EC2257 WEIN BRIDGE OSCILLATOR Anna University lab manual - Computer Programming

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Monday, April 18, 2011

EC2257 WEIN BRIDGE OSCILLATOR Anna University lab manual


WEIN BRIDGE OSCILLATOR

Aim : To Design and construct a Wein – Bridge Oscillator for a 
           given cut-off frequency .

APPARATUS REQUIRED:


S.NO
ITEM
RANGE
Q.TY
1
TRANSISTOR
BC107
2
2
RESISTOR


3
CAPACITOR



4
CRO
-
1
5
RPS
DUAL(0-30) V
1


























CIRCUIT DIAGRAM:

 


R1          RC!                                                  CC2
                                       R3                      Rc2            














 

                                 +    -


 


                                    Cc
R2
                                         R4                          RE2
                                                                                                CE
























 







                                                R

                                             C













 



                                                                                                DRB
                                                            R


 


                                          C                                                


                                                            GND


MODEL GRAPH:


 






Design
Given  :  Vcc = 12V ,  fo = 2 KHz, Ic1= Ic2 = 1mA.;  Stability factor = [0-10],
          fL = 100Hz
When the bridge is balanced,
                             fo = 1/ 2πRC
Assume,   C = 0.1μF
Find,          fo = ?
Given data :  Vcc = 15V ,  fL = 50Hz, Ic1= Ic2 = 1mA.; AvT = 3 ; Av1 =2; Av2 = 1;
                        Stability factor = [10]
                        Gain formula is given by
                        Av = -hfe RLeff / Zi   
                   RLeff = R c2 | | RL
                   hfe2 = 200 (from multimeter )
                   re2 = 26mV / IE2  = 26
                                hie2 = hfe2 re 2   = 200 x 26 = 5.2kW
From dc bias analysis , on applying KVL to the outer loop, we get
          Vcc = Ic2Rc2 + VCE2+VE2
                   VcE2  = Vcc/2 ;      VE2  = Vcc / 10 ; Ic2 = 1mA
                   Rc2 = ?
Since IB is very small when compared with Ic
                   Ic approximately equal to IE
                   Av2 = -hfe2 RLeff / Zi2   
Find  RL|| Rc2 from above equation
Since Rc2 is known , Calculate RL.
                             VE2  =  IE2RE2
Calculate RE2
S = 1+ RB2 /  RE2
                             RB 2 =?
                             RB 2 =R3 || R4
                             VB2 = VCC . R4 / R3 +  R4
                             VB2 = VBE2 + VE2
                                                R3 =?
   Find  R4
           Zi2 = (RB2 || hie2 )
          Zi2 = ?
          Rleff1 = Zi2|| Rc1
          Find Rleff1 from the gain formula given above
          Av1 = -hfe1 RLeff 1/ Zi1   
          RLeff1  = ?
On applying KVL to the first stage, we get
          Vcc = Ic1 Rc1 + VCE1 +VE1
          VCE1 = VCC / 2 ; VE1 = VCC / 10
          Rc1 = ?
Find   Ic1 approximately equal to IE1
                   R6 = RE1=?
S = 1+ RB1 /  RE1
                             RB 1 =?
                             RB 1 =R1 || R2
                             VB1 = VCC . R2 / R1 +  R2
                             VB1 = VBE2 +VE2
                   Find     R1 = ?
          Therefore find R2 = ?
 Zi1 = (RB1 || hie1 )
                              R5 = RL – R6
Coupling and bypass capacitors can be thus found out.
          Input coupling capacitor is given by , Xci = Z i / 10
                             Xci = 1/ 2pf Ci
                             Ci = ?
          output coupling capacitor is given by ,
                             X co=(Rc2 | | RL2) / 10
                             Xc0 = 1/ 2pf Co
                               Co =?
          By-pass capacitor is given by, XCE = RE2 / 10
                             XCE 1/ 2pf CE2
         CE =?

THEORY:

          In wein bridge oscillator, wein bridge circuit is connected between the amplifier input terminals and output terminals. The bridge has a series rc network in one arm and parallel network in the adjoining arm. In the remaining 2 arms of the bridge resistors R1and Rf are connected . To maintain oscillations total phase shift around the circuit must be zero and loop gain unity. First condition occurs only when the bridge is balanced . Assuming that the resistors and capacitors are equal in value, the resonant frequency of balanced bridge is given by
                                     
                                      Fo  =  0.159 RC








PROCEDURE:

1.           The circuit is constructed as per the given circuit diagram.
2.           Switch on the power supply and observe the output on the CRO( sine wave)
3.           Note down the practical frequency and compare it with the theoretical frequency.

RESULT :


Theoritical
Practical
Frequency
f = 1 / 2 P RC  



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