INDEX

Now let's start analyzing Where to start is one of the problem. See the capacitor first. The voltage at Vc=0 or <1/3Vcc because if you keep a charged capacitor for sometime, it's charge will be lost because of its internal impedance and we normally use capacitors of 1uF,10uF as C in the circuit, so they will lose their charge in seconds and if some charge exist, it will be so minute compared to 1/3Vcc. So we start with Vc=0. Now what about Q,it can be 0 or 1, we can't say surely that at the starting Q=0 or Q=1. Now let's start with Vc=0 and Q=0. Then voltage at pin2 become 0 and the
output of comparatorII is HIGH(1) and voltage at pin6 is 0 so the output of comparatorI is LOW(0), thus S=1 and R=0 which makes Q=1 and Q'=0. If in the starting Q=1, then no difference will come because Q becomes HIGH again. Now let's see what the limits 1/3Vcc and 2/3Vcc. So we have 3
states Vc>2/3Vcc, 1/3Vcc<Vc<2/3Vcc,Vc<1/3Vcc. Now let's see what will happen 1/3Vcc<Vc<2/3Vcc, which is the next state. Pin2 become Vc>1/3vcc makes comparatorII to LOW and Pin6<2/3VCC make comparatorI to LOW(0). That is R=S=0, then output will be previous state,
previous state was Q=0,Q'=1 thus present state become Q=0,Q'=1. Now Vc starts increasing and Vc>2/3Vcc then PIN2=2/3Vcc and ComparatorII becomes LOW(0) and PIN6=2/3VCC makes comparator I to HIGH ie, R=1 and S=0 makes Q=0 and Q'=1. Till this time capacitor was charging
because capacitor has not got a way to discharge because a capacitor discharge requires a closed circuit. Now when Q'=1 this makes a discharge path for the capacitor because C one lead is to ground and other to resistor and resistor's one end to C and other end to ground which forms a closed
circuit, see the discharging circuit of a capacitor. Now Vc<2/3Vcc and >1/3Vcc because capacitor is discharging, so voltage Vc will decrease. At this state R=0,S=0 makes Q=0 and Q'=1(ie previous state). So capacitor still got a discharge path and continue to discharge till Vc<1/3Vcc. At Vc<1/3Vcc R=0 and S=1 makes Q=1 and Q=0, so the charging path of capacitor is lost and capacitor starts charge again. Now it charges to Vc>1/3Vcc , Vc<2/3Vcc and continue till Vc>2/3 Vcc and starts discharge till Vc<1/3Vcc. This cycle repeats again. i.e capacitor charges and discharges between 2/3Vcc and 1/3Vcc. Only at the starting capacitor starts from 0 to 2/3Vcc, remaining time it will be from 1/3Vcc to 2/3Vcc. Now analyze the charging path, it consist of Ra,Rb and C and discharging path has only Rb,C. During charging time Q=1 and Q=0 when capacitor discharges.

Thus charging time
charging time= time required to charge from 0 to 2/3vcc – ( time required to charge from 0 to 1/3vcc )

discharging time = time required to discharge up to 1/3Vcc from 2/3Vcc
charging equation V=Vcc(1-exp(-t/RC))
charge from 0 to 2/3vcc
2/3vcc = vcc(1-exp(-t/RC)) -> 1/3=exp(-t/RC) t=RC ln 3
charging from 0 to 1/3vcc
1/3vcc = vcc(1-exp(-t/RC)) -> 2/3=exp(-t/RC) t=RC (ln 3 – ln 2)
charging time,t1= charge from 0 to 2/3vcc - charge from 0 to 1/3vcc
=ln 2 RC
=.693 (Ra+Rb)C because capacitor charges through Ra and Rb
discharging time
1/3Vcc=2/3Vcc exp(-t2/RC)
t2=ln 2 RC = .693 RbC
Duty cycle = t1/(t1+t2) = (Ra+Rb)/(Ra+2Rb)
frequency = 1.44/(Ra+2Rb)C