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I've decided to conquer my fear of switched-mode power supplies and build one. Right now, I have a L4960 regulator.

Its datasheet recommends use of a 150 μH inductor rated at 5 A. My problem is that I can't find such an inductor. I can either get inductors with lower current rating (for example 150 μH at 1 A) or inductors with smaller or higher inductance (for example 100 μH at 10 A).

So my question is: How is it calculated that 150 μH 5 A inductor is needed?

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I'm not totally clear on the question.

If the question is: "Why was 150 µH and 5A chosen?", it's a matter of preference.

Generally, the size of the inductor in a buck regulator is chosen to achieve a certain ripple voltage in the output capacitors at a given load, or to set the discontinuous-continuous threshold at a given load. The 5A saturation current threshold is most likely to ensure that the regulator itself will go into over-current protection before the inductor saturates out.

If the question is: how are the inductance and saturation current properties of an inductor calculated, then it's a matter of physics.

The inductance and saturation current of an inductor are primarily properties of numerous characteristics, the most important of which are:

  • the construction geometry (shape, number of turns, etc.)
  • the core material (air, ferrite, etc.)

Different numbers of turns will change the inductance for a given core. The core material (and air gap, if applicable) will determine how much DC current the inductor can carry before the permeability rolls off, which is generally what saturates out an inductor in a buck design (it carries more DC than ripple current, generally speaking).

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  • \$\begingroup\$ My question is the first one. So if I understand your answer correctly, the correct number has been determined experimentally? \$\endgroup\$ – AndrejaKo Apr 19 '11 at 13:39
  • \$\begingroup\$ The capacitor ripple voltage is fundamentally due to the superimposed inductor current, which is calculated as a function of the inductance, duty cycle, input voltage and load. If you look up buck converters you can easily find the formulae. \$\endgroup\$ – Adam Lawrence Apr 20 '11 at 3:15
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If you can't find one, it's quite easy to wind your own with a suitable toroidal ferrite core. 150 uH is quite a low value and won't require many turns. You need to ensure that you use a wire gauge that can handle 5A, of course. Suitable cores are available from most component suppliers, and the manufacturer's web site will have design details. I've wound larger inductors using a pot core for an MCU-controlled switching power supply without any problems.

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  • \$\begingroup\$ I was considering that idea too, but there's a but. The datasheet recommends a way to wind the inductor, but I can't find any information on the recommended ferrite cores. To make things even better, catalogs of the stores in my area don't have any information on relative permeability of the cores. There are markings such as "FM 452 50" or "45010", but Internet searches bring up tons of unrelated information. \$\endgroup\$ – AndrejaKo Apr 19 '11 at 12:04
  • \$\begingroup\$ Try Amidon: amidoncorp.com \$\endgroup\$ – Leon Heller Apr 19 '11 at 13:07
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The inductance determines how much energy you put in the magnetic field. Lower switching frequencies require larger inductance. The wire diameter is determined by the maximum current the coil will see. Too thin a wire will have a higher resistance and reduce efficiency.

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I don't think the coil's value is found experimentally, but rather that it's a complex calculation that the manufacturer has done for us. Same, by the way, for the other components. Take National's Webench, for instance. You supply the SMPS's black box parameters, and it returns a full schematic including BOM. If you're interested in the details of the calculation, however, have a look at a Linear Technology datasheet.

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  • \$\begingroup\$ I'm aware of general use of the coil and of its significant characteristics. Here I'm more looking for a way to understand how the number itself was derived. \$\endgroup\$ – AndrejaKo Apr 19 '11 at 13:21
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When designing a SMPS, the inductor is the most (like numero uno) import part of the whole circuit. The inductor will decide how nice your output rail is. The actual choice of inductor will change based on application, load, etc.

When selecting an inductor, you'll need to make sure your max supported amperage on the SMPS is lower than the inductors inherit saturation current. Check out this tool from Vishay. They have very high quality inductors and are great with their documentation.

You'll also want to work with the package size of the inductor, as this will affect its inherit DCR (resistance) as well as its power dissipation capabilities. A larger package will generally have better power dissipation properties although its DCR will be higher as well (which is bad)

Many of these things are found through calculations, simulations, and such although a good place to start is:

  1. Pick your target output current - based on what it is powering.
  2. Find your switching controller IC
  3. Calculate your feedback loop (the feedback resistors)
  4. Find your mosfets (if they are needed), otherwise skip this
  5. Pick your inductor

Use the vishay tool to decide certain factors. Play with the parts they offer. You'll want to keep an eye on your core losses and fiddle with the switching frequency. Fiddle with the parts they offer some more. It is a good idea to de-rate the current capabilities in the back of your mind just to give your application some headroom.

  1. Pick your output capacitance

Your output capacitors will help filter out noise on the output rail. You'll need some low capacitance ones like .1uF to filter out high-frequency noise, as well as some larger bulk capacitors too. Bulk capacitors can be 330uF or 470uF or anything large, but it really depends on your application. Make sure these have a high enough voltage rating to handle your output voltage.

And there you have it, that is a really really really general, non-scientific way to make an SMPS.

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  • \$\begingroup\$ Oh, there is already an accepted answer....oh well \$\endgroup\$ – Funkyguy Jul 9 '14 at 21:16

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