A class A power amplifier is defined as a power amplifier in which output current flows for the full-cycle (360°) of the input signal. In other words, the transistor remains forward biased throughout the input cycle.

A schematic circuit of a series fed class A large signal amplifier using resistive load R_c is shown below. The term “series fed” is derived from the fact that the load R_c is connected in series with the transistor output. The only difference between this circuit and the small-signal amplifier circuits considered previously is that the signals handled by the large-signal circuit are in the range of volts and the transistor used is a power transistor capable of operating in the range of a few watts. This circuit is seldom used for power amplification because of its poor collector efficiency but will give clear understanding of class A operation to the readers. The output characteristics with operating point Q are also shown. I_CQ and V_CEQ represent no signal collector current and collector-emitter voltage respectively. When ac input signal is applied, the operating point Q shifts up and down causing output current and voltage to vary about it. The output current increases to I_{c max} and falls to I_{c min}. Similarly, the collector-emitter voltage increases to V_{ce max} and falls to V_{ce min}.

Power Distribution in Class A power amplifiers.

Input power from the collector supply V_CC, P_in(dc) = V_CC I_CQ

The power drawn from the collector supply is used in the following two components

• Power dissipated in collector load as heat, P_{RC (dc)} = (I_CQ)² R_C

• Power supplied to the transistor, P_{tr (dc)} = P_in(dc) – P_{RC (dc)}

Power supplied to the transistor, P_{tr (dc)} is further subdivided into

ac power developed across the load resistor constituting ac power output and is given as

P _{out (ac)} = (I_C)² R_C = (V_ce)²/ Rc

where Ic and V_ce are the rms values of collector current and

Collector- emitter votage = {(I_{c max})²÷ √2}R_{C ­} = V² _{CE max} / 2 Rc

(I² c (peak-to-peak)  Rc)/ 8 = V²_CC (peak-to-peak) / 8 R_C

Power dissipated, in the form of heat, by the transistor itself. The cause of power dissipation   in transistor is explained below :

Consider an N-P-N transistor. The potential difference across the depletion layer formed near the collector junction is called the barrier potential. This potential gives the P-region (base) slightly more energy than N-region (collector). Thus when electronics emitted from emitter cross the base junction and enter the collector region, they give up energy in the form of heat and it is this energy that the transistor has to dissipate to the surrounding.
With zero signal applied at the input of the class A power amplifier, ac power devel­oped across the load reduces to zero and therefore all the power fed to the transistor is wasted in the form of heat. Thus, a transistor dissipates maximum power under zero-signal condition. Thus the device is cooler when delivering power to a load than with zero-signal condition.

Since in class A operation, maximum power dissipation in the transistor occurs under zero-signal condition, the power dissipation capacity of a power transistor, for class A  operation, must be at least equal to the zero-signal rating.

Collector Efficiency: The collector efficiency of a transistor is given as

Efficiency = Average ac power output, P_{out (ac)} /  Average dc power input to the transistor P_{tr (dc)},

Power Efficiency: A measure of the ability of an active device to convert the dc power of supply into the ac (signal) power delivered to the load is called the power or conversion or theoretical efficiency. By definition the efficiency is 

Efficiency = AC power delivered to the load, P_{out (ac)} / Total power drawn from dc supply P_{in (dc)},

Now ac power delivered to the load,

_{Pout (ac) = (Ic (peak-to-peak) *Vce (peak-to-peak)) ÷ 8}

Maximum Power and Efficiency. If the operating point Q is set at the midpoint of the

maximum signal swing, the resulting maximum power  condition may be achieved.

Maximum V_{CE(peak-to-peak)} = V_cc

Maximum I_{CE(peak-to-peak)} = V_cc ÷ R_C

we have maximum ac power developed across the load resistor,.

P_{out (ac) max} = 1/8 * Vcc/Rc  * Vcc = V²cc/8 Rc

For the quiescent point Q, I_CQ = (Vcc/Rc) ÷2

and dc power drawn from dc supply, P_{in (dc) max} = V_cc I_CQ = V²cc/2Rc

So maximum efficiency of an amplifier (class A power) is given as

Efficiency = P_{out (ac) max} / P_{in (dc) max} = (V²cc/8 Rc) ÷ (V²cc/2Rc)

This is the maximum percent efficiency for a series-fed class A power amplifier. Since this maximum efficiency will occur only under ideal conditions and for the maxi­mum ac signal swings, most series-fed class A power amplifiers have power efficiencies much less than 25%.