**1 Answer**

written 6.6 years ago by | • modified 6.6 years ago |

• A power amplifier is referred to as class A power amplifier if the transistor used for amplification conducts for the full cycle of the ac input signal. The transistor is biased in such a way that the q point is present at the center of DC load line. Due to this the output signal is obtained for full cycle of the input ac signal.

Diagram of transformer coupled class A power amplifier is given by:

output characteristics of class A given by following diagram:

• Now when we apply an ac input signal to base of the transistor, The base current changes sinusoidally above and below the quiescent base current IBQ.

• As base current and collector current is related to each other so in response to base current the collector current Ic also changes below and above its quiescent value i.e ICQ.The collector and base current are in phase with each other.

• Due to change in Ic, the voltage VCE will also changes but Ic and VCE are 180 degree out of phase with respect each other. The transistor is remains in active region for whole input signal cycle and never enters the saturation and cutoff region. Faithful amplification takes place without any harmonic distortion. Thus the harmonic contents in the output is very low.

• As the transistor is continuously operates in active region the voltage VCE across it and current IC through it both are simulteously high. Hence large power is dissipated in the form of heat from the transistor. Therefore the efficiency of class A power amplifier is very low.

• Typically the efficiency of class A power amplifier is lies between 25% to 50%.

**Expression for the efficiency of class A power amplifier:**

• **Obtain the Q-points of the circuit** (Ic , Vce) :Q point is situated at the center of DC load line.from the given circuit we can write the value of IBQ (Quiescent point base current ) as:

```
IBQ = (VCC – VBE) / RB =(VCC – 0.7) / RB ………….1
```

Note that at Q-point ac current considered to be zero and the only dc voltages and currents considered .

Collector current is given by ICQ = β IBQ Apply KVL to the output side of the circuit or collector loop to write

```
VCC - ICQ RL – VCEQ =0
```

Therefore VCEQ = VCC - ICQ RL.

Thus we can obtain the Q point.

**DC input power:**DC input power supplied by the dc voltage source VCC is given by

```
Pdc = VCC × average current supplied by the source .
```

But we know the Ic(AVG) = ICQ as collector current varies above and below ICQ.

Pdc = VCC × ICQ……………………………….2

**AC output power :** due to application of sinusoidal signal at the base ,sinusoidal collector current will flow through the transistor. This current will vary above and below the Q-pint value of ICQ.

Ac output power is given by: Pac = VLrms × ILrms

VLrms = Vm/√2 and ILrms = Im/√2

Therefore

Pac = (Vm× Im)/2 =[ (Vp-p/2)(Ip-p/2)]/2 = (Vpp × Ipp)/8………………….3

From the figure the Vp-p =[Vmax - Vmin] and IP-P =[ Imax - Imin]

Equation 3 can be written as : Pac = [( Vmax - Vmin)( Imax - Imin)]/8 …………4

**Efficiency:**The expression for the efficiency of the transformer coupled class A power amplifier is as follow:

% ɳ = (Pac/Pdc )/100 ={[( Vmax - Vmin)( Imax - Imin)]/[8VCC × ICQ]}×100………5

**Maximum efficiency :**
The maximum efficiency is given by : Vmax = 2 vCC and Imax = 2 ICQ

Substitute the values in the above equation and solving for the maximum efficiency we get

% ɳmax = ={[(2 VCC - 0)( 2 ICQ - 0)]/[8VCC × ICQ]}×100

% ɳmax ={[( 4VCC × ICQ)]/[8VCC × ICQ]}×100

```
= 50%
```

Maximum efficiency = 50 %