Question

using the circuit above as refernce solve problem 3

. Construct Figure 2.3, using VCC-5 V, RI -6.8k2, R2 3.3k, C10.luF, and C2 -0.0luF. R1 6.sk R2 3.3k THRES OUT SV x1 C2 0.01u o.1u Figure 2.3 2. Enter your results below. Duty Cycle Turn ON Time ToN Total Time Period | Turn OFF Time Quantity Measured Theoretical Value Design a circuit for a duty cycle 75%, and a frequency of 1 KHz. Show all calculations and results. 3. 4.

Answer 1

f = 1 khZ or Time period, T = 1/f = 1 ms Duty cycly = T_ON/T = 75% or T_ON = 0.75 times T T_ON = 0.75 times 1 ms = 0.75 ms T_OFF = T - T_0 N = 1ms - 0.75 ms = 0.25 ms T_ON = 0.69 (R_1 + R_2) C_1 & T_OFF = 0.69 T_z C_L Assume C_1 = 0.1 mu F Since T_OFF = 0.69 R_2 C_1 = 0.25 ms R_2 = 0.25 ms/0.69 times 0.1 mu F = 3.62 K Ohm Also T_ON = 0.69 (R_1 + R_2) C_1 = 0.75 ms R_1 = 0.75 ms - (0.69 times 3.62 k Ohm times 0.1 mu F)/0.69 times 0.1 mu F R_1 = 0.50 ms/0.69 times .1 mu F = 7.25 k Ohm Hence R_1 = 7.25 k Ohm, R_2 = 3.62 k Ohm & C_1 = 0.1 mu F Design is verified on MultiSim

Design is verified on MultiSim