I have been trying to implement a current limiting circuit using a high power NMOS and high side gate driver.

My idea is to adjust the PWM duty-cycle of the gate driver to give an limited (average) current to the load. So far, I had good luck with a small load (8 Ohm at 24V). However, when I try a bigger load (2 Ohm at 24V), the gate driver stops working.

My original guess is that the NMOS gate is taking too much current form the gate driver output pin, but the simulations tell me that the TG pin is only outputting 2.4A max (LTC4440 can do 4A max for 1us). Another guess is that the inductance in the wire make the peak current higher than what the simulations show. Does anyone have a recommendation ?

The schematic below shows my setup. The plot shows the output voltage in green, and the current at at the gate of the NMOS in blue.


Simulation results

Here is more information about the experiment I conducted: The circuit (shown in the schematic) seemed to work great when using a 8 Ohm resistor as a load (at 24V). I was controlling the PWM duty-cycle with a potentiometer, and I was able to vary the (average) load current from 0 to 4A. Then, I changed the load for a 2 Ohm resistor (2000W), which should theoretically draw 12A everytime the FET is on. JHowever, as soon as I started adjusting the duty-cycle of the PWM, the gate driver stopped driving the gate. I went back to the smaller load (8 Ohm), but the gate driver does not drive the gate anymore. The chip seems to have an issue. I had this issue happened to me three times (gate driver fails at higher load, but works well at small loads).

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    \$\begingroup\$ Where is the current limiting? Or how is this supposed to work? I admit, I'm having trouble finding that out. You want current limiting with a PWM driver that has no filter at the output, yet it worked? Could you explain how? \$\endgroup\$ – a concerned citizen May 7 '18 at 19:37
  • \$\begingroup\$ Please explain how you are bootstrapping it. \$\endgroup\$ – winny May 7 '18 at 19:39
  • \$\begingroup\$ Apologies for the confusion: a micro-controller adjusts the duty-cycle of the PWM according to the load current (a Hall current sensor monitors the load current at all time). For the bootstrap circuitry, I tried to match what was recommended in the LTC4440 datasheet. \$\endgroup\$ – user188096 May 7 '18 at 19:51
  • \$\begingroup\$ To me, the circuit looks completely wrong, pretty much everything. \$\endgroup\$ – Ale..chenski May 7 '18 at 19:57
  • \$\begingroup\$ Where is your energy storage inductor? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 7 '18 at 20:08

The gate driver and the MOSFET are working just fine.

However, with a source impedance (R2) of 0.6Ω, you're simply not going to get full voltage with a 2Ω load (R1).

These two resistors form a voltage divider that gives you the 18.46 volts that you see at the output.

What exactly is the purpose of R2?

| improve this answer | |
  • \$\begingroup\$ R2 is just to simulate the output impedance of the 24V (lithium-ion battery). If I remove it, I get the same waveform, but with an amplitude of 24V (instead 18V). As you said, the driver and the MOSFET work just fine on the simulation. That's why I'm puzzled that it does not work on the testbench. \$\endgroup\$ – user188096 May 7 '18 at 20:48
  • \$\begingroup\$ If you're trying to debug a prototype, why does your question talk exclusively about a simulation? What exactly is going wrong with the prototype? Edit the information into your question. \$\endgroup\$ – Dave Tweed May 7 '18 at 21:06
  • \$\begingroup\$ OP may need a more robust MOSFET with a much lower ON resistance. 600 miliohms is actually a bit high-not meant to be a high current switch. \$\endgroup\$ – VTNCaGNtdDVNalUy May 7 '18 at 21:29
  • \$\begingroup\$ Dave, you are 100% right. I edited my question and added more info on my experiment. \$\endgroup\$ – user188096 May 7 '18 at 21:43

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