I need the solution quickly with in 15 minutes Give reason(s) for the followings:a. Normally a transformer is found in the input supply of a rectifier.b. The use of center tapped full wave rectifier.c. It is preferable to apply a train of pulses instead of continuous signal on the gate of athyristor.d. The study of harmonics of rectifier input current.e. The efficiency of any power electronics converter is not 100%.f. IGBT or MOSFET are normally used in digitally controlled power electronicsconverters.g. In voltage control, the use of semiconductor control instead of variable resistancecontrol.h. It is not usual to the use of thyristors in de/ac converters.

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Answered: Give reason(s) for the followings: a.…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Discussion and calculations 1. Compare between the simulation and theoretical results for input and output voltages 2. What is delay angle control of three-phase half- wave converter? 3. When you design high power rectifier, what type of converter must be use? Why?
1. The full-wave rectifier circuit shown in following Figure is to deliver 0.2 A and 12 V (peak values) to a load. The ripple voltage is to be limited to 0.25 V. The input signal is 120 V (rms) at 60 Hz. Assume diode cut-in voltages of 0.7 V. (a) Determine the required turns ratio of the transformer. (b) Find the required value of the capacitor. (c) What is the PIV rating of the diode? vo D2
Create a rectifier circuit that provides an output of 100vdc by using the following circuit: a. Half Wave Rectifier In each circuit, calculate the following if the power is taken from a 220V line AC supply a. transformer peak voltage rating b. diode PIV ratings c. transformer turns ratio
Create this using LTSpice B. Simulation showing each required output/waveform.a. Output of transformerb. Output of rectifierc. Output of filterd. Zener diode or load output
Full Wave Rectifier Circuit The circuit of the full-wave rectifier can be constructed in two ways. The first method uses a centre tapped transformer and two diodes. This arrangement is known as a centre tapped full-wave rectifier. The second method uses a standard transformer with four diodes arranged as a bridge. This is known as a bridge rectifier. Advantages of Full Wave Rectifier The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The efficiency of half wave rectifiers is 40.6% while the rectification efficiency of full wave rectifiers is 81.2%. The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple factor in full wave rectifier is 0.482 while in half wave rectifier it is about 1.21. The output voltage and the output power obtained in full wave rectifiers are higher than that obtained using half wave rectifiers. Question: What is the main disadvantage of a conventional full-wave…
A load is connected in the Full Bridge rectifier circuit (Four Diodes). Usedtransformer primary voltage value is 220 V (rms). The transformer winding ratio is 1:10.Using the 0.7 V model of the diode,Load Rating: 7 K Ohmsa) How many % of conduction does each diode branch stay on?b) Find the average values of the load voltage.c) Find the average values of the load current.d) Find the effective value of the load voltage.e) Find the effective value of the load current.
A three phase full wave rectifier is shown below along with the peak input 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 Vpc=280.7V and. The RMS transformer secondary current is equal to 81.72A, the output RMS voltage is equal to VRMS=280.93V and the output resistance is equal to 2.8070. The TUF would be equal to: Secondary a平a字a辛 R 太D。太D。太D. Select one: 0.875 Ob. 0.955 O C. 0.551 O d. 0.432 Compared with power transistors, thyristors have the following characteristic
1.) Compare the DC output voltage of half-wave and full-wave rectifier circuit. 2.) Compare the ripples and DC output of half-wave and full-wave rectifiers without filter capacitors. 3.) Make a Conclusion for Bridge Rectifier.
The main advantage of a bridge rectifier using tha same input transformer as a full-ave rectifier is that _______________. a) İt will double the output voltage. b) less current is required fort he diodes to conduct. c) the ripple frequency is twice that of the full wave rectifier. d) the ripple frequency is on-half that of the full wave rectifier.
Facts Full Wave Rectifier Circuit The circuit of the full-wave rectifier can be constructed in two ways. The first method uses a centre tapped transformer and two diodes. This arrangement is known as a centre tapped full-wave rectifier. The second method uses a standard transformer with four diodes arranged as a bridge. This is known as a bridge rectifier. Advantages of Full Wave Rectifier • The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The efficiency of half wave rectifiers is 40.6% while the rectification efficiency of full wave rectifiers is 81.2%. • The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple factor in full wave rectifier is 0.482 while in half wave rectifier it is about 1.21. • The output voltage and the output power obtained in full wave rectifiers are higher than that obtained using half wave rectifiers. Question: What is the main disadvantage of a conventional full-wave…
A three-phase rectifier has a purely resistive load of R. If the rectifier delivers Idc = 50A at an out voltage of Vdc = 20V, determine a. The peak inverse voltage of each diode b. Peak current through a diode c. The value of R d. The power factor e. The VA rating of the transformer.
For a Full wave rectifier circuit, the AC voltage input to transformer primary is 115V. Transformer secondary voltage is 50V. Load resistor is 25 ohms. Determine Peak and Average DC Component of load voltage. (Assume ideal diodes) a.VDC (peak) = 25.4 V; VDC (average) = 18.5V b.VDC (peak) = 35.4 V; VDC (average) = 22.5V c.VDC (peak) = 38.4 V; VDC (average) = 23.5V d.VDC (peak) = 50 V; VDC (average) = 34.5V

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