Electronic Devices and Amplifier Circuits

7.9 Solutions to End−of−Chapter Exercises1.v ( V)5T 1T0 1 2T 23 4Solutions to End−of−Chapter Exercisest ( μs)The period is T = T 1 + T 2 = 3 + 1 = 4 μsand the duty cycle isFrom relation (7.6)T 1⁄ T = 3⁄ 4 = 0.75 = 75 %T = 4× 10 – 6= 0.69( R A + 2R B )C = 0.69× 10 – 9( R A + 2R B )R A+ 2R B4 × 10 – 6= -------------------------- =0.69 × 10 5.77 KΩ– 9(1)The duty cycle isT 1⁄ Tand from relations (7.3) and (7.6) we obtainTDuty cycle 0.75 1 0.69( R-----A + R B )C R------------------------------------------ A + R= = = = ---------------------- B=T 0.69( R A + 2R B )C R A + 2R BR A+ R--------------------- B5.77 KΩR A+ = 0.75× 5.77 = 4.33 KΩR B(2)Subtraction of (2) from (1) yieldsand from (2)R B = 5.77 – 4.33 = 3.44 KΩR A = 4.33 – 3.44 = 890 Ω2.v ( V)5TT 1 T 20 1 23 4 5t ( μs)The period is T = T 1 + T 2 = 1 + 3 = 4 μs . The duty cycle is T 1 ⁄ T = 1⁄ 4 = 0.25 = 25 % . Toachieve a duty cycle at 25 % , we must make T 2 = 3T 1 and this condition will be satisfied if wemultiply relation (7.11) by 3 and equate it to relation (7.14) subject to the constraint of (7.16).Then,Electronic Devices and Amplifier Circuits with MATLAB® / Simulink® / SimElectronics® Examples, Third EditionCopyright © Orchard Publications7−39