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Transcript
Example
A dc battery of constant emf E is being charged through a resistor using halfwave diode rectifier. For source voltage of 230 V, 50 Hz and for R =8Ξ©, 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
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
rectifier efficiency
π‘ƒπ‘œπ‘€π‘’π‘Ÿ π‘‘π‘’π‘™π‘–π‘£π‘’π‘Ÿπ‘’π‘‘ π‘‘π‘œ π‘π‘Žπ‘‘π‘‘π‘’π‘Ÿπ‘¦
π‘‡π‘œπ‘‘π‘Žπ‘™ 𝑖𝑛𝑝𝑒𝑑 π‘π‘œπ‘€π‘’π‘Ÿ
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
Solution
Heater resistance:
For Ξ±=450, RMS output voltage:
π‘‰π‘œπ‘Ÿ =
2.230
2 πœ‹
πœ‹βˆ’
πœ‹
4
1
2
+ 𝑠𝑖𝑛900 = 155.071 V
absorbed by heater element for Ξ±=450
Dr. Oday A. Ahmed
Power
Power absorbed by heater element for Ξ±=450
2
π‘‰π‘œπ‘Ÿ
155.071
=
𝑅
230
2
× 1000 = 454.57 π‘€π‘Žπ‘‘π‘‘
For Ξ±=900, RMS output voltage:
2. 230
π‘‰π‘œπ‘Ÿ =
2 πœ‹
πœ‹
1
πœ‹βˆ’
+ 𝑠𝑖𝑛1800 = 115 V
2
2
Power absorbed by heater element for Ξ±=900
2
π‘‰π‘œπ‘Ÿ
115
=
𝑅
230
2
× 1000 = 250 π‘€π‘Žπ‘‘π‘‘
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 Ο‰t =Ο€ but will continue to conduct until t=( Ο€/Ο‰ )+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
(a) For f= 2.5k Hz, the reduction in the average output voltage,
Percentage reduction in average output voltage
Dr. Oday A. Ahmed