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