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I need to design a three-phase inverter for motor control and isolate microcotnroller from power stage (gate drivers + transistors) .. look at the image

If I use LM5018 as flybuck, at the output there are two 12V ... one is not isolated (and I don't need it) while the other is isolated. The latter isolated one I connect to the microcontroller. In fact, I have indicated two different symbols for the two "isolated" grounds.

However, the controller produces the PWM signals (about 30k or 60kHz, I don't know yet) that need to get to the gate drivers .. does this mean that I need digital isolators (marked in red)?

If the answer is yes ... do you recommend any? (I will probably use TMS320F28069 and LMG2100 power stage)

Did I come up with the design well? Or have I made mistakes that currently escape me?

EDIT: I did not draw gate drivers! I connected uC directly with mosfets .. for simplicity!

enter image description here

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    \$\begingroup\$ Why use an LM5018 if you're not going to use the non-isolated output? The isolated output is not regulated, and I'm not sure if it will work at all without a load on the non-isolated output. Why not use a conventional flyback controller with optocoupler feedback? \$\endgroup\$ Commented May 22 at 15:22
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    \$\begingroup\$ The LM5018 is a buck converter that TI advertises as suitable for their "flybuck" topology that gets an extra output from a normal buck converter in a way that looks a little bit like a flyback's output. A flyback (not "flybuck") controller is designed for a specific well-established isolated topology that has one or more outputs, all of which are isolated. The LM5158 might be suitable. \$\endgroup\$ Commented May 23 at 14:16
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    \$\begingroup\$ Basically, the 5018 is a non-isolated buck converter, so anything reasonably "normal" that you do with it is going to have a non-isolated output as the primary output, and the only output that's accurately regulated. You can add an additional output or two by replacing the filter inductor with a flyback transformer (note the "a" in "flyback"), a technique that TI refers to as "flybuck" (note the "u" in "flybuck"). This gets you an additional, poorly-regulated but isolated output. This additional output relies on there being some nonzero load current on the primary output to function. \$\endgroup\$ Commented May 23 at 14:56
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    \$\begingroup\$ Yes, exactly. You can actually add an isolated output like that to any regular buck converter, the TI one just recommends it. \$\endgroup\$ Commented May 24 at 16:33
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    \$\begingroup\$ Yes, it is. (not the same chip, but it serves the same purpose.) If it says "fly-buck" or "flybuck" it's a buck converter that TI's advertising a specific use for, not an actual flyback converter. \$\endgroup\$ Commented May 24 at 16:34

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Yes, you do need isolation here. Lacking isolation means you no longer have an isolation barrier between your MCU and inverter input or output.

This exact situation is very common, which has led to there being many options on the market for isolated gate drivers, which combine a gate driver (or two in many cases) with an isolator.

You could use a separate non-isolated gate driver and digital isolator, but isolated gate drivers make things easier. TI, Maxim, Analog Devices, Infineon, and several others make them, and there's an entire product category for them on Digikey.

Some of these are dual drivers intended to drive both sides of a half-bridge (and may include functionality specific to that situation, like enforcing dead time), which may be useful for your case.


In light of the comment below:

I don't need gate drivers, because I use a LMG2100 power stage which contains both gate drivers and GaN transistors [...]

While it would be possible to still use an isolated gate driver to drive the LMG2100's inputs, you're right that it would be redundant.

The LMG2100 half-bridge power stage (datasheet) does not include any form of isolation, though it does have a level shifter so both of its inputs are ground-referenced. This means that all inputs of all phases, unless this is being used for a multilevel converter or other unusual topology, are referenced to the same ground. This means you can use a single multi-channel digital isolator for all inputs. An ISO7760 might be suitable to drive a three-phase bridge, for instance.

Especially if operating at high speeds, be sure to check the pulse width distortion and channel-to-channel or part-to-part matching of your isolator of choice; these specifications tell you how much the isolator will affect the timing of your switching. Unless you're switching at inadvisably high frequencies, most general-purpose digital isolators should be capable of your switching frequency (though some old optocouplers may not be).

Remember that you will need a separate power supply for both sides of the isolator; there are very few isolators that integrate an isolated power supply into the part, and those that do tend to be more expensive than just using a separate isolated power supply. In all likelihood, the input side will share a power supply with your MCU, and the output side will share a power supply with the Vcc of your LMG2100. (Note, though, that some isolators may not function at the 5 V the LMG2100 requires; if this is the case, a linear regulator from that down to 3.3 V would be simple enough. An isolator simply doesn't consume enough power for the poor efficiency of a linear regulator to be a concern here.)

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  • \$\begingroup\$ thanks for the answer. I don't need gate drivers, because I use a LMG2100 power stage which contains both gate drivers and GaN transistors ... in fact the PWM input can be connected directly to the uC. What I need (I still need to do a thorough research) is just a buffer that isolates GPIO of the uC from input of the LMG2100 \$\endgroup\$ Commented May 23 at 6:57
  • \$\begingroup\$ @KaleM Then any isolator should do, as long as it can work fast enough. Toshiba makes some pretty good optocouplers that might work. \$\endgroup\$ Commented May 23 at 14:08
  • \$\begingroup\$ @KaleM I've gone into more detail on selecting an isolator, since using an isolated gate driver to drive the input of a gate driver isn't exactly necessary. (It would work, but there's no point.) \$\endgroup\$ Commented May 23 at 15:23

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