> as junction temperature rises so does Ron, which creates more heat What's interesting about this is that bipolar transistor has the opposite property -- as i

> as junction temperature rises so does Ron, which creates more heat What's interesting about this is that bipolar transistor has the opposite property -- as it heats up, resistance decreases -- and yet this opposite property causes the same problems with thermal runaway. (Okay, when talking about bipolar transistors we don't really talk about "resistance", but if you know that then you probably know what I'm going to say anyway...) When hooked up to a load with a relatively constant voltage, there will also be a relatively constant voltage on the transistor. As the bipolar transistor heats up, the resistance decreases and more current flows through the transistor, and the transistor will dissipate power according to the law P=V^2/R. So once a transistor gets hot, it gets hotter until it blows. The question is, "what kind of load looks like a voltage source?" There's actually a quite common load -- any amplifier with parallel output transistors will look like it's driving a constant voltage load, from the perspective of one of the parallel transistors. Basically, hooking up bipolar transistors in parallel does NOT multiply the power rating as you would think, because thermal runaway might cause one of the transistors to dissipate all the power.