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Barkhausen criteria and relaxation oscillator

Short answer

Barkhausen: |Aβ| = 1 and loop phase = 0° or 360°. A relaxation oscillator uses a Schmitt trigger (R1‑R2) and an RC timing network. The capacitor charges/discharges between ±βV_sat, producing a square wave. Frequency f = 1/[2RC ln[(1+β)/(1−β)]]; with R2 ≈ 1.16 R1, f ≈ 1/(2RC).

State Barkhausen criteria. Draw and explain circuit diagram of Relaxation oscillator (square wave generator). Write its frequency of oscillation.

Barkhausen criteria

For sustained oscillations in a feedback circuit with amplifier gain A and feedback factor β:

  1. The loop gain magnitude must be unity: |Aβ| = 1.
  2. The total phase shift around the loop must be 0° or 360° (positive feedback).

(In practice |Aβ| is made slightly greater than 1 to start oscillations.)

Relaxation oscillator (op-amp square wave generator)

A relaxation oscillator produces a non-sinusoidal output by repeatedly charging and discharging a capacitor. The op-amp works as a comparator (Schmitt trigger) with positive feedback through R1–R2 and an RC timing circuit on the inverting input.

              R
        +---/\/\/\-----+
        |              |
        |    |\        |
 Vc  +--+----|-\       |
     |       |  >------+---o Vo (±Vsat)
    C    +---|+/       |
     |   |   |/        |
    GND  +----/\/\/\---+
         |     R2
         R1
         |
        GND

Operation

  • Positive feedback gives V+ = β Vo, where β = R1/(R1 + R2). The output is always at +Vsat or −Vsat.
  • Suppose Vo = +Vsat. Then V+ = +β Vsat. C charges through R towards +Vsat.
  • When Vc just exceeds +β Vsat, V− > V+ and the output switches to −Vsat. Now V+ = −β Vsat.
  • C now discharges and charges in the negative direction through R towards −Vsat.
  • When Vc falls below −β Vsat, the output switches back to +Vsat. The cycle repeats.

The output is a square wave (±Vsat) and the capacitor voltage is an exponential triangle-like wave between ±β Vsat.

 Vo  +Vsat |‾‾‾‾|    |‾‾‾‾|
         0 +----+----+----+---- t
     -Vsat      |____|    |___
 Vc  +βVsat   /\        /\
     -βVsat  /  \__/   /  \__/

Frequency of oscillation

T = 2RC ln[(1 + β)/(1 − β)]
f = 1/T = 1/(2RC ln[(1 + β)/(1 − β)])

If R2 = 1.16 R1 (β ≈ 0.462), ln[(1 + β)/(1 − β)] ≈ 1, so f ≈ 1/(2RC). If R1 = R2 (β = 0.5), T = 2RC ln 3 ≈ 2.2RC.

✓ Answer checked by Claude Opus 5.5 (high effort)

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