constant current or constant limit - Page 1
Forum section: electronics / beginners

author: VEGETA
author_rank: Super Contributor
timestamp: October 21, 2016, 09:48:54 pm
content: Hello,I have a question about the CC mode in lab power supplies, if I put it to 1A for example, will it be a constant current source or just current limit?I want to make my own power supply using LTC3649 device and I already have a circuit ready to be made but I want to ask about this CC mode. I used the same circuit everyone uses which is putting an op-amp comparator of 0-2v (=0-2A) vs the measured current... and the output is fed into a transistor which is tied to the "set" pin of the regulator... if transistor is on, it pulls the set pin to ground making it 0v... so continuous operation of making it on and off will regulate the current, as Dave's circuit on his usupply.However, I don't think this will be a current source (where it outputs say 1A regardless of the load) but instead it will be a current limit where it allows the load to get any lower value of current but not more. I want the previous one.is there any way I can implement this? especially without using any shunt resistor to measure current because the device (LTC3649) has a unique way of measuring current. It has a feature of regulating current but I couldn't fully understand it and won't be a software controlled solution.Thanks!

author: BobsURuncle
author_rank: Regular Contributor
timestamp: October 21, 2016, 10:20:24 pm
content: On a lab supply the current limiter behaves just like a CC source if you turn up the supply voltage and voltage limiter all the way. As long as the IR drop across the load doesn't exceed the max supply voltage and the current doesn't exceed the max current the supply can handle it is a CC source.

author: timb
author_rank: Super Contributor
timestamp: October 21, 2016, 10:26:26 pm
content: Well, they're sort of the same thing. Most good lab power supplies have an adjustable current limit and will go into CC mode when that limit is reached. So, in that case you would get the current limit to the desired setting and then set the voltage to the maximum compliance voltage you want.You can't arbitrarily force 1A through something, because current is a function of the load. Think about Ohm's Law.So, an adjustable current limit will essentially turn a power supply into a current source (assuming it's implemented correctly), as long as you set the current limit below whatever the max current drawn by the load is.Does that make sense?

author: VEGETA
author_rank: Super Contributor
timestamp: October 21, 2016, 10:31:28 pm
content: Quote from: BobsURuncle on October 21, 2016, 10:20:24 pmMy supply is meant to be 0-20v/0-2A one. So if it is put to CC, software must output the full 20v to the set pin immediately. This way if i put it at 1A will it output 1A regardless of the load?My concern is not when the load draws more, because limiter can act as current source here without problems. My concern is when the load doesn't really ask for much current... let us say it requires 0.5A only, how can my supply deliver 1A to it? This is where current limiting feature becomes useless simply because we haven't yet exceeded the set current.Delivering full output voltage while being of CC mode is good, and I actually thought of it. However, it doesn't guarantee delivering higher current than load requirements.QuoteI already know about that, but you just said I can not force 1A through anything... Ok, what is current source? when you get a current source IC and put it to output 1A... then connect an LED on the output, what will happen? the 1A will flow regardless of the LED requirements and will burn it up right?thanks

author: IanB
author_rank: Super Contributor
timestamp: October 22, 2016, 01:19:33 am
content: Quote from: VEGETA on October 21, 2016, 10:31:28 pmYou are thinking about this the wrong way. A "constant current" supply means "vary the voltage until the current equals the set point". If the current is below the set point, the only thing the supply can do is to increase the voltage up and up until the current reaches the required value. This is in fact what a power supply does, subject to the upper limit on the voltage, which is the "compliance voltage".You might imagine what would happen if you had an "ideal" supply where the upper limit on the voltage was really large, say 1 000 000 volts. You try to feed 1 A through a load with a resistance of 1 M\$\Omega\$. The supply will increase the voltage to 1 MV, the power will be 1 MW, and something will explode in very spectacular fashionObviously there are realistic limits to what a practical power supply can do or should do.Also, you should realize that even when a power supply is in constant voltage mode, this is not actually true. It is really in "voltage limited" mode. What the power supply does is to increase the current until the voltage equals the set point and then it stops. You can verify this by setting your power supply to some voltage like 10 V and then shorting out the terminals. You can be sure that you won't see 10 V between the terminals any more. (Only do this test with a proper, safe power supply. Don't do it if the supply is able to provide a huge amount of power or the wire will melt with danger to yourself and anything it touches.)

author: Brumby
author_rank: Unknown
timestamp: October 22, 2016, 01:54:29 am
content: (1)Timb gave you the most important piece of information you need to understand what can and CANNOT be achieved here:Quote from: timb on October 21, 2016, 10:26:26 pmThis willalwaysbe true. Plug in any two parameters and the third will be immediately determined. There is no room for discussion on this. It is simple arithmetic that cannot be 'got around'.(2)The next most important thing is to realise that you are taking the term 'constant current' far too literally - and you arecompletely ignoring the limiting conditions. You would be far better off thinking of such a supply as acurrent limitingone.(3)The regulation capability of any power supply has limits (upper and lower ... but we will just look at the maximums for this discussion). There is a maximum voltage that it can deliver and there is a maximum current it can deliver. Any values in excess of these indicate an area of operation that the supply simply cannot achieve - and any attempt to implement these values will result in the supply falling out of (the desired) regulation.With these three things in mind, we can look at a couple of examples...Let's say we want to push a constant 1A through a load - which is, say, 10 ohms. (I will stick to simple resistive loads for this part of the discussion.) Using Ohm's Law, we need the terminal voltage to be 10V. This is great - your supply can do this without any worries. Set the voltage to 10V and have the current control turned up to the maximum so it doesn't interfere. The actual current that flows will be a function of the voltage and load according to Ohm's Law.Now, let's set the voltage to 20V.Using the same load of 10 ohms, you are going to get 2A flowing - simple Ohm's Law (again)! To get the current we want, we are now going to have to wind back the current adjustment control until we get only 1A flowing. So , now that we've done that - what voltage is coming out of the supply? 20V you say? Wrong. It's actually 10V - it cannot be anything else.What has happened is that the supply has altered the voltage it delivers to make sure the current limit is not exceeded. If you change the load to 5 ohms or 15 ohms or 18.65 ohms, the supply will limit the current to 1A by changing the voltage. This is how a constant current source works.Now, what happens if we have a load of 50 ohms and we still want that 1A to flow?Ohm's Law shows us that the supply needs to deliver 50V. That is a problem for a supply that can only go up to 20V. So what happens? It's really very simple - and Ohm's Law gives the answer (yet again).Your supply puts out 20V and the 50 ohm load will draw 400mA.That's it.The current limit setting doesn't come into play, because the current is not trying to exceed that limit.In theory, a truly 'constant current' source must have the capability to deliver an infinite voltage ... which is kinda difficult.

author: IanB
author_rank: Super Contributor
timestamp: October 22, 2016, 02:25:11 am
content: Quote from: Brumby on October 22, 2016, 01:54:29 amEven more particularly, the voltageincreases to the upper limitof 20 V and then stops. The power supply ends up in a voltage limited mode where the voltage hits the upper limit before the current hits the upper limit. Whichever limit is encountered first is the one that becomes controlling.(It is an interesting control problem if both limits are hit at the same time. The power supply designer has to take special steps to avoid the supply "hunting" between CC mode and CV mode.)

author: jitter
author_rank: Frequent Contributor
timestamp: October 22, 2016, 07:41:34 am
content: All good answers given here.If you want to visualize this stuff start by drawing a simple graph.On the X-axis, you place the voltage, on the Y-axis the current.Next draw in the limits for your supply, i.e. a vertical line at 20 V and a horizontal line at 2 A. The resulting box represents the ability of your supply.Now draw in a line starting at the origin (0,0) and have it go all the way up to the upper right corner (20 V, 2 A).Can you tell me what that line represents? Hint: use Ohm's Law U = I * R.Answer:The graph intersects at 10 V and 1 A, and also at 20 V and 2 A. So R = U/I = 10/1 = 10 Ohms, double-check: 20 V /2 A = yep, also 10 Ohms.It represents the load curve of a 10 Ohm resistor.With that load connected to your supply, the only possible outcomes are on that line.Now manipulate the PSU's voltage and current settings, basically you only alter the shape of the box, but the load curve remains unchanged.E.g. set the voltage to 10 V, and that load curve stays the same, it just runs out at 10 V, 1 A. Set the voltage back to 20 V but lower the max. current to 1.5 A and it will again run out, but now at 15 V, 1.5 A. But again, the curve stays the same line.You can also hook up a different load, say a 20 Ohm resistor.Now draw its load curve, it will be shallower. It also starts at 0 V, 0 A, but it will hit the 20 V limit at 1 A (I = U/R = 20/20 = 1 A). Again, with this load,only the points on that curve can exist, and only one point at a given voltage/current combination.Now I'll let you do the math for a 5 Ohms resistor.But yeah, "constant limit" seems to be the better term.

author: VEGETA
author_rank: Super Contributor
timestamp: October 22, 2016, 08:11:50 am
content: So if I use CC mode, should I alter the voltage? let us say I want 5v output voltage while I want it to deliver 1A only (or a limit of), I put it as 5v at set pin and activate CC mode at 1A, right?While if I put it at 20v maximum and then set it to 1A it will be like a current source where it is guaranteed (to some extent) to deliver 1A.So my question is, if the user activated CC mode, should I make the supply automatically deliver the maximum voltage? or keep at at the previously set voltage of say 5v?thanks for the info.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 22, 2016, 08:16:54 am
content: Plot the power supply's output V(I) on graph paper.You will get a square shaped curve, rising at constant voltage from 0 to Ilim amps, then falling from Vnom to 0 volts at Ilim amps.DC.At AC, meaning for any changing, dynamic load, the capacitor on the supply's output will dominate. This is why you can't connect the supply (while open circuit) to LEDs, willy nilly, without blowing them up that is. (You can, if you use the power supply as a proper current source should be used: alwaysshort the outputbefore connecting another load!)Very few power supplies also respond quickly when going from voltage to current regulation: that is, if you sweep back and forth around the corner of the V(I) curve, it won't be square, but will over- and under-shoot as the control circuit responds in a limited time frame.Tim

author: Kleinstein
author_rank: Super Contributor
timestamp: October 22, 2016, 09:09:09 am
content: An a lob supply you are not manually choosing between CC and CV mode. You set limit for the voltage and current. If the current is less than the limit, the voltage will be at is limit - thus the supply is working in CV mode. If the current reaches it's limit the voltage will be reduced so that the current will stay at that limit - thus the supply is working in CC mode.So there is no switch to choose CC or CV mode, but usually an LED that shows you which mode is active.By itself a switched mode regulator does not make a good lab supply. There are two reasons for this:1) There is quite some HF ripple left - one can filter this, but this adds output impedance and thus no more precise regulation.2) The control loop in switched mode controller is usually quite slow. So reaction to load changes will be slow. To make it not that slow the adjustment is often so that it will not be stable with any load. So an unusual load (like a large low ESR capacitor) could make the supply oscillate in maybe the 10 kHz range.

author: timb
author_rank: Super Contributor
timestamp: October 22, 2016, 09:21:46 am
content: Quote from: VEGETA on October 22, 2016, 08:11:50 amSure, you can do 5V at 1A, so long as you've got a 5 ohm load. However, as soon as the load changes, the current or voltage will change. For example, if the load changed to 10 ohms, the voltage limit (compliance voltage) would have to be increased to >10 volts in order to maintain 1 amps of current flow; or you could decrease the current to 500mA if you want to maintain the 5V. That's the difference between CC and CV modes.Continuing with the 10 ohm load example from above: You're in CC mode, delivering 1A at 10V; now you set the supply's voltage limit to 9V and the current limit stays at 1A, now you're delivering 9V@900mA to the load and the supply is in CV mode. This is because of Ohm's Law.Is this making sense? Here's some illustrations to help you out:First, the 5 ohm load with a 1A current limit and 10V voltage limit. Notice how the load is only seeing 5V, because the supply is in CC mode.Now, if we change the load to 10 ohm, we hit both the current and voltage limits, moving the supply into CC/CV mode.Now, if we change the load to 20 ohms, we hit the voltage limit and the supply moves into CV mode. Notice how the current flowing into the load decreases to 500mA.

author: VEGETA
author_rank: Super Contributor
timestamp: October 22, 2016, 09:41:19 am
content: Quote from: Kleinstein on October 22, 2016, 09:09:09 amBut there are lots of power supplies that uses switching technique. Main gain is very high efficiency, and to me this is critical because I want this supply to be battery powered (portable using 2 18650 batteries), linear supplies are not perfect for this. My choice after many iterations is this one (http://www.linear.com/product/LTC3649). It is rail-to-rail which can reach up to 0v which is one of my requirements as well. It offers a great feature of monitoring the output current without any sense resistor which is just awesome! All I need is a resistor to its II_mon pin to read out a 0-2v voltage. It offers current regulation via the same pin but I couldn't figure how to use it properly so I will not use it.I thought of putting a MOSFET as a pass transistor acting as a linear supply part, however this will force me to use a heatsink which is out of the question. Is there a way of making it like 0.1v drop voltage only? I mean making the input voltage - output voltage of the output transistor = 0.1v or anything small to achieve linear supply but I couldn't figure it out yet, did not think of it.Timbthanks for the illustrations, I do understand that. My misconception was about current source chips that are available on the market and how they supply constant current.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 22, 2016, 09:57:07 am
content: The good way to use a switched mode regulator in a lab supply is to have a two stage regulation. The switched mode regulation first, than a filter and than a linear stage. The drop in the linear stage can be relatively small, like 1-3 V, but one still needs a little to compensate for ripple and fast transients. At 1 -2 V the heat sink does not need to be that large and efficiency can be still good, as the switched mode stage does not need to be extra fast (which reduces efficiency).I don't think a 0.1 V drop a feasible, as the error from the switched mode part can be larger on transients. I would consider about 0.5 V as an low limit.

author: VEGETA
author_rank: Super Contributor
timestamp: October 22, 2016, 10:40:58 am
content: I don't want to add extra regulator... so how to do that linear stage using only one MOSFET? what will the efficiency be?

author: Kleinstein
author_rank: Super Contributor
timestamp: October 22, 2016, 10:59:15 am
content: Using only a MOSFET will not help. One would still need something to control MOSFET, like an OP. (or two).It depends on the transients you want to catch with the linear stage how much drop is needed. Transients from the switched mode stage can only be compensated up to the drop you have before. So with a .5 V drop at the MOSFETs one may not be able to fully surpress large transients, but this would be enough to reduce ripple quite a lot. How much loss in efficiency this is depends on the output voltage.

author: VEGETA
author_rank: Super Contributor
timestamp: October 22, 2016, 11:07:32 am
content: Ok, I know I must use op-amps, actually one is enough. But how to do the actual chop of the voltage using the mosfet? and most importantly the alteration of the switching supply voltage accordingly.current control is via the switching supply while the final output is via the mosfet. the usage of this linear stage is mainly for eliminating the ripple and noise so it will be enough.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 22, 2016, 11:29:28 am
content: For a combined regulator usually the switched mode regulator follows the actual output voltage. This does not have to be very precise, so one can use something like a simple transistor circuit to modify the voltage feedback. So instead of the usual resistor to the output, there is a (usually PNP) transistor with collector toward the feedback node that senses the voltage difference. So fast feedback might still need to come from switcher output as well.Usually the current limiting is done in the linear stage and the switched mode stage than will just follows the output voltage. So the switched mode stage does not need adjustable current limiting but would only act as a kind of backup.In addition to reducing ripple, the linear stage can also be faster. So transients on load changes can be smaller and faster.

author: VEGETA
author_rank: Super Contributor
timestamp: October 22, 2016, 01:42:42 pm
content: I think someone already built an example of that... I will search for it since dave's psu uses lt3080 not a simple transistor.Anyway, this will further increase the parts count and cost... I already have a boost converter to convert the 8.4v nominal battery voltage to 25v and this 25v is the input to the buck which is the main regulator. I need to use a simpler solution than an already made chip... meaning, another transistor switching stuff with an inductor.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 22, 2016, 02:11:19 pm
content: There are combined buck - boost converters. So you can go up or down from the 8.4 V. It still takes two MOSFETs (and 2 diodes or 2 more FETs) to do the switching, but at least you only need one inductor.The ready made chips are usually quite good - doing it by hand makes things only more complicated or not that good. It is only that less common regulators can get expensive. If you can use common types the ready made chips are hard to beat.

author: tggzzz
author_rank: Super Contributor
timestamp: October 22, 2016, 03:11:58 pm
content: Quote from: VEGETA on October 21, 2016, 09:48:54 pmThat's a sensible question, and one where the terminology is often wrong - even with decent bench power supplies.Most bench power supplies have separate controls to limit the voltage and current.All too often they call the "current limit" control a "constant current" control, which is simply wrong. So, what is what?the voltage control sets the voltage the supply applies to the load, and the supply delivers whatever current is required - up to the current specifiedacurrent limitcontrol (the common case) sets the maximum current the supply is allowed to deliver. The constant voltage is applied to the load, and if the current limit is exceeded, two actions are possible:with foldback-limited supplies, the power supply is turned off, thus stopping power dissipation in the loadmore commonly, the current continues to flow but the supply voltage falls. Power continues to be dissipated in the load (and in the supply)aconstant currentcontrol (less common) is present on some bench power supplies, where the specified current is forced through the load, using whatever voltage is necessary - up to the voltage limit. If the voltage limit is reached then the current cannot reach the specified valueSo your confusion is unsurprising, and is compounded by incorrect terminology in the manufacturers products

author: jitter
author_rank: Frequent Contributor
timestamp: October 22, 2016, 05:47:45 pm
content: Agreed. A lab supply can be set to a certain voltage and a certain (max.) current.If that max. current is not reached, it is said to be in CV mode. If the max. current has been reached, it is said to be in CC mode.Maybe it's better to take a constant current LED driver as an example, say a 350 mA one. E.g.this(obsolete) one by Recom.On it you will see two specs, the current (obviously) and the voltage range in which it can achieve this current. In this case 350 mA and 3-36 V.This is a true CC power source in that it will always go for the 350 mA current. It can do this as long as the load is such that the voltage across it results in a value between 3 V and 36 V.So, to go back to the 10 Ohm load resistor: 350 mA through 10 Ohms results in 3.5 V across it, just within the capability of this driver. 20 Ohms, and the driver would put 7 V across it to reach that 350 mA. 40 Ohms and we're at 14 V to reach 350 mA...I guess you see the picture.But you will also notice that a lab supply set to 350 mA and a voltage of 36 V (let's forget about peaks and output caps etc. for a moment) will result in exactly the same values if those 10 Ohm, 20 Ohm and 40 Ohms loads were connected to it... Ohm's Law...So basically, they're not diffrent in behaviour after all...

author: tggzzz
author_rank: Super Contributor
timestamp: October 22, 2016, 06:28:25 pm
content: Quote from: jitter on October 22, 2016, 05:47:45 pmIt may be sad, but only incorrectly. In the situation you describe, it could be in constant current or current limited mode. You can only distinguish by chaging the set current/voltage, and observing how the load current/voltage changes.QuoteOhms law is a useful convenient fiction. Only a subset of components (and therefore systems) have I-V relationships that approximate V=IR, and then only under a limited range of conditions.

author: rob77
author_rank: Super Contributor
timestamp: October 22, 2016, 06:45:03 pm
content: btw... what's the difference between:- 100mA current source with compliance voltage of 40V- lab supply set to 40V CV and 100mA CCthey both behave the very same (as long as the lab supply is a REAL lab supply)

author: timb
author_rank: Super Contributor
timestamp: October 22, 2016, 08:22:42 pm
content: Quote from: rob77 on October 22, 2016, 06:45:03 pmFor most practical purposes, there is no difference.

author: timb
author_rank: Super Contributor
timestamp: October 22, 2016, 08:30:43 pm
content: Quote from: VEGETA on October 22, 2016, 09:41:19 amRight, well the simple answer to that question is this: They put out whatever the set current is until they reach their compliance voltage, then they move into Constant Voltage mode.So if you're putting 30V into the current source chip (assuming it's a linear *or* buck mode), it can never put out more than that. That's the compliance voltage. (Now a boost mode source would have a higher compliance voltage than what you put in, but it's not infinite; you're looking at 5-10V higher than the input voltage at moderate output currents.)

author: rob77
author_rank: Super Contributor
timestamp: October 22, 2016, 08:46:30 pm
content: Quote from: timb on October 22, 2016, 08:22:42 pmexactly that was my point, so in fact it answers the question in the opening postQuote

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 23, 2016, 04:01:07 am
content: Quote from: rob77 on October 22, 2016, 06:45:03 pmThe compliance range of the CCS will probably be "squishy", whereas the regulator will saturate sharply at a constant voltage.A current source with relatively large emitter resistors (dropping a useful fraction of a volt) will exhibit a saturation with the same "on resistance". So, at ~0A, it's very nearly +V (maybe 20mV shy of it). At Iccs/2, it'll be half the nominal voltage drop, and just below Iccs, it'll be the nominal voltage drop (very slightly more than this, and the voltage swings down suddenly, showing that two CCS's fighting gives huge voltage gain, as well it should).You could emulate a sharp knee by supplying a TL431 with a CCS. The voltage rises to a constant 2.50V, but if you pull more than (CCS) amps, you get constant current. (This is a good example, but it doesn't generalize very well, actually!)Accurate supply voltages are very rarely required. The most important aspect of a voltage regulator is simply to remove low frequency ripple. Constant voltage output is a bonus you simply get on the side!Tim

author: tggzzz
author_rank: Super Contributor
timestamp: October 23, 2016, 09:07:54 am
content: Quote from: T3sl4co1l on October 23, 2016, 04:01:07 amAny design that requires an accurate supply is probably going to experience problems when used.OTOH, I have a Farnell L30AT (0-50V, 0-500mA) where the fine control has a span of ~0.25V, and is a 10-turn pot. I have difficulty finding a use-case for that.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: October 23, 2016, 09:35:45 am
content: This is the graph I think people were trying to describe.Together the load and the supply reach an operating point which on the graph is where their functions intersect.http://powersupply.blogs.keysight.com/2012/03/what-is-going-on-when-my-power-supply.html

author: jitter
author_rank: Frequent Contributor
timestamp: October 23, 2016, 09:55:57 am
content: Exactly what I meant to describe, albeit with X and Y axis reversed.In this case the steeper line is the lower load (which operates in CV mode), the shallower line is the higher load (and operates in CC mode). Reversing the axis results in a more intuitive steeper curve = higher load.