Op-amp astable multivibrator

Multivibrator

A multivibrator is a regenerative circuit with two states in which the output switches between them. It has three types: astable (no stable state, free-running square wave), monostable (one stable state, gives a single pulse when triggered) and bistable (two stable states, flip-flop).

Op-amp astable (square-wave) multivibrator

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

The inverting input is connected to the output through R and to ground through C. The non-inverting input receives a fraction of the output from the divider R1, R2 (positive feedback).

β = R1/(R1 + R2)

Operation

1. Suppose the output is at +Vsat. The voltage at the (+) input is +βVsat. The capacitor charges toward +Vsat through R.
2. When vC reaches +βVsat, the (−) input becomes higher than (+), and the output switches to −Vsat. Now the (+) input is −βVsat.
3. The capacitor discharges and then charges toward −Vsat, until vC reaches −βVsat; then the output switches to +Vsat again.
4. The cycle repeats, giving a square wave while the capacitor voltage is an exponential triangular-like wave between ±βVsat.

 Vo    +Vsat ____      ____
             |    |    |    |
 ------------|    |____|    |____
                -Vsat
 vC    +βVsat  /\    /\
 ------------ /  \  /  \  <- exponential
                \/    \/
               -βVsat

Frequency of oscillation

During charging, vC rises from −βVsat to +βVsat, towards Vsat:

vC(t) = Vsat − (Vsat + βVsat) e^(−t/RC)

At t = T/2, vC = +βVsat:

βVsat = Vsat − Vsat (1 + β) e^(−T/2RC) → e^(−T/2RC) = (1 − β)/(1 + β)

T/2 = RC ln[(1 + β)/(1 − β)]

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

With β = R1/(R1 + R2): the expression becomes T = 2RC ln(1 + 2R1/R2).

Special case: R2 = 1.16 R1 (β = 0.462) gives ln(...) = 1, so f = 1/(2RC).

Example: R = 10 kΩ, C = 0.1 µF, R1 = R2 → β = 0.5, ln(3) = 1.0986, T = 2 × 10⁴ × 10⁻⁷ × 1.0986 = 2.2 ms, f = 455 Hz.