The IC 555 timer is connected for astable operation. Only one resistor is used as timing resistor in monostable multivibrator using 555 and here the timing resistor is split into two sections as Ra and Rb, with the discharge pin (pin 7) is connected to the junction of Ra and Rb. When the power supply is switched on, the timing capacitor C charges towards 2/3Vcc through Ra and Rb. When the capacitor voltage reaches 2/3 of Vcc, the upper comparator triggers the flip-flop and the capacitor starts to discharge towards ground through Rb. When the discharge reaches 1/3 of Vcc, the lower comparator is triggered and a new cycle is started.

The capacitor is charged and discharged periodically between 1/3 of Vcc and 2/3 of Vcc.

The output stage is High during the charging cycle for a time period t1, so

$$t_{1}=(R_{a}+R_{b})C text{ } ln{frac{V_{cc}-frac{1}{3}V_{cc}}{V_{cc}-frac{2}{3}V_{cc}}}$$

$$t_{1}=(R_{a}+R_{b})C text{ } ln{frac{3V_{cc}-V_{cc}}{3V_{cc}-{2}V_{cc}}}$$

$$t_{1}=(R_{a}+R_{b})C text{ } ln{frac{2V_{cc}}{V_{cc}}}$$

$$t_{1}=(R_{a}+R_{b})C text{ } ln{2}$$

$$therefore t_{1}=0.693(R_{a}+R_{b})C —- (1)$$

The output stage is low during the discharge cycle for a time period t2

$$therefore t_{2}=0.693(R_{b})C —- (2)$$

So, the total time period for charging and discharging is,

$$T=t_{1}+t_{2}$$

$$T=0.693(R_{a}+2R_{b})C$$

Output frequency is given as,

$$f=frac{1}{T}=frac{1.443}{(R_{a}+2R_{b})C}$$

**DUTY CYCLE**