Using a current transformer to harvest energy - Page 1
Forum section: electronics / projects

author: splin
author_rank: Frequent Contributor
timestamp: May 14, 2016, 07:16:52 pm
content: I'd like to be able to scavenge approximately 3.3V, 20mW from a conductor carrying 415V, 50Hz up to 500A and ideally it would work down to 1A or preferably lower. A current transformer is an obvious solution but 500A split core transformers are big and heavy as they need to avoid saturation to preserve accuracy.I assume that it would be possible to use a small core which saturates at say 1A. I guess the waveform would get pretty messy at 500A depending on the burden - i.e. whether it is a simple burden resistor + diode and capacitor or a more sophisticated adaptive or MPPT type load. Saturating at a low level should simplify the secondary side protection needed in the event of some failure by restricting the maximum energy that could be transferred (but a high dI/dt transient on the input could produce high voltages on the secondary that would have to be clamped).Is this reasonable? Would an iron or ferrite core be best? Can it be easily modelled in Ltspice or Matlab? changed 'harvest power' to 'harvest energy' corrected in. to i.e.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 14, 2016, 07:50:29 pm
content: A TVS clamp diode would be good for handling spikes, yes, and you should want a choke-input filter to get a fairly constant output voltage: you'll get a constant average because the flux is limited and the line frequency is constant. You still need a burden resistor, otherwise there's no fixed, linear ratio between current and flux.This can be modeled, yes. You need a saturable core model, such as:Code:* Saturable Core Model, copied from:* _SPICE Models For Power Electronics_, Meares and Hymowitz.*.SUBCKT INDSAT 1 2 PARAMS: VSEC=1e-4 LMAG=1e-5 LSAT=1e-7 FEDDY=1e6F1 1 2 VM1 1G2 2 3 1 2 1E1 4 2 3 2 1VM1 4 5 0RX 3 2 1E12CB 3 2 {VSEC/500} IC=0RB 5 2 {LMAG*500/VSEC}RS 5 6 {LSAT*500/VSEC}VP 7 2 250VN 2 8 250D1 6 7 DCLAMPD2 8 6 DCLAMP.MODEL DCLAMP D(CJO={3*VSEC/(6.28*FEDDY*500*LMAG)} VJ=25).ENDSThe complete model, of a nonlinear, nonideal transformer, consists of a primary self-inductance (using the above nonlinear inductor), a series inductor (leakage), another parallel inductor (secondary self-inductance, can be assumed negligible i.e. absent), and an ideal transformer for the ratio (use an E and F source to do that).Probably, since you don't need much flux, a ferrite core would be okay, but you'll have to run some numbers to find out, first.Tim

author: splin
author_rank: Frequent Contributor
timestamp: May 14, 2016, 09:54:06 pm
content: That was quick, thanks - I found satcore.pdf which I assume is your source. It's going to take a bit of digesting though as magnetics really aren't my thing. Where did the numbers you posted come from - are they selected as 'reasonable' for my requirements or an unrelated example?When you say input filter I assume you mean post - burden resistor to smooth the spikes from the secondary?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 14, 2016, 11:49:08 pm
content: The parameters in the model are defaults. Typically the simulator asks you for parameters for the instance. Or for a plain text netlist, you'd haveCode:xcore 1 2 INDSAT PARAMS: VSEC=1m LMAG=50m FEDDY=5kOr something like that.I have no idea how much flux you'd need, I'd have to run the numbers.By input filter, I was imagining you'd probably have:Transformer -- burden resistor -- rectifier (schottky FWB?) -- filter choke -- cap and TVS and additional regulators.Tim

author: Zeranin
author_rank: Regular Contributor
timestamp: May 16, 2016, 01:32:01 am
content: An interesting question. As you have 415V available, then in principle the easiest way to get your power is to use a conventional, small xformer with a 415V (or 240V P-N) primary. I presume that you either can’t or do not wish to make direct electrical connection, hence your interesting idea of using a ‘current xformer’.It should be possible, given you only need 20mW of power. I haven’t thought through in detail, but it seems that the first step is to produce a design that can provide your desired voltage (3.3V) and power (20mW) at the minimum available ‘primary’ current, say 1 amp or even less of you can do it. Then, you need to think about what needs to be done to keep things under control at 500A primary.Presumably you will use a slit core of some sort,or can you thread the 500A conductor through a non-split core?Either way, we are presumably talking about a toroid, or ‘rectangular toroid’ kind of core shape, with the 500A primary passing through the hole in the middle of the core. I would use the smallest core that gives enough space for the primary conductor, plus your secondary wnding. For example, if we talk toroid shape, maybe something like a 30mm hole in the middle, with a cross sectional are of magnetic material of around 15x15 mm. In principle it could be laminated iron or ferrite, but for now I’ll talk as if it was iron.Next part of the design process it to estimate how many turn-per-volt you will be operating at, and this is determined by the cross sectional are of the magnetic circuit, the frequency, and saturation flux density of the magnetic material, according to the formula :-N/V = 1.414 / (6.28 x f x A x Bpeak)Plugging in f=50Hz, Bpeak=1.8 Tesla for steel and A=20x20mm = 0.000225m^2 gives :-N/V = 11.1 turns/voltAs N=1 on the primary, this sets the primary voltage at 0.091V, above which the core will saturate. The saturation flux density of ferrite is lower than for steel, maybe 0.5 to 1 Tesla depending on the ferrite, but just plug it into the formula above to get your turns/volt.If, for example, you want 5V RMS on the secondary, then the required number of turns in this example is 5 x 11.1 = 55 turns.As the xformer goes into saturation, it will provide a bit more that your design voltage, and for fast spikes/transients it can produce MUCH more voltage, so it is up to you to provide suitable voltage clamping on the secondary, by way of back-to-back zeners, transient voltage suppressors or whatever.As with any current xformer, the secondary voltage will be whatever your circuit across the secondary defines it to be, but ultimately limited by core saturation at about (1/6.25) volts/turn, for this example. If Ip=1.0A, then a resistive load of 0.091 ohms on the primary will give Vpri=0.091V, as per our design target. In this example, the turns ratio is 55:1, so the impedance transformation ratio is 55^2 = 3025. Therefore, a resistance of 3025x0.091 = 275 ohms on the secondary will ‘look’ like 0.091 ohms on the primary, and will limit the primary voltage to 0.091V, and the secondary voltage to 5.0V. The maximum power that can be input to the primary is 1.0A x 0.091 = 91mW, enough for your stated need of 21mW. Losses aside, this must of course be the same as the power delivered to the secondary, namely 5x5/275 = 91mW. The numbers have to add up, and they do.You will be relying on core saturation to limit the maximum voltage that is developed on the secondary for primary current above 1A. In theory, a current transformer produces an infinite secondary voltage if the secondary is opened.If it was me, I would grab the proposed core, pass a 1A AC current through a primary conductor passing through the core, and see exactly what comes out of the 31 turn (or whatever) secondary. First make sure you can get the voltage and power that you need on the secondary, at 1 amp primary Then crank up the primary current to 5A, 10A etc to whatever AC current you can reasonably produce, and see what happens.The actual behavior is complicated by the large and unknown leakage inductance of the single turn primary, that may limit the power you can actually draw from the secondary, at low primary currents. Probably this can be minimized by looping the primary conductor in such a way that it completes a nice, proper single turn.Because you don’t really know your leakage inductance, and you probably don’t know the exact saturation characteristics of your core, my gut feeling is that modelling, and/or using the formula above will only be a guide to actual performance and behavior, which is why I recommend trying things out on the bench. Even so, I reckon that the formula and general approach that I have described will make a good starting point.I have many cores lying around, plus can easily produce variable AC current up to 20A or so, so might ‘play around’ and see what happens, see if what I’ve said above is nonsense or basically right.

author: coppice
author_rank: Super Contributor
timestamp: May 16, 2016, 02:10:30 am
content: Current transformers are widely used to scavenge power where only a single conductor is available, such as in breakers. That means you might well find suitable a suitable transformer as an off the shelf part.The problem with these things is tolerating the worst that the world might throw at you. For a 500A line you might see currents of 10,000A or more under fault conditions, for the few milliseconds before the breaker breaks. If you need to operate normally at just 1A that's a pretty wide dynamic range. If you want to scavenge 20mA from 1A your transformer will be 50:1. Getting hit by 1/50th of 10,000A and surviving takes some beefy components. A saturating core might help, but as the core saturates it looks less like a current transformer and more like a Rogowski coil. You have mitigated the high worst case current, but you have a horrible worst case di/dt voltage to deal with instead.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 16, 2016, 04:02:26 am
content: Quote from: coppice on May 16, 2016, 02:10:30 amAgreed that the potential problem is very large current spikes/transients (large di/dt) in the primary. If you get a short somewhere in your 500A primary circuit, it would not surprise me if the peak current even went above 10kA. Agreed that the current transformer looks more like a Roglowski coil for large current transients that saturate the magnetic core, and that you could getVERYlarge secondary voltages if the secondary was unloaded, or loaded in to a high impedance when such a transient occurred. However, given that we can easily place zeners or transient voltage suppressors across the secondary, the real question is what magnitude of CURRENT would be induced into the secondary, for that is what dictates the difficulty of suppression. My guess is that the magnitude of induced CURRENT spikes in the secondary will be manageable, given that TVS’s can easily handle 100’s or even 1000’s of amps for short periods of time. In fact, what actually damages a TVS is more likely to be the total ENERGY dumped into it, rather than the peak current/power as such. So the question becomes, if we get a 10kA current transient in the primary, how many joules of energy can be induced into the secondary. That could be difficult to calculate, so I’ll just wave my hands and say I suspect it’s manageable. Once the core saturates, the Pri-Sec coupling becomes very loose indeed, AKA high leakage inductance, and in conjunction with transient suppression on the secondary, that’s what will save you. If I find time to play with this, perhaps I’ll include some experiments with short-circuiting a large cap across the primary, and see how well the secondary suppression handles it.BTW, the OP specified 20mW at the secondary, not 20mA.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 16, 2016, 05:09:58 am
content: Thinking more about my Reply #4, I did not consider the primary magnetizing current, and this changes things for the worse as far as extracting the required power at 1A of primary current. In the particular example that I gave of a 30mm ID toroid with an iron cross section of 15x15, the magnetizing current at 1.8 Tesla would be more than 1 ampere-turn, maybe more like 4 AT. What that means, is that you would need (for example) maybe 4 amps just to get to 1.8T, and to achieve the calculated 0.091 V across the primary. This can be compensated for by adding more turns to the secondary, but it does mean that it is harder than you might think to extract even quite small powers from the secondary. Current xformers are easy to analyse when the secondary is shorted, which is close to how they are normally operated when measuring currents, but it's a different ballgame when you want to get currentandvoltage (ie power) from the secondary.I'll do the full calculation later.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 16, 2016, 11:18:29 am
content: Quote from: Zeranin on May 16, 2016, 05:09:58 amI'm starting to think that the OPs apparently simple request might not even be possible, or at least practical. Never mind for a moment what happens at 500A (almost certainly OK), or what happens for large transients of di/dt, which 'may be' OK with suitable transient suppression.As I said earlier on, the very first step is just to see if we can extract the very modest amount of power required (20mW) at a primary current of 1.0 amps, and this appears increasingly difficult to achieve. Sure, you can extract useful power at 10A or 100A or 500A, but how about at 1A? My posting #4 was all very well, but assumed an ideal transformer, with infinite primary inductance and therefore zero (or very small) primary magnetizing current. Fundamentally, it is not magnetizing current, but magnetizing amp-turns. If for example you have a 240V primary, then the number of turns is large, so the magnetizing current is correspondingly small. However, when you have only asingleprimary turn, then the magnetizing current needed to get the iron core flux density up to a normal operating value of 1.6 Tesla or so suddenly becomes uncomfortably large.My post #4 was based on the conventional wisdom for a power xformer, where the aim is to operate just below saturation flux density, say at 1.6 Tesla. Normally, this is done for reasons of economy, to minimize the size of core needed to get the desired power output. In this unusual application, we don't care about core size as such, but it would have been nice to operate somewhere near core saturation with a secondary voltage of say 5V. That way, core saturation will limit the available secondary voltage to say around 7.5V at 2.5 Tesla, as almost no amount of primary current will get the flux density in iron higher than that. As a simple example illustrating this point, take a conventional 100VA 240V power xformer, rated at 5V on the secondary. If you start raising the primary voltage above 240V, then at first you will increase the secondary voltage, but soon enough the core will saturate, leading to massive primary current, but little increase in secondary voltage. Clearly this is a good thing for the OPs application, all else equal, which is why I proceeded along those lines in my design procedure in Posting #4.However, I had not realized the magnitude of the magnetizing current, with only a single turn on the primary, for the size of core that might be suitable. I figure that if you have a 500A conductor passing through the center of the toroidal core, then you need a core of roughly similar dimensions to a 100VA. 50Hz, toroidal power xformer, which is typically 92mm OD x 45mm deep, with windings. A 240V toroidal xformer of that size will have around 1600 primary turns (6.7 turns-per-volt), and a 'low' magnetizing current of only 8.5mA, corresponding to 1600 x 0.0085 = 13.6 ampere turns. No big dealexceptthat if we instead have a single turn primary, then the magnetizing current will be 1600 times higher, or 13.7 amps! So much for operating such an xformer at only 1.0A, for we'll need 13.7A just to obtain the magnetizing current necessary to operate the core at 1.6 Tesla. Bugger! That is not compatible with operating at a minimum primary current of 1.0 A. Sure, it would work at a primary current of >15A, but not at 1 amp, damn it.I have 'played around' with a couple of very-high-quality grain-oriented, tape-wound, steel toroidal cores, and this is indeed what happens. No success at all trying to operate at 1A of primary current. The best I was able to achieve was 100mW of secondary power at 4A single-turn-primary current, with the secondary voltage rising by about a factor of x2 when increasing primary current to 30A, with no discernible further increase in secondary voltage when increasing Ipri from 30A up to 60A, because by then the core is totally saturated. The other thing that 'gets you' is the very high leakage inductance that you get with just a single primary turn, with the result that as soon as you start loading the secondary, the voltage sags alarmingly. In this respect you could do better by having multiple, parallel-connected single-turn primaries, but then you can't just thread a single 500A conductor through the centre of the toroid, defeating the original purpose. Of course, the 4A minimum primary current that I achieved could be reduced to 1A by using 4 primary turns rather than one, but there are all sorts of reasons why that would not be an acceptable solution either.In summary, I have demonstrated a working solution, that gives 100mW of secondary power for 4A primary current, that saturates nicely at only double the design secondary voltage, but to extend that down to 1A as requested by the OP appears very difficult, unless others can see some other way forward that I have missed.If it's true that scavenging power from 'current xformers' is a common practice, then you can bet that others have optimized the way to do it. Whether such devices work at 1 amp may be another matter.Getting something-for-free is always an interesting, irresistible challenge, so I'm keen to hear more comments and opinions about how to do it! I may well be missing something.

author: coppice
author_rank: Super Contributor
timestamp: May 16, 2016, 11:54:46 am
content: Quote from: Zeranin on May 16, 2016, 11:18:29 amReally? That's going to come as a big surprise to the people shipping millions of units that do EXACTLY what the original poster asked for, using a parasitic CT power supply.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 16, 2016, 12:43:58 pm
content: Quote from: coppice on May 16, 2016, 11:54:46 amWell, tell us how it's done then! Can you point us to a link giving the specs for these units? It's not that I disbelieve you, but without some sort of backup your words are empty.Here is a link for someone that has done this before.http://www.discovercircuits.com/DJ-Circuits/energy-harvesting.htmObserve that for a single primary turn at 1A, the DC output power is 3V x 86uA = 0.26 mWObserve that for a single primary turn at 4A, the DC output power is 3V x 800uA = 2.4 mWAs I recall, the OP needed 20 mW, and preferably for 1A of primary current, so this magic device is off the pace by a factor of 20/0.26 = 77 times less than needed.Note also that with around 30 minutes of development time, for a single turn primary of 4A, I achieved ~100mW of power, around 100/2.4 = 42 times better than this device.The ball is in your court.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 17, 2016, 05:49:08 am
content: Quote from: coppice on May 16, 2016, 11:54:46 amWho are these 'people' shipping units that can scavenge 20mW of power using a Current Transformer, with only 1A of primary current? I want one! Are yoursurethese things exist? I can't find them anywhere on the net, so please tell us where we can buy them, and point us to the specifications.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 17, 2016, 07:03:16 am
content: This topic tickled me from the outset, because at first glance it seems like it should be easy to harvest a mere 20mW of power from a ‘current xformer’, with a single-turn primary current of 1 amp, and yet when you dig deeper you realize that it can’t practically be done.I mentioned previously that the high leakage inductantance of a single turn primary (ie loose magnetic coupling to the core and secondary winding)doesn’t help. However, after further measurement and thought I realize that leakage inductance, while present, is not a significant player. So why is it impossible in practice to harvest any decent amount of power from a CT, at say 1A of primary current. Leakage inductance is not a player.I explained previously that what mucks things up is the magnetization current of a real-world xformer, and this is true. Here I attempt to explain qualitatively what is going on, in a slightly different way.A perfect xformer would have infinite winding inductance, and thus draw zero no-load current. If we had such an xformer, we could easily harvest the required power, and then some, but I need to make the case that that is true.In any ‘normal’ xformer application, the magnetization current is not an important player. For example, take a 100VA toroidal power xformer with 240V primary. The no-load magnetization current Imag would be typically a mere 8.5mA, and does not represent a loss anyway, as it is a purely reactive load. Further, Imag=8.5mA is negligible compared to the primary load current for a 240V 100VA xformer and can therefore be neglected.Our example xformer has a no-load primary impedance of 240/0.0085 = 28.2 kohms. A typical 100VA xformer has about 1600 primary turns. Inductance scales as the square of the turns, so a single turn primary will have a smaller inductance, and therefore inductive reactance, by a factor of 1600^2. Specifically then, the impedance of a single-turn primary winding on the same xformer will be 28.2kohms / (1600^2) = 0.011 ohms!So what are the implications of such a low no-load primary impedance, if the same 100VA toroidal core is used as a current xformer? For simplicity, let’s consider the case of trying to extract power on the primary side. You can step up the voltage to the secondary, but you won’t get any more power, actually a whisker less from losses. Also for simplicity, let’s consider a resistive load, again the best possible case, for there are slight losses converting to DC.Therefore, our no load model is the primary current, passing through the single-turn primary impedance of 0.011 ohms, and we have the option of placing a resistive load of our choosing across this primary winding. If we choose the optimum value of load resistor, then how much power can we get delivered to this resistive load, for a given primary current of 1.0A? That is the question.So what is the optimum load resistance, meaning the resistance for which the delivered power is greatest? Let’s try the extreme cases first. If the load resistance is chosen to be infinite, then the voltage will be maximized (=1.0A x 0.011 ohms), but the current will be zero, and therefore the power will be zero. If we make the load resistance zero, then the current will be maximized (= 1.0A), but the voltage will be dragged down to zero, so again the power delivered to the load resistance will be zero. For simplicity, I’m going to assume that the no load impedance of 0.011 ohms is resistive. Of course, actually it’s an inductive reactance, but assuming it to be resistive won’t introduce a great error, and I want to keep the analysis simple, so we can see what is going on, rather than getting bogged down in complex math. OK. So the power delivered to the load is zero for both infinite and zero load resistance. I’ll state without proof that the maximum power will be delivered to the load, when we choose a load resistance of 0.011 ohms, equal to the primary resistance that we are placing the load across. So the model is a 1A current, passing through two 0.011 ohm resistances in parallel. Clearly the current in each resistor will be 0.5A, and the power dissipated in the load will be 0.5x0.5x0.011 = 2.75 mW. If you like math, then you can do the calculation with load resistances a little higher and lower than 0.011 ohms, and you’ll find that the load power is less. The very best we can do, is to harvest 2.75 mW, using an optimized load resistance of 0.011 ohms. As I said before, you can transform the voltage up and the current down on the secondary side, but you won’t get any more power. If you love math then you can do the calculation ‘properly’ with the primary impedance being inductive rather than resistive, but the answer will be similar. For our present purposes we don’t care if the precise power that can be delivered is 2mW or 3.5mW, the key point is that the maximum power available is TINY, and that nothing can be done to improve matters, because the limiting factor is the low core inductance with only a single primary turn. Understanding beats modelling, IMHO.For me, that’s a fascinating and counterintuitive result. With a 100VA toroidal core, and a 1A single-turn primary current, we fundamentally will not harvest more than about 1.8 mW of power, and this is consistent with my own measurements. At Ipri=4A, things look much better. In this case, the load current would be 2A, so the load power is 2 x 2 x 0.011 = 44mW.The best that I have managed to get experimentally, using a very-high-quality grain-oriented toroidal core of similar dimension, is 54 mW at Ipri=4A.For me, the topic is done and dusted, and I enjoyed it greatly. Many thanks to the OP.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 17, 2016, 07:35:17 am
content: Yes, magnetization sets the minimum amp turns at a given frequency.So use more turns! There's also permalloy and nanocrystalline materials, which have much higher mu than the best GOSS.More turns, of course, incur greater winding area, and thus larger cores. Saturation may not be an acceptable power limiting method, as well, though I don't think so (a VITROPERM 500 core is pretty well saturated at less than 1At, so as long as magnetizing current is swinging to a similar amplitude, saturation will follow soon after).Tim

author: Zeranin
author_rank: Regular Contributor
timestamp: May 17, 2016, 08:16:05 am
content: Quote from: T3sl4co1l on May 17, 2016, 07:35:17 amOf course, if you double the number of turns, then you halve the primary current for the same secondary yield power. The problem with more turns, is that the OP has a 500A conductor running thru the core. The best solution of all is to use a conventional voltage xformer across the phases, but I gather the OP can't do that because he wants a non-intrusive way of harvesting his 20 mW. My guess is that threading 4 turns of his 500A cable or busbar through his CT core would also be overly intrusive, not to mention messy and needing a large core.You are spot on in saying that a very-high-permeability material would help, such as some of the nickel based alloys related to mu-metal, as this would raise the inductance and thus the primary no-load impedance, which is at the heart of the problem. Such cores may be overly exotic and expensive for the OP though. Ironically, the lower saturation flux density of these alloys may actually be an advantage here, as we won't necessarily be operating at very high flux density, in which case the earlier saturation provides greater protection at 500A.I'm fairly sure that core saturation will keep the secondary voltage under control at a nice, steady 500A of primary, and would be more concerned about multi-kA current transients from switch-on-off or accidental short in the system.One can also increase yield by using a larger core, such as stacking multiple toroids, but this is only a linear improvement, and will quickly get out of hand with issues of practical size. Stacking toroids increases the cross sectional area of the magnetic path, thus increasing primary inductance and no-load impedance, which is what is required.