# Monostable Multivibrators using Op-amp

The monostable multivibrator is also called as the one-shot multivibrator. The circuit produces a single pulse of specified duration in response to each external trigger signal. For such a circuit, only one stable state exists. When an external trigger is applied, the output changes its state. The new state is called as a quasi-stable state. The circuit remains in this state for a fixed interval of time. After some time it returns back to its original stable state. In fact, an internal trigger signal is generated which drives the circuit back to its original stable state. Usually, the charging and discharging of a capacitor provide this internal trigger signal.

• The diode $D_1$ connected across the capacitor is called clamping diode. It clamps the capacitor voltage to $\displaystyle 0.7 V$ when the output is at $\displaystyle +V_{sat}$.
• A negative triggering pulse is applied to the Non-inverting terminal of Op-amp through RC differentiator circuit and diode $\displaystyle D_2$.

Operation of the Circuit

(i). To understand the operation of the circuit, let us assume that the output $\displaystyle V_0$ is at $\displaystyle +V_{sat}$ i.e. in its stable state.

(ii). The diode $\displaystyle D_1$ (Connected across Capacitor) conducts and the voltage across the capacitor $\displaystyle C$  $\displaystyle \Rightarrow V_C$ gets clamped to $\displaystyle 0.7$ Volts.

(iii). The voltage at the non-inverting terminal is controlled by voltage divider circuit of $\displaystyle R_1$ and $\displaystyle R_2$

Voltage at non-inverting terminal $(V_2) = +\beta V_0$

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

(iv). If $\displaystyle V_{T}$, a negative trigger of amplitude $\displaystyle V_{T}$ is applied to the non-inverting terminal, so that the effective voltage at this terminal is less than $\displaystyle 0.7 V \Rightarrow$ then the output of the Op-amp changes its state from $\displaystyle +V_{sat}$ to $\displaystyle -V_{sat}$.

(v). The diode is now reverse biased and the capacitor starts charging exponentially to $\displaystyle -V_{sat}$ through resistance $\displaystyle R$.

(vi). The voltage at the non-inverting terminal is now $\displaystyle -\beta V_{sat}$. When the capacitor voltage becomes just slightly more negative than $\displaystyle -\beta V_{sat}$, the output of the Op-amp changes its state back to $\displaystyle +V_{sat}$

(vii). The capacitor now starts charging towards $\displaystyle +V_{sat}$ through $\displaystyle R$ until $\displaystyle V_{C}$ reaches $\displaystyle 0.7 V$ as capacitor gets clamped to the voltage.

Expression for pulse width $\displaystyle T$

Initial voltage ( at t=0 ) across Capacitor = $\displaystyle V_{CO} = V_{D1}$

Here Role of supply Voltage will play output voltage $\displaystyle \Rightarrow V_{S} = V_{0} = - V_{sat}$

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

$\displaystyle V_{C}(t) = V_{D1} e^{-\frac{t}{RC}} + V_0 (1 - e^{-\frac {t}{RC}})$

At time $\displaystyle t = T$

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

$\displaystyle \Rightarrow V_C(T) = -\beta V_{sat}$

$\displaystyle V_{C}(t) = (V_{D1} - V_{0}) e^{-\frac {t}{RC}} + V_{0}$

$\displaystyle - \beta V_{sat} = (V_{D1} + V_{sat}) e^{-\frac {T}{RC}} - V_{sat}$

$\displaystyle T = RC \frac {ln (1 + \frac {V_{D1}}{V_{sat}})}{1-\beta}$

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

If $\displaystyle V_{sat} >> V_{D1}$  and $\displaystyle R_1 = R_2$

Then $\Rightarrow T = 0.69 RC$

For Monostable operation, the trigger pulse width $T_{p}$ should be much less than $T$.

The diode $D_2$ is not essential but it is used to avoid malfunctioning if any positive noise spikes are present in triggering line.

It can be seen from the waveform that the voltage $V_{C}$ does not reach its quiescent value $V_{D1}$ until time $T^{'} > T$ . Hence it is necessary that a recovery time ($T^{'} - T$) be allowed to elapse before the next triggering signal is applied.