# Astable Multivibrator using Op-amp

The astable multivibrator is also called as a free running multivibrator. It has two quasi-stable states i.e. no stable state such. No external signal is required to produce the changes in state. The component values used to decide the time for which circuit remains in each state. Usually, as the astable multivibrator oscillates between two states, is used to produce a square wave.

The circuit looks like a Schmitt trigger except that the input voltage is replaced by a capacitor. As shown in fig. 6.15 the comparator and positive feedback resistors $R_1$ and $\displaystyle R_2$ form an inverting Schmitt trigger.

When $\displaystyle V_{0}$ is at $\displaystyle V_{sat}$ , the feedback voltage is called the upper threshold voltage $\displaystyle V_{UT}$ and is given as

$\displaystyle V_{UT} = R_1 \frac {V_{sat}}{R_1+R_2}$

When $V_{0}$ is at $-V_{sat}$, the feedback voltage is called the lower threshold voltage $V_{LT}$  and is given as

$\displaystyle V_{LT} = - \frac {R_1 V_{sat}}{R_1+R_2}$

Circuit operation

(i). When power is turned ON, $V_{0}$ automatically swings either $V_{sat}$ or to $- V_{sat}$ since these are the only stable states allowed by Schmitt trigger. Assume it swings to $+V_{sat}$.

(ii). Now capacitor starts charging towards $+ V_{sat}$ through the feedback path provided by the resistor $R_f$ to the inverting input. As long as the capacitor voltage $V_{C}$ is less than $\displaystyle V_{UT}$, the output voltage remains at $V_{sat}$

(iii). As soon as $V_{C}$ charges to a value slightly greater than $V_{UT}$ , the input goes positive with respect to the input. This switches the output voltage from $+ V_{sat}$  to $-V_{sat}$

(iv). As $V_{0}$ switches to $-V_{sat}$ , capacitor starts discharging via . The current $I$ discharges capacitor to 0 V and recharges capacitor to $V_{LT}$ When $V_{C}$ becomes slightly more negative than the feedback voltage $V_{LT}$ output voltage switches back to $+V_{sat}$.

• Frequency of Oscillation: The frequency of oscillation is determined by the time it takes the capacitor to charge from $V_{LT}$ to $V_{UT}$ and vice versa.

Initial voltage ( at t=0 ) across Capacitor

$\displaystyle \Rightarrow V_{CO} = -\beta V_{sat}$

Here Role of supply Voltage will play, the output voltage

$\displaystyle V_{s} = V_{0} = + V_{sat}$

Voltage across Capacitor at time $t$ is given by the eq

$\displaystyle V_{C}(t) = V_{CO} e^{-\frac {t}{RC}} + V_{0} \biggl ( 1 - e^{-\frac {t}{RC}} \biggr )$

At time $t = T_{C}$

$\displaystyle V_{0} = - V_{sat}$

$\displaystyle V_{C} (T_{C}) = + \beta V_{sat}$

$\displaystyle V_{C} (t) = \biggl ( V_{CO} - V_{0} \biggr ) e^{-\frac {t}{RC}} + V_{0}$

$\displaystyle +\beta V_{sat} = \biggl ( - \beta V_{sat} + V_{sat} \biggr ) e^{-\frac {T_{C}}{RC}} + V_{sat}$

$\displaystyle T_{C} = RC ln \biggl (\frac {1+\beta}{1 - \beta} \biggr )$

$\displaystyle \beta = \frac {R_2}{R_1 + R_2}$

$\displaystyle T_{C} = T_{d}$

Total time period ($T$ )

$\displaystyle T = T_{C} + T_{d}$

$\displaystyle T = 2RC ln \biggl (\frac {1+\beta}{1- \beta} \biggr )$