# Monostable Amplifier (OPAMP) / Op-amp Monostable

Single stable circuits can be easily made using discrete components or digital logic gates, but single stable circuits can also be created using operational amplifiers.

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Op-amp Monostable Multivibrator circuits are positive feedback (or regenerative) switching circuits with only one stable state, producing an output pulse with a certain T-time.

An external trigger signal is applied to change its state, and after a certain period of time, in microseconds, milliseconds or seconds, at a time period specified by the RC components, the monostable circuit then returns to its original stable state. The next trigger remains until the input signal arrives.

The basic single stable multivibrator block scheme is given as follows:

## Op-amp Monostable Block Schema

The block diagram above shows the creation of a single stable multivibrator by adding an external resistance (R) and capacitor (C) throughout a switching circuit. Switching circuitry can be done using transistors, digital logic gates or general purpose operational amplifiers. The time constant of the resistance-capacitor combination determines the length of the τ pulse, T.

In this lesson, we will create a single stable multivibrator circuit using a comparative amplifier comparator circuit with a positive feedback pathway. Since the feedback is positive, the circuit is regenerative, that is, it is added to the differential input signal.

## Op-amp Monostable Circuit

First of all, let's consider the Inverting Amplifier circuit as follows.

In this inverted operational amplifier configuration, part of the output signal (called the feedback fraction) is fed back to the inverted input of the transactional amplifier through the resistant network.

In this basic invert configuration, the feedback fraction is negative because it feeds back into the inverting input. This negative feedback configuration between the output and the inverted input terminal forces the differential input voltage towards zero.

The result of this negative feedback is that the op-amp produces a reinforced output signal that is out of phase 180° with the input signal. Thus, an increase in the inverted terminal voltage, the -V, which feeds back from the output, causes a decrease in the output voltage, VO produces a balanced and stable amplifier that operates in the linear zone.

Now consider the same transactional amplifier circuit in which the inverter and non-inverting inputs of the op-amp are replaced. That is, the feedback signal feeds back into the inverted input, and the feedback process is now positive and produces a basic op-amp comparator circuit with built-in hysteresis.

The op-amp monostable multivibrator circuit is built around a transactional amplifier configured as a closed loop Schmitt Trigger circuit that uses positive feedback provided by R1 and R2 resistors to create the necessary hysteresis. The use of positive feedback means that feedback is refreshing and provides the necessary state dependency that turns op-amp into a two-state memory device.

Consider the basic op-amp voltage comparator circuit below.

## Op-amp Schmitt Comparator

A resistant network is connected between the op-amp output and the inverted (+) input. When the Vout positive feed rail (+Vcc) becomes saturated, a positive voltage is applied relative to the soil to the non-inverted input of op-amps. Similarly, when the Vout is saturated with negative feed rail (-Vcc), a negative voltage is applied relative to the soil to the non-inverted input of op-amps.

Since the two resistors are configured as a voltage dividing network along the output, the voltage VB at the inverted input will depend on the fraction of the output voltage, which is fed back by the ratio of the two resistances. This feedback fraction is given β as follows:

Keep in mind that by replacing the resistances of R1 and R2 with a ponciometer, in which the posiometer wiper directly depends on the non-inverted input of op-amps, we can make the value of the β variable so that we can change the feedback fraction.

Since the amount of hysteresis is directly related to the amount of feedback fraction, it is best to avoid creating a Schmitt trigger op-amp (regenerative comparator) with very small amounts of hysteresis (small β), since this can result in op-amp. oscillating between the top and bottom points when switching.

Now, if we place a feedback network on the Schmitt trigger between the output and the inverted (-) input, we can check the time it took for the Schmitt op-amp to change its status. By doing so, the signal to the inverting input of op-amps is provided by the op-amp itself through the external RC feedback network, as shown now.

## Basic Op-amp Monostable Circuit

At the first opening (i.e. t = 0), the output (VOUT) will be saturated with either positive rail (+Vcc) or negative rail (-Vcc), because these are only two stable conditions. op-amp. For now, let's say the output turns towards the positive feed rail, +Vcc. Next, the voltage on the nonverting input will be equal to VB, +Vcc*β. Β feedback here is the fraction.

The inverting input is held at 0.7 volts. The forward voltage drop of the diode prevents it from going more positive by compressing to 0v (soil) by D1 and diode. Thus, the potential in the VA is much less than in VB, and the output remains constant in +Vcc. At the same time, the capacitor (C) charges up to the same 0.7 volt potential and is held there by the forward voltage drop of the diode.

If we apply a negative pulse to the inverted input, the voltage of 0.7v in the VA will now be larger than the voltage in VB, since VB is now negative. Thus, the output of the op-amp switches configured by Schmitt becomes saturated and saturated towards the negative feed rail, -Vcc. The result is that the potential in VB is now equal to -Vcc*β.

This temporary meta-stable state causes the capacitor to charge exponentially in the opposite direction through feedback resistance, from R +0.7 volts to the saturated output it replaced just before the R+0.7 volt, -Vcc. Diode, D1 becomes inverted, so it has no effect. The capacitor will be emptied in a time constant τ = RC.

As soon as the capacitor voltage in the VA reaches the same potential as VB, that is, -Vcc*β, the op-amp returns to its original permanent stable state, once again saturated in the output +Vcc.

Note that when the timing period is complete and the op-amp output returns to its stable state and the positive feed rail is saturated, the capacitor tries to reverse-charge to +Vcc but can only charge to a maximum value of 0.7v. forward voltage drop by diodes. We can graphically show this effect as follows:

## Op-amp Monostable WaveForms

Next, we can see that a negative outgoing trigger input will temporarily destabilly transfer the op-amp monostable circuit. After a time delay, capacitor C charges over feedback resistance R, while the circuit returns to its normal stable state when the capacitor voltage reaches the required potential.

This latency (T) of the rectangular pulse on the output, which is unstable state time, is given as follows:

## Op-amp Monostable Scheduling Period

If the two operational amplifier feedback resistors are of the same value, that is: R1 = R2, then the equation above is also simplistic:

Obviously, there is a certain amount of time it takes for the capacitor to recharge from -Vcc*β to VD (0.7v), and therefore a second negative impact during this period may not start a new timing period.

Then, to ensure the correct operation of the op-amp monostable circuit upon the application of the next trigger pulse, the time between the trigger pulses (Ttoplam) must be greater than the timing time, the time required for the T-plus capacitor. charging, (Charging).

The charging recovery time is given as follows:

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

We can add an RC differential circuit to the input to ensure that the op-amp monostable circuit has a good negative trigger signal that initiates the timing period on the front edge of the negative outgoing pulse, as well as to stop any incorrect triggering of the circuit while in stable condition.

A differentiator circuit is useful in generating a negative output increment from a square or rectangular input waveform. The sharp and sudden drop of the comparator threshold voltage below the feedback fraction pushes the β value, op-amp monostabilini into the timing period. A differentiator circuit is created using a resistance-capacitor (RC network as shown).

## RC Differentiator Circuit

The basic differentiator circuit above uses another resistance-capacitor (RC) network, which is a time-based derivative of the output voltage input voltage. 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, creating a negative rise. Then the capacitor decreases exponentially while charging.

Usually for an RC differentiator circuit, the peak value of negative elevation is approximately equal to the size of the trigger waveform. Also, as a general rule, in order for an RC differentiator circuit to produce good sharp narrow spikes, the time constant ( τ ) must be at least ten times smaller than the input pulse width. For example, if the input pulse width is 10 ms, the 5RC time constant must be less than 1 ms (10%).

The advantage of using a differentiator circuit is that it blocks any constant DC voltage or slow-changing signal and allows only rapidly changing trigger pulses to start the single stable timing period. The diode ensures that the trigger pulse to the input that does not reverse the D, op-amps is always negative.

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