- 2075 Baisakh · 2+2+2 marks
Barkhausen criteria; Wien bridge frequency
Short answer
Barkhausen criteria: |Aβ|=1 and loop phase shift 0°. The Wien bridge oscillator uses a non-inverting amplifier with gain A=3 (Rf=2R1) and a lead-lag RC network; setting imaginary part of feedback factor to zero gives ω=1/(RC), so oscillation frequency f=1/(2πRC).
State Barkhausen criteria for oscillation. Draw Wien bridge oscillator circuit to generate sine wave and derive the frequency of the generated sine wave.
Barkhausen criteria
- Loop gain magnitude |Aβ| = 1.
- Total loop phase shift = 0° (or 360°).
Wien bridge oscillator
+------[ Rf ]-------+
| |
GND--[R1]-+ |\ |
+--------|- \ |
| >------+----o Vo
+--------|+ / |
| |/ |
| |
+--[ R ]--||--------+
| C (Zs: series R and C)
+-+-+
| |
[R] === C (Zp: R parallel C)
| |
+-+-+
|
GND
The amplifier is a non-inverting op-amp with gain A = 1 + Rf/R1. The positive feedback is taken through a lead-lag network: a series R and C from the output to the non-inverting input, and a parallel R and C from that input to ground. Rf and R1 form the negative feedback and fix the gain.
Derivation of frequency
Series arm: Zs = R + 1/(jωC) Parallel arm: Zp = R/(1 + jωRC)
Feedback factor: β = Zp/(Zs + Zp) = 1/(1 + Zs/Zp)
β = 1 / [ (1 + jωRC)(R + 1/jωC)/R + 1 ]
= 1 / [ 1 + (1 + jωRC)(1 + 1/jωRC) ]
= 1 / [ 3 + j(ωRC − 1/ωRC) ]
For a 0° phase shift, the imaginary part must be zero:
ωRC = 1/(ωRC), so ω² = 1/(RC)², i.e. ω = 1/(RC)
f = 1/(2πRC)
At this frequency β = 1/3. For |Aβ| = 1 we need A = 3, i.e. 1 + Rf/R1 = 3, so Rf = 2R1 (in practice slightly more than 2R1, so that oscillation can start).
✓ Answer checked by Claude Opus 5.5 (high effort)
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