The circuit shown in Fig. Q.8(c) is a variable frequency Wien bridge oscillator.

Question

The circuit shown in Fig. Q.8(c) is a variable frequency Wien bridge oscillator. The resistors R are ganged and different frequency ranges are obtained by switching the three capacitors. To start oscillations the voltage gain of the non-inverting amplifier Av = (1 + R2/R1) must be greater than +3 and to maintain constant amplitude Av must be +3. Briefly explain how replacing R2 by a thermistor can meet these conditions. Determine the frequency range when resistance R can vary from 1 kohm to 10 kohm and C = 0.15 muF. Note: That the frequency of oscillation is f = 1/(2piCR) Hz. Calculate the different frequency ranges that are obtained by switching the capacitors. Use the results of part for the middle capacitor, and assume that adjacent capacitors in the circuit have values: 1.5 pF, 0.15 muF and 15 muF.

Explanation / Answer

1)

Av= (1+R2 /R1 )

for oscillations Av = 3

R2 = 2 R1

2)

f0 = 1/(2 *pi* R*C)

if R=1 kilo ohm

c=0.15 pF

f0 = 10.610 MHz

if R=10 kilo ohm

c=0.15 pF

f0 = 1.0610 MHz

3)

f0 = 1/(2 *pi* R*C)

if R=1 kilo ohm

c=1.5 uF

f0 = 106 Hz

c=0.15uF

f0 = 1.061 kHz

if c = 15nF

f0 = 10.610 kHz

if R=10 kilo ohm

c=1.5uF

f0 = 10.610 Hz

C= 0.15uF

f0 = 106 Hz

c=15nF

f0 = 1.061 kHz

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