Hey friends, welcome to the YouTube channel ALL ABOUT ELECTRONICS. So, in this video, we will learn, how the monostable multivibrator can be designed using the BJT. So, the monostable multivibrator is the circuit where the output of the circuit remains in the stable state but whenever the external trigger signal is applied then momentarily the output goes into the unstable state. And once again, it comes back into the stable state. So, the duration for which it remains in this unstable state depends on the values of the resistor as well as the capacitor. So, now let's see, how this monostable multivibrator can be designed using the BJT. And this is the circuit of the monostable multivibrator. So, as you can see, it consists of two transistors. So, this is the first transistor, while this is the second transistor. So, here this resistor R2 is the base resistor, while this resistor Rc2 is the collector resistor. And as you can see, both transistors are cross-coupled. That means here, the collector of the first transistor is connected to the base of the second transistor via this capacitor C. While the collector of the second transistor is connected to the base of the first transistor via this resistor R1. And through this portion of the circuit, the trigger signal is applied to the circuit. So, this portion of the circuit will come into the picture whenever we apply the trigger signal. And later on, we will talk about it in detail. Now, here during the operation of the circuit, the transistor operates in the cut-off and the saturation region. So, here when the transistor operates in the cut-off region, then we will assume that it will act as an open circuit. So, at that time, voltage Vce is approximately equal to Vcc. Similarly, when it operates in the saturation region, then we will assume that it acts as a closed switch. So, at that time, this voltage Vce is approximately equal to 0V. So, now let's see, how this monostable multivibrator circuit works. So, when the circuit is powered up, then the transistor Q1 will be OFF, while the transistor Q2 will be ON. So, let's understand, how this circuit works whenever it is powered up. So, after the circuit is powered up, then both transistors try to conduct. But if you see over here, the collector of this transistor Q2 is connected to the base of this transistor Q1 via this resistor R1. So, as this transistor Q2 conducts more and more, and goes into the saturation, then it will act as a closed switch. And due to that, the voltage at the base of this transistor Q1 will be approximately equal to zero volt. So, due to that, this transistor Q1 will operate in the cut-off region and it will act as an open circuit. That means initially after the power-up, the transistor Q1 will be OFF, while the transistor Q2 will be ON. And here, at this node, or at the collector of this transistor Q2 we will get the output. So, initially, after the power-up, the voltage at the transistor Q2 will be approximately equal to 0V, while the voltage at the collector of this transistor Q1 will be approximately equal to Vcc. And this is the stable state of the monostable. That means in this state, the output of the monostable will be low. Now, as soon as we apply the trigger signal, then momentarily, the output goes into the unstable state and then after it comes back into the stable state. So, now let's see, when we apply this trigger signal, then how this circuit works. Now, as as I said, in the stable state, the transistor Q2 is ON and the Q1 is OFF. So, in this condition, this capacitor C charges through this path. And here, as this transistor Q2 is ON, so the voltage between this base and the emitter terminal will be equal to 0.7V. That means in the stable state, this capacitor C will charge up to the voltage Vcc - Vbe. Now, in this stable condition, if a very narrow triggering pulse is readily available then through the resistor it can be directly applied to the base of this transistor Q1. But if it is not available, then the pulse of any finite duration can be applied to this circuit. So, basically, this circuit is the differentiator circuit. And whenever we apply the pulse to this differentiator circuit then it detects the rising and the falling edge of the pulse. And by passing it through the diode, we can get the only positive edge. So, when we apply the pulse of finite duration to this differentiator circuit, then we will get a very narrow triggering pulse at the base of this transistor Q1. And as soon as the trigger pulse is applied to the base of this transistor Q1, then it will go into saturation. And due to that, it will act as a short circuit. So, now if you see, the left end side of the capacitor will get grounded. And due to that, the voltage -(Vcc - Vbe), will directly get applied to the base of this transistor Q2. So, due to this, large negative voltage, the transistor Q2 will go into the cut-off region and it will act as an open circuit. That means now, the voltage at the collector of this transistor Q2 will be high and the same voltage is also applied to the base of this transistor Q1. So, Q1 will remain in the saturation region, and it will act as a closed switch. And now in this condition, this capacitor C reverse charges or starts charging in the reverse direction. So, basically, it will start charging towards the voltage Vcc. And during the charging as soon as the voltage across the capacitor reaches the 0.7V, or the voltage Vbe, the transistor Q2 starts conducting and eventually it will go into the saturation. And due to that, the voltage at the collector of this transistor Q2 will become low. Or we can say that the output of the circuit will become low. And as the same voltage is also applied to the base of this transistor Q1, so this transistor Q1 will also become off. And due to that, it will act as an open circuit. That means once again, the circuit will come back into the stable state. And the time for which it remains in this unstable state depends on the value of this capacitor and the resistor R2. So, let's say, the time for which it remains in this unstable state is equal to T1. So, now let's find the expression of this time T1 in terms of the RC time constant. And to find that, let's write down the general expression of the capacitor charging. So, the voltage across the capacitor Vc(t) can be given as Vc(∞) + [ Vc(0) - Vc(∞)]e^(-t/RC) Now, as I said earlier, the output goes into the unstable state when the Q1 becomes ON and the Q2 goes OFF. But before that, the voltage across the capacitor is equal to Vcc - Vbe. But after that, the capacitor starts charging in the reverse direction. So, we can say that the initial voltage across the capacitor or the Vc(0) is equal to - (Vcc - Vbe) And in the unstable state, when this Q1 becomes ON then the capacitor starts charging towards the voltage Vcc. That means Vc (∞) = Vcc. That means if the capacitor charges fully, then it will get charged up to the voltage Vcc. But here we are interested in finding the time T1 when the capacitor reaches the 0.7 V. Because at that voltage, there will be a transition in the output. And once again, at that time, the output will come back into the stable state. So, let's say, at time t = T1, the voltage across the capacitor, that is the Vc(T1) = Vbe So, at time t= T1, if we write this expression, then it will be equal to voltage Vbe = Vc (∞) + [ Vc (0) - Vc(∞)] e^(-T1/RC) Now, here the Vc (∞) = Vcc, while the Vc(0) = Vbe - Vcc And here this RC time constant is equal to R2C. That means we can say that, the voltage Vbe = Vcc + [ Vbe - Vcc - Vcc] * e^(-T1/R2C) Or we can say that, e^(-T1/R2C) = ( Vbe - Vcc)/ (Vbe - 2 Vcc) Now, here for this circuit to work, the voltage Vcc should be much larger than the voltage Vbe. And usually, that is the case in the circuit. So, we can neglect this term voltage Vbe and we can write this expression as e ^ (-T1/ R2C) = (-Vcc/ -2Vcc) That is equal to 1/2. Or if we take the natural log on both sides, then we can write it as, T1 = -R2C ln (1/2) That is equal to R2Cln (2) Or we can say that it is equal to 0.693R2C. That means time T1 = 0.693R2*C. That means once we apply this trigger pulse, then this is the time T1, for which the circuit will go into the unstable state. And here the voltage in the unstable state will depend on the value of this Rc2 and R1. So, if we want the value of the output voltage close to Vcc, then the value of the R1 should be at least 10 times higher than the Rc2. Because when the circuit operates in the unstable state, then this transistor Q1 is ON. And for a moment, if we neglect this voltage Vbe then the collector voltage Vc can be found by applying the voltage divider rule. Or for more precise value, we can apply the KVL in this loop. And by applying the KVL, we can find the voltage at this node. So, in short, to get the output very close to Vcc, the value of R1 should be at least 10 times higher than the Rc2. So, anyway, I hope in this video, you understood how the monostable multivibrator can be designed using the BJT. So, if you have any questions or suggestions, do let me know here in the comment section below. If you like this video, hit the like button and subscribe to the channel for more such videos.