Editing Multivibrator (section)

==Monostable==
[[Image:Transistor Monostable.svg|thumbnail|220px|Figure 2: Basic [[Bipolar junction transistor|BJT]] monostable multivibrator]]{{unreferenced section|date=February 2022}}

In the monostable multivibrator, one resistive-capacitive network (C<sub>2</sub>-R<sub>3</sub> in Figure 1) is replaced by a resistive network (just a resistor). The circuit can be thought as a 1/2 [[Multivibrator#Astable|astable multivibrator]]. Q2 collector voltage is the output of the circuit (in contrast to the astable circuit, it has a perfect square waveform since the output is not loaded by the capacitor).

When triggered by an input pulse, a monostable multivibrator will switch to its unstable position for a period of time, and then return to its stable state. The time period monostable multivibrator remains in unstable state is given by ''t''&nbsp;= ln(2)''R''<sub>2</sub>''C''<sub>1</sub>. If repeated application of the input pulse maintains the circuit in the unstable state, it is called a ''retriggerable'' monostable. If further trigger pulses do not affect the period, the circuit is a ''non-retriggerable'' multivibrator.

For the circuit in Figure 2, in the stable state Q1 is turned off and Q2 is turned on. It is triggered by zero or negative input signal applied to Q2 base (with the same success it can be triggered by applying a positive input signal through a resistor to Q1 base). As a result, the circuit goes in [[#State 1|State 1]] described above. After elapsing the time, it returns to its stable initial state.

=== Monostable using op-amp ===
[[File:Monostable.jpg|thumb|monostable multivibrator using op-amp]]
The circuit is useful for generating single output pulse of adjustable time duration in response to a triggering signal. The width of the output pulse
depends only on external components connected to the op-amp. A diode D1 clamps the capacitor voltage to 0.7 V when the output is at +Vsat. Let us assume that in the stable
state the output Vo = +Vsat. The diode D1 clamps the capacitor to 0.7 V. The voltage at the non-inverting terminal through the potential divider will be + βVsat. Now a negative trigger of magnitude V1 is applied to the non-inverting terminal so that the effective signal at this terminal is less than 0.7 V. Then the output voltage switches from +Vsat to -Vsat. The diode will now get reverse biased and the capacitor starts charging exponentially to -Vsat through R. The voltage at the non-inverting terminal through the potential divider will be - βVsat. After some time the capacitor charges to a voltage more than - βVsat. The voltage on the non-inverting input is now greater than on the inverting input and the output of the op-amp switches again to +Vsat. The capacitor discharges through resistor R and charges again to 0.7 V.

The pulse width T of a monostable multivibrator is calculated as follows:
The general solution for a  low pass RC circuit is
:<math>V_o = V_f +(V_i - V_f)e^{-t/RC}</math>
where <math>V_f = -V_\text{sat}</math> and <math>V_i = V_d</math>, the diode forward voltage.  Therefore,
:<math>V_c = -V_\text{sat} +(V_d + V_\text{sat})e^{-t/RC}</math>
at <math>t = T</math>,
:<math>V_c = -\beta V_\text{sat}</math>
:<math>-\beta V_\text{sat} = -V_\text{sat} +(V_d + V_\text{sat})e^{-T/RC}</math>
after simplification,
:<math>T = RC\ln\left({1+V_d/V_\text{sat} \over 1 - \beta}\right)</math>
where <math>\beta = {R2 \over R1+R2}</math>
           
If <math>V_\text{sat} >> V_d</math> and <math>R1 = R2</math> so that <math>\beta=0.5</math>, then
<math>T = 0.69RC</math>