30°, the average value of output current ir A single-phase full-wave transistor rectifien feeds power to motor load. The source voltage is 230V, 50 Hz, eloadR = 2N, L = 10mH and E = 100V. For a firing angle or 30, the average value of output current ir case conduction is stopped at 170° is (a) 2.8 A (c) 5.69 A (b) 28.46 A (d) 56.92 A
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- What is CEMF?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 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 diode
- 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.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…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)
- 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 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 referenceA 10 V Zener diode is used to regulate the voltage across a variable loadresistor. The input voltage Vs varies between 13 and16 V. The loadcurrent (??) varies between 10 and 85 mA. The minimum Zener current is15 mA. Findi. The maximum value of Rsii. The maximum power dissipated by the Zener diode, using the value ofRs.
- 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?Half-wave 60Hz sinusoidal uncontrolled rectifier circuit with a peak voltage value of 100V and a source inductance of 8mH is feeding a highly inductive load of 8A. The average value of the output voltage and power delivered are, respectively and what is the communication angle (u) (Power Electronics)Design rectifier circuits to provide an output of100 VDC using a.) HWR b.) Center-tappedFWR and c.) Bridge-Type FWR circuits. For each circuit, calculate the transformer peakvoltage rating, the diode PIV ratings, and thetransformer turns ratio, if power is taken fromthe 220V line ac supply