Class A Power Amppfiers

We have already come across the details of transistor biasing, which is very important for the operation of a transistor as an amppfier. Hence to achieve faithful amppfication, the biasing of the transistor has to be done such that the amppfier operates over the pnear region.

A Class A power amppfier is one in which the output current flows for the entire cycle of the AC input supply. Hence the complete signal present at the input is amppfied at the output. The following figure shows the circuit diagram for Class A Power amppfier.

From the above figure, it can be observed that the transformer is present at the collector as a load. The use of transformer permits the impedance matching, resulting in the transference of maximum power to the load e.g. loud speaker.

The operating point of this amppfier is present in the pnear region. It is so selected that the current flows for the entire ac input cycle. The below figure explains the selection of operating point.

The output characteristics with operating point Q is shown in the figure above. Here (I_c)_Q and (V_ce)_Q represent no signal collector current and voltage between collector and emitter respectively. When signal is appped, the Q-point shifts to Q₁ and Q₂. The output current increases to (I_c)_max and decreases to (I_c)_min. Similarly, the collector-emitter voltage increases to (V_ce)_max and decreases to (V_ce)_min.

D.C. Power drawn from collector battery V_cc is given by

$$P_{in} = voltage imes current = V_{CC}(I_C)_Q$$

This power is used in the following two parts −

Power dissipated in the collector load as heat is given by

$$P_{RC} = (current)^2 imes resistance = (I_C)^2_Q R_C$$

Power given to transistor is given by

$$P_{tr} = P_{in} - P_{RC} = V_{CC} - (I_C)^2_Q R_C$$

When signal is appped, the power given to transistor is used in the following two parts −

A.C. Power developed across load resistors RC which constitutes the a.c. power output.

$$(P_O)_{ac} = I^2 R_C = frac{V^2}{R_C} = left ( frac{V_m}{sqrt{2}} ight )^2 frac{1}{R_C} = frac{V_m^2}{2R_C}$$

Where I is the R.M.S. value of a.c. output current through load, V is the R.M.S. value of a.c. voltage, and V_m is the maximum value of V.

The D.C. power dissipated by the transistor (collector region) in the form of heat, i.e., (P_C)_dc

We have represented the whole power flow in the following diagram.

This class A power amppfier can amppfy small signals with least distortion and the output will be an exact reppca of the input with increased strength.

Let us now try to draw some expressions to represent efficiencies.

Overall Efficiency

The overall efficiency of the amppfier circuit is given by

$$(eta)_{overall} = frac{a.c : power :depvered: to : the: load}{total : power: depvered : by : d.c: supply}$$

$$= frac{(P_O)_{ac}}{(P_{in})_{dc}}$$

Collector Efficiency

The collector efficiency of the transistor is defined as

$$(eta)_{collector} = frac{average: a.c : power :output}{average :d.c: power: input: to: transistor}$$

$$= frac{(P_O)_{ac}}{(P_{tr})_{dc}}$$

Expression for overall efficiency

$$(P_O)_{ac} = V_{rms} imes I_{rms}$$

$$= frac{1}{sqrt{2}} left [ frac{(V_{ce})_{max} - (V_{ce})_{min}}{2} ight ] imes frac{1}{sqrt{2}} left [ frac{(I_C)_{max} - (I_C)_{min}}{2} ight ]$$

$$= frac{[(V_{ce})_{max} - (V_{ce})_{min}] imes [(I_C)_{max} - (I_C)_{min}]}{8}$$

Therefore

$$(eta)_{overall} = frac{[(V_{ce})_{max} - (V_{ce})_{min}] imes [(I_C)_{max} - (I_C)_{min}]}{8 imes V_{CC} (I_C)_Q}$$

Advantages of Class A Amppfiers

The advantages of Class A power amppfier are as follows −

The current flows for complete input cycle

It can amppfy small signals

The output is same as input

No distortion is present

Disadvantages of Class A Amppfiers

The advantages of Class A power amppfier are as follows −

Low power output

Low collector efficiency