Calculating Magnetic Flux Density in Transformer/Inductor? - Page 1
Forum section: electronics / beginners

author: sean0118
author_rank: Frequent Contributor
timestamp: September 28, 2014, 09:09:21 am
content: Hi everyone,Sorry for asking something that's so trivial and covered in a million books etc. But I have confused myself and just want to check the basics.When calculating the magnetic flux density Bmax I have come across two different formulas: Bmax = (Vrms*0.25*T)/(N*Ae) Bmax = (L*delta_I)/(2*Ae*N)The problem is they give me different answers! Not only that, as N increases  decreases while  increases... not so good, which way is Bmax meant to go with a change of N?Either I broke maths/physics or I'm doing something wrong.edit: Just to be clearer, I'm using Vrms as the voltage applied over the winding (square wave), T as the time period, N as number of turns, Ae as cross sectional area of core, L as magnetising inductance and delta_I as the peak-to-peak of the current.

author: IanB
author_rank: Super Contributor
timestamp: September 28, 2014, 05:00:39 pm
content: Quote from: sean0118 on September 28, 2014, 09:09:21 amI'm not an expert on this, but a couple of thoughts.1. When you see a formula with Vrms in it, that formula is most likely designed for use with regular sinusoidal voltages.2. You can know the applied voltage across the transformer winding as it is a given specification, but you cannot know the current flowing through the transformer without further calculation. So formulas involving current like that are probably not very useful.In short, you should use a variant of formula 1, but go back to the derivation of that equation and find a specific form of it that is applicable to square waves. (In general the change in flux density depends on the voltage applied to the coil and the length of time it is applied.)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 28, 2014, 07:23:56 pm
content: If nothing else, you can go back to the basics:- B is flux/area, and being a spacial quantity, is /turn.- H is amps/length, and being a spacial quantity, is *turn (i.e., amp-turns).- mu is the conversion factor between B and H. mu_0 ~= 1.257 nH mm^-1 t^-2 is the permeability of free space, and mu_r is the relative permeability (mu/mu_0) of the core (while it's unsaturated, anyway).- Flux comes in two flavors, the circuit value (no turns) and the geometrical form (/turn). Flux is volts*time, the integral of voltage (EMF, technically) through time. So you need to do an integral to find the flux of an arbitrary waveform, but straight-line waveforms (like square waves) you can do this implicitly.- V is the voltage (technically EMF) applied to the winding, and I is the current through it.- L is the inductance of the winding. Henries are flux per amp.- N is the number of turns. Important to keep sorted. Circuit parameters (V, I, L..) never use turns, because they reflect what appears at the terminals of some component, don't care what's inside it. The spacial fields have turns associated with them, because it matters how many times the coil wraps around those fields.- A_e and l_e are core parameters, derived from the geometry of the core (as seen by the fields inside).- A_L is the inductivity of a given core geometry (material, shape, mu, gap, etc.), usually in nH t^-2 or some power-of-10 multiplier (uH t^-2 is sometimes given for ferrites, while some powdered iron materials are given as "uH for 100 turns" or something silly like that).Now... Bmax. You're talking transformers or inductors, just going up to some Bmax (which may be saturation Bsat, or some fraction thereof to provide a safety margin, or to reduce power losses). The inductance doesn't matter, so the gap and mu and everything also don't matter! Yes,  contains inductance, but it's factored out immediately by current. So the two equations are saying the same thing.Closer look: has Vrms * T which is flux (at the terminals -- circuit flux).  has L*delta_I, but inductance is flux per amp, so this is just applied flux again. And I think you'll agree that N*A_e in the denominator is the same thing. So, that leaves everything up to a constant.One nice thing about inductors is, as long as you know the current flowing at some instant, you knowexactlyhow much flux has been delivered to it to put it in that state. This is true whether it's DC bias, peak AC or whatever. Transformers you don't usually care about inductance and magnetizing current (so long as it's enough not to care), so you have to go from the voltage waveforms. It matters whether it's single ended or full wave.So what is it?If we consider the voltage and current waveforms on a winding, then for a square wave voltage, the current will be a triangle wave. If I_DC is zero, then the triangle rises above and below symmetrically, with the peaks coinciding with edges of the square wave, and the zero crossings occurring in the middle between edges. Now, notice... it takes exactly 1/4 of a wave to go from zero to peak, then back to zero, then to negative peak, and back to zero. So the peak current is I = (Vpk * T/4) / L. Or, to put it another way, the peak flux is Vpk*T/4. It would seem,  is written for a zero-DC condition (typical of the transformer in a full wave forward converter).If we write it in terms of current instead, we might be inclined to take Ip-p (aka delta_I), since that's easy to measure, putting cursors on the waveform. For a 50% duty cycle, this is half the wave. So to find the 1/4 wave quantity, divide by 2. Which looks like .These are both assuming zero DC, so Bmax = Bpk = -Bmin. This won't work for a single-ended condition, where B never returns to zero. For that, you need to double either equation.Regarding waveform, Vrms = Vpk for a square wave, so this is correct, but a little misleading because square waves aren't usually referred to by RMS value. Curiously, the equation doesn't change much for sine waves: you get 1 / (sqrt(2)*pi) (1 / 4.44...) instead of 1/4.Tim

author: sean0118
author_rank: Frequent Contributor
timestamp: September 29, 2014, 02:51:00 am
content: Thanks for your replies.Quote from: T3sl4co1l on September 28, 2014, 07:23:56 pmyep, so just to be crystal clear, your saying as N increases, B decreases while H increases? So more turns would help if a transformer were saturating?Quote from: T3sl4co1l on September 28, 2014, 07:23:56 pmI agree with this for most cases, but for this specific case I'm aiming for a specific magnetising inductance, which is why I'm adding an air gap in an attempt to allow more turns and a lower B. That's one of the uses of an air gap right?Quote from: T3sl4co1l on September 28, 2014, 07:23:56 pmI think this might have been where I was going wrong, I used the pk-pk of the current, I didn't divide by 2. But if L is the magnetising inductance, doesn't that increase rapidly with an increase of N? Wouldn't that make B increase with N?Quote from: T3sl4co1l on September 28, 2014, 07:23:56 pmYes, your right, I probably should have used the peak voltage. I also should have mentioned that the waveform is not quite square wave (image attached), but I approximated it to be squarewave to simplify the B calculation, that's why I used RMS. Would I be better using the average voltage of a half cycle instead of RMS?edit: I just tested this, the difference between the RMS of the total cycle and average of half a cycle is only a couple of volts in this case.

author: sean0118
author_rank: Frequent Contributor
timestamp: September 29, 2014, 02:55:43 am
content: Quote from: IanB on September 28, 2014, 05:00:39 pmI think I'm getting some success with formula , however as Tim mentioned I should not be using RMS.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 29, 2014, 03:48:31 am
content: Quote from: sean0118 on September 29, 2014, 02:51:00 amYes and no. In agiven core and geometry, holding all else constant, B and H are always proportional (B = mu_r * mu_0 * H), so if B reduces, H reduces as well.An in-circuit explanation of that would be, more turns means more flux handling (before saturation) and more inductance, so the current drops, so B and H both drop.If you're controlling something else at the same time, like the gap (to keep the same desired inductance), then H must rise (because you're asking the same amps, but more turns = more amp-turns = more H). B and H are then related effectively by a different mu_r.B is generally better to use, because B is equal throughout a loop, but H depends on the material it's measured in (H is discontinuous at the surface of a permeable material).To figure this, you can imagine the core, with magnetic path length l_e, being shrunk by a factor of mu_r (of the core material), to represent an equivalent air gap. Consider the purpose, why magnetic cores are used for in the first place: free space isn't very inductive, so to get higher inductivity (and thus, higher inductance from a given pile of wire), we want to concentrate the magnetic field into a smaller (thinner, wider) volume than the space naturally around the coil.So, if you take l_e / mu_r, you get the air gap equivalent of the core. Add to this the actual air gap you've used (which is probably a lot larger, anyway) to get l_eff, the equivalent air gap of the assembly.A_L = mu_0 * A_e / l_effSo, the equivalent air gap of almost all ferrites is vanishingly small: a very average EE core set might have l_e around 100mm, but with mu_r = 2000 or so, it amounts to an equivalent 0.05mm. This is handy to know, since any accidental gap on the order of l_eff will have a dramatic impact on the apparent permeability of the core -- if the core halves meet with more than, say, 20um RMS surface finish, or tilt or general out-of-touching-ness, mu_eff will be much lower than you were expecting.And, it follows, anywhere you require an extremely high mu (e.g., common mode chokes), you need very well fitting faces -- often, such cores aren't made from cut pieces, but completely solid pieces instead (toroids, figure-8 shapes, etc.).QuoteYes, precisely the use.Note that energy density goes as B^2 / (2*mu), so the energy storage in ferrite is abysmal (almost nil, because mu > 1000 for most), and most filter chokes are wound on gapped, or distributed gap (e.g., powdered iron) materials. An air core choke has the highest energy density, because it won't saturate and because mu = mu_0, as low as it can go. But it isn't very practical, because the copper resistance gets in the way, and it gets really bulky. Practical inductors are a matter of optimization for the engineer, rather than theoretical formulations derived by the physicist.Though this is worth keeping in mind if you ever have very peaky situations. Ferrite beads are completely and utterly futile for ESD and EFT, but air core coils can stand a chance (assuming they're big enough to do anything at all, which is kind of unlikely for casual use?). Snubbers ranging from switching circuits (especially using SCRs) up to photoflash and more can benefit, because although the peak power sucks, the pulsed operation means the average dissipation isn't a thermal problem. (A photoflash application might have peak currents in the 100s of A, even for a compact DSLR application. Jjust a few microhenry is enough to have an impact on the microsecond range transients, and a permeable cored inductor would be required to handle each and every ampere of that transient without saturating to do the job -- there's no way you could get the core small enough to matter at that current level!).QuoteNo, B decreases with N. If current is constant, then for rising L, you must be driving it with much more voltage or much lower frequency! Remember, Phi = L * I, and B = Phi / (N * A_e).QuoteYeah, a waveform like that, I'd approximate with a worst-case square wave of peak amplitudetimes worst case input variationand full duty cycle. You'll probably end up with 10-30% overhead, which is fine; all the more to help prevent it from saturating at high temperature (where Bsat is lower).Tim