√ 21 4 Home Work: Drive the form factor and ripple factor and efficiency of single phase half wave controlled rectifier. I lood
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- For bridge full wave rectifier with Vs = 5sin(wt), ripple factor = ? +Va O a. 1.59 Ob. 0.483 O C. 1,21 o d. 2.01Determine the voltage across the resistor, the current in the resistor, and the output voltage in figure 1 and figure 2. Plot these waveforms with proper ac cycle phase alignment. 1SR154-400 is a silicon rectifier 1 ampere diode. Indicate the peak values as per calculation.What condition should be followed to achieve discontinuous current mode operation in fully controlled rectifiers ? a. large L/R value b. large inductance value c. rated inductance value d. small L/R value
- Draw the waveform Vs , VL, ID ((each)) , is for single phase full wave rectifier with RL load for the following firing angle a) 0° b)45° c)90° d)135° f)180°A full wave bridge rectifier is supplied from 230V, 50Hz and uses a transformer of turns ration 15:1. It uses load resistance of 50 ohms. Calculate load voltage and ripple voltage. Assume ideal diode and transformer. Assume ripple factor equals to 0.482. a.Load Voltage= 13.8 V; Ripple Voltage=6.652 V b.Load Voltage= 15.6 V; Ripple Voltage=7.611 V c.Load Voltage= 21.3 V; Ripple Voltage=8.410 V d.Load Voltage= 25.1 V; Ripple Voltage=10.442 VDetermine the voltage across the diode, the output voltage, and the current in the resistor in the figure below. Plot these waveforms with proper ac cycle phase alignment. 1SR154-400 is a silicon rectifier 1 ampere diode. Indicate the peak values as per calculation.
- N H.W4 The diode in the single-phase half wave rectifier has a reverse recovery time of tr 150usec and the source voltage Vs = 200V at frequency = 5 kHz. Calculate the average output voltage. IC L.C. the HoNo locor3 Given the bridge-type full-wave rectifier circuit shown 220V 60Hz 220V:Es Es D1 D2 D3 D4 The load resistor is 220 and the transformer average current is 300mA. Determine: (a) Average value, RMS value, and frequency of load voltage VRL. (b) Average current and PIV of each diode. RLA single-phase center-tapped rectifier has an R-L of R= 20 0, L = 50mH, and a 240 V ac 50 Hz source voltage. Determine: i. The average and rms currents of the diodes and the load. ii. The rms and average 50 Hz source currents. iii. The power absorbed by the load. iv. The supply power factor.
- Create a Half-Wave Rectifier circuit. Use a resistor of 1000 Ohms. Input Voltage should be 95VAC - 65Hz Plot both the Input voltage and output voltage. Determine the Voltage drop across the diode by subtracting the output voltage to the input voltage peaks. Create a Bridge-Type Full Wave Rectifier Circuit. Repeat the steps done on the Half-Wave Rectifier circuit.Given the half-wave rectifier circuit shown 220V 60Hz 220V:Es Es D RL The load resistor is 2292 and it dissipates 2W. Determine: (a) Average value, RMS value, and frequency of load voltage VRL. (b) Average diode current and PIV of the diode.Maximum voltage applied at single phase full wave rectifier diode is 440 Volt. load is resistive find value of average output voltage and rms output voltage.
