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From Texas Instruments snoaa15, which is an Application Report about overcurrent protection of GaN FETs, in the first page, you can read the following quote:

Traditional approaches using current sense transformer, shunt resistors, or de-saturation detection circuits remain ineffective due to slow response time.

So, I stumbled on the matter de-saturation detection (which is widely used in IGBT converters for protection and works very well) is ineffective for GaN technology due to slow response time, and that's why we should use their LMG341X driver, which includes fast and reliable short-circuit and overcurrent protection.

My problem is I am designing a pcb in which I'll make use of GaN in order to shrink as much as possible the pcb size and I don't want to buy the driver since it's really expensive compared to other GaN FET drivers. So to sum it up, is there another effective and cheaper way I can protect my GaN against short-circuit without needing to buy this specific driver?

The circuit I want to design is a simple synchronous buck converter, and for a size cutback I thought of using GaN FETs. The circuit as an overall view looks like this:

enter image description here

Additionally, the footprints are shown in the following figure:

enter image description here

$U_{11}$ is the LMG3410 driver, I would need two of these to make a synchronous buck, which is comprised to the right with $U_{12}$ and both $Q_{5}$, a driver without short-circuit/overcurrent protection (PE29102) and the GaN FETs respectively.

Moreover, the circuit I am designing should not surpass 6A, so this driver becomes kind of useless (not totally since it features short-circuit protection) in this situation since it is capable of 100A of drain current.

Thus, is there another way to protect a GaN without the need of this driver and can be at the same way as effective?

Thank you in advance.

Iron Maiden
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  • The LMG341x isn't just a driver, it's a driver co-packaged with a GAN FET. That helps minimize parasitics much more than you could do with a PCB layout of a driver and GAN FET. – John D Jan 15 '20 at 05:22
  • It includes a 600 (800) V 100A GaAN FET. Look at the prices of bare FETs that have the same specs. The added driver cost does not seem to be vast for what it does. || The driver intro says sub 100 ns protection. That may not be too hard to match. – Russell McMahon Jan 15 '20 at 09:08
  • The TI application report is bound to use a little fear in order to sell the benefits of the device. Show your circuit and explain where and how you think it might become vulnerable. – Andy aka Jan 15 '20 at 09:27
  • I'll show what I intend as my circuit in order to discuss the possibilities. – Iron Maiden Jan 15 '20 at 14:58

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