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In EET 109 Midterm Problems 7 and 8, calculations for voltage trip points are detailed. The scenarios involve setting Vout max to Vout min equal to the power supply rail, diode characteristics, and current flow considerations. Find explanations and results for each step.
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Problems 7 and 8 Assume Vout,max = Vout, min = power supply rail 7. AV (V+ - V-) = Vout Vout = 10000 * (6 – 5) = 50 kV > 10 V = Vmax Vout,max = 10V Diode off 8. V- = 12 [3.3k/(2.2k + 3.3k)] = 7.2V Vout = ILED(R) + VLED = 10mA (200W) + 2V = 4V Av (V+ - V-) = Vout = 10000 (Vin – 7.2V) = 4V Vin = 7.2004 V In practice, setting Vin to this level is impossible. Vin will be set to drive Vout to the rail Iout = (15-2)/200 = 6.5mA EET 109 Midterm
Problem 15c,d Assume Vout,max = Vout,min = power supply rail 15c) LTP = -12 [22k / (3.3k + 2.2k)] = -4.8 UTP = +12[22k / (3.3k + 2.2k)] = +4.8 15d) Each trip point is Voffset + VR1. UTP = 1 + [(12 – 1)/(3.3k + 2.2k)](2.2k) = 5.4 VR1 LTP = 1+ [(-12 -1)/(3.3k + 2.2k)](2.2k) = -4.2 EET 109 Midterm