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)

where 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?

  • 1
    \$\begingroup\$ Square wave drive voltage? \$\endgroup\$ – jonk Dec 4 '16 at 6:39
  • 1
    \$\begingroup\$ is this a transformer, where you are driving a 100kHz square wave with +/- 15v, and the core flux is swinging +/- 0.49T, or is this a flyback where a 15v pulse is going on for 5uS, and the flux is swinging between 0 and 0.49T? \$\endgroup\$ – Neil_UK Dec 4 '16 at 6:39
  • 2
    \$\begingroup\$ Let me second that, voff must be 15 v in the other direction, in other words, ZERO DC voltage on average. For a gate drive transformer, this is OK, as FET gates can tolerate -15v for off. \$\endgroup\$ – Neil_UK Dec 4 '16 at 7:53
  • 2
    \$\begingroup\$ B field is not the voltage, B field is integral of voltage. dB/dt is voltage. B can be offset by a constant, and still get the same voltage. B will walk off to saturation if there is net DC voltage applied. \$\endgroup\$ – Neil_UK Dec 4 '16 at 7:54
  • 2
    \$\begingroup\$ No. You will go from \$B_{sat}\$ to 0. Just think. Neil correctly points out that this is about integrals. So it integrates up to \$B_{sat}\$ and then de-integrates (if I may use that term) back down to 0 -- assuming you allow it to do so, of course. That's the problem -- you must allow it to get back to 0 or else it will just walk upwards. \$\endgroup\$ – jonk Dec 4 '16 at 7:58

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.

  • \$\begingroup\$ excellent way to approach the problem!! \$\endgroup\$ – Miguel Sanchez Dec 4 '16 at 8:55
  • 1
    \$\begingroup\$ @MiguelSanchez Then the whole world starts making a lot more sense and you can then focus on what's important, I think. However, you will still need to balance that equation. Webers must return to 0 every cycle. So a reversing voltage is still required. \$\endgroup\$ – jonk Dec 4 '16 at 8:57
  • 1
    \$\begingroup\$ @jonk webers don't need to return to zero every cycle, they just need not to saturate, so the reversing voltage is needed to bring them back to somewhere where they can grow again next cycle. \$\endgroup\$ – Neil_UK Dec 4 '16 at 9:14
  • 1
    \$\begingroup\$ @MiguelSanchez Yes. I didn't disagree with your equation, before. If I had, I would have said so. I was just worried you didn't understand and I wanted to elaborate a few details. You need to spend time thinking more about magnetic fields. There's an excellent textbook called "Matter & Interactions" which gives you a seriously good and intuitive understanding of physics -- I haven't seen a better one. Another might be volume 2 of the Feynman Lecture series (now available on the web.) But it is not nearly so intuitive, yet still pretty good in a classic kind of way. \$\endgroup\$ – jonk Dec 4 '16 at 19:44
  • 1
    \$\begingroup\$ @MiguelSanchez You really need to read "Matter & Interactions". And also spend time thinking about classical mechanical physics, to start. There isn't space here to write chapter and verse. But try and think about the fact that the electrons circulating in a loop of wire must be under constant acceleration to keep changing direction. Since the magnitude doesn't change, the acceleration must be perpendicular to motion. But the force must be \$\frac{m\cdot v^2}{r}\$. Given same amps, you know same \$v\$. Double \$r\$ you must double electron mass. So applied force is dependent only on amps. \$\endgroup\$ – jonk Dec 5 '16 at 20:49

TL;DR you can use a 15v unidirectional supply, but only in flyback mode

You state in a comment to Jonk's answer that you do not want to use a negative V. There is no recipe for driving a transformer, and keeping it out of saturation (ie working for more than one cycle), without a negative V from somewhere.

The rate of change of flux is proportional to, and the same sign as, the voltage. No negative V, no drop in flux, therefore continuous walk of the flux to saturation.

In a conventional transformer, we drive with an AC waveform, this provides the negative V.

In a flyback, we use a unidirectional voltage drive, so there's no negative V supplied. How does that work?

a) Assume we start from zero flux. We apply a voltage, by closing a switch to a power supply. The current, and the flux, grow at some rate.

b) Before we reach saturation in the core, we open the switch, to try to stop the current. Let's model the open switch as a small capacitor.

c) The current continues to flow, driven by the stored energy in the inductor, and charges the capacitor, very quickly.

d) How high does the voltage go? If you don't actively design for this phase, usually higher than you want. If it's a relay you've switched with a transistor, it will probably break over the transistor, unless you've put a catch-diode across it, to provide a path for the current to stay flowing at a low voltage drop. If it's mechanical contacts, they will probably arc, unless you've added an extra capacitor across them to slow the rate of voltage rise until the contacts have opened far enough, as in an old style car ignition system.

You could say that this flyback voltage reduces the flux in the core. Or you could say that breaking the current to reduce the core flux has created a large negative voltage. It doesn't matter, they are equivalent, and you can't do one without the other.

So you can drive your transformer with a unidirectional 15v. But you can only do this in flyback mode, and you have to make allowances for the voltage spike that will occur when you switch the primary drive off. This negative voltage spike will break over your FET gates on the secondary, and your drive transistor on the primary, unless you add circuit components to limit both to the voltages that the data sheets say they can take. Hint, the higher the voltage you can tolerate safely, the faster the flux will fall in the core, ready for the next cycle.

Note that driving it as a flyback, and driving it with 0 to 15v are not the same thing. If you use an amplifier, or an H-bridge or something, to force 0 and 15v onto the terminals, then for the 15v phase, flux will increase, and for the 0v phase flux will stay the same and current will continue to flow.

Driving as a flyback means during the 15v phase, flux increases. During the 'off' phase however, you open circuit the drive to allow the inductor to create a negative voltage at its terminals, which drops the flux back, eventually to zero.

  • \$\begingroup\$ where the equation N = (V * t) / (B * Ae) fit in or is it just not relevant for this use case? \$\endgroup\$ – Miguel Sanchez Dec 4 '16 at 10:31
  • \$\begingroup\$ Could you elaborate your earlier statement B field is not the voltage, B field is integral of voltage. I need to get my head right on this, I'm still confused on the BH curve and this statement any reference's to web pages or books would help \$\endgroup\$ – Miguel Sanchez Dec 4 '16 at 10:34
  • \$\begingroup\$ as I also wrote to jonk I am confused because I always thought that magnetic flux was related to current (amps) I mean propotional to current but the concepts that you and jonk turned my world around! \$\endgroup\$ – Miguel Sanchez Dec 4 '16 at 10:40
  • \$\begingroup\$ Flux is proportional to current. But in an inductor, current is not proportional to voltage! Current is proprtional to flux, voltage is proportional to rate of change of flux. If you understand capacitors, think of it as the dual of that. Voltage is proportional to charge (the dual of flux), current is proprtional to the rate of change of charge. \$\endgroup\$ – Neil_UK Dec 4 '16 at 12:16
  • \$\begingroup\$ N=(vt)/(BAe) is just details. It's probably right. But it's unimportant until you get your head straight on how flux, current, voltage etc work. N=(... is like polishing your front crawl. Voltage and flux is like drowning or not. \$\endgroup\$ – Neil_UK Dec 4 '16 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.