 a session on biostable multivibrator using transistor. Learning outcomes are at the end of session students will be able to explain the operation of biostable multivibrator and its output waveforms. They will able to list the applications of biostable multivibrator. Students are like this. So, what is the biostable multivibrator? As you know it is a two stage amplifier and it has two stable states, but it requires two trigger pulses to generate these two stable states. It continues in the same state that is stable state unless external trigger is given to the circuit and it will generate the second stable state when the second trigger pulse is applied. So, it switches in between the two stable states that is logic 1 and logic 0 which will be obtained by the switching circuit and here we are using the transistor as our switching circuit. So, we are using two transistor which provides the total phase shift of 360 degree so that oscillations will be provided and coupling is provided through the capacitors. Figure 1 shows the circuit diagram for biostable multivibrator. It has two low resistances nothing but collector resistance. These two resistances will keep the value of collector current in the permissible range nothing but it will able to get the IC max and with the help of this RL1 and RL2 we can control the collector current. It has two transistor Q1 and Q2 which are identical, but it has two different characteristics nothing but due to its doping level will have the switching time different. Collector of transistor Q1 is connected to base of transistor Q2. Same collector of transistor Q2 is connected to base of transistor Q1. Now, we will assume one condition. Suppose transistor Q1 is in the on state and Q2 is in the off state. So, what happens when Q2 is off? Q2 is off therefore its collector circuit will be open therefore you will get the logic one at its output. We are assuming transistor Q2 as the output therefore we are having here logic one when we assume Q1 is on that is Vc2 is equal to Vcc and that voltage is provided to the base of transistor Q1. So, large voltage will appear at the base of transistor Q1. When this voltage is large then transistor Q1 will be in the on condition and now as Q1 is on nothing but it is in the saturation region its operating point is in the saturation region due to that its collector voltage will be very, very small. So, this small or this decreased amount of voltage is applied again back to the base of transistor Q2 and due to that transistor Q2 remains in the off condition. So, this is your first stable state. Now, we will assume another stable state for that we will apply trigger pulse at base of transistor Q1. Suppose we are applying trigger at this base of transistor Q1 through C3. So, what was the previous state? Q2 was off, Q1 was on. Now, we apply the negative trigger pulse which is of large amplitude and very small duration. When you apply that trigger pulse over here transistor Q1 will be now in the cutoff region or transistor Q1 will be now off. When Q1 is off, what happens to its collector voltage? Collector voltage will increase. This increased voltage will be applied to the base of transistor Q2. So, due to that this increased voltage means Q2 will be now on due to the large voltage at its base which is greater than 0.7. So, due to that Q2 is now on as Q2 is on it will be in the saturation its operating point will be in the saturation region and due to that its collector voltage will reduce and that gives the logic 0 signal at its output. So, this is your another stable state. To obtain this stable state either you can start with the negative signal or negative trigger pulse at the base of transistor Q1 or positive trigger pulse at the base of transistor Q2. So, what was our original state that you have understood and to obtain the next stable state you have to apply here negative pulse or here positive pulse. But again when you switch back or when you want the original stable state you have to apply here negative trigger pulse and here positive trigger pulse. In this way you can switch back to the original stable state. So, you have the two stable state that can be obtained by keeping the either transistor on and another is off. So, this is the working of biostable multivibrator. See the output waveforms for biostable multivibrator as I told you collector voltage is low that is logic 0 due to its on condition at that time transistor Q2 is off which is our original stable state. So, this is logic 1. Now when you apply the trigger pulse then you can see the change in the output. Now you will be from logic 1 to logic 0 and logic 0 to logic 1. So, this is for transistor Q1 this is for transistor Q2. So, you can see Q1 is off and therefore its collector voltage is large approximately equal to Vcc and that is nothing but logic 1 signal and this is logic 0 signal because Q2 is now on and its collector voltage now decrease and that will be very very less and that will be Vc e sat. Practically it will be from 0.1 volt to 0.3 volt. Same again to come back to its original stable state you can apply the another trigger pulse and you can see the variation in the output. This will goes on continuous process provided that you must apply the trigger pulse and you can obtain the two stable states. What are the conditions and equations for biostable multivibrator? As I told you collector resistance nothing but RL1 and RL2 at the transistor Q1 and Q2 must be selected in such a way that it must control the value of collector current to the permissible value. So, this is the first condition. Second thing you have to select R1, R2, R3, R4 and Vbe to keep the transistor in the cutoff and saturation region respectively. Means Ibe must be large to drive this transistor in the saturation region. This is one stable state and another to get the another stable state Vbe must be below its cutoff value. So, in this way these components will make the transistor as a on and off state and due to that you will obtain the two stable states. And at the same time to avoid the loading effect on the amplifier R1, R2, R3, R4 must be very very large than collector resistance. What will be the stable voltage and current values for that? We will analyze this. So, as we assumed original stable state has transistor Q1 in the on condition and Q2 is in the off condition. Here IC2 is approximately equal to 0 Vbe1 and Vce1 will be small with respect to the signal provided at the input nothing but Vbeb. So, at the base of transistor Q2 you will obtain the value Vbeb into R1 divided by R1 plus R2. Ibe1, Ibe1 will be calculated as Vcc minus Vbeb1 total divided by R2 minus Vbeb minus Vbe1 total divided by R3. If we consider Vbe1 is very very small then Ibe1 will be Vcc by R2 minus Vbe1 divided by R3. We want transistor to operate as a switch to keep the transistor Q1 in the on condition. So, Ibe1 should be greater than or equal to Vcc minus Vce sat divided by Hf into Rl1. So, we have to satisfy the equation that is Ibe1 should be greater than or equal to IC1 by beta. We call the circuit diagram and find out what is the use of capacitors C1 and C2 in the biostable multivibrators. It will definitely increase the speed of response. Biostable multivibrators flips between the two states therefore it is also used as a flip flop in the digital operation which is very important operation for the computers and in the digital communication circuits. It will act as a electronic switch because it toggle between the two states. It is also used as a counter and store for the binary information. It can be used as a frequency divider in the timing circuit and also used for the generation of clock pulses and useful in the relay controllers. So, biostable multivibrator find wide applications in the industrial automation references are like this. Thank you.