We saw in our tutorial about Monostable Multivibrators (MMVs), that monostable’s are electronic circuits which produces a single timed rectangular output pulse when triggered.

Monostable circuits can easily be made using discrete components or digital logic gates but monostable circuits can also be constructed using operational amplifiers.

**Op-amp Monostable Multivibrator** (one-shot multivibrator) circuits are positive-feedback (or regenerative) switching circuits that have only one stable state, producing an output pulse of a specified duration T. An external trigger signal is applied for it to change state and after a set period of time, either in microseconds, milliseconds or seconds, a time period which is determined by RC components, the monostable circuit then returns back to its original stable state were it remains until the next trigger input signal arrives.

The basic *monostable multivibrator* block diagram is given as:

The above block diagram shows that a monostable multivibrator is constructed by adding an external resistor, (R) and capacitor, (C) across a switching circuit. The switching circuit can be made using transistors, digital logic gates or general purpose operational amplifiers. The time constant, τ of the resistor-capacitor combination determines the length of the pulse, T.

In this tutorial we will construct a monostable multivibrator circuit using an operational amplifier comparator circuit with a positive feedback path. As the feedback is positive the circuit is regenerative, that is it adds to the differential input signal.

Firstly let’s consider the Inverting Amplifier circuit as shown.

In this inverting operational amplifier configuration, some of the output signal (called the feedback fraction) is fed back to the inverting input of the operational amplifier via the resistive network.

In this basic inverting configuration the feedback fraction is therefore negative as it is fed back to the inverting input. This negative feedback configuration between the output and the inverting input terminal forces the differential input voltage towards zero.

The result of this negative feedback is that the op-amp produces an amplified output signal which is 180^{o} out-of-phase with the input signal. So an increase in the inverting terminal voltage, -V fed back from the output causes a decrease in the output voltage, Vo producing a balanced and stable amplifier operating within its linear region.

Consider now the same identical operational amplifier circuit in which the inverting and non-inverting inputs of the op-amp have been interchanged. That is the feedback signal is fed back to the non-inverting input and the feedback process is now positive producing a basic op-amp comparator circuit with built-in hysteresis.

The op-amp monostable multivibrator circuit is constructed around an operational amplifier configured as a closed-loop *Schmitt Trigger* circuit that uses positive feedback provided by resistors R1 and R2 to generate the required hysteresis. The use of positive feedback means that the feedback is regenerative and provides the required state dependence which in effect changes the op-amp into a bistable memory device.

Consider the basic **op-amp voltage comparator** circuit below.

A resistive network is connected between the op-amps output and non-inverting (+) input. When Vout is saturated towards the positive supply rail, (+Vcc), a positive voltage, with respect to ground is applied to the op-amps non-inverting input. Likewise, when Vout is saturated towards the negative supply rail, (-Vcc), a negative voltage, with respect to ground is applied to the op-amps non-inverting input.

Since the two resistors are configured across the output in the form of a voltage divider network, the voltage V_{B} present at the non-inverting input will therefore be dependant on the fraction of the output voltage fed back by the ratio of the two resistors. This feedback fraction, β is given as:

Note that we can make the value of β variable by replacing resistors R_{1} and R_{2} with a potentiometer in which the potentiometers wiper is connected directly to the op-amps non-inverting input thereby allowing us to vary the feedback fraction.

As the amount of hysteresis is directly related to the amount of feedback fraction, it is best to avoid constructing a Schmitt trigger op-amp (regenerative comparator) with very small amounts of hysteresis (small β) as this may result in the op-amp oscillating between the upper and lower points when switching.

If we now place a feedback network across the Schmitt trigger between the output and the inverting (-) input, we can control the amount of time that it takes for the Schmitt op-amp to change state. By doing this, the signal to the op-amps inverting input is now provided by the op-amp itself via the external RC feedback network as shown.

At initial power on (that is t = 0), the output (V_{OUT}) will saturate towards either the positive rail, (+Vcc) or to the negative rail, -Vcc, since these are the only two stable states allowed by the op-amp. Lets assume for now that the output has swung towards the positive supply rail, +Vcc. Then the voltage at the non-inverting input, V_{B} will be equal to +Vcc.β where β is the feedback fraction.

The inverting input is held at 0.7 volts, the forward volt drop of diode, D_{1} and clamped to 0v (ground) by the diode, preventing it from going any more positive. Thus the potential at V_{A} is much less than that at V_{B} and the output remains stable at +Vcc. At the same time, the capacitor, (C) charges up to the same 0.7 volts potential and is held there by the forward-biased voltage drop of the diode.

If we were to apply a negative pulse to the non-inverting input, the 0.7v voltage at V_{A} now becomes greater than the voltage at V_{B} since V_{B} is now negative. Thus the output of the Schmitt configured op-amp switches state and saturates towards the negative supply rail, -Vcc. The result is that the potential at V_{B} is now equal to -Vcc.β.

This temporary meta-stable state causes the capacitor to charge up exponentially in the opposite direction through the feedback resistor, R from +0.7 volts down to the saturated output which it has just switched too, -Vcc. Diode, D_{1} becomes reverse-biased so has no effect. The capacitor, C will discharge at a time constant τ = RC.

As soon as the capacitor voltage at V_{A} reaches the same potential as V_{B}, that is -Vcc.β, the op-amp switches back to its original permanent stable state with the output saturated once again at +Vcc.

Note that once the timing period is complete and the op-amps output changes back to its stable state and saturates towards the positive supply rail, the capacitor tries to charge up in reverse to +Vcc but can only charge to a maximum value of 0.7v given by the diodes forward voltage drop. We can show this effect graphically as:

Then we can see that a negative-going trigger input, will switch the op-amp monostable circuit into its temporary unstable state. After a time delay, T while the capacitor, C charges up through the feedback resistor, R, the circuit switches back to its normal stable state once the capacitor voltage reaches the required potential.

This time delay period (T) of the rectangular pulse at the output, the unstable state time, is given as:

If the two operational amplifiers feedback resistors are of the same value, that is: R_{1} = R_{2}, then the above equation simplifies down too:

Obviously there is a a certain amount of time that the capacitor takes to charge again from -Vcc.β to V_{D} (0.7v) and therefore during this period a second negative pulse may not start a new timing period. Then in order to ensure the correct operation of the op-amp monostable circuit upon the application of the next trigger pulse, the time period between trigger pulses, (T_{total}) must be greater than the timing period, T plus the time required for the capacitor to recharge, (T_{charging}).

The charging recovery time is given as:

Where: Vcc is the supply voltage, V_{D} is the diodes forward voltage drop, (usually about 0.6 to 0.7 volts) and β is the feedback fraction.

In order to ensure that the op-amp monostable circuit has a good negative trigger signal which starts the timing period on the leading edge of the negative going pulse, and also to stop any false triggering of the circuit when it is in its stable state, we can add a RC differentiating circuit to the input.

A differentiator circuit is useful in producing a negative output spike from a square or rectangular input waveform. The sharp and abrupt reduction of the comparators threshold voltage below its feedback fraction, β value drives the op-amp monostable into its timing period. A differentiator circuit is formed using a resistor-capacitor (RC network as shown.

The differentiator circuit above uses another resistor-capacitor (RC) network whose output voltage is the derivative of the input voltage, with respect to time. When the input voltage changes from 0 to -Vcc, the capacitor begins to charge exponentially. Since the capacitor voltage, Vc is initially zero, the differentiator output voltage suddenly jumps from 0 to -Vcc producing a negative spike and then decays exponentially as the capacitor charges up.

Generally for a RC differentiator circuit, the peak value of the negative spike is approximately equal to the magnitude of the trigger waveform. Also, as a general rule-of-thumb, for an RC differentiator circuit to produce good sharp narrow spikes, the time constant, ( τ ) should be at least ten times smaller than the input pulse width. So for example, if the input pulse width is 10 ms, then the 5RC time constant should be less than 1 ms (10%).

The advantage of using a differentiator circuit is that any constant DC voltage or slowly varying signal will be blocked allowing only rapidly varying trigger pulses to initiate the monostable timing period. Diode, D ensures that the trigger pulse arriving at the op-amps non-inverting input is always negative.

Adding the RC differential circuit to the basic op-amp monostable gives:

An **op-amp monostable circuit** is constructed using the following components. R1 = 30kΩ, R2 = 30kΩ, R = 150kΩ and C = 1.0uF. If the op-amp monostable is supplied from a ±12V supply and the timing period is initiated with a 10ms pulse.

Calculate the circuits timing period, capacitor recovery time, total time between trigger pulses and the differentiator network values. Draw the completed circuit.

Data given: R1 = R2 = 30kΩ, R = 150kΩ, C = 1.0uF and pulse width equals ten milliseconds, (10ms).

1. Timing Period, T:

2. Capacitor recovery time:

3. Total time between trigger pulses:

4. The input pulse is given as 10ms, therefore the negative spike duration will be 1ms (10%). If we assume a capacitance value of 0.1uF, then the differentiator RC values are calculated as :

This gives the final **Op-amp Monostable circuit** for our example as:

We have seen in this tutorial that an Op-amp Monostable circuit can be constructed using a general purpose operational amplifier, such as the 741, and a few additional components. While it may be easier to construct monostable (one-shot) circuits using discrete components, digital logic gates or the common 555 IC chip, sometimes it is required to construct monostable’s using op-amps for use in analogue circuits.

By configuring an op-amp as a Schmitt trigger with positive feedback, the duration of the output pulse is determined by the time constant of the RC timing circuit, as well as the by the ratio value of the resistive voltage divider network providing the positive feedback which helps make the circuit unstable.

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Isn’t that circuit near the beginning a NON inverting amplifier?

can i install in my laptop