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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' = Rth,j-a ∙ dP

So an amplifier will be thermally stable if: Rth,j-a∙dP < dT, or: Rth,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=VCE∙IC.


dP/dIC=VS - 2∙IC∙RL. => dP=(VS - 2∙IC∙RL)∙dIC

The amplifier will therefore be stable if:

Rth,j-a∙(VS - 2∙IC∙RL)∙dIC/dT < 1

This will always be true if: VS < 2∙IC∙RL => IC∙RL > VS/2

IC∙RL = VS - VCE => VS - VCE > VS/2 => VCE < VS/2

So, the amplifier will be stable if VCE is smaller than VS/2. For maximum output amplitude, the collector voltage must be VS/2. If the emitter is grounded, VCE will be equal to VS/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 VS/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.

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