 Session on monostable multivibrator using transistor. Hello, welcome to the session on monostable multivibrator using transistor. Learning outcomes are, at the end of session students will be able to explain the operation of monostable multivibrator as well as they can draw the output waveforms of monostable multivibrator. And I can also identify the time period for output waveforms. Contents are like this. So, what is a monostable multivibrator? It is a two-stage amplifier with two stable state out of which one is stable state which is zero that is logic zero and another is temporary state nothing but logic one which is your cosy stable state. So, one is stable state another is cosy stable state. So, this is obtained by transistor and the transistor remains in the cosy stable state and the time period is determined by the circuit elements. Then it again returns to its initial stable state nothing but logic zero. So, cosy stable state time period is determined by register and capacitor connected at the base of transistor. Here we have two transistors. So, which transistor you are considering as the output transistor to the base of that transistor R and C are connected and that will decide the time period of monostable multivibrator's cosy stable state. Figure one shows the circuit diagram for monostable multivibrator transistor Q1 and Q2 are wired as a switch which provides total 360 degree phase shift. R1 and C1 are the components which will decide the time period of output nothing but cosy stable states time. Collector of Q2 is connected to base of Q1 through register R2 and collector of Q1 is connected to transistor base of transistor Q2 through C1. DC coupling is provided through capacitor C2 and AC coupling is provided through capacitor C1. So, both AC and DC coupling are provided in this circuit. Now, assume VCC is connected that is switch on and PBB is also connected through register R3 to base of Q1. At base of Q1 you have negative voltage. As you know this is your NPN transistor this is your NPN transistor. So, at the base you have the negative voltage and that is why transistor will be in the cutoff region. When it is in the cutoff region it will be off and when transistor is off it will be like a disconnected from the VCC and ground and therefore you will be obtained the voltage VCC across this transistor that is open circuit condition. As this is open circuit condition collector voltage will rise as the collector voltage rises and that collector voltage is fed to the base of Q2. So, increased value of collector voltage will make the transistor Q2 on nothing but now transistor Q2 is in the saturation region and therefore it will act as a short circuit and when it is short circuit its collector voltage will decrease. So, this decreased value of collector voltage will make the transistor Q1 again in the off condition. So, this is your stable state. So, continuously you will obtain the negative nothing but 0 volt at the output of Q2. So, we are considering V0 across this transistor Q2. This is your stable state. Now, if you apply trigger at capacitor C2 which is a large amplitude short duration pulse at the base of transistor Q1. So, at base of transistor Q1 you have the high voltage and due to this high voltage transistor will be on which goes to the saturation region and due to that it will reduce the collector voltage at transistor Q1. This decreased value of voltage is fed to transistor Q2. Therefore, this voltage at base terminal will cause this transistor Q2 in the off state. When transistor Q2 is off its collector voltage will rise and therefore it will go to the logic 1 level. So, you will obtain the cause is stable state of logic 1 level due to this trigger pulse and therefore this is also called as one short or mono short multivibrator. We are applying the trigger pulse at once only. So, this time period is determined by R1 and C1 and after this you will obtain the logic 0 level. In this way you will obtain one stable state which is logic 0 and one cause is stable state which is logic 1. In this way you will get the only one stable state out of this multivibrator. Have a look at the output waveforms of mono stable multivibrator. This is the negative trigger pulse applied at the base of Q2 and due to that you will reach to the cause is stable state of logic 1. If the value of R and C is less then you will get the short time constant signal. If you have the larger value of R and C you will get longer duration time pulse that is logic 1 signal. Now we will see how this time period can be calculated with the help of circuit diagram. So, again have a look at the base of transistor Q2. So, what it consists of? It has the voltage that is capacitor will discharge through transistor Q2 and therefore you will get the negatively exponential signal as well as it will first reach to its Vcc. So, now we will see the derivation for time period of mono stable multivibrator. So, we must concentrate on the voltage at base of transistor Q2 as this voltage is not constant. We assume it as V of t that is function of time period and that is equal to Vi plus Vcc minus Vi into bracket e to the power minus t divided by t. So, what this Vi is? We will see Vi is nothing but Vb saturation minus Vcc plus Vce sat. So, this is the value at the base of transistor Q2 due to the either state that is due to the either side of the multivibrator and this is the capacitor discharged voltage. Value of V of t will be 0.5 due to its cut in voltage. Next Vb sat will be 0.7 volt and Vc sat as we know that is 0.3 volt practically. If you put the value of Vi in the equation of Vi V of t you can easily calculate the time period of Cauchy stable state. So, you can get V of t minus Vi divided by Vcc minus Vi equal to e to the power minus t divided by t. So, put the value of Vi again put the values of these quantity and then you can obtain the time period of your Cauchy stable state nothing but logic one level that is 0.69 into RC. With this you are able to identify the time period of your signal. Now recall the concept we have learnt and identify out of which of this following multivibrator that is more stable multivibrator will work. Is it a pulse generator? Is it a pulse stretcher or is it a pulse loser? So, what is a pulse generator? It is a circuit which continuously generate the pulses but monostable multivibrator will not generate the pulses unless it has a trigger pulse. But again after applying a trigger pulse it will give you only one pulse at the output. So, it is not a pulse generator. So, pulse stretcher. So, will it stretch the pulse? Yes, based on the value of R and C it can stretch the time period of the Cauchy stable state. So, it can be a pulse stretcher. It is not a pulse loser because as soon as it will get the pulse it will give you the output. So, it is not a pulse loser circuit. So, correct answer is pulse stretcher. So, what are the applications of monostable multivibrator? It is a circuit which provides the delay. So, it is very useful for the timing circuits and for the digital circuit it will used as a temporary memories. It is very useful to trigger any kind of circuit most of the time for any kind of oscillator circuit and it will also regenerate old and worn out pulses. Therefore, it is very useful to use the monostable multivibrator to regenerate old and worn out pulses. Most of the time it is used in the television circuit and control system circuit to obtain its application to obtain the pulse at the output. References are like this. Thank you.