Magnetic circuit formulas - Page 1
Forum section: electronics / beginners

author: Picuino
author_rank: Super Contributor
timestamp: May 07, 2023, 09:45:27 am
content: Compilation of useful formulas for calculatingmagnetic circuits.Interesting links on the subject*https://en.wikipedia.org/wiki/Magnetic_circuit*https://www.seventransistorlabs.com/tmoranwms/Elec_Magnetics.html#magnet*Calculating a Coil*Calculating a Transformer

author: Picuino
author_rank: Super Contributor
timestamp: May 07, 2023, 09:46:11 am
content: Magnetic formulasℜ = λ / (μ0 · μr · A)ℜ = Magnetic reluctance in Henri-1[H-1]λ = Magnetic circuit length in meters μ0 = 1.2566 x 10-6Magnetic permeability of vacuumμr = Relative magnetic permeability of core materialA = Effective area of the magnetic circuit in meters2If you have an air gap, the reluctances of the core without air gap and of the air gap must be calculated separately. They are then added together to find the total reluctance:ℜtotal= ℜcore+ ℜgapCore datasheets often give the value of permeance of the core (inverse of reluctance ℜ) with the letters ALand in units of Henries.ℜ = 1 / ALℜ = Magnetic reluctance in Henri-1[H-1]AL = Magnetic permeance in Henri B = N · I / (A · ℜ)B = Magnetic field in Tesla N = Number of turns of the solenoidI = Electric current in amperes A = Effective area of the magnetic circuit in meters2ℜ = Magnetic reluctance in Henri-1[H-1]Calculation of the maximum saturation current of a coil:Imax= Bmax· Amin· ℜ / NImax= Bmax· Amin/ (N · AL)

author: Picuino
author_rank: Super Contributor
timestamp: May 07, 2023, 09:47:11 am
content: L = N2/ ℜL = Inductance in henry N = Number of turns of the solenoidℜ = Magnetic reluctance Henri-1[H-1]E = 0.5 · L · I2E = Energy stored in the core in Joules L = Inductance of the coil in Henry I = Electric current in Amperes E = B2· A · λgap/ (2 · μ0)E = Energy stored in the air gap in Joules B = Magnetic field in Tesla A = Cross section of the air gap in meters2λgap= Length of the air gap in meters μ0 = 1.2566 x 10-6 Magnetic permeability of vacuum

author: Picuino
author_rank: Super Contributor
timestamp: May 07, 2023, 09:50:46 am
content: Φ = N · I / ℜΦ = Magnetic flux in Weber N = Number of turns of the solenoidI = Electric current in amperes ℜ = Magnetic reluctance Henri-1[H-1]Copper resistivity:ρCu = 1/58 - (1 + 0.00393 · (T - 20)) [Ω·mm2/m].T = Copper temperature in ºCρCu = Copper resistivity in Ω·mm2/mTemp ρCu [Ω·mm2/m]200,01724300,01792400,01860500,01927600,01995700,02063800,02131900,021981000,022661100,023341200,024021300,024691400,02537R = ρCu · L / SR = copper wire resistance in ohms ρCu = Copper resistivity in Ω·mm2/mL = Copper wire length in meters S = Copper wire cross section in milimeters2

author: Picuino
author_rank: Super Contributor
timestamp: May 07, 2023, 09:55:29 am
content: Maximum AC voltage allowed by a coil:In sinusoidal alternating current:Vrms_max= Imax· sqrt(2) · π · f · LVrms_max= Maximum rms sinusoidal voltage applied to the coil, in volts Imax= Maximum peak electric current admitted by the coil, in amperes according to the formula: Imax= Bmax· A · ℜ / Nf = Frequency of applied voltage, in Hertzs L = Inductance of the coil, in henry Vrms_max= Bmax· sqrt(2) · π · N · f · AminVrms_max= Maximum sinusoidal rms voltage applied to the coil, in volts Bmax= Maximum magnetic field admitted by the coil, in Tesla N = Number of turns of the coilf = Frequency of applied voltage, in Hertzs Amin= Minimum cross section of the core in meters2Variation of current with time in an inductance:dI = dt · V / LdI = Variation of electric current through inductance (I1- I0) in Amperes dt = Time differential (t1- t0) in seconds [ s ]V = Constant voltage between inductance terminals in Volts L = Inductance in Henrys

author: MrAl
author_rank: Super Contributor
timestamp: May 07, 2023, 01:23:15 pm
content: Quote from: Picuino on May 07, 2023, 09:55:29 amThis doesn't look right because of the sqrt(2). This is the relationship between AC current and voltage in an inductor:V=I*w*Land in the time domain:v=L*di/dt

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 07, 2023, 01:29:15 pm
content: Wrote this many years ago, has aged well enough I think?https://www.seventransistorlabs.com/tmoranwms/Elec_Magnetics.html#magnetPersonally I don't find much use of reluctance, mainly because transformer and inductor designs are largely simple, or mu_eff or A_L is as provided, but it's a perfectly valid way to reason through things of course.Quote from: MrAl on May 07, 2023, 01:23:15 pmYes, be careful which kinds you're using, RMS or peak. Saturation flux density for example only cares about peak, so you must use peak in its formula; the sqrt(2) and pi combine to give the traditional "4.44" coefficient in Bpk = Vrms / (4.44 N F Ae).Just explaining this one already is probably worthwhile. Dimensional analysis suffices. Flux density is, well, a density: per m^2. Flux is webers (volt-seconds). Vpk / F gives the correct units for flux, and dividing by effective area Ae gives flux density. Simple. Just don't forget to include N, the conversion factor between fields (one turn, as it were) and circuit/terminals (the winding is N times the flux).To be clear, you can rub any units together, in these magnetics problems, and get something meaningful in some contrivance. The coefficients are usually either 1, simple factors like sqrt(2) (for RMS/peak), or pi (for area under a sine); or horrendously complicated geometry factors that you'll never solve analytically so just look up in a table, simulate, or measure.Tim

author: Picuino
author_rank: Super Contributor
timestamp: May 07, 2023, 02:20:03 pm
content: Quote from: MrAl on May 07, 2023, 01:23:15 pmComes from:Vpeak = Ipeak · XL = Ipeak · (2·pi·f·L)Vpeak = Vrms · sqrt(2)so:Vrms · sqrt(2) = Ipeak · (2·pi·f·L)Vrms = Ipeak · sqrt(2) · pi · f · LQuote from: T3sl4co1l on May 07, 2023, 01:29:15 pmGood text. Added to the list in the first post.

author: Picuino
author_rank: Super Contributor
timestamp: May 07, 2023, 06:05:33 pm
content: Example: calculating a coilCore:E 30/15/7gaps: 0.5mm in one core and 0mm in other core (ungapped)Material:N87Parameters:A = 60 mm2= 60 · 10-6m2Amin= 49 mm2= 49 · 10-6m2AL = 145nH = 0.145 · 10-6Hℜ = 1/AL = 6.897 · 106H-1Bmax= 350 mTCalculations and results:N = 1 -> L = 12· AL = 0.145 uH -> Imax= 118.3 AN = 5 -> L = 52· AL = 3.63 uH -> Imax= 23.66 AN = 10 -> L = 102· AL = 14.5 uH -> Imax= 11.8 AN = 15 -> L = 152· AL = 32.6 uH -> Imax= 7.88 AN = 20 -> L = 202· AL = 58 uH -> Imax= 5.9 AN = 25 -> L = 252· AL = 90.6 uH -> Imax= 4.73 AN = 30 -> L = 302· AL = 131 uH -> Imax= 3.94 AN = 50 -> L = 502· AL = 363 uH -> Imax= 2.37 AImax= Bmax· Amin/ (N · AL) = 0.350 · 49 · 10-6/ (1 · 0.145 · 10-6) = 118.3 AThese are the maximum values, but should not be reached because a maximum magnetic field of 200mT is recommended for N87 material. At this value, the peak current results should drop by almost half.Magnetic energy stored:Another way of looking at it is that the air gap always stores the same amount of maximum energy, which is given by the formula:Emax= 0.5 · L · Imax2Emax= Maximum energy stored in the core in Joules L = Inductance of the coil in Henry Imax= Maximum electric current in Amperes Emax= 1.015 mJ (Constant for this core and gap)

author: Picuino
author_rank: Super Contributor
timestamp: May 07, 2023, 07:05:59 pm
content: Example: calculating a transformerFlyback transformers are designed like coils. They must store energy in the primary coil during the first part of the cycle, to be returned through the secondary in the second part of the cycle. They are actually coils with two separate windings.For this reason flyback transformers need a relatively large air gap, which is the area of the magnetic circuit that stores the most energy.All other transformers operate by making a direct transfer of energy, so they do not need to store it. The air gap in these cases will be very small or not exist at all.This is the type of transformer that we are going to calculate in this post.Core:E 30/15/7gaps: 0mm in one core and 0mm in other core (ungapped)Material:N87Parameters:A = 60 mm2= 60 · 10-6m2Amin= 49 mm2= 49 · 10-6m2AL = 1900 nH = 1.9 · 10-6HBmax= 200 mT (below saturation)Copper window (between former and core): 17mm x 5mm = 85mm2Current density of copper: 5 A/mm2rmsFrecuency of operation: 20 kHzCalculations and results:We are going to try to fill with copper wire all the window between the former and the core.Half of the window with the primary coil and the other half with the secondary coil.We will start with a round wire of 0.5mm diameter, which will fill the hole of the window as if it were a square section wire.d1= 0.5 mm (diameter of primary wire)N1= 0.5 · Window / d12= 0.5 · 85mm2/ 0.52= 170 Turnsd2= 0.5 mm (diameter of secundary wire)N2= 0.5 · Window / d22= 0.5 · 85mm2/ 0.52= 170 TurnsLenght of one turn of wire (median):LTurn= 4 · (9mm + 19mm) / 2 = 56mmLenght of primary wire:LN1= N1 · Lturn= 170 · 0.056m = 9.52mCross section of primary wire:SN1= 0.52· π / 4 = 0.1963 mm2Estimated resistance of primary wire:RN1= ρCu · LN1/ SN1= 0.020 · 9.52 / 0.1963 = 0.97 ΩMaximum current allowed by copper:Irms_max_N1= 5A/mm2· SN1= 5 · 0.1963 = 0.982 ArmsMaximum power losses in copper:PN1= Irms_max_N12· RN1= 0.9822· 0.97 = 0.935 WattPN2= PN1= 0.935 WattPCu_N1+N2= 1.87 WTransformer excitation inductance:Lexcitation= N12· AL = 1702· 1.9 uH = 54.91 mHMaximum excitation current allowed by magnetic core:Imax= Bmax· Amin/ (N · AL) = 0.2 · 49 · 10-6/ (170 · 1.9 · 10-6) = 30.34 mAMaximum primary supply voltage:Vrms_max= Imax· sqrt(2) · π · f · L = 0.03034 · 4.44 · 20000 · 0.05491 = 148 VoltsPower constant:Now, if I change the number of turns of the primary by half and the wire section by double (for example, by using double parallel wires to make the winding) what is achieved is to double the maximum admissible current and halve the maximum admissible voltage.Here again we can see a constant which is the power that the transformer can handle with this core at calculated frequency:P = I · V = 148 · 0.982 = 145 Watt (Constant for this core @ f = 20kHz)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 07, 2023, 07:56:51 pm
content: Quote from: Picuino on May 07, 2023, 06:05:33 pmThe Maxwell stress:$$ σ = \frac{B^2}{2 \mu} $$This gives a pressure (e.g., T / (H/m) == Pa), which is also an energy density (J/m^3)! Multiply by the volume of the air gap (\$A_e l_g\$) to get energy stored in it at a given peak flux density.This doesn't help you determine theoverallsize of core required for a given transformer or inductor -- that also depends on copper resistance / current density and winding area available to put it in -- but it's a step along the way!Tim

author: MathWizard
author_rank: Super Contributor
timestamp: May 07, 2023, 11:05:58 pm
content: Hmmm, I was expecting there to be a bunch of terms or coefficient's that would be hard to find or calculate, relating to the material or something. Those eqn's look quite doable, especially if you have caliper's.Soon I want to try making a metal detector again, I'll have to try a few of these eqn's, thanks. I have all this and more in books, but I haven't gotten around to any amount of magnetics or detailed more use of inductors yet.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 08, 2023, 01:15:27 am
content: Yes, you can measure cores!Identifying material type isn't so easy, but you can at least measure mu_r from dimensions and A_L. Give or take air gap; if you have the tools to lap the core faces, you can minimize air gap and get most purely the core inductivity by itself. (They measure material properties on a toroid of course, which solves that problem.)There are some odd powers (exponents) that come up in core loss models, which I imagine we'll get to here sooner or later; but they're also approximate models at best. (Turns out magnetic materials are really, really hard physics.So, empirical models reign.)Tim

author: Picuino
author_rank: Super Contributor
timestamp: May 08, 2023, 04:20:34 pm
content: Yes, in some cases I have had to measure with the caliper the dimensions of a toroidal iron powder core to know its parameters.If you don't have the dimensions of the core and its air gap values (it is frequent in toroids), they can be estimated from measurements.When measuring the inductance of a wire wound to a core, I prefer an indirect method which consists in switching with a mosfet a known voltage (5V) for 1us every 1ms and measure in each case the peak current absorbed by the coil. From there the inductance can be estimated. You can also measure directly on the oscilloscope how the sawtooth current increases and from there estimate the inductance.

author: Picuino
author_rank: Super Contributor
timestamp: May 08, 2023, 04:30:06 pm
content: This is the circuit used.1º Measure power supply current (I_mean) with 1us on and 999us off in mosfet gate2º Calculate peak current in inductance:I_L_peak = 2 · 1000 · I_mean3º Calculate the value of LL = 1us · 5V / I_L_peakIf the peak current is small because the inductance is relatively large, it is easy to increase the ON time and recalculate.It is convenient to put good capacitors in parallel with the power supply, close to the mosfet.

author: Picuino
author_rank: Super Contributor
timestamp: May 08, 2023, 05:28:35 pm
content: Quote from: T3sl4co1l on May 07, 2023, 07:56:51 pmAdded like energy stored in air gap:Quote from: Picuino on May 07, 2023, 09:47:11 am

author: MrAl
author_rank: Super Contributor
timestamp: May 08, 2023, 07:53:06 pm
content: Quote from: Picuino on May 07, 2023, 02:20:03 pmHello again,I was referring to this formula you wrote:"Vrms_max = Imax · sqrt(2) · π · f · L"which when rewritten with w=2*pi*f would look like this:"Vrms_max = Imax · w · L /sqrt(2)"which implies that Imax is really Ipeak, but Imax is used for RMS quantities also so it does not look concise enough.Really we dont need to play around with sqrt(2) at all because if we just write it out as normal it would look very simple:V=I*w*Land note we dont need sqrt(2) anywhere because if we are working in RMS then both V and I are in RMS units, and if we are working in Peak units then both V and I are in Peak units. It doesnt make too much sense to write something like:Vrms=Ipeak*w*L/sqrt(2)because that is a mix of two different basic concepts it would be better to write it this way:V=I*w*LVrms=Vpeak/sqrt(2)and now the two concepts are completely independent, although you dont even need the second because it's elementary that Vpeak=Vrms*sqrt(2).In definitions like this we try to stay as simple as possible and keep to the most basic concepts when possible. If we didnt do that we would end up having to write out a LOT of extra formulas such as:Vrms=Ipeak/sqrt(2)*w*fVpeak=Irms*sqrt(2)*w*tIt's better to keep the inductor relationship apart from the RMS calculations. That way we have just one RMS calculation for everything and we can describe the other relationships apart from that so that they are clearer and more concise with less clutter.I hope this makes sense to you.

author: MrAl
author_rank: Super Contributor
timestamp: May 08, 2023, 08:01:18 pm
content: Quote from: Picuino on May 08, 2023, 04:20:34 pmHi,For power inductor measurements one of the best ways to measure the inductance is to insert the unknown inductor into a buck circuit with perhaps variable frequency, or a buck-like circuit.The reason for this is because in a buck circuit we can observe both the dynamics of the current and voltage and also the response to the DC current. The DC current is important in power circuits because there is always a DC current in a power supply circuit and could be a net DC current in just about any circuit. We can then vary the pulse width and watch the current ramp up and calculate the inductance, and then increase the DC current and repeat and that tells us how much the inductance changed as the DC current went up. Since the permeability of the core can change substantially with DC current, this turns out to be a very good test for a power inductor.Something like this also works for the capacitor. We can observe the capacitance by the way the voltage and current changes and we can estimate the ESR by observing the faster change in voltage compared to the current. ESR is an important parameter in power circuits also.

author: Picuino
author_rank: Super Contributor
timestamp: May 09, 2023, 03:41:58 pm
content: Quote from: MrAl on May 08, 2023, 07:53:06 pmI understand what you mean. Also sine Vrms is not useful for transformers with square waveforms because they are part of switching power supplies.But right now I don't know how to change the formula to be useful in all cases.At least I have defined more precisely what each term in the formula refers to, to try to avoid mistakes.