Tutorial 11  Astables
We need to consider some definitions:
The mark time [t_{(H)}] is the time at which the output is a 1.
t_{H}= 0.7(R_{A} + R_{B})C
The space time [t_{(L)}] is the time at which the output is a 0.
t_{L} = 0.7 R_{B}C
The mark to space ratio = mark time ÷ space time.
Ratio = t_{H} ÷ t_{L}
The astable period T is the time taken for one complete cycle, the mark and the space times added together.
T = mark + space = t_{L} + t_{H}
The frequency = 1 ÷ period.
The time t_{H} will be longer than t_{L}, unless R_{1} is very small compared to R_{2}. If this is the case, then t_{H} will be approximately equal to t_{L}, but not quite equal. We can say to a first approximation that the mark to space ratio is 1. This will result in a square wave output.
The duty cycle = mark time ÷ period.
Worked Example What is the frequency of the square wave output from the circuit?
Use the formula

Answer From the diagram we can see that:
We need to substitute these values into the formula: f = ____1.4_____ = __________________ 1.4____________ (R_{1} + 2R_{2})C [4700 W + (2 × 2200 W)] × 2.0 × 10^{6 }F
= ____ 1.4_______ = 77 Hz 9100 × 2.0 × 10^{6} 
An astable 555 timer has the following external component values: R_{1} = 100 kilohms; R_{2} = 47 kilohms; C_{1} = 10 microfarads. What is the mark and space? 

Question 2 
An astable has a 2.2
F
capacitor, R_{1}
value of 10 k, and R_{2}
value 20 k. 
A NAND gate astable has a capacitor of capacitance 20 mF with a resistor of resistance 150 kW. What is the period of the astable? 