Power amplifiers

Power amplifiers
It is used to deliver a large amount of power to the load it contain bulky component .A power transistor of large surface area and metal case is suitable for power amplifier .A power amplifier is used as the large stage of a communication system .It is widely used in audio component radios,TV receivers etc

Efficiency /Conversion efficiency/collector efficiency
$η=\frac{P_{oac}}{P_{odc}}$
$=\frac{V_{rms}I_{rms}}{V_{cc}I_c}$
$=\frac{\frac{V_m}{\sqrt{2}}\frac{I_m}{\sqrt{2}}}{V_{cc}I_c}$ 
$=\frac{\frac{V_{pp}}{2\sqrt{2}}\frac{I_{pp}}{2\sqrt{2}}}{V_{cc}I_c}$  
$=\frac{V_{pp}I_{pp}}{8V_{cc}I_c}$

Class A Power amplifiers(series fed)

An amplifier is of type class A if its output remain in the active region during a complete cycle of sine wave input signal .It is an amplifier under normal condition i.e the output never saturates or cut off.If the input is 360°,then output is also 360° ,i.e distortion is very low

Efficiency

$V_{pp}=V_{cc}$
$I_{pp}=\frac{V_{cc}}{R_c}$ 
$I_c=\frac{V_{cc}}{2R_c}$
$η =\frac{V_{pp}I_{pp}}{8V_{cc}I_c}$
$=\frac{V_{cc}\frac{V_{cc}}{R_c}}{8V_{cc}\frac{V_{cc}}{2R_c}}$ 
$=\frac{1}{4}$=25%
η is low .So class A is never used as a power amplifier 

Transformer coupled(Class A power amplifier)
Instead of resistive coupling transformer coupling is used

$R^{,}_{L}=(\frac{N_1}{N_2})^2R_L$ 

$N_2=2N_1$
$\frac{N_1}{N_2}=\frac{V_1}{V_2}$  =>
$V_2=2V_1$

$R^{,}_{L}$ is the resistance reflected to $1°$ winding.It acts like $R_c$ in series fed class A
$V_{pp}=2V_{cc}$ 
$I_{pp}=\frac{2V_{cc}}{R^{,}_{L}}$  
$I_c=\frac{V_{cc}}{R^{,}_{L}}$
$η =\frac{V_{pp}I_{pp}}{8V_{cc}I_c}$
$=\frac{2V_{cc}\frac{2V_{cc}}{R^{,}_{L}}}{8V_{cc}\frac{V_{cc}}{R^{,}_{L}}}=\frac{1}{2}$=50%    

Harmonic Distortion/Non-linear /amplitude distortion

The presence of unwanted frequency components in the output which are harmonics of the input frequency is called harmonic distortion .When a sinusoidal signal is applied to a transistor ,non-linearity occurs.Some portion of the signal is amplified more than the other portion


$I_c=K_1I_b$(linear circuit)
with harmonic distortion $I_c=K_1I_b+K_2I_B^2+K_3I_B^3....$ 
if $I_b$ is sinusoidal  $I_b=I_bcosωt$
$I_c=K_1I_bcosωt+K_2I_B^2I_bcos^2ωt+K_3I_B^3cos^3ωt....$ 
$=K_1I_bcosωt+K_2I_B^2[\frac{1+cos2ωt}{2}]....$ 
$=K_1I_bcosωt+\frac{1}{2}K_2I_B^2+\frac{1}{2}K_2I_B^2[cos2ωt]....$
  $=B_1cosωt+B_0+B_2cos2ωt....$
$D_2=\frac{B_2}{B_1}(2^{nd})$      $D_3=\frac{B_3}{B_1}(3^{rd})$     $D_4=\frac{B_4}{B_1}(4^{th})$
Total harmonic distortion=$\sqrt{D^2_2+D^2_3+D^2_4....}$

Class A Push-pull power amplifier

During positive half cycle $Q_1$ conducts,So $I_{c1}$ flows
During negative half cycle $Q_2$ conducts,So $I_{c2}$ flows
 $R^{,}_{L}=\left( \frac{\frac{N_1}{2}}{N_2}\right)^2R_L$
$=\left( \frac{\frac{N_1}{2}}{N_1}\right)^2R_L=\frac{R_L}{4}$ 

$V_{cc}$ is center tapped to $N_1$ in first half cycle ,only $Q_1$ conduct.So effective primary winding is $\frac{N_1}{2}$ same for $Q_2$
 Output current
 $I_{c1}=B_0+B_1cosωt+B_2cos2ωt+....$ 
$I_{c2}=B_0+B_1cos(ωt+180°)+B_2cos2(ωt+180°)+....$ 
$I_{c2}=B_0-B_1cos(ωt)+B_2cos(2ωt)+....$
Total current $I_{c}=k[I_{c1}-I_{c2}]$ 
$=2kB_1cos(ωt)+2B_3cos(3ωt)+....$
Thus even harmonics are eliminated

Class B push-pull (Transformer coupled)

In normal class B amplifier output current flows only for one half cycle i.e conduction angle is 180° to ensure 360° operation class B push-pull is used

$Q_1$ conducts during positive half cycle of input ,so $I_{c1}$ flows
During negative half cycle of input ,$Q_2$ conduct $I_{c2}$ flows.these 2 current are combined at output
Eficiency
$V_{pp}=2V_{cc}$

$I_{pp}=\frac{2V_{cc}}{R^{,}_{L}}$ 
This circuit resembles action of a fullwave rectifier therfore
$I_c=\frac{2I_{m}}{Π}=\frac{2V_{cc}}{ΠR^{,}_{L}}$

$η =\frac{V_{pp}I_{pp}}{8V_{cc}I_c}$

$=\frac{2V_{cc}\frac{2V_{cc}}{R^{,}_{L}}}{8V_{cc}\frac{2V_{cc}}{ΠR^{,}_{L}}}=\frac{Π}{4}$=78.5%  

Complementary Class B push-pull power amplifier

or

transformerless power amplifier

Class B push-pull amplifier uses 2 transfomer which make it bulky and costly .To avoid using transformer complementary symmetry class B is used

It consist of  NPN and PNP transistor >During positive  half cycle of input ,$Q_1$ conducts and $I_{c1}$ flows from $V_{cc1}$ through $R_L$ .During negative half cycle of input $Q_2$ conducts and $I_{c2}$ flows from $V_{cc2}$ through $R_L$ .Thus we get a complete amplified waveform of input across $R_L$

Cross over distortion

Since the transistors are biased at cutt-off no current flows through the load till the amplitude of input signal exceeds cut in voltage .It means that the amplifier cannot reproduce or amplify the input signal at zero crossing point ,This type of distortion is called cross over distortion


Class AB push-pull(Circuit same as  Class A push-pull)

Class AB is a compromise between class A and class B .Class B amplifiers are highly efficient but their output waveform is distorted due to cross over,This occur in class B because of the absence of current at zero crossing point of input signal .This limitation can be overcome by biasing the transistor just at cut in (0.7V for Si).This resulting configuration is Class AB. To reduce harmonic distortion also class A is used the value of $R_2$ is chosen in such a way that only 0.7v is allowed to drop across $R_2$ .Performance of class AB is between class A and class B ,less efficient than class B and more than class A(0.5 & 0.785) .Distortion in Class AB is less than class B

Large signal tuned amplifier
Class C power amplifier

Class C amplifier is tuned  amplifier which can amplify only a narrow band of frequencies around the center frequency.The conduction angle is less than 180°.The output  contain lots of harmonics and these are eliminated by using tuned circuit as load
$C_{c}$,R and Base emitter diode acts as a clamping circuits which clamps the input signal towards negative .Hence the transistor conduct only for a short duration during each positive peak of input signal .the output current $I_{c}$ appears in the form of pulses .But the CE voltage will sinosuidal because of tank circuit $V_{CE}$ will be 180° out of phase with $V_{BE}$
Power Dissipation is less