I have wound my first inductor and I have verified the inductance with 2 methods.

However, when I test its saturation current, its much lower than the formula gave me:

\$B_{peak} = \dfrac{V\cdot T_{on}}{A_e\cdot N}\$ (units: volts, microseconds, mm2, turns)

I set \$B_{peak}\$ at 0.2 Tesla and I'm using N87 material in my core.

I admit my windings were sloppy, but other than that I'm not sure what could cause such low saturation current. This has been causing my boost converter to blow up every time.

Here is my test circuit for measuring both the saturation current, where I increase pulse width til it saturates and also use for method 2 inductance measurement.


simulate this circuit – Schematic created using CircuitLab

enter image description here

  • 1
    \$\begingroup\$ Which methods have you used to verify the inductance? What geometry core are you using? \$\endgroup\$ – user36129 Aug 15 '13 at 13:53
  • \$\begingroup\$ First I used a 1.5kohm resistor in series with the 6.8mH inductor and verified half amplitude at ~61 kHz 1vpp sine wave. Second, I measured the volt across a current sense resistor and divided by that resistance to give me current into my inductor with a DC voltage pulsed across it with a known pulse width. I then used the equation L = V dt/di to calculate L. I'm using a toroid: B64290L0651X087 epcos.com/inf/80/db/fer_07/r_22.pdf It is the one with Ae = 51mm^2 and N87 material (Bsat is .39T) \$\endgroup\$ – EwokNightmares Aug 15 '13 at 14:00
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    \$\begingroup\$ Because \$V_L\$ is 90 degrees to \$V_R\$ measuring half amplitude won't necessarily be giving you the answer that you think. What is the \$A_L\$ of the toroid? Also I'd check your formula for \$B_{MAX}\$ \$\endgroup\$ – Andy aka Aug 15 '13 at 14:17
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    \$\begingroup\$ You don't say how many turns of wire or what pulse width you are using. From the figures you give, I would expect 51 turns for 6.8μH. For 0.2T maximum field density, your maximum voltage-time product would be around 520μVs ie 104μs at 5V. \$\endgroup\$ – MikeJ-UK Aug 16 '13 at 9:05
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    \$\begingroup\$ Yes, it would seem to be about 50mA - certainly less than the predicted 77mA. It should apply to all reasonable Vt combinations. \$\endgroup\$ – MikeJ-UK Aug 16 '13 at 15:10
  1. N87 is straight-up ferrite material, not distributed air gap like powder iron types of material. Just because it's in toroidal form doesn't mean it's distributed-gap material - N87 in a toroid will saturate the same way as N87 in an E-core. There's nothing wrong with using straight-up ferrite for a boost inductor, so long as you gap it (more on this later). The fact that it's in toroidal form means you can't gap it. You might want to switch to Kool-Mu if you want to stick with a toroidal form factor.

  2. Deriving the core inductance from the \$A_L\$ factor is reasonably good, so long as you have sufficient turns on the core and can keep the winding reasonably uniform and in as few layers as possible. Bear in mind that \$A_L\$ can vary by +/- 25% on some cores.

  3. Boost inductors carry both the magnetizing current and energy for the load (that will be stored magnetically and delivered during the off-time.) Once the converter starts operating in continuous conduction mode (when the inductor current never goes to zero), it's even worse since you start operating on a B-H curve that doesn't reset to zero. (Bmax is still Bmax, but you now have a DC offset on which Bpeak is riding.) These are the reasons that the inductor needs air gapping - the core won't be able to handle any significant DC current without saturation otherwise.

  4. I'm not sure I understand your test circuit. Both ends of the inductor are essentially clamped to 5V, meaning that the two capacitors (C1 and C2) contribute nothing to the simulation. If your real boost converter is arranged in this manner, it isn't a boost converter and will never work. L1 needs to release its stored energy through D1 to the load, which can never happen when D1 and the load is connected as shown. The only connection between input and output must be through L1 and D1. I would also put R1 in the source of Q1 and do a single ground-referenced measurement instead of a mathematical construction. (L1 will only saturate when Q1 is on, so measuring it when Q1 is off is irrelevant.)


Answer changed to suit changed question

This answer has been edited because the focus of the question has changed. My original answer is still below because it was relevant to the original question.

In any inductor, B (Magnetic flux density) and H (Magnetic field strength) form the B-H curve and from that curve you can see that B doesn't increase linearly with H - this is called saturation: -

enter image description here

H is the ampere-turns driving force behind creating flux and is dimensioned in units of ampere's per metre. It's formula is:

H = \$\frac{I N}{l_e}\$ where I is current, N is number of turns and \$l_e\$ is the magnetic path length and for a toroid \$l_e\$ is the average diameter of the core x \$\pi\$. You don't need to calculate it - all toroids will have this specified in the data sheet.

B, flux density is related to H in following formula:

\$\frac{B}{H} = \mu_0 \mu_r\$

where \$\mu_o\$ and \$\mu_r\$ are the magnetic constant (\$4\pi\times10^{-7}\$) and the relative permeability of your core material respectively.

So, if you know what your current peaks at (or is expected to) and you know how many turns you have (and what material and core size you are using) you can calculate B, flux density.

From the specification for the toroid \$l_e\$ is 54.15mm, and the OP suggests that 77mA is the peak current and, that the toroid is wound with 51 turns. From this we can calculate H: -

\$H = \frac{0.077 \times 51}{0.05415} = 72.5\$ amperes per metre

If we plug this into the B/H formula and use a relative permeability (2200) from the data sheets of N87 we get: -

\$B = 4\pi\times10^{-7}\times 72.5\times 2200\$ = 200.4 mT and this is what the OP states in his question.

This can only mean that the core is saturating because:

  • Not all the magnetic energy has been removed by the time the inductor is pulsed again
  • Remanence flux + new flux (pulse) is causing saturation (see BH curve diagram)
  • For whatever reason, there is more current going into the inductor
  • Unlikely as it seems, the ferrite is not N87

Personally I'd look at the Remanence flux density to see how high this might be. Just taken a look and the coercive field strength in the spec for N87 is 21 A/m. Because you are not getting rid of the Remanence flux there is an equivalent magnetic field strength of 21 A/m which adds to the 72.5 A/m you are applying meaning you are actually applying 93.5A/m and this results in a flux density of more like 260mT.

Added to this, if you are not reducing the inductor current to zero you will be compounding the problem. Given also that the inductor value may be a little lower than you think (\$A_L\$ can be 25% low) these may be enough reasons to account for your problem.

On a different tack, 6.8mH is a mighty big value of inductance to be using in a switcher for what I can perceive is your application. To get the same energy from an inductor of 3400uH only requires current to rise to \$0.077\times\sqrt{2}\$ = 109mA. What is preventing you using a much smaller inductor?

Original Answer

Below was taken from a comment by the OP and my explanation further down is to explain how his method is faulty: -

First I used a 1.5kohm resistor in series with the 6.8mH inductor and verified half amplitude at ~61 kHz 1vpp sine wave

Firstly, if you calculated \$X_L\$ based on it equalling 1500 ohms at 61kHz, you'd get an inductance of \$\frac{1500}{2\Pi F}\$ = 3.9mH. Now look at the phasor diagram below:

enter image description here

In reality if there is 1Vp-p across the inductor, this will be when it has a reactance of more like 1060 ohms and at 61kHz, this is when L = 2.8mH.

If you are nearly 2.5x out in your actual inductance, it is likely that the current through it at \$T_{ON}\$ is 2.5x greater and this of course is going to push a "close-to-saturation" inductor fully into saturation.

  • \$\begingroup\$ When I run the circuit in LTSpice, I get half voltage (-6dB) at 61 kHz. I originally used laplace analysis to determine the same result. \$\endgroup\$ – EwokNightmares Aug 15 '13 at 18:26
  • \$\begingroup\$ I meant when I run it with 6.8mH and 1.5kohm, I get half voltage at 61kHz. When I run it at 2.8mH instead, it shifts the cutoff to 150kHz. \$\endgroup\$ – EwokNightmares Aug 15 '13 at 18:45
  • \$\begingroup\$ Also, my circuit is in the LPF configuration, I should have mentioned that originally. \$\endgroup\$ – EwokNightmares Aug 15 '13 at 18:54
  • \$\begingroup\$ Ok, the above comments are with AC analysis (bode plot). When I do a transient analysis, your value works. I am confused why AC analysis tells me half voltage is at my calculated frequency, but time domain disagrees. \$\endgroup\$ – EwokNightmares Aug 15 '13 at 19:11
  • \$\begingroup\$ I have found that swapping the resistor and inductor make a big difference. In HPF configuration as you draw it, you are correct. In LPF configuration as I calculated and tested it, I am correct. \$\endgroup\$ – EwokNightmares Aug 15 '13 at 19:18

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