(a) Find the value of average charging current, Example A dc battery of constant emf E is being charged through a resistor using half- wave diode rectifier. For source voltage of 230 V, 50 Hz and for R =8Ω, E =150 V, (a) Find the value of average charging current, (b) Find the power supplied to battery and that dissipated in the resistor, (c) Calculate the supply pf, (d) Find the charging time in case battery capacity is 1000 Wh and (e) Find rectifier efficiency and PIV of the diode. Dr. Oday A. Ahmed
the value of average charging current Dr. Oday A. Ahmed
the power supplied to battery the power dissipated in the resistor Dr. Oday A. Ahmed
Find the charging time in case battery capacity is 1000 Wh Calculate the supply pf, Find the charging time in case battery capacity is 1000 Wh (Power delivered to battery) ×(charging time in hours)= Battery capacity in Wh Hence, charging time is Dr. Oday A. Ahmed
𝑃𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 rectifier efficiency 𝑃𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 PIV of the diode What is the PIV of the single-phase diode Rectifier Supply R-Load Dr. Oday A. Ahmed
A single-phase 230V: 1kW heater is connected across I-phase 230V, 50Hz supply through an SCR. For firing angle delays of 45° and 90°, calculate the power absorbed in the heater element. 230V, 1kW Heater Dr. Oday A. Ahmed
For α=450, RMS output voltage: Solution Heater resistance: For α=450, RMS output voltage: 𝑉 𝑜𝑟 = 2 .230 2 𝜋 𝜋− 𝜋 4 + 1 2 𝑠𝑖𝑛 90 0 =155.071 VPower absorbed by heater element for α=450 Dr. Oday A. Ahmed
For α=900, RMS output voltage: Power absorbed by heater element for α=450 𝑉 𝑜𝑟 2 𝑅 = 155.071 230 2 ×1000=454.57 𝑤𝑎𝑡𝑡 For α=900, RMS output voltage: 𝑉 𝑜𝑟 = 2 .230 2 𝜋 𝜋− 𝜋 2 + 1 2 𝑠𝑖𝑛 180 0 =115 V Power absorbed by heater element for α=900 𝑉 𝑜𝑟 2 𝑅 = 115 230 2 ×1000=250 𝑤𝑎𝑡𝑡 Dr. Oday A. Ahmed
In a single-phase full-wave diode bridge rectifier, the diodes have a reverse recovery time of 40 µs. For an AC input voltage of 230 V, determine the effect of reverse recovery time on the average output voltage for a supply frequency of (a) 50 Hz and (b) 2.5 kHz. Solution D 1 and D2 will not be off at ωt =π but will continue to conduct until t=( π/ω )+trr Dr. Oday A. Ahmed
With zero reverse recovery time, average output voltage, Dr. Oday A. Ahmed
For f= 50 Hz and trr = 40 µs, the reduction in the average output voltage, Percentage reduction in average outpace voltage Dr. Oday A. Ahmed
For f= 2.5k Hz, the reduction in the average output voltage, Percentage reduction in average output voltage Dr. Oday A. Ahmed