Why class A amplifier is cooler with load

A class A amplifier draws a constant dc power Pi = VCC · ICQ from the supply, whether or not a signal is present. This power is shared between the transistor, the load and any resistors in the circuit.

Without a signal and without a load (no ac power delivered)

Po = 0, so all the input power that is not lost as dc in resistors is dissipated in the transistor. In a transformer-coupled stage (negligible dc resistance in the primary):

PD = Pi − Po = Pi (maximum)

The transistor is at its hottest. In a class A stage the dissipation is maximum under no-signal conditions.

With a load and a signal

Part of the input power is converted to useful ac power Po delivered to the load, so the heat in the transistor falls:

PD = Pi − Po (minus any dc power lost in resistors such as RC)

Example 1 (transformer-coupled): VCC = 12 V, ICQ = 60 mA, so Pi = 0.72 W. With no signal, PD = 0.72 W. At full swing (50 % efficiency) Po = 0.36 W, so PD falls to 0.36 W, half the no-signal value.

Example 2 (series-fed): VCC = 12 V, RC = 100 Ω, Q point at the centre: ICQ = 60 mA, VCEQ = 6 V, Pi = 0.72 W. Here RC is the load, and it also takes dc power ICQ² RC = 0.36 W, so with no signal the transistor dissipates VCEQ · ICQ = 0.36 W. At full swing RC also receives Po = 0.18 W of ac power (25 % efficiency), so the transistor dissipation falls to 0.36 − 0.18 = 0.18 W.

Explanation in points:

1. Input power is constant (Pi = VCC ICQ), independent of the signal.
2. With the load connected, the signal power Po leaves the transistor and goes to the load.
3. So the heat in the transistor is lower, and it runs cooler.
4. Without a load (or without a signal) Po = 0, so the transistor dissipates the most (PD = Pi in a transformer-coupled stage): it is at its hottest.

This is why a class A amplifier must be designed for the no-signal worst case, and the heat sink is chosen for the no-signal dissipation PD = VCEQ · ICQ.