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Consider the folLowing half bridge circuit: enter image description here

I'm designing a half-bridge circuit to be controlled by a microcontroller. Ultimately, there will be 3 copies of this circuit for an ESC. I am hoping to have reasonably fast switching times for the power MOSFETs M8 and M9.

V_C is the battery voltage, up to 50V V_L and V_H are 12V above GND and V_C respectively and will be used to drive M8 and M9. M1, M2, M3, and M4 can be driven with logic levels. V1, V2, V3, and V4 are signals from the microcontroller and can be controlled independently to enforce dead-time, preventing shoot-through.

The low side is easy. To turn on M8, V1 is LOW and V2 is HIGH, pulling the M5's gate to GND, which in turn, pulls M8's gate to V_L. To turn off M8, V2 goes LOW and V1 goes HIGH. R1 turns off M5, and M1+R2 pulls M8's gate LOW.

The high side is where things get tricky. To turn on M9, V4 goes HIGH, pulling M7's gate to GND. However, with V_C ~50V, pulling M7's gate to GND is beyond the MOSFET's max Vgs. D3 prevents this by "clamping" this Vgs, with the remaining voltage across R8. The problem is that in this state, M7's gate is ~50V, meaning R8 is dissipating ~2.5W. I can increase the resistor values, but I want to avoid weakening the drive characteristics to keep switching times short.

My instinct for turning off M9 is to use the same kind of setup as the above. V1 goes HIGH, pulling M6's gate to GND, with D1 clamping the gate voltage. In this state, V_load would also be near GND or floating, so I wouldn't even have the same resistor dissipation issue as above. However, this wouldn't work at all. In this schematic, there wouldn't even be a current path through D1.

Am I correct about my assumptions here? Is there a better schematic I can use to drive the high side without wasting so much power in R8? Is there a simple change to introduce a current path for D1 without messing anything else up? Is there a different way I can actually turn off M9 actively? Are M1 and M6 overkill, and I can still have fast enough turn-off times for my purposes relying only on R2 and R3?

Any help/advice you can give me would be greatly appreciated! I don't have any specific part numbers chosen, but I'd like to keep it from breaking the bank. (Also the reason I don't just put a bunch of high-speed optocouplers everywhere). Thanks so much!

Levi Applebaum
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    Why not use a half-bridge FET driver IC? Fewer components, some are available with protection features like UVLO, desat, cross-conduction prevention, etc. There are several issues with your schematic that I don't have time to elaborate on now. – John D Aug 30 '23 at 15:46
  • Mostly, I don't seem to find what I'm looking for. I want to be able to have M8 & M9 turned off. I want to have M9 ON for long periods of time without having to charge a bootstrap cap. And I want all of this to be able to work up to 50V, making the bootstrap issue even worse (MOSFET Vgs limits) and limiting parts in general. I'd also like to be able to experiment with dead-times etc, and I value the "close to the metal" control. Partially, I want to experiment and enjoy the engineering spirit. That said, do you have any parts you recommend? Thanks for the response! – Levi Applebaum Aug 30 '23 at 15:59
  • Use an isolating dc-dc converter whose negative output is connected to M9's source, Then use an optically coupled (and quick) driver. Let the silicon chips do the heavy lifting for you. ADuM4146 springs to mind as do several Murata dc-dc converters. If you are looking for a bargain basement BoM cost then I'm not the man to make recommendations that would suit your budget. I will add this though... use a simulation tool (maybe you do). – Andy aka Aug 30 '23 at 16:07
  • You are trying to generate a voltage referenced to `V_load` lane with using a voltage referenced to local ground. This is beyond level translation so it seems to require a floating supply which can be either a bootstrap cap (although can't be used for your long duration requirement) or, as Andy stated in his answer. – Rohat Kılıç Aug 30 '23 at 16:14
  • Take a look at something like this, with 100% duty cycle support for the high side: https://www.ti.com/lit/ds/symlink/drv8302.pdf?ts=1693412085650&ref_url=https%253A%252F%252Fwww.ti.com%252Fmotor-drivers%252Fbrushless-dc-bldc-drivers%252Fbldc-drivers%252Fproducts.html It might be overkill for your application, but there are other options out there. – John D Aug 30 '23 at 16:16
  • @Andyaka V_L & V_H were to be created by such isolated DC-DC supplies, but V_H relative to V_C, not V_load. Might be better your way, but does that still leave me with the same overpowered R8? I may have to invest in some OPTOs, but avoiding for now. From a quick look at ADu.., actually looks pretty good. I didn't mention in the post that I'm hoping to be able to use down to 7V, but 8V may be agreeable. I plan on simulating as much as I can before designing anything for real! – Levi Applebaum Aug 30 '23 at 16:25
  • @RohatKılıç For the low side, M8's source is at GND, so I figured 12V referenced to GND was fine. For the high side when M9 is active, M9's source is ~V_C. So I figured 12V referenced to V_C was fine. That being said, my plan was to use an isolated converter anyway, so referencing to V_load is probably a better idea! – Levi Applebaum Aug 30 '23 at 16:28
  • @JohnD Thanks! From what I know about bootstrapping, I have no idea how the 100% duty cycle is actually implemented. Now I'm desperately curious. Either way, looks like a promising IC! – Levi Applebaum Aug 30 '23 at 16:30
  • @LeviApplebaum no it doesn't; it's the standard way of driving higher powered MOSFETs using a floating supply. – Andy aka Aug 30 '23 at 16:33
  • @LeviApplebaum There's a supplementary charge pump that allows 100% duty cycle: "The half-bridge drivers use a bootstrap configuration with a trickle charge pump to support 100% duty cycle" – John D Aug 30 '23 at 16:42
  • @Andyaka Do you mean for the R8 question? If I switch to OPTOs, that eliminated this issue entirely. Otherwise, the V_H relative to V_load still has M9's gate at ~50V, dissipating the same ~2.5W through R8. Am I misunderstanding? I know the standard way is with ICs and OPTOs somewhere. I was just trying to use discretes and no OPTOs because some low- power MOSFETs, resistors, zeners are pennies vs $5-10 for a driver IC and maybe the same for a decent high-speed optocoupler. – Levi Applebaum Aug 30 '23 at 16:42
  • @JohnD Thanks again! I'm coming across this info now looking around. – Levi Applebaum Aug 30 '23 at 16:43
  • Ditch the entire driver circuit and use opto-coupled drivers and an isolated DC converter <-- that's what I'm saying @LeviApplebaum – Andy aka Aug 30 '23 at 17:10
  • @Andyaka Gotcha! Looking more and more like I'll have to, but not giving up quite yet. – Levi Applebaum Aug 30 '23 at 17:17

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