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I'm using a LM2576 switching voltage regulator and I see that an inductor and a Schottky diode are required at the output.

The schematics is below:

enter image description here

I understand why the capacitors are there, but why is the inductor required? Is it used for filtering purposes?

Why is the diode there? Can I put a normal diode instead of a Schottky one?

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    \$\begingroup\$ You might want to take a look at the theory of operation: en.wikipedia.org/wiki/Buck%E2%80%93boost_converter \$\endgroup\$
    – Eugene Sh.
    Commented Mar 2, 2021 at 22:29
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    \$\begingroup\$ Because they (together with a switch in the chip) are the very core of a buck convertor. \$\endgroup\$
    – user16324
    Commented Mar 2, 2021 at 22:29
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    \$\begingroup\$ ... and you can't fit them inside the IC because they would be too big and or too hot. \$\endgroup\$ Commented Mar 2, 2021 at 22:47
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    \$\begingroup\$ Monolithic Power makes a whole business of integrating inductors inside their packages. Some are very small - 2x2mm and 1.6mm thick. Example: monolithicpower.com/en/products/power-modules/step-down/… \$\endgroup\$ Commented Mar 2, 2021 at 23:53
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    \$\begingroup\$ Martel What is the startup power from 40V to a 1000uF cap with 10 mOhm ESR? 1kW? 100W? 10W? \$\endgroup\$ Commented Mar 3, 2021 at 0:42

3 Answers 3

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The inductor and diode are integral parts of a buck converter (this type of regulator). Without them, you cannot make a buck converter. In fact, most switching regulators need an inductor in them (charge pumps are the exception).

You seem to be asking why you need your own diode and inductor, and why they aren't just part of the regulator chip. Well, actually you can get regulators that are all-inclusive. This one isn't.

They make regulator modules with different levels of integration - this one includes the switching element and the control circuitry, but not the inductor (a.k.a. storage element). Why not? Well there are different reasons - having not everything inside the regulator module means you can customize the parts that aren't on the chip; it makes the chip smaller and cheaper; it allows for different thermal management strategies.

Chips can't contain inductors, so when the module includes an inductor, they just glue it to the chip and then put the whole thing in a plastic box. Convenient, but not good for heat.

At the other end of the spectrum, there are "build your own regulator" chips that contain all the needed parts of the control circuit, which you have to wire together yourself, and then add your own switching transistor and inductor and diode. Obviously very flexible - you could make a tiny converter or a giant one with these chips - but you have to actually design the converter yourself.

Somewhere in the middle, we have the LM2576.

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  • \$\begingroup\$ I've been using MPS integrated-inductor modules for years. They even have versions that support up to 6A in a 4x6mm x 1.6mm thick package (MPM3860, MPM3650). What makes them possible is using higher frequency switching, which permits smaller inductance values to be used, and thus, smaller inductors. \$\endgroup\$ Commented Mar 3, 2021 at 19:20
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Preface: we're talking about 'buck' (step-down) regulators here. Unless stated otherwise all comments apply to that topology only.

tl; dr: The LM2576 is a very old chip. Switching regulator technology has evolved over time, with further integration happening (including that pesky inductor). More about that below.

Why isn't the inductor part of the silicon?

In theory it's possible using MEMS techniques to make inductors (and some RF systems do exactly that), but power inductor values and currents are too large to be practical for silicon. So, the inductor lives outside.

Why the inductor at all?

In all switching regulators, the inductor is the energy storage element that converts a pulsed source into a smooth DC (along with the filter cap) to the load. In a buck regulator it’s basically working as a low pass filter (integrator), but its role is more subtle than that. In short, it's a way to transfer energy in one voltage domain to another, without the losses that happen with linear regulators.

The inductor is in one of two states:

  • charging: high-side FET is on, flux is increasing (storing energy)
  • discharging: high-side FET is off, flux is collapsing (releasing energy)

So the regulator IC's job is to steer that energy at the right time from supply to load, by steering the current path to/from the inductor during the regulator switching cycle.

For the buck regulator, the inductor input (left) side swings positive during charge, then negative during discharge. Why does it swing negative? Because of that collapsing flux resulting in cutting off the current from the high-side FET. That current has to go somewhere, otherwise that negative voltage gets very large.

Which brings us to the diode. Why?

The diode catches the inductor's collapsing-flux reverse voltage and forms a path for the inductor to dump its energy into the load. Unsurprisingly, it’s literally called a catch diode in some documents; and I'll keep calling it that here, because it's catchy...

And why Schottky?

Schottky diodes are preferred for catch diodes because they have a lower forward voltage (Vf) of 0.3V vs. 0.7V for normal diodes, and they have a much shorter recovery time (the time to switch from forward to reverse. More here: What is the reverse recovery time in a diode?) These traits improve efficiency.

Is there anything better?

You bet.

Some switching regulators include an internal low-side FET to catch the discharge, and so don't need a catch diode. These are called synchronous regulators, and they're more efficient since there is no diode forward voltage drop during the discharge-and-catch state. This type is popular and inexpensive now.

For high-power and high-voltage applications there are also switching regulator controllers that use external FETs, because the FETs are so huge that they too aren't practical to combine with the control circuit silicon due to the special process and packaging needed. Some controllers even support multiple phases for extremely high currents (hundreds of Amps for power-hungry CPU / GPU chips.)

Can I find a chip that has all this stuff in one package?

¡Sí, se puede!

For low-to-medium power applications, there are modules that integrate the inductor as part of the package. What makes this possible is ever-increasing switching frequency. The higher the frequency, the smaller the inductor value needs to be. So now it's possible to actually fabricate the inductor as part of the package leadframe (clever, huh?)

Several companies offer very compact solutions in the 500mA - 6A range that only need external capacitors for filtering. They're super easy to use, save board space, and in my experience, work really well. I've used them on M.2 cards for example, which have a strict height limit.

Here's a 6A, 17V-in buck DC-DC module that's only 4x6mm, that would replace the LM2576 in most applications: https://www.monolithicpower.com/en/products/power-modules/step-down/mpm3650.html

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    \$\begingroup\$ This is a great description. I would add that the current measured in the inductor is more or less a triangle wave at the switching frequency with an average value equal to the load current. The voltage at the left side of the inductor looks like a square wave, with varying duty cycle. You may recall the voltage and current relationship of an inductor: v = Ldi/dt. The voltage across the inductor is being switched between 2 values: Vin-Vload (positive) and 0-Vload (negative). The duty cycle is varied (PWM) as needed to raise and lower the output voltage (measured at feedback pin). \$\endgroup\$
    – Troutdog
    Commented Mar 3, 2021 at 1:03
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    \$\begingroup\$ I think that level of detail is beyond the scope of question.Others linked wiki pages, etc. I may link in a sim later though to illustrate the buck behavior. The key takeaway is that the inductor stores and releases energy, and that energy is in the form of flux. \$\endgroup\$ Commented Mar 3, 2021 at 1:16
  • \$\begingroup\$ Thanks, yeah you're right. \$\endgroup\$
    – Troutdog
    Commented Mar 3, 2021 at 1:52
  • \$\begingroup\$ As well as the lower forward voltage, isn't the faster recovery time of the Schottky diode important? \$\endgroup\$
    – nekomatic
    Commented Mar 3, 2021 at 10:43
  • \$\begingroup\$ Yes, it is, and I rolled that in as part of the rewrite. In fact the fast recovery is essential for increasing switching frequency, especially with high stepping ratios. \$\endgroup\$ Commented Mar 3, 2021 at 20:42
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Think of the left side of the inductor as wiggling very fast between the input voltage Vin, and ground. The inductor is like a big "mass" that evens this out with its inertia. The right side of the inductor is mostly fixed at the output voltage, assuming everything is already up and running.

The controller causes the the left side of the inductor to connect to Vin for a short while, then when it lets go, the inductor snaps its voltage down towards negative infinity, but the diode catches it, so the voltage at the left of the inductor only goes down to about -0.3 V or so (for Schottky type diode), until the controller decides to turn on the high side switch again. While the inductor is connected to Vin, the current increases through the inductor, then while it is disconnected and the inductor node is near ground, the current decreases. But it is always flowing to the right (at least in this simplified continuous model).

The rate of increase is proportional to the difference between Vin and Vout. The rate of decrease is proportional to the difference between Vout and -0.3 V. The rates are also inversely proportional to the inductance value of the inductor -- a larger inductor has more "mass", so its current changes more slowly.

The total voltage on the right side of the inductor is proportional to Vin and ratio of time the left side of the inductor spends at Vin vs. -0.3 V.

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