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
                                                     loading.
                                                   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



        Input Loading                                                                   Output Loading

                                      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
            I1                                             I2            Amplifier gain including loading effects
                                                     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




  Oscillator specifications: o=1x106 rad/s
                                                                     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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