COMPUTING

Chapter 7 Pulse Circuits and Waveform GeneratorsFor the discharging interval the circuit is as shown in figure 7.9.T 2V CCR AR BCv C () tFigure 7.9. Circuit for the discharging intervalFor this interval, = 0 V, V in = 2 ⁄ 3 V CC , and R eq = R B . Then, relation (7.1) becomesV ∞v C () t 0 ( 0 – 2 ⁄ 3 V CC )e t R B– – ⁄ C( 2⁄ 3 V CC )e t R B= =– ⁄ C(7.4)At t = T 2 , v C () t = 1⁄ 3 V CC and relation (7.4)becomes1⁄ 3 V CC ( 2 ⁄ 3 V CC )e T 2 R B=– ⁄ C( 2 ⁄ 3 )e – T 2 ⁄ R B C= 1 ⁄ 3e T 2 ⁄ R B C– = 1 ⁄ 2orThe period T– ⁄ R B C = ln( 1⁄2)= ln1 – ln2= 0 – ln2T 2T 2 = ln2⋅ ( R A + R B )C = 0.69R B Cis the summation of (7.3) and (7.5). Thus,T = T 1 + T 2 = 0.69( R A + R B )C + 0.69R B C = 0.69( R A + 2R B )C(7.5)(7.6)The right side of (7.6) cannot be negative; accordingly, the circuit of Figure 7.9 cannot achievethe condition T 2 > T 1 . Moreover, from (7.3) and (7.5) we observe that T 1 and T 2 cannot beequal. Therefore, with the circuit of Figure 7.9 T 1 will always be greater than T 2 and the dutycycle will always be greater than 0.5 of 50 percent. But we can achieve a duty cycle close to 50percent if we choose resistor to be much smaller than resistor .R A R B10 nFExample 7.1For the astable multivibrator of Figure 7.10 the capacitor has the value of . Determineappropriate values for the resistors and so that the circuit will produce a pulse repetitionR AR Bfrequency of 200 KHz with a duty cycle of 60%.7−6Electronic Devices and Amplifier Circuits with MATLAB Computing, Second EditionCopyright © Orchard Publications