Setup the Half Wave Rectifier using LiveWire Simulation software for the given below values. f=1HZ Vs=5V
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Setup the Half Wave Rectifier using LiveWire Simulation software for the given below values.
f=1HZ
Vs=5V
Step by step
Solved in 4 steps with 4 images
- How is a solid-state diode tested? Explain.For a full wave controlled rectifier with pure resistive load (R=100),the supply voltage is (V-50Sin314t). When the firing angle (a =0%) determine the followings: 1) The mean (dc) value of the load current. 2) The RMS value of the load voltage.A R-load half-wave rectifier used sinusoidal voltage source with a peak value (35.353kV). The firing angle is (30). The total resistance is 50k2 and frequency is 50HZ. Calculate (in details): load direct voltage, maximum direct voltage, normalized voltage and effective voltage, power delivered to the load, power factor, form factor, ripple factor?
- In the circuit in the figure, an AC voltmeter will be made with full wave rectifier circuit structure. R = 50ohm and diodesResistance values in the direction of transmission are Rd = 100ohm. Inside of the DC ammeter to be used as indicatorthe resistance is very, very small. The AC mark to be measured is Vs = 100 Sinwt Volts.a- Draw the shape of the current passing through the DC Ammeter and calculate its maximum value.b- Find the average value of the current passing through the DC Ammeter.c- Analyze the voltage seen at the ends of diode D1 for both alternans.d- Find the Rms value of the voltage seen at the ends of the diode D1.Calculate the rms load current for the single phase uncontrolled half-wave rectifier when the source voltage of 200V at a frequency of 50Hz, and load resistance is 10ohm.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.
- Q1 Figure Q1(a) is a full wave bridge circuit with a capacitor filter and the output waveform produced by the circuit is shown in Figure Q1(b). Assume all the diodes in the circuit have a forward voltage of 0.7 V. (a) Sketch the waveform in A, B, C and D as shown in Figure Q1(a).b. Calculate the following about the circuit in the figure (If silicon diodes are employed in the rectification);i. the peak value of the output voltage considering the drop across each diode, Vpk.ii. the average voltage, Vdc. iii. The current through the load resistor, IL.iv. The current diode, Id.v. The frequency of the output signal, Fout.vi. Calculate the efficiency of the full wave rectifier expressed in percentage. vii. Sketch a graph of the input and output voltage against time.Why is the ripple voltage inversely proportional to the Load Resistance and Capacitance in an rectifier circuit.
- A half-wave rectifier is to provide an averagevoltage of 50 V at its output.a. Draw a schematic diagram of the circuit.b. Sketch the output voltage waveshape.c. Determine the peak value of the output voltage. d. Sketch the input voltage waveshape.e. What is the rms voltage at the input?.Draw a full-wave rectifier circuit (using a diode bridge) and predict the output voltage signal when the input is a sinusoidal signal with amplitude 10 V (p p). Use the previous full wave rectifier to create a peak detector with a 2 V ripple.3. For the half-wave rectifier given in the figure below, find the average values ofthe output voltage and current, and the ripple factor of the output voltage.