Determine the peak output voltage for the bridge rectifier. Assuming ideal model, what PIV rating is required for the diodes? The transformer is specified to have a 12 VRMS secondary voltage for the standard 110V across the primary. Vout 10KΩ ANS. PIV = 16.97 V ll
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- 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 VA single-phase full-wave bridge rectifier circuit is fed from a 220 V, 50 Hz supply. It consists of four diodes, a load resistance 20 Q and a very large inductance so that the load current is constant. What is the average or dc output voltage?A full-wave rectifier uses 2 diodes. The internal resistance of each diode is 20 Q. The transformer RMS secondary voltage from centre tap to each end of the secondary is 50 V and the load resistance is 980 Q. Mean load current will be
- A certain unfiltered center-tapped full wave rectifier is powered by a 120 Vrms, 60 Hz power system. The peak value of the output voltage under loaded conditions is 30 V. The capacitance value is 2000 uF, load current of 2A, determine the following: 4. DC load voltage Ripple factor C.Setup the Half Wave Rectifier using LiveWire Simulation software for the given below values.f=1HZVs=5V3 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. RL
- In the circuit shown below. Determine the following a. Vs (at the secondary) b. Vout (across RL) c. Vrip (ripple votage) d. VDC e. PIV (Peak Inverse Voltage) 10:1 Output 115 V mis 60 Hz RL 2.2 k) D, 50μF- All diodes are IN4001. Tund AlMannai EENG261 Page 5/11Problem # 4 In the circuit shown below. Determine the following a. Vs (at the secondary) b. Vout (across RL) c. Vrip (ripple votage) d. VpC e. PIV (Peak Inverse Voltage) 10:1 Dy 1ISV ms Ouiput RL 2.2 k2 D, 50 pF All diodes are IN4001. alleeeWhen the rms output voltage of a full-wave bridge rectifier is 26.20 V, the peak inverse voltage (PIV, in V) across the diodes (neglecting the diode drop)
- Q5. For the following center tapped transformer, show the waveform across each half of the secondary winding and across Ri, when a 100 V peak sine wave is applied to the primarywinding. What is the PIV rating must the diode have? (Use constant voltage drop model for the Silicon diode) IN4001 Ry OV 10k) D IN40011) AC power in a load can be controlled by using a. Two SCR's in parallel opposition b. Two SCR's in series c. Three SCR's in series d. Four SCR's in series 2) The advantage of using free - wheeling diode in half controlled bridge converter is that a. There is always a path for the ac current independent of the ac line b. There is always a path for the dc current independent of the ac line c. There is always a path for the dc current dependent of the ac line d. There is always a path for the ac current dependent of the ac line 3) Silicon controlled rectifier can be turned on a. By applying a gate pulse and turned off only when current becomes zero b. And turned off by applying gate pulse c. By applying a gate pulse and turned off by removing the gate pulse d. By making current non zero and turned off by making current zeroA three phase full wave rectifier is shown below along with peak phase voltage Vm-169.7V. The load is purely resistive. The rectifier delivers Ipc = 100 A and the source frequency is 60 Hz. The DC output voltage is %3D VDc=280.7V and the output RMS voltage is equal to VRMS=280.93V. The efficiency, FF and RF are respectively equal to: Secondary D D D, R AD. Z D. D, Select one: O a. 99.83%, 100.08%, 4% O b. 99.83%, 55.08%, 4% O C. 99.83%, 100.08%, 16% O d. 87.83%, 100.08%, 4%