Hartley oscillator

Hartley oscillator

This is under the category of tuned oscillator or resonant circuit oscillator
Operation
When $V_{cc}$ is applied ,the collector current begins to flow and the drop in collector voltage is coupled through capacitor C and $L_{2}$ .Thus the capacitor charges to its minimum voltage ,this voltage acts as initial excitation for tank circuit ,causing a current to flow in the LC circuit  .This current induces damped harmonic oscillation across $L_{1}$ and this circuit acts as input to base of transistor
This damped signal is amplified and appears at collector which is coupled as feedback to the tank circuit  C and $L_{1}$. The feedback voltage across $L_{2}$ is in phase with the input voltage across $L_{1}$ results in sustained oscillation
The feedback voltage is in phase shift with input voltage since 180° phase shift produced by the transistor and another 180° being provided by the tank circuit
Capacitor block ,dc component of the collector circuit ,but coupled c signal
As a results of this dc is out of tank circuit this energy loss due to tank circuit  is reduced and hence the oscillation is more stable

WRITING NODE EQUATION
    $\frac{V_π}{r_π}= \frac{V_π}{SL_1}+ (V_π-V_c)S_c=0$
    $g_mV_π+\frac{V_c}{R}= \frac{V_c}{SL_2}+ (V_c-V_π)S_c=0$
\end{array}\right]

= $\left[\begin{array}{cc}V_π\\V_c\\ \end{array}\right]=0$

$\left[\begin{array}{ccc}   (\frac{1}{r_π}+\frac{1}{SL_1}+S_c)&-S_c\\  (g_m-S_c)&\frac{1}{R}+\frac{1}{SL_2}+S_c\\ \end{array}\right]=0$
will continue.......

TUNED COLLECTOR OSCILLATOR

TUNED COLLECTOR OSCILLATOR

It is used for high frequency application .It is called tuned -tuned collector oscillation because  the tuned circuit is connected to the collector .Capacitor  C and primary coil L form the tuned circuits it forms the load impedance and determine frequency of oscillation
the output voltage developed across the tuned circuit is inductively coupled to base through secondary coil  $L_{1}$
the feedback appears across base emitter junction .transistor amplifier provides 180° phase shift,and tuned circuit provides another 180° so total 360° phase shift is obtained , ie there is a positive feedback

Working
  
       When   $V_{cc}$ is applied a transient current is developed in tuned L-c circuit -this transient current start natural oscillations in the tank circuit . these natural oscillation induce sonic voltage into $L_{1}$ due to mutual induction which causes corresponding variation in base current these variation are amplified ß times and appear in the collector circuit .      

  $\frac{1}{2π\sqrt{LC}}$ is the frequencies oscillation 

Crystal oscillator

Crystal oscillator


Crystal oscillator is used for high frequency application .It is basically a tuned circuit using a pizzo electric crystals as the resonant tank circuits.
A piezzo electrical crystal displays piezo electronic property.It is the ability to transform mechanical deformation into electricals charge and vice versa .Whwn the crystal is squazed it develope voltage and if a voltage is applied across  it a change in materials dimension results.
The most common piezzo electric materials are Rochelle salt and quartz
the resonant frequency and Q of a crystal depend upon the crystal dimension ,ie how the surface are oriented with respect to it axis and how the devices are m=mounted thinner and more frogile crystal are needed for high frequency oscillations
 
If the  resistance R is regulated the impedance of the crystal is purely reluctance

$iX=\frac{(jωL+\frac{1}{jωc} ) ( \frac{1}{jωc} )}{jωL+\frac{1}{jωc}+\frac{1}{jωc^`}}$

$=\frac{1-ω^2LC}{jωc^`-jω^3Lcc^`+jωc)}$
$=\frac{1-ω^2LC}{jω(c^`+c)-jω^3Lcc^`}$

let the reactance be zero at $ω=ω_s$ where $ω_s$ is series resonant frequency

$1-ω^2LC=0$ =>

$ω^2_s=\frac{1}{LC}$
$ω_s=\sqrt{\frac{1}{LC}}$
$f_s=\frac{1}{2π\sqrt{LC}}$
let the reactance become ∞ at $ω=ω_p$ where
$ω_p=$paralel resonant frequency
at $ω=ω_p$

 $=jω_p(c^`+c)=jω^3Lcc^`$
$c^`+c=jω^2_pLcc^`$
$ω^2_p=\frac{Lcc^`}{c^`+c}$
$ω_p=\sqrt{\frac{Lcc^`}{c^`+c}}$   
$f_p=\frac{1}{2π\sqrt{\frac{Lcc^`}{c^`+c}}}$  
 The circuits can be oscillate at a frequency between $ω_s$ and $ω_p$ .The oscillate frequency is determined by the crystal and not by rest of the circuit

RC wein bridge oscillator

RC wein bridge oscillator

RC network does not produce any phase shift .therefore to obtain total shift of 0° or 360° ,a two stage CE amplifier is required
$R_1,C_1$ and $R_2,C_2$ acts as forward network ,voltage across the parallel combination of $R_2,C_2$ is fed to the input of the amplifier .The frequency of oscillation is determined by $R_1,C_2 and R_2,C_2$ .The desired frequency of oscillation can be obtained by varying 2 capacitors and resistors
The feedback network provides the positive feed back .In addition to this the resistors $R_3 and R_4$ provide negative feedback.Hence this oscillator has better amplitude stability $R_4$ is often a temperature sensitive resistor with positive temperature co-efficient
If the amplitude of oscillation increases the resistance $R_2$ increases .This reduces the negative feedback which reduces the amplitude of the gain and the amplitude of oscillation is restored to stable value

Advantages
low distortion
better stability
adjustable frequency


Disadvantages
Costlier
used only in low frequency

$V_f=V_o\frac{R_2||X_{c2}}{R_2||X_{c2}+R_1+X_{c1}}$

let $R_1=R_2=R$

$V_f=V_o\frac{\frac{R\frac{1}{jωc}}{R+\frac{1}{jωc}}}{\frac{R\frac{1}{jωc}}{R+\frac{1}{jωc}}+R+\frac{1}{jωc}}$
$=V_o\frac{\frac{R\frac{1}{jωc}}{R+\frac{1}{jωc}}}{\frac{R\frac{1}{jωc}+(R+\frac{1}{jωc})(R+\frac{1}{jωc})}{(R+\frac{1}{jωc})}}=V_o\frac{R\frac{1}{jωc}}{R\frac{1}{jωc}+(R+\frac{1}{jωc})(R+\frac{1}{jωc})}$       
 $=V_o\frac{\frac{R}{jωc}}{\frac{R}{jωc}+(R+\frac{1}{jωc})^2}=V_o\frac{\frac{R}{jωc}}{\frac{R}{jωc}+\frac{(Rjωc+1)}{(jωc)^2}^2}$
$=V_o\frac{R}{R+\frac{(Rjωc+1)}{jωc}^2}=\frac{V_oRjωc}{Rjωc+(Rjωc+1)^2}$ 
$=\frac{V_oRjωc}{Rjωc-R^2ω^2c^2+2Rjωc+1}=\frac{V_oRjωc}{3Rjωc-R^2jω^2c^2+1}$

 $\frac{V_o}{V_f}=\frac{3Rjωc-R^2jω^2c^2+1}{Rjωc}$




 $\frac{V_o}{V_f}=\frac{3Rjωc-R^2ω^2c^2+1}{Rjωc}=3+\frac{1-R^2ω^2c^2}{Rjωc}$

Equating imaginary part 

 $\frac{1-R^2ω^2c^2}{Rjωc}=0$
 $=\frac{1-R^2ω^2c^2}{Rωc}=0$
 $1=R^2ω^2c^2$
$ω=\frac{1}{Rc}$

 $ω=2πf$

$f=\frac{1}{2πRc}$
Now from real part
 $\frac{V_o}{V_f}=3$=> $ß=\frac{1}{3}$
$|Aß|=1$=>
$A=3$
therfore gain of 2 amplifier required is 3

RC phase shift oscillator

RC phase shift oscillator

In this case the CE amplifiers is followed by a frequency determining network ,$R_1,R_2$ combination provides dc potential divides bias and $R-E,C_E$ provides temperature stability and provids AC signal degeneration RC-RC-RC network may be used as positive feedback between input and output.The frequency determining network provides 180°  ie each RC produces 60°,amplifier produces another 180° phase shift.thus the total phase shift is 180+180=360° and satisfy the bark criterion ,.When $V_CC$ is applied ,the current through $R_L$ increases beacuase of biasing .this change capacitor C an induces voltages across $R_1$

Advantages
1) not bulky and expensive
2) pure sine wave output
Disadvantages
1) suited for low frequency
Replacing transistor by its approximate small signal model

writing loop equations
loop 1
$ßI_3R_L+I_1R_L+\frac{I_1}{jωc}+(I_1-I_2)R=0$
    $$\bbox[5px,border:2px solid black]{     +I_1(R_L+\frac{1}{jωc}+R)-I_2R+ßI_3R_L=0}$$
Loop 2
$(I_2-I_1)R+\frac{I_2}{jωc}+(I_2-I_3)R=0$

$-I_1R+I_2(R+\frac{1}{jωc}+R)-I_3R=0$
$$\bbox[5px,border:2px solid black]{ -I_1R+I_2(\frac{1}{jωc}+2R)-I_3R=0}$$
loop 3

$(I_3-I_2)R+\frac{I_3}{jωc}+I_3R=0$

$-I_1R+I_2(R+\frac{1}{jωc}+R)-I_3R=0$
$-I_2R+I_3(R+\frac{1}{jωc}+R)=0$
$$\bbox[5px,border:2px solid black]{ 0-I_2R+I_3(\frac{1}{jωc}+2R)=0 }$$

this can be put in determinent form by

$\left[\begin{array}{ccc}   (R_L+\frac{1}{jωc}+R)&-R&ßR_L\\   -R&(\frac{1}{jωc}+2R)&-R\\ 0&-R&(\frac{1}{jωc}+2R) \end{array}\right]\left[\begin{array}{ccc}   I_1\\   I_2\\ I_3 \end{array}\right]=0$
loop current  $I_1, I_2,I_3$ cannot be zero

$\left[\begin{array}{ccc}   (R_L+\frac{1}{jωc}+R)&-R&ßR_L\\   -R&(\frac{1}{jωc}+2R)&-R\\ 0&-R&(\frac{1}{jωc}+2R) \end{array}\right]=0$
=>

$(R_L+\frac{1}{jωc}+R)\left[(\frac{1}{jωc}+2R)^2-R^2 \right]+R\left[-R(\frac{1}{jωc}+2R)+0 \right]+ßR_LR^2=0$


$(R_L+\frac{1}{jωc}+R)\left[\frac{1}{j^2ω^2c^2}+4R\frac{1}{jωc} +4R^2-R^2 \right]-\frac{R^2}{jωc}-2R^3+ßR_LR^2=0$
-------------------------------------------------------------
$(R_L+\frac{1}{jωc}+R)\left[\frac{1}{j^2ω^2c^2}+4R\frac{1}{jωc} +3R^2 \right]-\frac{R^2}{jωc}-2R^3+ßR_LR^2=0$


$-\frac{R_L}{ω^2c^2}+4R\frac{R_L}{jωc} +3R^2R_L-\frac{1}{jω^3c^3}-4R\frac{1}{ω^2c^2} +3R^2\frac{1}{jωc}-\frac{R}{ω^2c^2}+4R^2\frac{1}{jωc} +3R^3-\frac{R^2}{jωc}-2R^3+ßR_LR^2=0$


 
$ßR_LR^2+3R^2R_L +3R^3-2R^3+4R\frac{R_L}{jωc}+3R^2\frac{1}{jωc}+4R^2\frac{1}{jωc}- \frac{R^2}{jωc}-\frac{R_L}{ω^2c^2}-4R\frac{1}{ω^2c^2} -\frac{R}{ω^2c^2} -\frac{1}{jω^3c^3}=0$
$ßR_LR^2+3R^2R_L +R^3+\frac{4RR_L+3R^2+4R^2-R^2}{jωc}+\frac{-R_L-4R-R}{ω^2c^2} -\frac{1}{jω^3c^3}=0$

$ßR_LR^2+3R^2R_L +R^3+\frac{4RR_L+6R^2}{jωc}-\frac{(R_L+5R)}{ω^2c^2} -\frac{1}{jω^3c^3}=0$
$$\bbox[5px,border:2px solid red]{    ßR_LR^2+3R^2R_L +R^3+\frac{4RR_L+6R^2}{jωc}-\frac{(R_L+5R)}{ω^2c^2} -\frac{1}{jω^3c^3}=0          }$$       
to make total phase shift around loop zero 
Equating imaginary part to zero

$\frac{4RR_L+6R^2}{jωc}-\frac{1}{jω^3c^3}=0$        

$\frac{4RR_L+6R^2}{jωc}=\frac{1}{jω^3c^3}$ 
         
$4RR_L+6R^2=\frac{1}{ω^2c^2}$

$\frac{1}{4RR_L+6R^2}=ω^2c^2$
$ωc=\sqrt{\frac{1}{4RR_L+6R^2}}$
$2πfc=\sqrt{\frac{1}{4RR_L+6R^2}}$
$$\bbox[5px,border:2px solid black]{  f=\frac{1}{2πc\sqrt{4RR_L+6R^2}}}$$
is the expression for frequency of oscillation
Equating real part
$ßR_LR^2+3R^2R_L +R^3-\frac{(R_L+5R)}{ω^2c^2} =0$
here we have→
$ω^2c^2=\frac{1}{4RR_L+6R^2}$
$ßR_LR^2+3R^2R_L +R^3-(R_L+5R)(4RR_L+6R^2) =0$
$ßR_LR^2+3R^2R_L +R^3-4RR_L^2-20R^2R_L-6R_LR^2-30R^3 =0$ 
$ßR_LR^2 +R^3-30R^3-4RR_L^2-20R^2R_L-6R_LR^2+3R^2R_L =0$ 
$ßR_LR^2 -29R^3-4RR_L^2-23R^2R_L =0$   
$ßR_LR^2  =29R^3-4RR_L^2+23R^2R_L$
$ß  =\frac{29R^3}{R_LR^2}+\frac{4RR_L^2}{R_LR^2}+\frac{23R^2R_L}{R_LR^2}$
  $ß  =\frac{29R}{R_L}+\frac{4R_L}{R}+23$
$ß  =\frac{29R}{R_L}+\frac{4R_L}{R}+23$
$$\bbox[5px,border:2px solid black]{ let  k=\frac{R_L}{R} }$$
$ß  =\frac{29}{k}+4k+23$
the min value of k is obtained by differentiating with respect to k  
 $\frac{dß}{dk}=4-\frac{29}{k^2}$
Equating to zero
$4-\frac{29}{k^2}=0$
     
$4=\frac{29}{k^2}$
$k=\sqrt{\frac{29}{4}}=2.69$
The min value of ß required by the transistor can be obtained by

$ß  =\frac{29}{2.69}+4*2.69+23$
$ß  =44.5$
The phase sift oscillator using BJT will not oscillate if transmitter is chosen whose ß<44.5

RC and LC oscillator

RC and LC oscillator
Based on the feedback circuits used ,oscillator are classified into RC and & LC ocillator .The time period of RC and LC oscillator are 2ΠRC and $2Π\sqrt{LC}$ respecyively.For low frequency or Audio Frequency (AF range) the time periods are large .Since larger values of L&C are bulky and rare ,we use RC oscillator for AF. For high frequency or radion frequencies ,the time periods are smaller , The R &C become compatible with the internal parameter of transistor and hence LC are used for high frequency RF oscillator

Class A Power amplifiers(series fed)

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)

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

 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

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)

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

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

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 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

Class C power amplifier (Large signal tuned amplifier )

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

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

Common gain ,Relation between alpha α beta β gamma γ

Common gain

Current gain or current amplification factor is defined by the ratio of output current to input current when output voltage is kept constant

CB Configuration
$\left. α=\frac{ΔI_{C}}{ΔI_{E}}  \right |_{V_{CB}=constant }$
CE Configuration
$\left. β=\frac{ΔI_{C}}{ΔI_{B}}  \right |_{V_{CE}=constant }$
CC Configuration 

$\left. γ=\frac{ΔI_{E}}{ΔI_{B}}  \right |_{V_{CE}=constant }$

Relation between alpha α beta β gamma γ

We know that 

$I_E=I_B+I_C$→1

dividing it by  $I_C$ gives

$\frac{I_E}{I_C}=\frac{I_B}{I_C}+\frac{I_C}{I_C}$
$\frac{1}{α }=\frac{1}{β }+\frac{1}{1}=\frac{1}{β }+1$
$α =\frac{β}{β+1 }$  

$\frac{1}{β }=\frac{1}{α }-1$ 
$\frac{1}{β }=\frac{1-α}{α }$ 
$β =\frac{α }{1-α}$ 

dividing it by  $I_B$ gives

 $\frac{I_E}{I_B}=\frac{I_B}{I_B}+\frac{I_C}{I_B}$
$γ=1+β$
$β=γ-1$

Transistor Configuration

Transistor Configuration
if base is free→input
if collector is free→output
The transistor can be operated in 3 region
1)cut off voltage
the region for $I_E=0$ is called cut off region .This is a high voltage low current region in which the transistor is perfectly off. Hence the both input and output junction are reverse biased
2)Saturation region
The region to left of $V_{CB}=0$ is called saturation region .this is low voltage high current region in which the transistor is perfectly ON .Hence both input and Output junction are forward biased
3)Active region
It is the normal region of operation of a transistor .In active region input junction is forward biased and output junction is reversed biased .The transistor can be used as a switch  operating in saturation and cut off region ,In the active region the transistor can be used as as amplifier


Common Base Configuration
In this,base is common to both input(Emitter) and output(collector)

Input charcteristics

Input current $I_E$ VS Input voltage $(V_{EB})$ for different values of output voltage $(V_{CB})$

After the cut in voltage ,normally 0.7V for silicon and 0.3v Germanium ,$I_E$ increases rapidly with very small increase in $V_{EB}$



$$\bbox[5px,border:2px solid red] { Dynamic input resistance \left. r_i=\frac{ΔV_{EB}}{ΔI_{E}}  \right |_{V_{CB}=constant }  }$$

Output characteristics

It is the curve betweeen collector current $I_c$ and collector base voltage $V_{VB}$ at constant Emitter current $I_E$
      
$$\bbox[5px,border:2px solid red] { The dynamic output resistance \left. r_o=\frac{ΔV_{CB}}{ΔI_{C}}  \right |_{I_{E}=constant }  }$$

Common collector configuration

 Here the emitter is common to both input (base) and Output (collector)
Input charcteristics

$$\bbox[5px,border:2px solid red] { Dynamic input characteristics \left. r_i=\frac{ΔV_{BE}}{ΔI_{B}}  \right |_{V_{CE}=constant }  }$$
Output Characteristics

$$\bbox[5px,border:2px solid red] { The dynamic output resistance \left. r_o=\frac{ΔV_{CE}}{ΔI_{C}}  \right |_{I_{B}=constant }  }$$

Common collector configuration(Emitter flower buffer)

Here the collector is made common to both input (base) and output (Emitter)
Common collector configuration has very high input impedence and very low output impedance
Due to this it is coomonly used for impedance matching
As the current gain is high and voltage gain is nearly unity ,it is also used as buffer
$\frac{v_o}{V_i}=1 ==> V_o=V_i$

Frequency response of RC Coupled amplifier

Frequency response of RC Coupled amplifier

Characteristics of a RC coupled amplifier ii low ,middle and  high frequency range is called frequency response curve
This also called log magnitude plot
$f_H→$ upper 3db frequency
$f_L→$ lower 3db frequency
$f_H-f_L→$3db bandwidth
At lower drequency the reactance of $C_c$ is high and gain reduced due to in sufficient coupling .
At high frequency the reactance of $C_o$is small and act as short circuit and reduce gain
At mid frequency $C_c$ reduces and $C_o$ increases ,so gain is constant
GBW(gain bandwidt product)→Amid($F_2-F_1$)

RC Phase shift Amplifier

RC- Phaseshift Amplifier

Output of  $1^{st}$ stage is coupled to the $2^{nd}$ stage via RC network . So the the name coulpling capacitors
 $C_{c1},C_{c2},C_{c3}$  blocks dc components.
 $R_{c1},R_{c2}$→ collector resistance
$R_{11},R_{12},R_{21},R_{22}$ →are biasing resistors
$R_{e1},,R_{e2}$→emitter resistors
$C_{e1},C_{e2}$→bypass capacitors prevents loss of amplification
These are 3 different frequency ranges
1) low frequency range
2) highr frequency range
3)  mid frequency range

Equivalent circuit (Single stage)

Middle frequency range

Current gain
$A_{IM}=\frac{I_o}{I_b}$
$I_{o}=-g_mV_{b^,e}\frac{R_{co}}{R_{co}+R_{bc}}$
$I_{o}=-g_mI_br_{b^,e}\frac{R_{co}}{R_{co}+R_{bc}}$
$\frac{I_0}{I_b}=-g_mr_{b^,e}\frac{R_{co}}{R_{co}+R_{bc}}$
$g_mr_{b^,e}=I_c\frac{r_{b^,e}}{V_{b^,e}}=\frac{I_c}{I_b}=h_{fe}$
$\frac{I_0}{I_b}=A_{IM}=-h_{fe}\frac{R_{co}}{R_{co}+R_{bc}}$
$$\bbox[5px,border:2px solid red]{ A_{IM}=-h_{fe}\frac{R_{co}}{R_{co}+R_{bc}} }$$  

Voltage Gain
$A_{VM}=\frac{V_o}{V_i}$
$V_{o}=I_oR_{bc}=-g_mV_{b^,e}\frac{R_{co}R_{bc}}{R_{co}+R_{bc}}=-g_mI_br_{b^,e}R_{cobc}$
$V_i=I_b(r_{b^,b}+{r_{b^,e}})=I_bh_{ie}$
$\frac{V_{o}}{V_i}=\frac{-g_mI_br_{b^,e}R_{cobc}}{I_bh_{ie}}=\frac{-g_mr_{b^,e}R_{cobc}}{h_{ie}}=\frac{-h_{fe}R_{cobc}}{h_{ie}}$ 
$$\bbox[5px,border:2px solid red]{ A_{VM}=\frac{-h_{fe}R_{cobc}}{h_{ie}} }$$
Low frequence range

Current Gain
  
$A_{IL}=\frac{I_o}{I_b}$
$I_{o}=-g_mV_{b^,e}\frac{R_{co}}{R_{co}+R_{bc}+\frac{1}{jωc_b}}$  
$I_{o}=-g_mI_br_{b^,e}\frac{R_{co}}{R_{co}+R_{bc}+\frac{1}{jωc_b}}$
$\frac{I_0}{I_b}=-g_mr_{b^,e}\frac{R_{co}}{R_{co}+R_{bc}+\frac{1}{jωc_b}}$
$g_mr_{b^,e}=I_c\frac{r_{b^,e}}{V_{b^,e}}=\frac{I_c}{I_b}=h_{fe}$
$\frac{I_0}{I_b}=A_{IL}=-h_{fe}\frac{R_{co}}{R_{co}+R_{bc}+\frac{1}{jωc_b}}$ 
$$\bbox[5px,border:2px solid red]{ A_{IL}=-h_{fe}\frac{R_{co}}{R_{co}+R_{bc}+\frac{1}{jωc_b}} }$$  

Voltage Gain

 $A_{VL}=\frac{V_o}{V_i}$   
$V_{o}=I_oR_{bc}=-g_mV_{b^,e}\frac{R_{co}R_{bc}}{R_{co}+R_{bc}+\frac{1}{jωc_b}}=-g_mI_br_{b^,e}\frac{R_{co}R_{bc}}{R_{co}+R_{bc}+\frac{1}{jωc_b}}$
$V_i=I_b(r_{b^,b}+{r_{b^,e}})=I_bh_{ie}$ 
$\frac{V_{o}}{V_i}=\frac{-g_mI_br_{b^,e}\frac{R_{co}R_{bc}}{R_{co}+R_{bc}+\frac{1}{jωc_b}}}{I_bh_{ie}}=\frac{-g_mr_{b^,e}R_{co}R_{bc}}{h_{ie}(R_{co}+R_{bc}+\frac{1}{jωc_b})}=\frac{-h_{fe}R_{co}R_{bc}}{h_{ie}(R_{co}+R_{bc}+\frac{1}{jωc_b})}$ 
$$\bbox[5px,border:2px solid red]{ A_{VL}=\frac{-h_{fe}R_{co}R_{bc}}{h_{ie}(R_{co}+R_{bc}+\frac{1}{jωc_b})} }$$  

High Frequency Range

Current Gain

$A_{IH}=\frac{I_o}{I_b}$
$I_{o}=-g_mV_{b^,e}\frac{R_{co}}{R_{co}+R_{bc}}$
$V_{b^,e}=\frac{I_br_{b^,e}\frac{1}{jωc}}{r_{b^,e}+\frac{1}{jωc}}=\frac{I_br_{b^,e}}{r_{b^,e}jωc+1}$ 

$I_{o}=-g_m\frac{I_br_{b^,e}}{r_{b^,e}jωc+1}\frac{R_{co}}{R_{co}+R_{bc}}$

$\frac{I_o}{I_b}=\frac{-g_mr_b^,e}{r_{b^,e}jωc+1}\frac{R_{co}}{R_{co}+R_{bc}}=\frac{-h_{fe}R_{co}}{(r_{b^,e}jωc+1)(R_{co}+R_{bc})}$ 

$$\bbox[5px,border:2px solid red]{ A_{IH}=\frac{-h_{fe}R_{co}}{(r_{b^,e}jωc+1)(R_{co}+R_{bc})} }$$  

Volatage Gain

 $A_{VH}=\frac{V_o}{V_i}$
$V_{o}=I_oR_{bc}=\frac{-I_bh_{fe}R_{co}R_{bc}}{(r_{b^,e}jωc+1)(R_{co}+R_{bc})}$
$V_i=(r_{b^,b}+{\frac{r_{b^,e}}{r_{b^,e}jωc+1}})I_b=\frac{I_b(r_{b^,b}+r_{b^,b}r_{b^,e}jωc+r_{b^,e})}{r_{b^,e}jωc+1}$ 
 $\frac{V_o}{V_i}=\frac{\frac{-I_bh_{fe}R_{co}R_{bc}}{(r_{b^,e}jωc+1)(R_{co}+R_{bc})}}{\frac{I_b(r_{b^,b}+r_{b^,b}r_{b^,e}jωc+r_{b^,e})}{r_{b^,e}jωc+1}}= \frac{\frac{-I_bh_{fe}R_{cobc}}{(r_{b^,e}jωc+1)}}{\frac{I_b(r_{b^,b}+r_{b^,b}r_{b^,e}jωc+r_{b^,e})}{r_{b^,e}jωc+1}}=\frac{-h_{fe}R_{cobc}}{r_{b^,b}+r_{b^,b}r_{b^,e}jωc+r_{b^,e}}$
$\frac{V_o}{V_i}=\frac{-h_{fe}R_{cobc}}{r_{b^,b}+r_{b^,b}r_{b^,e}jωc+r_{b^,e}}=$ 
$$\bbox[5px,border:2px solid red]{ A_{VH}=\frac{-h_{fe}R_{cobc}}{r_{b^,b}+r_{b^,b}r_{b^,e}jωc+r_{b^,e}} }$$

Hybrid Parameter or h parameter

Transistors is a 2 part network.Input and Output of a transistor are related through 2 equation
$$V_1=h_{11}I_1+h_{12}V_2$$

$$I_2=h_{21}I_1+h_{22}V_2$$


$\left . \frac{V_1}{I_1}  \right |_{V_2=0}$ derived point impedance when output is short circuited $$h_{i},h_{ie},h_{ib}h_{ic}$$
$\left . \frac{V_1}{V_2}  \right |_{I_1=0}$ Reverse transfer voltage ratio when input is open circuit
  $$h_{r},h_{re},h_{rb}h_{rc}$$
$\left . \frac{I_2}{I_1}  \right |_{V_2=0}$ forward current ratio when output is short circuited $$h_{f},h_{fe},h_{fb}h_{fc}$$
$\left . \frac{I_2}{V_2}  \right |_{I_1=0}$ driving point output admittance when input is open circuit $$h_{o},h_{oe},h_{ob}h_{oc}$$

Equivalent circuit

For CE

$$V_b=h_{ie}I_b+h_{re}V_c$$

$$I_c=h_{fe}I_b+h_{oe}V_c$$

For CC

$$V_b=h_{ic}I_b+h_{rc}V_e$$
$$I_e=h_{fc}I_b+h_{oc}V_e$$

For CB

$$V_e=h_{ib}I_e+h_{rb}V_c$$
$$I_c=h_{fb}I_e+h_{ob}V_c$$


Analysis (using without considering source)

Current Gain
$A_i=\frac {I_L}{I_I}=\frac {-I_2}{I_1}$

$I_2=h_fI_1+h_oV_2$
$= h_fI_1+h_oI_LR_L$
$I_2 = h_fI_1-h_oI_LR_L$
$I_2(1+h+oR_L) = h_fI_1$
${\frac{I_2}{I_1} }= {\frac{h_f}{1+h_oR_L}}$
$A_I={\frac{-I_2}{I_1} }= {\frac{-h_f}{1+h_oR_L}}$
cc
$A_I= {\frac{-h_{fc}}{1+h_{oc}R_L}}$
ce
$A_I= {\frac{-h_{fe}}{1+h_{oe}R_L}}$
cb
$A_I= {\frac{-h_{fb}}{1+h_{ob}R_L}}$
$2)Input Resistance(R_i)$
$R_i= \frac{v_1}{I_1}$
$V_1=h_iI_1+h_rV_2$
$V_1=h_iI_1-h_rI_2R_L$
$\frac{V_1}{I_1}=h_i-\frac{h_rI_2R_L}{I_1}$
$$\bbox[5px,border:2px solid red] {  R_i=h_i+h_rA_IR_L }$$

ce
$R_i=h_{ie}+h_{re}A_IR_L$
cc
$R_i=h_{ic}+h_{rc}A_IR_L$
cb
$R_i=h_{ib}+h_{rb}A_IR_L$

3) $Voltage Gain (A_v)$
$A_v=\frac{V_2}{V_1}$
$V_2=-I_2R_L$
$A_I=\frac{-I_2}{I_1}$
$-I_2={A_II_1}$
=>$V_2=A_II_1R_L$
$A_v=\frac{V_2}{V_1}=\frac{A_II_1R_L}{V_1}=\frac{A_IR_L}{R_i}$

$$\bbox[5px,border:2px solid red] {  A_v=\frac{A_II_1R_L}{V_1}=\frac{A_IR_L}{R_i} }$$

4 )  $Output Resistance(R_o)$
$R_o=\frac{V_2}{I_2}$
$I_2= h_fI_1+h_oV_2$
$\frac{I_2}{V_2} = \frac{h_fI_1}{V_2}+h_o$
$Y_o= \frac{h_fI_1}{V_2}+h_o$

when $V_s=0$

$(R_s+h_i)I_1+h_rV_2=0$
$\frac{I_1}{V_2}=\frac{-h_r}{h_i+R_s}$
$Y_o= \frac{-h_fh_r}{h_i+R_s}+h_o$
$Y_o= h_o-\frac{h_fh_r}{h_i+R_s}$   
$R_o=\frac{1}{Y_o}$ 
Power Gain
$A_p=A_vA_I$   

Analysis (using with considering source)

Voltage gain

$A_{vs}=\frac{V_2}{V_s}=\frac{V_2}{V_1}\frac{V_1}{V_s}$
$A_{vs}=A_v\frac{V_1}{V_s}$

$V_1=\frac{V_sR_i}{R_i+R_s}$
=>$\frac{V_1}{V_s}=\frac{R_i}{R_i+R_s}$

$$\bbox[5px,border:2px solid red]{  A_{vs}=\frac{A_vR_i}{R_i+R_s}}$$

Current Gain


$A_{is}=\frac {I_L}{I_s}=\frac {-I_2}{I_s}=\frac {-I_2}{I_1}\frac {I_1}{I_s}$
$A_{is}=A_I\frac {I_1}{I_s}$

$I_1=I_s\frac {R_s}{R_s+R_L}$
$\frac {I_1}{I_s} = \frac {R_s}{R_s+R_L}$
$$\bbox[5px,border:2px solid red]{ A_{is} = A_I \frac {R_s}{R_s+R_L} }$$