise continuously if there is no load attached. This is why boost converters can 'boost' - the inductor is 'charged' and 'discharged' into the output cap pumping current into it, so its voltage rises. A boost also has one node of the inductor at the input voltage, so when voltage develops across it, it is added to the input voltage - that is why boost converters cannot output a voltage lower than the input (well, it can, but only by the forward voltage of the free-wheeling diode).A SMPS supply would reduce probably both the on-time of the inductor switch (how long it is connected to the input supply, ergo: how much energy is stored) and the frequency (how often it gets charged) if the load is light - the load isn't drawing much power. But also the inductor can only store so much energy in one go, so its 'on-time' has a limit, and because it takes time for this energy to be dumped in to the output cap/load that is another limit.Therefore the PWM you speak of, which I assume is the control of the switch in our SMPS has upper and lower limits. So as you draw more and more current from the supply, its output voltage starts to sag because you're drawing the same amount of current from the output cap, than is being pushed in, so its voltage can't rise. In that instance, assuming the SMPS controller has limits on duty cycle, the current limit is actually 'built in'.You can blow things up with short circuits in non-isolated supplies. Like the boost and the buck converter. In the boost the output is constantly connected to the input via the inductor and diode - short the output to ground and the full input voltage is now across the inductor (likely to have a very low resistance) and the diode (voltage drop of ~0.3V for a schottky). So the basic boost converter has no inherent current limit. The buck converter uses a high side switch - switching the inductor to the input power. This means if you short the output to ground, the input voltage appears across the switch and inductor. Often (but not always!) dedicated IC's will have measures in place to cope with shorted outputs and over-current - often a current sense resistor and an amplifier. In the case of the boost converter, that is pretty much in series with the output, in a buck converter this is in series with the switch.Really only meant to write a few lines... probably didn't really add anything >.<

author: Rick Law
author_rank: Super Contributor
timestamp: May 06, 2016, 10:46:48 pm
content: Quote from: Red Squirrel on May 04, 2016, 07:12:24 amNow that we got the confusion of "constant current" vs "current limit" out of the way, back to "how is that done?"In my own quest to understand, I decided to add "constant current" to a cheap Chinese adjustable voltage buck board. I used a XL4051E (LM2596 like) based board. In doing that, and in looking at how some other boards did it, now I feel I have a better grasp of how that is typically done - at least with the cheap boards.You may want to look at this thread, you can see how limit current can be done. Of course, there are many ways to skin a cat and this is but just one of the many ways.https://www.eevblog.com/forum/beginners/adding-cc-to-a-cv-buck-to-make-it-cccv/