# E3 Solution Essay

1653 Words7 Pages
1 SEE3263 09102 Question 1 a) Two factors • The ac line voltage is not constant. The so-called 115 volts ac can vary from about 105 volts ac to 125 volts ac. This means that the peak ac voltage to which the rectifier responds can vary from about 148 volts to 177 volts. The ac line voltage alone can be percent change in the dc output voltage. • The changingof the load resistance. If the load resistance decreases, the internal resistance of the power supply drops more voltage. This causes a decrease in the voltage across the load. responsible for nearly a 20 b) i. ii. R4gives enough current to Q2 so that it will operate in active region. Q2 acts as a comparator or feedback amplifier used to compare the voltage at R2 and reference voltage where any voltage error can be amplified to trigger the Q1. iii. By applying voltage divider, R R R R R R Vo = 5.2 + 0.7 9 = 5.9 R RA2 + 10k = 655.55m (RA2 + 13.3k ) (10k – 8.718k) = 655.55mRA2 –RA2 - 0.3444RA2 Therefore; RA2 = = 3.722k 2 SEE3263 09102 iv. v. Active region. From the figure shown, total resistance is RA= RA1 5k = RA1 + 3.722k RA1=1.278k Therefore, the current of RA is A + RA2 =R V = =0.493mA. The load current could be obtained; L =R V = =212.12mA By applying Kirchhoff Current Law (KCL); L= A+ o o =211.63mA Hence; =(211.63m)(10 ) =2.116V. =(211.63m)(100k ) =21.163kV. 3 SEE3263 09102 c) i) As a current detector. The value of R1 and R2 will determine the current value. ii) R4 = 0 The output formula could be obtained. Where; Vo=VREF = (1.2) =1.2V When R4 = 5k ; Vo=VREF = (1.2) =26.2V Therefore the range of output voltage is 1.2 V Vo iii) Vo = 26.2 V When RL= 1k ; Io = =26.2mA I1 = 0A When RL = 20 Io = =1.31A ; 26.2 V. R R + IADJ R2 +0 + IADJ R2 +0 4 SEE3263 09102 The total output current is; Io = I1 + IN Where; I1=Io-IN =1.31-233.33m =1.077A 5