TPS546D24A: How to estimate the Power dissipation of an Rboot resistor?

The BOOT pin of a synchronous BUCK converter like the TPS546D24A (and many others) powers the circuitry of a "floating" gate driver for the high-side FET and is connected to the SW pin with a capacitor to maintain the BOOT to SW voltage when the high-side FET is turned on.  This "Boot Strap" power supply maintains a loosely regulated voltage from BOOT to SW for driving N-channel MOSEFTs that need a gate drive voltage that exceeds the power input voltage PVIN.

The BOOT pin draws high peak currents that can be a few Amperes, but only during the short duration when the high-side FET is turning on.  The Boot resistor, Rboot, limits this current to control the slew-rate, ringing, and peak voltage stresses during the turn-on of the high-side FET, which can complicate estimating power dissipation.

The approximate power dissipation is given by:

Pdiss = R * I^2 * Sqrt ( D)

Rboot * [ (Vdrive - Vplat) / (Rdrive + Rboot) ] ^2 * [ Qg * (Rdrive + Rboot) / { (Vdrive - Vplat) * Tsw } ] 

{Typo in the above equation was corrected 8/21/2020 to replace the incorrect (Rdrive + Vboot) with the correct (Rdrive + Rboot) }

Unfortunately, more integrated FET devices, like the TPS546D24A, do not include factors like Qg, Vdrive, Rdrive or Vplat, because the power dissipation Rboot is generally neglectable, even for an 0402 resistor.

For the TPS546D24A, the approximate values are:

Vdrive = 4.7V, Vplat = 2.0V, Qg = 6.5nC and Rdrive = 4.2-Ohms

Solving this for a couple of different resistor values at 1.5MHz:

1-Ohm:  1 * (2.7V / 5.2-Ohms)^2 * ( 6.5nC * 5.2-Ohm / (2.7V * 667ns)) = 5mW

4.7-Ohm: 14mW

10-Ohm (Not recommended) 18.5mW