Assume that a dT increase in temperature results in a dP increase in the transistor's power dissipation, and that a power increase of dP results in a temperature increase of dT'. Everything will be fine as long as dT' < dT.

We can calculate dT' by multiplying the power increase (dP) with the thermal resistance from junction to ambient, so:

dT' = R_th,j-a ∙ dP

So an amplifier will be thermally stable if: R_th,j-a∙dP < dT, or: R_th,j-a∙dP/dT < 1

Let's take a look at a very simple amplifier:

Without input signal the transistor (NPN or PNP) will dissipate: P=V_CE∙I_C.

V_CE=V_S - I_C∙RL => P=(V_S - I_C∙RL)∙I_C=V_S∙I_C - I_C²∙RL.

dP/dI_C=V_S - 2∙I_C∙RL. => dP=(V_S - 2∙I_C∙RL)∙dI_C

The amplifier will therefore be stable if:

R_th,j-a∙(V_S - 2∙I_C∙RL)∙dI_C/dT < 1

This will always be true if: V_S < 2∙I_C∙RL => I_C∙RL > V_S/2

I_C∙RL = V_S - V_CE => V_S - V_CE > V_S/2 => V_CE < V_S/2

So, the amplifier will be stable if V_CE is smaller than V_S/2. For maximum output amplitude, the collector voltage must be V_S/2. If the emitter is grounded, V_CE will be equal to V_S/2 and the amplifier may be thermally instable. Therefore a small resistor is connected between emitter and ground. The voltage across this resistor must be about V_S/5. This is a compromise; less voltage may result into instability due to component spread and aging. A higher voltage will decrease the maximum output amplitude and thus the efficiency.