A single-phase full-wave transistor rectifier power to motor load. The source feeds voltage is 230V, 50 Hz, eloadR = 22, L 10mH and E = 100V. For a firing angle of 30", the average value of output current in case conduction is stopped at 170° is (a) 2.8 A (c) 5.69 A %3D %3D %3D (b) 28.46 A (d) 56.92 A
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- A 60-Hz full-wave rectifier is built with a transformer having an rms secondary voltage of 20 V and filter capacitance C =150,000 μF. What is the largest current that can be supplied by the rectifier circuit if the ripple must be less than 0.3 V?A schematic design of a full-wave bridge type power supply is given below. Provide the appropriate values for the inductors, diodes, capacitor and resistors, such that the output DC voltage is 6.2 V and the max. output power for the load resistance is 20 mW (percentage error for output voltage and output power is 2%). AC voltage source: Amplitude=311.13, Freq=60, DC offset= 0. For the rectifier part, choose diodes with an appropriate PIV rating. Show all the formulas and computations involved in acquiring the values for the inductors, diodes, capacitor and resistors. In choosing your Zener diode, consider the output voltage and the output current. Note that the output current should be the minimum Zener current. For the value of the series resistor RS, choose a lower value for less ripple but always consider the required output voltage. The ripple voltage peak-to-peak value should be less than or equal to 1% of the required output voltage. Use the formula below for choosing the value…In the figure, SCR semiconductor switched rectifier to a 220 Vrms ac source at 50 Hz, at load Vdc = 100 V, R = 8 Ω and has an inductor large enough to supply continuous current. (a) Trigger angle α = 25o determine the total power absorbed by the load. (b) If the inductance value is 30 mH, the change value from the top Determine.
- Design a CT-FWR to supply a load of (50) with a waveform of the following specifications: - Vdc = 12 V Ripple factor = 0.1 % the main power supply is (220 Vrms, 50 Hz). Determine the following values: - 1- The value of capacitor filter. 2- The maximum load voltage (VmR) 3- The transformer turns ratio (a). 4- The RMS value of the load voltage. 5- Draw the output waveform. (assume ideal diodes)A step-down transformer supplies 25Vrms to a simple half wave rectifier power supply which is connected to a load resistance of 912ohm. The diode breakdown voltage, Vf is 0.7 V. Calculate the peak signal voltage received by the diodeDesign a full-wave bridge type power supply. Design the transformer, filter and regulator section. For the rectifier part, choose diode with appropriate PIV rating. Use standard values for your design. The output dc voltage should be 7.5 V and the Maximum output power be 20 mW. Use the topology below as reference
- A half-wave rectifier is needed to supply 15-V dc to a load that draws an average current of 250 mA. The peak-to-peak ripple is required to be 0.2 V or less. What is the minimum value allowed for the smoothing capacitance? If a full-wave rectifier is needed?The voltage waveforms shown below are obtained from a three-phase power rectifier used in an ESKOM application.The rectifier output before the inductive filter is denoted by VL and the load voltage after the inductive filter by Vo. Ignore any diode losses and consider a load resistance of 12.7 Ω. Analyse the waveforms and specify / calculate,11) the supply frequency;12) the rectifier pulse number;13) the rectifier average output voltage;14) the average output current;15) the r.m.s. value of the output current;16) the r.m.s. value of the harmonics present in the output current;17) the dc current rating of the rectifier diodes;18) the r.m.s. current rating of the rectifier diodes;19) the peak repetitive reverse voltage rating of the rectifier diodes;20) the peak repetitive forward current rating of the rectifier diodes;21) the line-line r.m.s. voltage of the three-phase supply;22) the rectification efficiency of the converter;23) the converter transformer utilisation factor;During output power measurement DC gives a deflection whereas AC gives properoutput. Why? Explain it. (Voltmeter, Ammeter and Isolator Connection of a Simple Half Wave Rectifier Circuit.)
- Design a circuit for an uncontrolled half-wave Rectifier based transformer step down containing a capacitor filter circuit, with writing the laws of the circuit and the shape of the output wave, with a comparison between the current output wave and the voltage output wave in terms of being affected by the placement of the filter.4- What is the effect of adding capacitor in parallel to the load at the rectifier output? Note: Please do not handwriting and write the answer briefly. What value of filter capacitor is required to produce a 1% ripple factor for a full-wave rectifier havinga load resistance of 1.5 kΩ? Assume the rectifier produces a peak output of 18 V.