## 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