# Oscillators by 8D89xPT

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```									                                      Oscillators

* Feedback amplifier but frequency dependent feedback
As 
A f s  
1   f (s) A(s)

* Positive feedback, i.e. βf () A () < 0
As 
* Oscillator gain defined by
A f s  
1   f (s) A(s)
* Oscillation condition at ω = ωo (Barkhausen’s criterion) Af (ωo) = 
L( o )   f ( o ) A( o )   f ( o ) A( o ) e j ( o )  1
 ( o )  phase of  f ( o ) A( o )

ECES 352 Winter 2007                         Ch 13 Oscillators                             1
Wien Bridge Oscillator
* Based on op amp
* Combination of R’s and C’s in
feedback loop so feedback factor
βf has a frequency dependence.
* Analysis assumes op amp is ideal.
 Gain A is very large
 Input currents are negligibly
small (I+  I_  0).
R2
 Input terminals are virtually
R1                                         shorted (V+  V_ ).
V0
* Analyze like a normal feedback
amplifier.
Vi
If               ZS                 Determine input and output
 Determine feedback factor.
ZP
 Determine gain with feedback.
* Shunt-shunt configuration.

ECES 352 Winter 2007            Ch 13 Oscillators                        2
Wien Bridge Oscillator
Define
R1                                                                                                1 1  sRC
R2                        Z S  R  ZC  R       
sC    sC
V0                                       1               1
1  1            1     
R Z 
Z P  R ZC                  sC 
Vi                                                            ZS                                                       R     
If                                                                            C 

R

ZP                                                                      1  sCR

ZS                                                                 ZS
Z1                                         V0 = 0
Vi = 0
Z2
ZP                                                                             ZP

1
 1   1 
Z1  Z P Z S                                                                                      1  sRC
 ZP ZS                                                         Z 2  Z S  R  ZC 
sC
R1  sCR 
1
1  sCR sC 
 R  1  sCR 
             

sCR  (1  sCR ) 2

ECES 352 Winter 2007                                          Ch 13 Oscillators                                         3
Wien Bridge Oscillator
R2                               V0 V0 Vi
R1                                                                 Ar       
I S Vi I S
Vi                                                                V0
V0                       Vo
IS                                                 To get        , we use I1  I 2          and
IS                                         Z2                                  Vi                     R1  R2
IS                 Z1                                                                               Vo
Vi  V  V  I1 R1            R1 so
R1  R2
V0 R1  R2     R
        1 2
Vi    R1       R1
Feedback factor
ZS                                                           Vi
If                                                                       Since I   0,          Z1 and
IS
V0 Vi        R 
V0                             Ar         Z1  1  2 
 R 
ZP                                                         Vi I S           1 

R1  sCR 
where Z1                      so
Xf        If        1                                               sCR  (1  sCR ) 2
f                  
 R  R1  sCR 
Xo        Vo        ZS
sC                                                       Ar  1  2 
 R  sCR  (1  sCR ) 2
                                                                      1 
1  sRC

ECES 352 Winter 2007                                       Ch 13 Oscillators                                      4
Wien Bridge Oscillator
Oscillation condition
Phase of  f Ar equal to 180o. It already is since  f Ar  0.
   R        sCR
Then need only  f Ar  1  2 
       sCR  (1  sCR ) 2  1
   R1 
Rewriting
   R        sCR
 f Ar  1  2 

   R1  sCR  (1  sCR ) 2

   R              sCR
 1  2 

   R1            
 sCR  1  2 sCR  s 2C 2 R 2   
   R            sCR                R2       1
Loop Gain                                                   1  3sCR  s 2C 2 R 2  1  R 
 1  2 
                                       
   R1                               1 3
1
 sCR
sCR
    sC 
 f Ar          Ar                                   R 
 1  2 
1
 1  sCR                                    


R1                 1 
3  j  CR         
    sC  R2  R1  sCR                                              CR 
         1  
                                   Then imaginary term  0 at the oscillatio n frequency
 1  sCR  R1  sCR  (1  sCR )
2
1
  o 
 R          sCR                                        RC
 1  2 
 R  sCR  (1  sCR ) 2
      1 
Then, we can get  f Ar  1 by selecting the resistors R1 and R2

Gain with feedback is                               appropriately using
   R 1        R
Ar                                        1  2   1 or 2  2
      3
Arf                                                   R1         R1
1   f Ar
ECES 352 Winter 2007                       Ch 13 Oscillators                                                    5
Wien Bridge Oscillator - Example

1
Selecting for convenience C  10 nF , then from o 
RC
1          1
R                            100 
oC 10nF (1x106 rad / s)
Choosing R1  10K , then since R 2  2 R1 we get
R2  2(10K )  20K

ECES 352 Winter 2007                   Ch 13 Oscillators                 6
Wien Bridge Oscillator

Final note: No input signal is needed. Noise at the desired oscillation frequency
will likely be present at the input and when picked up by the oscillator when
the DC power is turned on, it will start the oscillator and the output will
quickly buildup to an acceptable level.

ECES 352 Winter 2007                 Ch 13 Oscillators                                7
Wien Bridge Oscillator
*    Once oscillations start, a limiting circuit is needed to prevent
them from growing too large in amplitude

ECES 352 Winter 2007                 Ch 13 Oscillators                            8
Phase Shift Oscillator

If     Rf                                    Vo   V   V      1 
IC3          IC2          IC1                            I C 2  I R1  I C1           o  o 1    
V2            V1                                                           sCRR f R f R f  sCR 
Vo   V      1  1
VX                 C                                          V2  V1  I C 2 Z C           o 1    
C                      C                                                          sCR f R f  sCR  sC
R            R IR1                           V0
IR2
Vo       1 
         2    
sCR f  sCR 
*     Based on op amp using inverting input         V          Vo         1 
I R2  2              2       
*     Combination of R’s and C’s in                   R      sCRR f  sCR 
feedback loop so get additional phase                         Vo          1  Vo     1 
shift. Target 180 o to get oscillation. IC3  I R2  IC 2          2         1    
sCRR f  sCR  R f  sCR 
*     Analysis assumes op amp is ideal.
Vo    1  1      1  Vo    3     1 
     1       2       1         2
V  V  0 so I f 
Vo
 I C1                    Rf sCR  sCR  sCR  R f  sCR ( sCR ) 
Rf                       Finally
Vo                                                 V      1  Vo                  1 
V1  V  I C1Z C                                    V X  V2 
IC3
  o 2 
3
sCR f                                                                    1         2
sC     sCR f  sCR  sCR f      sCR ( sCR ) 
 V1  1  Vo


  Vo                       Vo         4    1 
I R1                                                
R    R  sCR f        sCRR f                            3         2
                                       sCR f    sCR ( sCR ) 

ECES 352 Winter 2007                              Ch 13 Oscillators                                          9
Phase Shift Oscillator
Rearranging
If       Rf                     Vo         4       1 
VX           3  sCR  ( sCR ) 2 
IC3           IC2          IC1                                      sCR f                     
V2           V1
we get for the loop gain
VX    C             C            C                                                      V             sCR f
V0      L( )   ( ) A( )  0                           1
R        IR2 R     IR1                                                        VX          4         1 
3  sCR  ( sCR ) 2 
                     
 jCR f                      2C 2 RR f
                         
        4         1                         1 
Example                                                           3 j                       4  j  3CR         
Oscillator specifications: o=1x106 rad/s                              CR (CR ) 2                       CR  
To get oscillatio ns, we need the imaginary term to go to zero.
Selecting for convenience C  10 nF ,                       We can achieve this at one frequency o so
1
then from  o                                              3CR 
1
so   0 
1
3RC                                            CR             3RC
1                 1
R                                        58             To get oscillatio ns, we also need L(ωo )  1 so
3 oC       3 10nF (1x106 rad / s)
0 2C 2 RR f
Then                                  L(ωo )                 1 and substituting for ωo we get
4
R f  12(58 )  0.67 K
0 2C 2 RR f C 2 RR f 1           Rf
             2 2
       1 so
Note: We get 180o phase shift from op amp      4             4 3R C        12 R
since input is to inverting terminal and  R f  12 R
another 180o from the RC ladder.
ECES 352 Winter 2007                                 Ch 13 Oscillators                                         10
Colpitts LC-Tuned Oscillator
* Feedback amplifier with inductor L and
capacitors C1 and C2 in feedback network.
 Feedback is frequency dependent.
 Aim to adjust components to get
CB                                     positive feedback and oscillation.
 Output taken at collector Vo.
V0
 No input needed, noise at oscillation
Vi                          CE              frequency o is picked up and
amplified.
* RB1 and RB2 are biasing resistors.
* RFC is RF Choke (inductor) to allow dc
V0     current flow for transistor biasing, but to
block ac current flow to ac ground.
Vi                              * Simplified circuit shown at midband
frequencies where large emitter bypass
capacitor CE and base capacitor CB are
shorts and transistor capacitances (C and
C) are opens.
ECES 352 Winter 2007              Ch 13 Oscillators                     11
Colpitts LC-Tuned Oscillator
*   Voltage across C2 is just V
V
IC 2       sC2V
ZC 2
*   Neglecting input current to transistor (I  0),
V
I L  I C 2    sC2V
ZC 2
*   Then, output voltage Vo is


Vo  V  I L ZL  V  (sC2V )(sL)  V 1  s2 LC2         
AC equivalent circuit          *   KCL at output node (C)
Assuming oscillations have started, then V ≠ 0 and Vo ≠ 0, so
sC2V                                                   1      
sC 2V  g mV    sC1 Vo  0
R      
V0
sC 2V  g mV    sC1 V 1  s 2 LC2   0
Iπ ≈ 0                                               1      
R      
sC2V                                                           LC                       1
s 3 LC1C2  s 2  2   sC1  C2    g m    0
 R                        R
*   Setting s = j

 gm  

1  2 LC2 
R 
                         
  j  C1  C2    3 LC1C2  0
     R         
ECES 352 Winter 2007         Ch 13 Oscillators                                        12
Colpitts LC-Tuned Oscillator
* To get oscillations, both the real and imaginary parts
of this equation must be set equal to zero.

 gm  

1  2 LC2 
R 
                     
  j  C1  C2    3 LC1C2  0
     R         
* From the imaginary part we get the expression for
the oscillation frequency
 o C1  C2    o LC1C2  0
3

C1  C2          1
o              
LC1C2          CC 
L 1 2 
C C 
 1  2

* From the real part, we get the condition on the ratio
of C2/C1        1  o LC2
2
g  m        0
R       R
 C  C2      C2
1  g m R   o LC2  LC2  1
2
 1 C
 LC1C2        1

C2
 gm R
C1
ECES 352 Winter 2007               Ch 13 Oscillators                                     13
Colpitts LC-Tuned Oscillator
Example              * Given:
 Design oscillator at 150 MHz

     
o  2f  2 150x106  9.4x108 rad / s

Transistor gm = 100 mA/V, R = 0.5 K

* Design:
C2
 g m R  (100 mA / V )( 0.5K )  50
C1
   Select L= 50 nH, then calculate C2, and then C1
C1  C2        1  C2 
o                     1     
LC1C2         LC2 
    C1 

1  C2              1
C2      1                        (1  50)  1.13x10 9 F  1,130 pF
L o 
2
    C1  50nH (9.4 x108 ) 2

C    1,130 pF
C1  2             23 pF
50       50

ECES 352 Winter 2007                      Ch 13 Oscillators                                14
Summary of Oscillator Design
Wien Bridge Oscillator            *    Shown how feedback can be used with
reactive components (capacitors) in the
feedback path.
*    Can be used to achieve positive feedback.
 With appropriate choice of the resistor
sizes, can get feedback signal in phase
with the input signal.
 Resulting circuit can produce large
amplitude sinusoidal oscillations.
Phase Shift Oscillator
*    Demonstrated three oscillator circuits:
 Wien Bridge oscillator
 Phase Shift oscillator
 Colpitts LC-Tuned oscillator
*    Derived equations for calculating resistor and
capacitor sizes to produce oscillations at the
Colpitts LC-Tuned Oscillator             desired oscillator frequency.
*    Key result: Oscillator design depends
primarily on components in feedback
network, i.e. not on the amplifier’s
characteristics.

ECES 352 Winter 2007             Ch 13 Oscillators                              15

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