Why does transformer flux need to be reset in every half cycle? - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 21, 2016, 07:04:14 am
content: I'm trying to understand Gate Drive Transformer's, I was reading the particular attached document and there is this statement in itQuoteBased on my understanding If I have a say for example a toroidal core and a coil wound on it and I supply some current into the coil, the coil will generate some amount of flux in the core which will be proportional to the current passed in the coil until a point that the core saturates.Now when I remove the current in the coil the magnetism should come to some point called the remanence of the core, based on the core's B/H curve, it might not go completely to zero even though the current in the coil is zero.But I do not understand why do we need to give a negative current into the coil to reset the core?Why cant' I operate the core with positive current pulse and the core magnetism oscillates between +Bremanence to +Bsaturation ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 21, 2016, 08:09:19 am
content: If you let it relax to remanence (or more likely, a little bit below that, because if the node is left "floating", circuit capacitance will hold up the voltage a little while longer), then you're only using half the flux capacity of the core.If that's enough, there's no problem.Single and two-switch forward converters do this, too (and flyback, but flyback has a gapped core so the remanence is small, and you get very nearly 50% usable flux).The important part is you must wait until enough negative flux has been delivered, before starting another cycle. This causes the positive duty cycle to be limited by the negative peak voltage limit.Failure to do so will cause the flux to ratchet upwards, until the core saturates, and you'll run out of drive voltage (and draw short-circuit current from the driver) at an inconvenient moment.If you have a full wave converter, then duty cycle of each half-cycle is necessarily under 50%, and the two half cycles balance each other (as long as they're timed and driven consistently, which should be the case), so you can use the full flux capacity of the GDT, which is nice.That appnote isn't talking about very effective gate driving methods, anyway. There are many ways you can assist drive while using a transformer: depending on needs, and how many parts you can afford to stick in there.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 21, 2016, 08:24:04 am
content: Quote from: T3sl4co1l on December 21, 2016, 08:09:19 amWould'nt the regular cores be optimized for as low remanance as possible, I mean I don't know the exact values but would the Brem be like under 10% of Bsat... and I am still trying understand your point...what you are trying to say is that the entire range of flux in the core is from -Bsat to +Bsat so if I'm using from 0 to +Bsat I'm using only 50% of the flux capacity of the core and lower than 50% if lets say Brem is 10% then I would actually using 40% of the total flux capacity, is this right?On another note if a core has a saturating flux density Bmax of 1T does it mean it can go from +0.5T to -0.5T on the BH curve or is it actually +1T to -1T?, because if its the latter then I would be using the fully flux density of the core from 0 to 1T right?

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 21, 2016, 05:36:28 pm
content: Quote from: T3sl4co1lI fail to understand why the flux would continue to build up towards saturation when the current is cycling from 0 to max should'nt the flux follow the current every cycle and also move between 0 to +Bmax or +Brem to +Bmax ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 21, 2016, 05:50:49 pm
content: Quote from: ZeroResistance on December 21, 2016, 08:24:04 amI don't think remenance is a big priority for them. Permeability and losses are the bigger concern.Remenance (and coercive force) are important for losses (indeed, the product of the two is the hysteresis loss per cycle!), but there's no pressure to reduce oneorthe other in particular.For instance, the hysteresis loop of #77 is something like 40%:http://www.fair-rite.com/77-material-data-sheet/A little gap can help a lot!QuotePrecisely!QuoteBmax is Bmax; by using full wave action, you get -Bmax to +Bmax, or a range of 2T in this case!QuoteNot current -- voltage. You're driving it with a constant voltage driver (usually).If the source impedance is changing (i.e., a switch that turns off, and no diode or switch turns on to 'catch' it), then current will decay to zero, and flux with it (well, towards remanence of course).For this reason, even with a balanced signal source (consider TL598 for instance), it's a good idea to include a coupling capacitor in series with the gate drive transformer; or for a single-ended case, to use a switch that turns off between cycles, so the "reset" flux can take care of itself.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 21, 2016, 06:34:35 pm
content: Quote from: T3sl4co1l on December 21, 2016, 05:50:49 pmI still did not understand the voltage part. How does voltage affect flux I always though that current affects flux.Now lets say I have a N mosfet and to the drain I have connected a ferrite core inductor and the other end is connected to +V of 15V and across the inductor I put a freewheeling diode.Now I drive the gate @ 100Khz with 50% duty.So the pulse is 5us ON and 5us OFF.Now when the mosfet is ON current flows through the inductor and lets say the flux builds up to +Bmax.Now when the mosfet turns off current flows through the catch diode and eventually falls to 0 lets say for arguments sake that it falls to 0 within the off time of 5us. Now even the flux has fallen to ether 0 or +Brem.And this cycle would continue wouldn't it and the flux would oscillate between +Brem to +Bsat.How does the flux inching towards saturation fit into all this?

author: Kleinstein
author_rank: Super Contributor
timestamp: December 21, 2016, 07:40:14 pm
content: The induction voltage is equal to number of turns time time derivative of the flux. So the flux is proportional to integrated voltage. As the flux is limited, the long time average of the voltage has to be zero, or there will be a DC current superimposed, that reduces the useful flux even more.A free wheeling diode slow's down the speed of field decay. A suitable resistor (or zener diode) in series to the diode can speed up the decay and still limit the voltage. With 50% PWM ratio and a free wheeling diode the current will usually not decay all the way to zero, unless the coil resistance is rather high. The negative voltage needs to be on average as high as the the positive when driving the FET on - so negative peak voltage should be even higher (e.g. 1.5 times).

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 21, 2016, 08:38:40 pm
content: Quote from: Kleinstein on December 21, 2016, 07:40:14 pmThanks Kleinstein,I think the issue I am getting confused with is the concept of Ampere turns where MMF = N * I, so current figures in this and also in quite of the other topics related to magnetism like right hand thumb rule where if I wrap my fingers around a coil the direction of the thumb shows the North pole, Also Oersted did a lot of work and found that a current flowing in a wire has a magnetic field associated with it.So even when I think of Voltage in my mind I tend to think that voltage applied to a circuit produces a current and it is this flow of current that produces a flux.I tend to see it this way, I still can't comprehend integral of voltage being related to flux directly.Is there an analogy that can help me solve this confusion.Thanks again!

author: Kleinstein
author_rank: Super Contributor
timestamp: December 21, 2016, 08:49:57 pm
content: It is Faradays induction law, or one of the Maxwell equations that gives the voltage for one turn as the derivative of the flux through that turn.Looking at transformers it is the area under the voltage curve that sets the flux. This is coming just from integrating the law of induction. There is no need to calculated the current and magnetic curve of the core. It is only winding resistance that adds a little voltage - so not something for an near ideal transformer.The way an inductor / transformer works it that the induction voltage compensated most of the external voltage and only a small fraction divided by coil resistance drives the magnetizing current. In the ideal transformer you neglect the magnetizing current and the voltage drop it causes.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 21, 2016, 08:58:24 pm
content: Quote from: Kleinstein on December 21, 2016, 08:49:57 pmThanks again Klienstein..Can these same principles be applied to coils and motors, for instance a stepper motor we always tend to give a unidirectional voltage / current to the coils, is there any flux reversal in these motors? why dont' they step towards saturation with each pulse we feed them?

author: Kleinstein
author_rank: Super Contributor
timestamp: December 21, 2016, 09:18:39 pm
content: Stepper motors are usually driving the coils in a + 0 - 0 + sequence. So the flux will reverse.For unipolar drive, there are usually two coupled coils like a center taped transformer. With unipolar drive it is a bad idea to have direct freewheeling diodes, as there will be a kind of transformer function from the other half coil.The same is true for many latching relays with 2 separate coils: with free-wheeling diodes more current / a longer pulse is needed. So at least add a series resistor to the diode.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2016, 12:11:43 am
content: Amps are unimportant. Amps depend on core properties -- you're putting flux in (B axis) and getting amps out (H axis). The B-H curve is very nonlinear (some materials are much worse than their curves dare to suggest..!), so the ratio between B and H isn't constant.When there's no material (air core!), or when the material contains so much air gap that the average permeability is relatively low, the curve straightens out and becomes more linear. Then it's more useful to speak of current, like saturation amps.It's practically by coincidence that zero amps is zero flux, but the fact that remanence and magnetization exist, shows that this isn't a necessary condition, at all.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 22, 2016, 06:24:01 pm
content: Thanks again Tim!I'm still trying to understand what you are trying to say and while searching on the topic on the web I found this imageIt show's that when the voltage goes to zero the flux tends to maintain to constant value, i thought this was wrong wouldn't the flux start going to zero when the voltage drops to 0...

author: Kleinstein
author_rank: Super Contributor
timestamp: December 22, 2016, 08:23:01 pm
content: The shown curve is an ideal case, ignoring coil and switch / diode resistance. In real world (and no superconductors), the flux would slowly decrease. Depending on the coil this could be in the ms to 10 µs time scale - so for the shown example one might hardly see the flux creeping down. With 50% PWM ration the flux would not at all reach zero before the next puls comes.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 22, 2016, 08:48:51 pm
content: Quote from: Kleinstein on December 22, 2016, 08:23:01 pmThanks Kleinstein!! Is there any software where I can simulate these kind of things I mean I want to play around with Volt, amps and flux and see those kind of waveforms, I don't think LTSpice will show me magnetic flux... but are you aware of anykind of simulator that can help visualize flux vs volts?If there is no such thing then I could do some tests in the real world just to improve my understanding, I mean I have a hall effect sensor can you suggest some basic experiments...

author: Kleinstein
author_rank: Super Contributor
timestamp: December 22, 2016, 08:59:24 pm
content: Spice can simulate nonlinear inductors (e.g. include are real B-H curve with saturation) - so there should be a way to look at the flux too.Using a hall effect sensor to watch the flux could be tricky: the sensor needs an air gap and will thus reduce the effect of the core quite a lot. So it should be used with a relatively large core, to get sufficient inductance.A simple experiment would be using just inductor, resistor and free-weeling diode and look at the current when sending pulse groups (e.g. 10 pulses with 5 or 50% PWM, and than a longer break). Once the current does not go to zero the current will add up over the pulses.It might be even enough to look at this in a simulation.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 22, 2016, 09:04:14 pm
content: Quote from: Kleinstein on December 22, 2016, 08:59:24 pmNow that's a superb idea... many thanks Kleinstein !!...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 23, 2016, 12:04:07 am
content: Or with more analytical correctness:The plot of V and Phi is true, when V == EMF. That is, the electromotive force (the part that comes from Faraday's Law) in the core itself.What you measure at the pins is "dirtied" by stuff like DC resistance and stray/leakage inductance.If the core acts more or less like a regular inductor (current proportional to flux), then nonzero flux will also have nonzero current, and that current will drop voltage across DCR. Thus, over time, flux decays to zero, following the L/R time constant of the source resistance + coil DCR, and winding inductance.Regardless of resistance, this much is true: to make flux move quickly, you must apply a large voltage. Doesn't matter whether it's a positive or negative voltage (with respect to the two terminals of the winding), for an increase or decrease in flux.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 23, 2016, 07:22:22 am
content: Quote from: T3sl4co1l on December 23, 2016, 12:04:07 amYou are trying to say that EMF as shown in the graph is the Back emf generated by the inductor and this would be proportional to the rate of change of flux.Another important question is who drives the flux when the Voltage drops to zero...I we take an analogy of an electrical circuit vs a magnetic circuitV = I * RMMF = Flux * Reluctance,is it that the current doesn't drop quickly to zero that so the flux stays on and follows the current ... since MMF = N * I (turns times amps) is it this current that is the driving force for the flux to still stay in the core even after the driving voltage is removed?Secondly from my initial simulations...If the offtime is sufficiently large the current seems to drop to 0 ... its strange to see that he current rises very fast in almost 1uS but drops in almost 50+ uS...Another interesting issue noticed is if I manually calculate the current it doesnt seem to match the simulated values...V = L (di/dt)di = 15/43 = 348mAwhereas LTSpice shows around 373mA..

author: albert22
author_rank: Regular Contributor
timestamp: December 23, 2016, 07:39:44 pm
content: Quote from: ZeroResistance on December 22, 2016, 06:24:01 pmI have been trying to understand this subject for a long time. I follow all the threads on this subject with attention and take the advantage to say thanks to TIM, T3sl4co1l, Kleinstein and others who have a solid understanding about it.My (perhaps naive) understanding of how the flux stays constant is that, once the applied voltage goes to zero, the inductor still keeps the current flowing and thus the flux. Unless there is ohmic resistance that dissipates the energy. The later case is evident in your ltspice model, where the diode and other components provide that ohmic looses. Please somebody correct me if what I saying is foolish.As was mentioned before it is possible to simulate in ltspice nonlinear cores. It is also possible to plot the flux because it is internally modeled by ltspice where a piece wise aprox of the core is made.I found find this article very useful.http://www.intusoft.com/articles/satcore.pdfThanks and Merry Christmas.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 23, 2016, 09:49:17 pm
content: I use this model: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 1.0G2 2 3 1 2 1.0E1 4 2 3 2 1.0VM1 4 5 0.0RX 3 2 1E9CB 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 RS={LSAT/VSEC} ).ENDSYou'll need to paste this into a subcircuit file, and load that file as a model or library. RTFMIt works by transforming terminal voltage into current in a capacitor; as the capacitor charges, flux increases. A diode clamps the flux, providing an exponential cutoff: flux cannot practically increase beyond this point. A second capacitor (the diode capacitance) even provides a lowpass filter characteristic, which has the effect of reducing flux at high frequencies: the permeability of the core drops. (The core also becomes lossy, i.e., the permeability has an imaginary component. Inductive reactance is already imaginary, so imaginary * imaginary = resistive loss!) A voltage source measures the current flow into the clamp diode section, relaying this back to the terminals; thus terminal current is controlled by flux, and attempting to push excessive flux into the core will draw a huge current (just as should be the case!).Which explains what the variables control:VSEC: saturation flux in volt-secondsLMAG: magnetizing inductance in henry (default, un-saturated inductance)LSAT: saturated inductance in henry (this should be mu_r times smaller than LMAG)FEDDY: cutoff frequency where permeability drops (as if due to eddy currents in a conductive core).This is equivalent to a ferrite core with cross-sectional area A_e, turns N, permeability mu_r, inductivity A_L, and:VSEC = A_e * NLMAG = A_L * N^2LSAT = A_L * N^2 / mu_rA typical #77 ferrite might be mu_r = 2000 and FEDDY = 1e6 or so. High mu (up to 10-15k) ferrites have lower FEDDY (down to about 0.3e6), and lower loss, lower permeability types have higher FEDDY. Laminated iron will have FEDDY inversely proportional with lamination thickness, with several kHz being typical. (More complicated core loss curves might simply be modeled with an RLC network in parallel with the core, rather than at the flux level. For instance, the diffusion characteristic that nanocrystalline steel has.)Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 23, 2016, 10:03:58 pm
content: Quote from: T3sl4co1l on December 23, 2016, 09:49:17 pmMany Thanks!Correct me if I'm wrong, do I need to use this model instead of the inductor in the previous simulation? Then what do I do to see the flux? Just watch the current through the inductor?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 23, 2016, 10:14:07 pm
content: Yes, the two terminals of the SUBCKT become the two terminals of the inductor.You could add a SUBCKT node to sense the internal flux value, but otherwise it will be invisible inside the SUBCKT.You could also take the contents of the SUBCKT, build up the rest of the circuit around it (connecting to the nodes as usual), and have everything in the top level simulation. This is, uh, easiest to do if you're comfortable with netlist syntax...Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 24, 2016, 03:30:19 am
content: Quote from: Kleinstein on December 21, 2016, 09:18:39 pmI just remembered that even relays and some solenoids are operated on DC... how do these devices reverse flux.. yes a relay is operated at very low frequencies ... and also the coil resistance is quite high... so do you think that has a bearing on resetting a the flux in the core.What if I operate a relay at say 50Hz or 100hz pulses will its core enter saturation on do they have some inbuilt protection against it?

author: MarginallyStable
author_rank: Regular Contributor
timestamp: December 24, 2016, 04:12:38 am
content: Relays and solenoids typically have a relatively large coil resistance. My guess is this would aid in dissipating the stored energy with the aid of a free-wheeling diode.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 24, 2016, 04:16:27 am
content: Relays turn off extremely slowly, especially with clamp diodes: milliseconds to tens of milliseconds.They do indeed go faster if you provide a higher voltage for freewheeling: often, a zener diode (from ground), or TVS diode (across the terminals).Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 24, 2016, 04:25:38 am
content: Quote from: MarginallyStable on December 24, 2016, 04:12:38 amThe current flowing in the inductor would eventually get dissipated by the high DC resistance of the winding.. but what about the property of the core itself .. if it can retain magnetism when current has long gone to zero... then that flux would stick around for some time...

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: December 25, 2016, 06:57:43 am
content: What about transformers with gapped cores or transformers using materials like powdered iron cores that have distributed air gaps, and these are known to store energy in air gaps (although I don't know how) ...So Would a transformer with air gap walk to saturation faster than a core without air gaps for the same Volt / sec applied?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 25, 2016, 08:20:43 am
content: Since flux depends on core area, not gap, it doesn't matter!Indeed, if you set up a saturation tester (typically by driving a pulse into the winding -- just as the simulation above), and you use an inductor on a gapped core, then you can adjust the gap manually, in real time, while testing. As you adjust the gap length, the slope of the current varies (because of the inductance -- more gap = less inductance = more magnetizing current), but the time where saturation begins will remain nearly constant. Because, for constant applied voltage, the same flux is reached at that moment, (almost) independent of gap (or inductance or current draw).Conversely, because the resistance of a real circuit is nonzero, a gapped core isless likely to saturatein a full-wave circuit. Why? For the same reason the current eventually reaches zero on the diode-clamped inductor, or in a short-circuited (non superconducting) inductor. The increased magnetizing current causes more voltage drop across circuit resistance, which drops a voltage, and thus a flux, that's proportional to current. It's a negative feedback process.When you solve the equation for that feedback process -- that is, where applied voltage is dependent on flux -- you get an exponential decay type solution: the inductor-resistor time constant.Test this in the simulator by connecting a voltage source to an inductor, and setting the inductor for a nonzero initial current, and a nonzero series resistance. You will see the average drifts back to zero, over time!The rate at which current decays is the L/R time constant. Written as a function, I(t) = I(0) exp(-t * R/L).Tim

author: chris_leyson
author_rank: Super Contributor
timestamp: December 25, 2016, 10:09:22 am
content: QuoteIf there was a large DC component in the inductor current, say for example in the case of a flyback transformer, than you would gap the core. A gapped core lets you use a higher MMF or higher current. Saturation flux density, Bsat, remains the same as this is a function of the number of turn and core area.