Am I calculating my shunt FET power dissipation correctly?

Question

^{simulate this circuit – Schematic created using CircuitLab}

I'm designing an LED driver circuit that requires very fast PWM switching at a high resolution (10bit at 30khz requiring a minimum pulse width ~30ns). In order to achieve the pulse timing required, I'm using a FET to shunt the LED driving current to ground during the "off" periods. I'm trying to determine the worst-case power dissipation for the FET, so I can plan heat mitigation strategies when designing my board.

One of the FETs I am considering is the BUK9M52-40E, because it works on logic-level voltages and on/off times in the single ns range. It is not a power MOSFET, however, and does not appear to have a substantial heat sink build in, so I would presumably need to rely on copper pours and thermal vias for heat management.

I will be driving 8 LEDs in series at 24V at a constant average current of 360mA with a peak current of 540mA.

The FET has a maximum RDSon of 52mO at 5V, though I hope to drive it with a 3.3V signal, which from what I can determine would result in an RDSon of closer to 100mO. So given a constant average current of 360mA, and assuming 100mO of resistance, (0.36A x 0.01O) the voltage across the drain-source will be 0.0036V. So the power dissipation in the FET will be 0.0036V x 0.36A = 0.001296W.

This number seems... really low. And given that the FET is rated to 31W, I assume that this is well within its capabilities, and I could probably get away with very little thermal management, and potentially a much smaller FET. Am I correct in this assumption?

Answer 1

Power dissipation is \$I^2R\$ so 12mW, plus switching losses. I didn’t check how realistic your 100m\$\Omega\$ looks. It apparently is not guaranteed, so you’re rolling the dice.

Note that average current is not as important as the average current squared (which is larger than the square of the average current if it’s switching at all).

Note that the switching speeds are specified with 1A peak gate current, so they are assuming a good gate driver. There is a lot of Miller capacitance in your situation due to the large change in Vds.