Low frequency power amplifier

Figure 1 is a Class A power amplifier. The load RL becomes the collector load RL=nRLo through the impedance transformer B. For DC, the primary DC resistance and Re of transformer B are very small, so the DC load line is close to a vertical line (see Figure 1 (b)). In order to make the amplifier output higher power, the AC load line can be placed at points a and b: Uce=UCM at point a, and the operating point Q is at the midpoint of straight line ab. Usually the saturation voltage drop and penetration current of the transistor are very small. In fact, it can be considered that Icmin=0 and Ucemin=0o. Therefore, the current and voltage amplitudes supplied to the load are:
Icm=IcM/2, Ucem=UCM/2 ------------------------------------------------Formula 1
The AC power of the load (or the output power of the amplifier) ​​is:
PL=（UceM/ ）×（IcM/ ）=(IcM/ )×(UcM/ )=(1/8)IcM×UcM-------Formula 2
The collector current ICQ and voltage UceQ at the operating point Q are:
ICQ=ICM/2, UceQ=Ec=UCM/2---------------------------------------------Equation 3
Therefore, the input power of the DC power supply:
PD=IcQ×UceQ=(ICM/2)×(UCM/2)=1/4IcMUcm-------------------------------Equation 4
The efficiency of the Class A power amplifier is:
η=PL/PD=50%---------------------------------------------------------Equation 5