COMPUTING

Astable Multivibrator with the 555 Timerand when it reaches the value v C = 1 ⁄ 3 V CC , the output of Comparator 2 goes High and Sets theSR flip−flop. The output Q goes High, Q goes Low, the transistor is turned OFF, the capacitorbegins to charge through the series combination of resistors R A and R B , and its voltage rises exponentiallywith time constant t r = ( R A + R B )C and when it reaches the value v C = 2⁄ 3 V CC itresets the flip−flop and the cycle is repeated.We are interested in the pulse repetition frequency f = 1⁄T and the duty cycle T 1 ⁄ T whereT = T 1 + T 2 and the desired values are dependent on appropriate values of resistors R A and R Band the capacitor C . As we know, the capacitor voltage as a function of time is given byv C () t V ∞ ( V ∞ – V in )e t R eq= –– ⁄ C(7.1)where is the final value and is the initial value of the voltage across the capacitor.V ∞V inFor the charging interval the circuit is as shown in figure 7.8.T 1V CCR AR BCv C () tFigure 7.8. Circuit for the charging intervalFor this interval, = V CC , V in = 1 ⁄ 3 V CC , and R eq = R A + R B . Then, relation (7.1) becomesV ∞– ⁄ ( + )C– ⁄ ( + )Cv C () t V CC ( V CC – 1⁄ 3 V CC )e t R A R–BV CC ( 2⁄ 3 V CC )e t R A R= = –B(7.2)At t = T 1 , v C () t = 2⁄ 3 V CC and relation (7.2)becomes⁄ ( + )C2 ⁄ 3 V CC V CC ( 2 ⁄ 3 V CC )e T 1 R A R= –B––⁄ ( + )C( 2 ⁄ 3 )e T 1 R A R B=1 ⁄ 3–e T 1 ⁄ ( R A + R B )C=1⁄2or– T 1 ⁄ ( R A + R B )C = ln( 1⁄2)= ln1 – ln2= 0 – ln2T 1 = ln2⋅ ( R A + R B )C = 0.69( R A + R B )C(7.3)Electronic Devices and Amplifier Circuits with MATLAB Computing, Second EditionCopyright © Orchard Publications7−5