Electronic – Turns for saturating a toroid at a particular frequency


I need to know the winding turns required to saturate a toroid for a particular drive frequency.

The toroid I have is having the following dimensions

  • OD = 19 mm
  • Thickness = 8 mm
  • Effective Area (Ae) = 32 sq. mm
  • Al (nH) = 1900
  • Bsat = 0.49T

My driving signal would be

  • f = 100kHz
  • Duty = 50%
  • Drive Voltage = 15V

From my initial research I have found that this relates to the following equation

N = (V * t) / (B * Ae)

V = drive voltage
t = on time of the drive voltage

Based on the above I am getting turns of 4.8

I need to know whether this is the right equation, I mean is it actually giving be the turns to drive the toroid to saturation?

Best Answer

It's unusual to specify a 50% duty cycle at the outset. This is kind of weird. Usually, you look at the input source situation and the output situation and work out the required duty cycle from there. The output is arranged so that the off-voltage and off-time of the inductor works out correctly. Starting with a duty cycle is an unusual approach.

The inductor volume is a useful metric (though not often directly considered in most write-ups.) You can compute it in your case (square wave) as:

$$l_e\cdot A_c \ge \frac{\mu_0\cdot\mu_r\cdot I_{peak}}{B_{max}^{~2}}\cdot V_{on}\cdot t_{on}$$

That can all be turned around to compute \$I_{peak}\$:

$$I_{peak} \le \frac{\left(l_e\cdot A_c\right)\cdot B_{max}^{~2}}{\mu_0\cdot\mu_r\cdot V_{on}\cdot t_{on}} $$

Plugging in your figures, I get:

$$I_{peak} \le \frac{\left(44.4\:\textrm{mm}\cdot 32\:\textrm{mm}^2\right)\cdot \left(0.49\:\textrm{T}\right)^{~2}}{\mu_0\cdot 2100 \cdot 15\:\textrm{V}\cdot 5\:\mu\textrm{s}} \approx 1.724\:\textrm{A}$$

And of course from your basic equation for an inductor, I can now compute:

$$L \ge \frac{V_{on}\cdot t_{on}}{I_{peak}} = \frac{15\:\textrm{V}\cdot 5\:\mu\textrm{s}}{1.724\:\textrm{A}}\approx 43.5\:\mu\textrm{H}$$

(I misspoke in a comment to you, earlier. I meant greater.)

From this, you can work out the windings needed on your toroid, of course. You have \$A_L\$.

The problem will now be that since you require 50% duty cycle, then you must provide for a reversed voltage across the inductor during the off-time that is at least as large as the applied voltage.

The reason for this is that the on-time volt-seconds (Webers) must be matched by off-time volt-seconds (except that the voltage polarity must be opposite.) It must be the case in each period that:

$$V_{on}\cdot t_{on} + V_{off}\cdot t_{off} = 0$$

Given enough cycles, there is no escaping that need. If there is even the smallest consistent deviation, each cycle, then it will build up and given enough cycle periods, walk itself so that it exceeds whatever limitations your core has (unless it is vacuum, which has no limitations -- with neutron stars perhaps demonstrating this fact.)

If \$t_{off}=t_{on}\$, then it must be the case that \$V_{off}=-V_{on}\$. Some circuits allow the inductor to find its own reverse voltage and provide for a long enough \$t_{off}\$ time so that in all circumstances it can return to zero. Then the inductor will automatically drop its own voltage back to zero, too, for whatever time remains in \$t_{off}\$ (until the next cycle starts.)

If \$\vert V_{off}\vert\ge \vert V_{on}\vert\$, then less time in the off period is required than you have provided. And that's fine if and only if the voltage across the inductor is zero for the remainder of the time. Not reversed. But exactly zero. Anything other than that will gradually accumulate Webers in one direction or the other until you exceed the core's ability to handle the flux (again, unless its vacuum.)

I started out talking about core volume. Yet far more often you will hear about core area. These are related, though. A Tesla is just flux divided by area. So 1 Weber divided by 1 Tesla comes out as area in \$m^2\$.

When you are winding a core, you are winding it around an area. So this is why the focus in most papers and documents discusses that concept a lot more. But to get a feel for magnetics, I think it's better to think about flux (Webers) than to think about flux density (Webers per meter squared.)

[Perhaps, just as it is often somewhat easier to think in terms of mass than to instead always force yourself into thinking of density. Density can be converted back to mass, of course. But it confuses a lot of equations and thinking if you aren't allowed to use mass, but always have to use density and volume, because now you have to keep two things in mind instead of one. Also, because of our biology, we have a more intuitive feel for the idea of "weight" (given gravity, proportional to mass) than we do of "density," which our biological sensors have no direct way to "observe."]

Sure, flux density is important. That's because matter acting as little magnetic dipoles that respond to an applied magnetic field does have its limitations. (In Tesla.) But flux is what must return to zero. Sure, you can also say that flux density must return to zero, too. Clearly, the area is non-zero, so if flux density goes to zero then flux also goes to zero. But again this forces you to imagine a slightly more complicated concept, the ratio of flux to area, instead of just focusing on the actual thing under consideration, which is just flux.

So if you have a material, such as some specific ferrite or powdered iron or whatever, it will have a limitation in flux density. Given a magnetic cross section (\$A_C\$), you can work out the allowable flux. But keep in mind that flux, itself, is only one of two orthogonal parts of the magnetic field. The total field energy isn't completely determined by just flux. It is determined by the magnetic flux AND the magnetic force, combined. You already know that the magnetic force is measured in amperes and the intensity (another paired dimensional unit thing) is the amperes per meter (H).

At this point you should be thinking: "Hmm. Flux times Force is Energy!! Wow! And Flux density times Force intensity then must be Energy per unit volume!! Incredible!!" Ah. See that 'volume' bit there, sneaking up on you??

Vacuum has no limitations here. But all matter does. And matter that can form into dipoles (by definition, a magnetic dipole opposes the applied field) will have some kind of limitations in the number of useful dipoles it can form up into, right? And this limitation is really throughout the entire material. Not just an area. But the entire VOLUME must exhibit this limitation. There's no reason to imagine that this is just a cross section behavior only. It's almost certainly a factor that affects all 3 dimensions!

So, now, if I know the energy I need stored, and if I know a particular material which can only support a certain 'energy per unit volume', then knowing the energy alone I can now compute the volume of the matter (material) I need to properly hold that energy and still be within its limitations of flux density and force intensity!

Another way of sweeping aside complications here is to imagine that energy MUST ONLY BE STORED in vacuum (which has no limitations) and that the magnetic dipoles formed up in matter are "short circuits" which cannot store energy of any kind (ideally; in practice, of course, it takes energy to cause them to rotate, which may also cause friction and heating and energy loss into the core, etc.) And that the value of \$\mu_r\$ is nothing more than a ratio of physical material volume to remaining magnetic vacuum volume in the matter itself. This is why the volume increases for a specific energy to be stored when you increase \$\mu_r\$. Nothing comes free. A point of a high value of \$\mu_r\$ (ignoring energy losses for the moment) is to concentrate flux lines and to keep them from spreading out into a huge volume of space around the inductor. To contain them, in short. You pay a price for that, which is the energy losses in a practical core and the limitations in energy storage given some volume of material to work with.