Ferrite core transformer  - Page 1
Forum section: electronics / projects

author: whatty
author_rank: Contributor
timestamp: February 18, 2023, 08:57:48 am
content: Hi,I am designing a smps in fly-back converter topology. I am using a EE core i needed to find the saturation current ISAT for my coreThe AL value is 800nh number turn N=63 Using EE 25/13/7 core Primary peak current is 1.25A how could i find the saturation current of the fer-rite core cor air gap=0.1mm operating frequency=65khz core cross sectional area=52mm square Bmax=0.3T.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 18, 2023, 09:59:22 am
content: Datasheet:https://www.tdk-electronics.tdk.com/inf/80/db/fer/e_25_13_7.pdfAssuming this is representative of course. E cores are... fairly standardized? But who knows.Ae = 52.5 mm^2, l_e = 57.5 mm.Interestingly, they give typical values including g = 0.1 mm, for which A_L = 489 nH/t^2. Evidently your 800 nH/t^2 is either the wrong gap (narrower than you think), or the core isn't the same shape.Even more precisely, they give formulas for air gap: A_L and NIsat. Well, parameters for the formulas; apparently those are in another document. If, of course, you have the same geometry and an equivalent ferrite material.For gaps much less than the width/thickness of the core leg, we can approximate:A_L ~= mu_0 Ae / (g + l_e / mu_r)Saturation is easier calculated from flux, which you already have from the switching waveforms in a flyback. Going backwards, we use inductance as the conversion factor between current and flux (H == Vs/A). Thus, Isat = Phi / L, or NIsat = Phi / A_L. And Phi = Ae Bsat. So all together,NIsat = (Ae Bsat) / (mu_0 Ae / (g + l_e / mu_r))= Bsat (g + l_e / mu_r) / mu_0Ae drops out because we are ignoring fringing fields (for which we could use an approximation, or empirical results). Inductance is proportional to Ae so varying Ae gives the same saturation current but proportional inductance, and therefore energy storage is proportional to Ae. Which is very reasonable. Likewise, energy is proportional to length: inductance is inverse with length but Isat is quadratic, cancelling out for a linear proportion overall. Together, volume is proportional to energy, which is most reasonable of all.We can go further and note that almost all the energy is stored in the airgap; its volume is proportional to energy storage. If you were expecting for example 3.1mH [63 turns at 0.8 uH/t^2] and 1.5A peak, that's 3.572mJ energy. The energy density of a magnetic field is the Maxwell stress, σ = B^2 / (2 mu_0 ). At 0.3T, that's 35.8 kPa == J/m^3. Evidently you need about 100 mm^3 of air gap. Which for example might be proportioned 12 x 12 x 0.7 mm (W x thk x gap). You'll need a... somewhat larger core for that.More likely, operation should be moved closer to DCM, and perhaps Fsw raised, to use a core like EE25. What are you doing, about 50W, offline flyback?Tim

author: CosteC
author_rank: Regular Contributor
timestamp: February 18, 2023, 10:10:58 am
content: I think you may find it useful:https://www.ferroxcube.com/en-global/design_tool/indexPC based version works on my contemporary windows, so it can be used still.EE core shapes are standardised, so shall not differer too much between suppliers.

author: whatty
author_rank: Contributor
timestamp: February 20, 2023, 04:45:04 am
content: Yep doing offline 30W as like you said its AL value is not 800nh i gave it as a example Thanks for the answer

author: whatty
author_rank: Contributor
timestamp: February 20, 2023, 04:52:41 am
content: Can i use this formula AL=B * Ae / N * IIsat = B * Ae / AL * NFor calculating the saturation current of ferrite

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 20, 2023, 06:07:08 am
content: In terms of A_L would be,Rearranged,(g + l_e / mu_r) = mu_0 Ae / A_LStarting with,NIsat = Bsat (g + l_e / mu_r) / mu_0substitute,NIsat = Bsat (mu_0 Ae / A_L) / mu_0Isat = Bsat Ae / (A_L N)Be careful where you associate the division. I should prettyprint to be clearest:\$I_\textrm{sat} = \dfrac{B_\textrm{sat} A_e}{A_L N} \$Tim

author: Doctorandus_P
author_rank: Super Contributor
timestamp: February 20, 2023, 06:51:28 am
content: It is quite easy to build a power inductor tester.1. Some power suppy (10 to 20V or so)2. The Inductorofcourse.3. Mosfet (preferably with a real gate driver)4. Some pulse generator, a function generator is good, but an NE555 is "acceptable' (this can be a half decent mosfet driver too).5. Extra flyback diode.6. Current measurement shunt.7. Oscilloscope.To measure:Short the inductor between the power supply and GND for a short time, and measure the current through the shunt resistor with your scope.The first section will be a straight line (you can calculate self induction from that), and when it nears saturation, the current will start increasing quicker.Turn off the MOSfet after a short time, to prevent over saturation and excessive currents.Do make sure to use a beefy MOSfet and big buffer elco's. With this setup it's quite easy to draw short 60A pulses from the buffer elco's, while the average load of the power supply is quite small (maybe 100mA or so).With this setup you can both get an (accurate enough) estimate of the induction of self wound inductors, and examine how the inductor behaves near and beyond saturation. Some have a sharp knee when going into saturation, while others have a much gentler increase of the slope. There are several of such schematics floating on the web.

author: james_s
author_rank: Super Contributor
timestamp: February 20, 2023, 08:11:05 am
content: A while back somebody posted this which is more or less what you describe.http://elm-chan.org/works/lchk/report.html

author: whatty
author_rank: Contributor
timestamp: February 21, 2023, 07:07:02 am
content: Hi there i am having an doubt consider the following dataI am having a ferrite core whose AL=1400 nhAe=52mm2Bmax=0.3TNumber of turns of primary N=25primary peak current to be 1AUpon this by applying it to the below formulaIsat = (Bmax*Ae)/(AL*N)the saturation occurs at 0.4 amps but my primary peak current will goes up to 1Aif i am adjusting the number of turns to 9 and keeping others as same then applying the above same formula i am getting saturation current at 1.23A (which is more than my peak primary current)so what i am concluding is it okay if i adjust just the number of turns instead of AL (without adjusting air gap) we could narrow the current saturation in ferrite core

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 21, 2023, 10:33:41 am
content: Sure, that works...But what's that going to do to your inductance?For example, if you need (25 t)^2 * 1.4 uH/t^2 = 875 uH, at say 1.1 A Isat, that's 530 uJ energy, which needs 4.4 mm^3, or 0.08 mm air gap at Ae = 52 mm^2. You then need 38 t primary to get the required inductance.Tim

author: jonpaul
author_rank: Super Contributor
timestamp: February 21, 2023, 11:55:11 am
content: see old app notes, textbooks on this.Best professional design software, Intusoft Magnetics Designer,http://www.intusoft.com/mag.htmA great transformers and inductors CAD program.Free versions were available after the firm's founder sadly passed.Jon

author: whatty
author_rank: Contributor
timestamp: February 21, 2023, 04:21:06 pm
content: yeah you are right as turns changes the inductance may change sorry i didn't notice