555 Timer IC | Monostable Mode

Monostable Mode of 555 Timer IC is discussed in this post. This is the part 3 of the series on 555 Timer IC. Part 1 discusses the pin configuration and working of the internal circuit of 555 Timer IC so you might want to have a look at it if you haven’t read it already: 555 Timer IC - Introduction. 

Introduction to Monostable Mode:

555 IC in its Monostable mode, also known as a monostable multivibrator has one stable state in which it can remain indefinitely. When a trigger pulse is applied to this monostable multivibrator, it switches to another quasi-stable state for a given time interval (let say T) and then reverts back to the original stable state. As it switches to the quasi-stable state once, for a single input trigger pulse, this circuit is also known as One-Shot. The duration ‘T’ for which the one-shot remains in the quasi-stable state is independent of the triggering pulse and can be adjusted using other parameters as we will see in the discussion below. This mode of 555 is generally used as a pulse-standardizer.

Circuit Diagram:

The circuit diagram below shows the connection of 555 in Monostable mode. This is a simple circuit which requires a 555 IC, a resistor and a capacitor. 

Circuit diagram of 555 in monostable mode

Working:

Now, let us look at the working of 555 IC in monostable mode. The flip-flop inside the 555 is in the reset state when the circuit is in the stable state. So, the output of the IC is LOW and the Q1 transistor is in saturation connecting the capacitor to ground, thus the capacitor is discharged initially (see the figure below for a better understanding). Now, if we keep the trigger pin HIGH, the input to comparator 1 and 2 will be LOW and the flip-flop will go into the memory state, keeping the circuit in this state forever. Thus, the trigger input is HIGH and the output is LOW in the stable state.

A negative input pulse is applied to the pin 2 of 555 to trigger it. The output of comparator 2 will be HIGH and that of comparator 1 will be still LOW. So the output of 555 will switch to HIGH, Q1 goes to cut off. Therefore the capacitor C starts charging through resistor R with a time constant T = R*C. Once the trigger pulse switches to HIGH again, the output of comparator 2 goes back to 0 and that of comparator 1 is still 0. So the output remains in the previous state and C still charges.

 As the voltage across capacitor crosses (2*VCC)/3, the output of comparator 1 switches to logic HIGH and that of comparator 2 still remains LOW. So 555's output is pulled to logic LOW and Q1 transistor comes back to saturation as it was initially and capacitor starts to discharge rapidly through transistor Q1. Now the voltage across the capacitor and trigger pulse both are same as they were initially. So the output remains in the same stable state until another trigger pulse is applied.

Thus, 555 produce an output pulse of a fixed duration ‘T’ which is very important to us as it characterizes a monostable multivibrator. we will now calculate ‘T’. 

Simulation of the output of the circuit given in the section on circuit diagram

Time Period:

If we notice in the graph of V_C above, we can see that the output pulse remains HIGH as long as the capacitor ‘C’ charges from zero volts to (2*VCC)/3. So, we can find the time T if we calculate the time it takes for C to charge from 0 to (2*VCC)/3. The capacitor charges with a time constant R*C, so its equation can be given by this equation below.

At, time T, voltage across the capacitor becomes 2*VCC/3, so we get,

So, depending on the pulse duration we require, we can choose for the corresponding values of R and C from the above equation. This completes our analysis of 555 Timer IC in Monostable mode. I hope you find this tutorial useful. Comment your opinions and topics you want me to discuss in future.

Thanks for reading!