ECD2601
PORTFOLIO
Examination
DUE: 7 October 2025
, PORTFOLIO EXAMINATION
07 OCTOBER 2025
SEMESTER COURSE
ENGINEERING GRAPHICS FOR ELECTRICAL DESIGN PROJECTS
Question 1: Rectifier Design (20 Marks)
1.1 Half-Wave Rectifier Design and Simulation (10 Marks)
Design and Explanation: A Half-wave rectifier is the simplest form of rectifier that
converts an AC voltage to a pulsating DC voltage. It uses a single diode to allow only
one half-cycle (either positive or negative) of the AC input voltage to pass through to the
load.
Components for an EasyEDA design:
• AC Voltage Source: To provide the input signal (e.g., a sin(𝑡) source with a
peak voltage 𝑉𝑝 ).
• Diode (D1): A standard signal or power diode (e.g., 1N4001).
• Load Resistor (𝑅𝐿 ): To represent the load (e.g., 1kΩ).
• Ground: Reference point.
Design Schematic: In EasyEDA, you would connect the components in series: AC
Source → Diode (anode to source, cathode to load) → Load Resistor → Ground.
Simulation Waveform: The output voltage (𝑉𝑜𝑢𝑡 ) across the load resistor will be:
• Positive half-cycle: 𝑉𝑜𝑢𝑡 ≈ 𝑉𝑖𝑛 − 0.7V (assuming a silicon diode) - a pulsating
positive voltage.
• Negative half-cycle: 𝑉𝑜𝑢𝑡 ≈ 0V (the diode is reverse-biased and acts as an open
circuit).
PORTFOLIO
Examination
DUE: 7 October 2025
, PORTFOLIO EXAMINATION
07 OCTOBER 2025
SEMESTER COURSE
ENGINEERING GRAPHICS FOR ELECTRICAL DESIGN PROJECTS
Question 1: Rectifier Design (20 Marks)
1.1 Half-Wave Rectifier Design and Simulation (10 Marks)
Design and Explanation: A Half-wave rectifier is the simplest form of rectifier that
converts an AC voltage to a pulsating DC voltage. It uses a single diode to allow only
one half-cycle (either positive or negative) of the AC input voltage to pass through to the
load.
Components for an EasyEDA design:
• AC Voltage Source: To provide the input signal (e.g., a sin(𝑡) source with a
peak voltage 𝑉𝑝 ).
• Diode (D1): A standard signal or power diode (e.g., 1N4001).
• Load Resistor (𝑅𝐿 ): To represent the load (e.g., 1kΩ).
• Ground: Reference point.
Design Schematic: In EasyEDA, you would connect the components in series: AC
Source → Diode (anode to source, cathode to load) → Load Resistor → Ground.
Simulation Waveform: The output voltage (𝑉𝑜𝑢𝑡 ) across the load resistor will be:
• Positive half-cycle: 𝑉𝑜𝑢𝑡 ≈ 𝑉𝑖𝑛 − 0.7V (assuming a silicon diode) - a pulsating
positive voltage.
• Negative half-cycle: 𝑉𝑜𝑢𝑡 ≈ 0V (the diode is reverse-biased and acts as an open
circuit).