19-EE-148 EDC Lab 4 - asdfg hjkl;\'asdfg hnjmk,l.;jhk cgncgb gsgsrfsf dvsvgfgd gdbdbb asdfg hjkl;\'asdfg PDF

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UNIVERSITY OF ENGINEERING AND TECHNOLOGY TAXILA ELECTRICAL ENGINEERING DEPARTMENT

LAB TASK # 04 For 3rd Semester Electrical Engineering Department (B.Sc. Electrical Engineering)

Submitted by: Muhammad Wasif Akhlaq 19-EE-148 (Section - D)

Submitted to: Prof. Hammad Shaukat

Task1: It is required to design an AC to DC rectifier. Following information is given: · Input source Vin (RMS) = 100 VRMS · Frequency of the source fin = 50Hz. · Load resistance RL = 100 Ω Following conditions must be satisfied: · Ripple voltage VR = 0.5V (it can vary from 0.45 to 0.55). · Suitable diodes should be selected, such that the Voltage/current/power rating of each diode must is not violated. · Select the diodes that meet the voltage/current/power requirement of the circuit, do not used over or under rated diodes & capacitor. Following results/outputs are required: · Calculate the value of the filter capacitor to achieve the required level of VR @ 100 Ω load. · Select suitable diode that can perform the desired task. · Justify your selection of a particular diode and why rejected other types. · Plot the graph for input VS output. 1. Show (using cursor) the peak value of the rectified output voltage 2. The value of the VR and compare confirm if it is close to 0.5V · Plot another graph that shows the diode current, diode reverse voltage and Load current. · Output frequency of the rectified waveform. Here we chose General Purpose Diode 1N4003 because it has an Vrms rating of 140V.

Using current = 983mA we calculate Filter capacitor value by C=(0.983)/2*50*0.5 Which gives us C=0.01966F or 19,666uF

0.5 V difference between min 136.1 and maximum 136.6

In the graph below we can see the reverse voltage of the Diodes, the diode current in the first cycle peaks due to Capacitor charging. The load current remains constant.

Task 2: Design a half wave rectifier, following information is provided: · Input source Vin (RMS) = 10 VRMS · Frequency of the source fin = 50 kHz. · Load resistance RL = 200 kΩ Following condition must be satisfied: · The output voltage must not go below -1.5V during the negative halfcycle. Following results/outputs are required: · Select suitable diode that can perform the desired task. · Justify your selection of a particular diode and why rejected other types. · Plot the graph of input vs output · Show using cursor the positive peak and the negative peak of the output waveform. Here we will choose a Switching diode 1N4148 as the current is low and the Frequency of the input AC current is high.

The negative peak is at -1.45V and the positive peak is at 13.61V

Task 3: It is required to design an AC to DC full wave rectifier. Following information is provided:

· Input source Vin (RMS) = 1 VRMS · Frequency of the source fin = 50kHz. · Load resistance RL = 100 Ω Following results/outputs are required: · Select suitable diode that can perform the desired task. · Justify your selection of a particular diode and why rejected other types. · Plot the graph of input VS full wave rectified output. (Output should be pulsating DC, no filter capacitor is required).

Here we will use A Schottky Diode because it is best suited for high frequency and low current environments. The circuit draws a maximum of 5.83mA for such use a Schottky Barrier Diode NSR05F40QNXT5G is used.

The Graph below shows us a pulsating output of which repeats between Maximum 906mV and Minimum 39uV.

Task 4: Design a voltage limiter circuit using Zener diode. Following information about the diode is given: · Input source peak voltage Vin(peak) = 10 Vp · Frequency of the source fin = 50 Hz.

Following condition must be satisfied: · Revers zener current should not exceed “IZM” for the selected Zener diode.

Following results/outputs are required: · Limit the output voltage at 5.0V. Plot the graph in Multisim to show your results similar to the plot shown below. · Calculate the value of current limiting resistor required in the circuit. · Show the current of Zener diode while it is limiting the output voltage Peak voltage of 10V Vp= Vrms *(2)^-2 Vrms = Vp/(2)^-2 Vrms= 7.07V...