DC-DC Converter Design and Magnetics--Do We Mean Inductor? Or Inductance?

Do We Mean Inductor? Or Inductance?

Note that in the previous section, we said nothing explicitly about what the inductance was - we just talked about the size of the inductor. We know that in theory, we can put almost any number of turns on a given core, and get almost any inductance. So inductance and size of inductor are not necessarily related. However, we will now see that in power conversion they often do turn out to be so, though rather indirectly.

Looking at Fgr. 6, we can see that a smaller r will require a higher energy-handling capability, and thus a larger inductor. Let us now formally go through all the possible ways of reducing r.

Since we are assuming our application conditions are fixed, the load current and input/output voltages are also fixed. Therefore, IDC is fixed too. The only way we can cause r to decrease under these circumstances is to make delta I smaller. However, delta Iis

delta I = voltseconds / inductance

(V-s/H)

But we know the applied voltseconds is fixed too (input and output voltages being fixed). So the only way to decrease r (for a given set of application conditions) is to increase the inductance. We can therefore conclude that if we choose a high inductance, we will invariably require a bigger inductor. It’s therefore no surprise that when power supply designers instinctively ask for a "large inductance," they might well mean a "large inductor." Therefore the designer is cautioned against being too "ripple-phobic" in their designs.

A certain amount of ripple is certainly "healthy."

However, we must not forget that if, for example, we increase the load current (i.e. a change in application conditions), we will clearly need to move to a larger inductor (with greater energy-handling capability). But simultaneously, we will need to decrease the inductance.

That's because IDC will increase, and so to keep to the "optimum" value of r, we will need to increase delta I in the same proportion as the increase in IDC. And to do this, we have to decrease, rather than increase, L.

Therefore the general statement that a "large inductance is equivalent to a large inductor" only applies to a given application.

How Inductance and Inductor Size Depend on Frequency

The following discussion applies to all the topologies.

If keeping everything else fixed (including D) we double the frequency, the voltseconds will halve, because the durations tON and tOFF have halved. But since delta I is "voltseconds per unit inductance," this too will halve. Further, since IDC has not changed, r = delta I/IDC will also halve. So if we started off with r = 0.4, we now have r = 0.2.

If we want to return the converter to the optimum value of r = 0.4, we will now need to somehow double the delta I we were left with at the end of the last step. The way to do that is to halve the inductance.

++ Therefore, we can generally state that inductance is inversely proportional to frequency.

Finally, having restored r to 0.4, the peak will still be 20% higher than the dc level. But the dc level has not changed. So the peak value is also unchanged (since r hasn't changed either, eventually). However, the energy-handling requirement (size of inductor) is

1/2 × L × IPK2.

So since L has halved, and IPK is unchanged, the required size of the inductor has halved.

++ Therefore, we can generally state that the size of the inductor is inversely proportional to frequency.

++ Note also that the required current rating of the inductor is independent of the frequency (since peak is unchanged).

How Inductance and Inductor Size Depend on Load Current

For all topologies, if we double the load current (keeping input/output voltages and D fixed), r will tend to halve since delta I has not changed but IDC has doubled. Therefore to restore r to its optimum value of 0.4, we need to get delta I to double too. But we know that delta I is simply "voltseconds per unit inductance," and in this case the voltseconds has not changed. So the only way to get delta I to double is to halve the inductance.

++ Therefore, we can generally state that inductance is inversely proportional to the load current.

What about the size? Since we doubled the load current, but still kept r at 0.4, the peak current IDC(1 + r/2) has also doubled. But the inductance has halved. So the energy-handling requirement (size of inductor), 1 / 2 × L × IPK2, will double.

++ Therefore, we can generally state that the size of the inductor is proportional to the load current.

How Vendors Specify the Current Rating of an Off-the-shelf Inductor and How to Select It

The "energy-handling capability" of an inductor, 1/2 × LI^2, is one way of picking the size of the inductor. But most vendors don’t provide this number upfront. However, they do provide one or more "current ratings." And if we interpret these current rating(s) correctly, that serves the purpose too.

The current rating may be expressed by the vendor either as a maximum rated IDC, or a maximum rated IRMS, or/and a maximum ISAT. The first two are usually considered synonymous, since the RMS and dc values of a typical inductor current waveform are almost equal (we had indicated previously that the RMS of the inductor current is not very "shape-dependent"). So the dc/RMS rating of an inductor is by definition basically the direct current we can pass through it, such that we get a specified temperature rise (typically 40 to 55fic depending on the vendor). The last rating, that is, the ISAT, is the maximum current we can pass, just before the core starts saturating. At that point, the inductor is considered close to the useful limit of its energy-storing capability.

We will also find that many, if not most, vendors have chosen the wire gauge in such a manner that the IDC and ISAT ratings of any inductor are also virtually the same. And by doing this, they can publish one (single) current rating - For example, "the inductor is rated for 5 A." Basically, having determined the ISAT of the inductor, the vendor has then consciously tweaked the wire gauge (at the saturation current level), so as to also get the specified temperature rise too.

The rationale for wanting to set IDC = ISAT is as follows - suppose the inductor had a dc rating of 3A and an ISAT of 5A. The 5-A rating is then likely to be superfluous, because users would probably never select this inductor for an application that required more than 3A anyway. Therefore, the excessive ISAT rating in this case essentially amounts to an unnecessarily over-sized core. Of course, if we do find an inductor with different IDC and ISAT ratings, it’s also possible the vendor may have (unsuccessfully) tried to exploit the larger size of the chosen core (by increasing the wire thickness), but the stumbling block was that the selected core geometry was somehow not conducive to doing so - maybe it just did not have enough window space for accommodating the thicker windings.

In general, an inductor with a "single" current rating is usually the most optimum/ cost-effective too.

However, in some rare off-the-shelf inductors, we may even find ISAT stated to be less than IDC. But what use is that? We can't operate beyond ISAT in any case! So the only advantage, if any, that can be gleaned from such an inductor is that the temperature rise in a real application will be less than the maximum specified. Automotive applications?

In general, for most practical purposes, the current rating of the inductor that we need to consider is the lowest rating of all the published current ratings. We can usually simply ignore all the rest.

There are some subtle considerations and exceptions to the argument for always preferring an inductor with IDC ˜ ISAT. For example, under transient/temporary conditions, the momentary current may exceed the normal steady operating current by a wide margin. So For example, suppose we are we using a switcher with an internally fixed current limit 'ICLIM' of 5A-in a 3-A application. Then under startup (or sudden line/load steps), the current is very likely to hit the limiting value of 5 A for several cycles in succession as the control circuitry struggles to bring up the output rail into regulation. We will discuss this issue in greater detail below - in particular, whether this is even a concern to start with! However, assuming for now that it is, it then seems that it may actually make sense to use an inductor rated for 3-A continuous current, with an ISAT rating of 5 A (provided such an inductor is freely available, and cheap). Of course, alternatively, we could just pick a standard "5 A inductor" (for the 3-A application), and thereby we would certainly avoid inductor saturation under all conditions (and the consequent likelihood of switch destruction). But we realize that in doing so, our inductor may be considered slightly over-designed from the viewpoint of its copper/temperature-rise - the wire being unnecessarily thicker. However, we should keep in mind that larger cores certainly affect cost, but a little more copper rarely does!