For improved thermal management, attach the heat-removal path to the region of the silicon where heat (I*V) is generated. Silicon thermal conductivity is 150 watts/meter * degree C. For a cubic meter of silicon. For a cubic centimeter, the heat travels only 1cm but the area to conduct the heat has dropped from one square meter to 0.01*0.01 square meter. Result is thermal conductivity drops to 1.5 watt/ degree C ----- if the heat has to travel that entire centimeter. Silicon wafers are 300 microns (0.3mm) thick, thus ThermCond is 5 watts/degree C. By removing heat nearest its generation region, we can cut that in half. But that results in floating heatsinks.
What happens when the heatsink is floating, and moving with some slewrate?
How severe is that transient current from heatsink to "ground"? and what deltaVoltage across a 10nH inductor? (1cm of return path)
Assume heatsink 10cm by 5cm; assume 20 mils between layers, or 500 microns. Assume Er = 5. Capacitance C = E0*Er*Area/Distance becomes
9e-12 * 5 * 10cm*5cm/0.5 millimeters = 45e-12 * 50 *0.0001 / 0.5 * 0.001 = 450pF
Assuming 200 volts in 200 nanoseconds transient on the FET drain.
I = C * dV/dT = 450 pF * 1e+9 = 0.45 amp, with Trise of 10 nanoseconds (the FET likely turns on in 10nS, though the slewing lasts for 200 ns).
What is the "ground" upset? V = L * dI/dT = 10nH * 0.45amp/10nSec = 450 milliVolts