Hey friends, welcome to the YouTube channel ALL ABOUT ELECTRONICS. In this video, we will learn how to design a A-stable multivibrator using the 555 timer IC. So, if you are not aware what is a-stable multivibrator then it is a circuit whose both states are unstable state and output continuously used to change between the two states.
And the most common application of this a-stable multivibrator is in the design of relaxation oscillator. So, in this video, let's see how we can design this astable multivibrator using the 555 timer IC. So, here is the circuit using which this 555 timer can be configured as a astable multivibrator.
So, if you see over here, this reset and control pin is not used in the circuit. So, this reset pin is connected to the supply voltage while this control pin is connected to the ground via this capacitor C1. And over here, this threshold and the trigger pins are connected together. Now to understand the working of this circuit, let me just draw the same circuit along with the internal block diagram.
Now, whenever the circuit is just turned on, at that time this capacitor will be fully uncharged. Or we can say that the voltage at this pin number 2 and 6 will be equal to 0V. Now, like we had seen in the previous videos, the voltage at these two nodes of the voltage divider will be equal to 2x3 Vcc and 1x3 Vcc respectively. So, initially if you see, the second comparator output will be equal to logic 1. While the first comparator output will be equal to logic zero.
Because initially, for the second comparator, the voltage at this inverting node is less than the voltage at this non-inverting node. While for the first comparator, the voltage at this inverting node is more than the voltage at the non-inverting node. So initially if you see, the S will be equal to 1 and R will be equal to zero.
Or we can say that output of the flip-flop will be equal to 1. And because of that, the output of the 555 timer will also equal to logic high value. Now whenever this Q is equal to 1, at that time this Q bar will be equal to 0. And because of that, this transistor will be in a off condition. So this capacitor C1 will start charging through this resistor R1 and R2. And gradually, the voltage at this pin number 2 and 6 will start increasing. So first of all, let's see whenever this voltage reaches the 1 by 3 Vcc voltage.
So, whenever this voltage just crosses this 1 by 3 Vcc voltage, the output of the second comparator will become logic low. Because at that time, the voltage at the inverting node will be slightly more than the voltage at the non-inverting node. While still the voltage is less than this 2 by 3 Vcc voltage, so the output of the second comparator will be equal to logic zero.
So, in this condition, both s and r equal to zero. So, the output of the flip-flop will retain the previous state. That means q will remain to logic 1 value. And hence, the output of the 555 timer will also remain to the logic high value.
Now, let's see what happens when this capacitor voltage reaches this 2x3VCC voltage. So now, whenever this capacitor voltage just crosses this 2x3VCC voltage, at that time the output of the first comparator will become logic high value. Because at that time, the voltage at this pin no. 6 is more than the reference voltage. So now if you see, S is equal to 0 and R is equal to 1. So, this flip-flop will get reset to the logic zero value. And whenever this capacitor voltage just reaches this 2x3 Vcc voltage, then you will see the transition in the output from logic high value to the logic low value.
Now whenever this Q is equal to zero, at that time the value of Q bar will be equal to 1. So, now this transistor will become ON. And once this transistor will become ON, this capacitor C will start discharging through this transistor. And gradually, the voltage across this capacitor will start reducing.
So as soon as the voltage just go below this 2x3VCC voltage, then the output of the first comparator will become logic 0. Because at that time, the voltage at this pin no. 6 is less than the reference voltage. And if you see the second comparator, then the output of the second comparator will also be equal to zero. Because the voltage at this pin no. 2 is less than the reference voltage. So in this condition, both S and R is equal to zero.
And the flip-flop will retain its previous state. That means Q will be equal to logic 0 value. And hence, the output of the 555 timer will also remain to the logic 0 value.
Now, let's see during the discharging process whenever the voltage across the capacitor just go below this 1 by 3 Vcc voltage. Now, once this voltage goes below this 1 by 3 Vcc voltage, then the output of the second comparator will become logic high value. Because at that time, the voltage at this non-inverting terminal is more than the voltage at this inverting terminal. And in this condition, s will be equal to 1 and r will be equal to 0. And now, the flip-flop will get set to the logic 1 value. And at the same time, you will see a transition in the output from logic low value to the logic high value.
Now, once this Q will become 1, the Q bar will be equal to logic 0. And as soon as this Q bar will become 0. this transistor will get OFF. And now this capacitor will again start charging through this resistor R1 and R2. So, in this way, this capacitor is charging between this 2x3 Vcc and 1x3 Vcc voltages. And because of the charging and discharging of this capacitor, we will see a transition in the output voltage.
So, during the charging process, whenever the voltage across the capacitor just crosses this 2 by 3 Vcc voltage, then there is a transition in the output from logic high value to the logic low value. And likewise, during the discharging process, whenever the voltage across the capacitor just go below this 1 by 3 Vcc voltage, then there is a transition from logic low value to the logic high value. And if we see the same sequence in a timing diagram. Then it will look like this. So, here this blue graph represents the charging and discharging of the capacitor.
While the yellow graph represents the output of the 555 timer. So, initially whenever the circuit is just turned on at that time, the output of the 555 timer will be high. And the capacitor will start charging towards the Vcc voltage.
Now as soon as the voltage across the capacitor just reaches this 2x3 Vcc voltage, then You will see a transition in the output from logic high value to the logic low value. And now the capacitor will start discharging towards the 0V. So now during the discharging process, as soon as it reaches this 1 by 3 Vcc voltage, then once again you will see a transition from logic low value to the logic high value.
And in this way, the capacitor is charging and discharging between these two voltages. So over here. T1 is the time for which the output of the timer is high. While T2 is the time for which the output of the timer remains the low value.
Or in other way, we can say that T1 is the time which capacitor requires to charge from 1 by 3 Vc voltage to the 2 by 3 Vc voltage. While T2 is the discharging time which is required by the capacitor to go from 1 to go from 2x3 Vcc voltage to the 1x3 Vcc voltage. So, if we represent this T1 and T2 mathematically, then mathematically it can be represented like this.
That means T1 will be equal to 0.693 times R1 plus R2 while T2 will be equal to 0.693 times R2. Now, one more thing if you observe over here, this T1 is greater than T2. Because, during the charging, the capacitor is charging through this resistor R1 and R2.
While during the discharging process, it is only discharging through this resistor R2. And that is why this T1 will be always more than T2. Or in other way, we can say that the duty cycle of this astable multivibrator will be always more than 50%. Now, over here the total time period t will be equal to the summation of this t1 and t2.
That is 0.693 times R1 plus 2R2. And the duty cycle can be represented as the on time divided by the total time period. That is t1 divided by t. Or we can say that, that is equal to 0.693 times R1 plus R2 divided by 0.693 times R1 plus 2R2.
That means the duty cycle will be equal to R1 plus R2 divided by R1 plus 2R2. So, from the expression you can see that for this particular circuit, the duty cycle will be always more than the 50%. And to get a 50% of duty cycle, the value of R1 should be equal to zero. So, in this particular circuit, whenever R1 is equal to zero, at that time we can achieve a 50% of duty cycle. Now, whenever this R1 is equal to zero, at that time the collector of this transistor will get directly exposed to the supply voltage.
And whenever this transistor is on, at that time it is possible that it will draw a more current. And because of that this internal transistor may get damaged. Now there is one more alternative using which we can also achieve the 50% of duty cycle. So, now in this circuit, the diode is connected between this pin number 7 and 2. So, using this circuit, it is possible to achieve a duty cycle of 50% or even less than the 50%.
So, now let's see the working of this circuit. So, now during the charging of this capacitor, this diode will become ON. And in the ON condition, the resistance offered by the diode will be less than this R2. So, during the charging time, this capacitor will charge through this resistor R1 and through this diode. So, the charging time T1 can be given as 0.693 times R1 plus Rd times C.
Where Rd is the forward resistance of this diode. And during the discharging time, this diode will remain OFF. So, this capacitor will get discharged through this resistor R2. So, the discharging time T2 can be given as 0.693 times R2 times C.
So, now if we adjust the value of R1 in such a way that this R1 plus Rd is equal to R2, then it is possible to achieve the 50% of duty cycle. Now what here? Suppose, if we assume this diode as an ideal diode, in that case, its forward resistance should be equal to zero. So, in that case, this T1 will be equal to 0.693 times R1 times C, while the T2 will be equal to 0.693 times R2 times C. And in that condition, the duty cycle will become R1 divided by R1 plus R2.
So, suppose if you want to achieve a 50% of duty cycle, then the value of R1 and R2 should be equal to same. And using this circuit, it is even possible to achieve either more or less than the 50% of duty cycle. So, if R1 is greater than R2, in that case the duty cycle will be more than the 50%. While whenever this R1 is less than R2, in that case the duty cycle will be less than 50%. So using this circuit, it is possible to achieve the desired duty cycle.
So now let's design one astable multivibrator whose frequency F is equal to 10 kHz and duty cycle is equal to 50%. So, now to get a duty cycle of 50%, the value of R1 and R2 should be equal to same. Let's say that is equal to R. And the total time period T will be equal to 0.693 times R1 plus R2 times C.
Now, over here as F is equal to 10 kHz, so the time period T will become 0.1 ms. Now, here let's assume the value of C as 0.1uF. So, we can write this expression as 0.1 ms that is equal to 0.693 times R1 plus R2 times 0.1uF. Or we can say that this R1 plus R2 will be equal to 0.1 ms divided by 0.693 times 0.1uF. Now, over here as this R1 and R2 are equal, so we can write it as 2R1.
And if we simplify it, then the value of R1 will roughly come around as 720 Ohm. So what we can do, we can use a potentiometer for this R1 and R2 and we can tune this potentiometer to the value of 720 Ohm. And in this way, it is possible to design a stable multivariate multivibrator which has a frequency F of 10 kHz and duty cycle of 50%. So now, let's derive the expressions that we have used in the design of this astable multivibrator. So, over here to find the time period T of this astable multivibrator, we need to find the value of this T1 and T2.
And like I said before, this T1 and T2 are the charging and discharging time of these capacitors. So first of all, let's find out the value of this T1. Now like we had seen in the earlier video of transient analysis, the capacitor voltage can be written as Vf plus Vin-Vf times e to the power of minus T divided by R'times C. Where here, R'is equal to R1 plus R2 because during the charging time, the capacitor is charging through the resistor R1 and R2.
Now initially, the voltage across the capacitor is equal to 1 by 3 Vcc voltage. So, in this expression, the V initial will be equal to Vcc by 3. And it is charging towards the supply voltage. So, Vf will be equal to the supply voltage.
So now, let's put these values in this expression. And over here, we want to find this time T1 whenever the voltage across the capacitor reaches this 2x3Vcc voltage. So, we can write this expression as Vc Will be equal to Vcc plus 1 by 3 Vcc minus Vcc times e to the power of minus T1 divided by R'c. And the value of this Vc of T1 is equal to 2 Vcc by 3. So, we can write this expression as 2 Vcc by 3 is equal to Vcc minus 2 Vcc by 3. times e to the power of minus T1 divided by R'c. And further if we simplify it, then we can write it as, Minus of Vcc by 3 is equal to minus 2 Vcc by 3 times e to the power of minus T1 divided by R'c.
can say that 1 by 2 is equal to e to the power of minus T1 divided by R'C. So, now if we take a natural log on both sides and if we simplify this expression, then the value of this time T1 will come out as natural log of 2 times R'C. Or we can say that this time T1 will be equal to 0.693 times T1.
R1 plus R2 times C. So, this is the time T1 which is required by the capacitor to charge from 1 by 3 Vcc voltage to the 2 by 3 Vcc voltage. So now, similarly let's find out this time T2. So once again, let's write down the expression of this capacitor voltage. That is Vc will be equal to Vf plus Vinitial minus Vf times e to the power of minus T divided by R2c.
Because during the discharging time, capacitor is discharging through this resistor R2. Now during this time, The initial voltage across the capacitor will be equal to 2x3 Vcc. So, we can say that the V initial will be equal to 2x3 Vcc.
And it is discharging towards the zero voltage. So, the final voltage across the capacitor should be equal to 0V. And we want to find the time T2 whenever the voltage across the capacitor just reaches this 1x3 Vcc voltage.
So, considering all this, we can write this As Vc will be equal to 0 plus 2 by 3 Vcc minus 0 times e to the power of minus T2 divided by R2 times C. And the value of this Vc is equal to 1 by 3 Vcc voltage. So we can write this expression as 1 by 3 Vcc is equal to 2 by Vcc times e to the power of minus T2 divided by R2 times C.
And if we further simplify it, then we can write it as 1 by 2 times e to the power of minus T2 divided by R2 times C. So, if we take the natural log on both sides and if we simplify it, then the value of T2 will come as R2 times C times natural log of 2. Or we can say that the value of T2 will be equal to 0.693 times R2 times C. So, this is the time required by the capacitor to discharge from this 2x3 Vcc voltage to the 1x3 Vcc voltage.
And the total time period T will be the summation of this time T1 and T2. That is equal to 0.693 times R1 plus 2R2 times C. So, using this expression, we can find the time period or the frequency of the astable multivibrator. And we can design a astable multivibrator of the desired frequency.
So, I hope in this video, you understood how we can design a astable multivibrator using this 555 timer IC. So, if you have any question or suggestion, do let me know in the comment section below. If you like this video, hit the like button and subscribe to the channel for more such videos.