Pin Diode Max Power Handling Receiver-Limiter

Hi Guys,
Wanted some advice on HSMP389C pin diode which i saw in receive design. Its was used as limiter, placed at the input of LNA. How can i calculate the maximum power handling of the limiter in that circuit before its damage due to high power transmited to receive path. I read thru the data sheet and no information was specified on the diode`s max power handling. Please help

well, if you make a few assumptions you can get a ballpark answer.

Assume the diode is turned-on enough to reach its data sheet 3.8 ohms series resistance.
Max junction temp is 150 deg C, and thermal resistance is 150 degC/Watt. So, the maximum power dissipated in the diode itself can be 1 watt.

Power dissipated = 1 Watt = I2 * 3.8 ohms. So the current flowing in the diode is Irms= √(1 watt/3.8 ohms) = 0.51 amps rms

Since the diode is acting very much like a short circuit, and since a short circuit is a boundary condition to travelling wave theory on the transmission line leading up to it, the diode current is twice the forward traveling wave current. (this is since a forward and equal reverse wave has to pass its current thru the diode boundary condition).

So the forward travelling wave current must have been I+=0.51/2 amps rms = 0.26 amps rms.

The forward travelling wave power then must be P = V+*I+=(I+*Z)*I+= I+2 * Z. (I+ and V+ mean forward traveling current and voltage waves)
Z is 50 ohms. So the max power that diode can handle is somewhere around Pmax= (0.262 * 50) = 3.4 watts.

OF COURSE, it all depends on if the diode is fully turned on. If it is only partially turned on, the diode series resistance will be higher, and the power it can take will be less!