# How does Class A power amplifier work?

Power amplifiers are electronic amplifiers used to amplify a signal voltage/current magnitude and therefore increase signal power. Power amplifier are used to amplify signal power in order to drive output actuators like loud speaker, RF transmitter transmitter, headphones etc.

Depending upon the percentage of the input cycle for which an amplifier operates in its linear region we can classify power amplifier as

1. Class A

2. Class B(see How does Class B power amplifier work?)

3. Class AB

4. Class C

5. Class D

Here we explain how BJT based class A power amplifier works.

### Class A power amplifier

A class A power amplifier is one which is biased such that it always operates in the linear region and the output signal is in phase with the input signal. The objective of class A power amplifier is to deliver high power rather than high voltage. The usual rated power dissipation of power amplifiers is 1W and heat dissipation problem is usually required. The collector terminal is where the heat appears and heat sinks maybe required.

#### Q-point(Quiescent-point) or Bias point

The Q-point of a transistor is a steady state operating condition for DC operation and AC operation. The following shows the Q-point on both DC load line and AC load line which intersects at the same Q-point.

When a transistor is operated with the Q-point the output signal is maximum. This is because the collector current, $$I_{C}$$, it's value at Q-point center, $$I_{CQ}$$, has maximum range to vary from 0 which is cutoff to $$I_{C(sat)}$$ which is saturation region. Similarly, value of the collector to emitter voltage $$V_{CE}$$, at the Q-point center $$V_{CEQ}$$, can vary maximally from 0 to $$V_{CE(cutoff)}$$. The following graph illustrates the excursion of input and output sine wave signal  over the range from the middle Q-point.

If the signal is not centered at Q-point, the output signal waveform is limited. For example, if Q-point is near the cutoff, the input signal amplitude excursion is limited as shown below.

Any further shift of Q-point towards cutoff, that is transistor is driven into cutoff region, the input and output signal waveform amplitude are clipped as shown below.

Similarly, if the Q-point is not in the middle and is moved towards saturation the input and output signal is limited by the transistor saturation point.

Any more shift towards saturation drives the transistor into saturation region and the input and output signal are clipped.

Power Gain

The power gain of an amplifier is the ratio of the power delivered to the load(output power) to the input power. It is given by,

$A_{p} = \frac{P_{L}}{P_{in}} ----------->(1)$

where $$A_{p}$$ is the power grain, $$P_{L}$$ is the power delivered to the load or the output power and $$P_{in}$$ is the input power.

There are several ways to calculate the power gain which depends upon what is known. In most cases the power gain is calculated using the input resistance, load resistance and voltage gain.

Power can expressed in terms of voltage and resistance as follows,

$P = \frac{V^2}{R}$

And so the output power to the load can be written as,

$P_{L} = \frac{V_{L}^2}{R_{L}}$

and the input power is,

$P_{in} = \frac{V_{in}^2}{R_{in}}$

Substituting these in equation(1),

$A_{p} = \frac{V_{L}^2 R_{in}}{V_{in}^2 R_{L}}$

Since, $A_{v} = \frac{V_{L}}{V_{in}}$

Therefore,

$A_{p} = A_{v} (\frac{R_{in}}{R_{L}})$

Power Dissipation/ DC Quiescent Power

The power dissipation or the DC quiescent power of a transistor is the product of its Q-point current and voltage and is given by,

$P_{DQ} = I_{CQ}V_{VEQ}$

This DC quiescent power is the maximum power that the class A amplifier must handle. The transistor used for class A amplifier must not exceed this value.

Output Power

The output signal power is the product of the rms load current and the rms load voltage. The maximum unclipped ac signal occurs when the Q-point is centered on the ac load line.

For a CE amplifier with a centered Q-point, the maximum peak voltage swing is given by,

$V_{C(max)} = I_{CQ} R_{C}$

The RMS voltage value is,

$V_{C(rms)} = 0.707 V_{C(max)}$

The maximum peak current swing is,

$I_{C(max)} = \frac{V_{CEQ}}{R_{C}}$

The RMS current value is,

$I_{C(rms)} = 0.707 I_{C(max)}$

The maximum output power calculated using the rms current and rms voltage as follows,

$$P_{out(max)} = I_{C(rms)} V_{C(rms)}$$

or,

$$P_{out(max)} = 0.707 I_{C} 0.707 V_{C}$$

that is,

$P_{out(max)} = 0.5 I_{CQ} V_{CEQ}$