ECD2601 Portfolio Examination 2025 (Exceptionally Crafted) Due on 07 October 2025

ECD2601
Portfolio
Examination
Due 07 October 2025

, Question 1

1.1 Design of a Half-Wave Rectifier in EasyEDA

The circuit consists of an AC source of 10 V peak at 50 Hz connected to a diode (1N4001)
and a load resistor of 10 kΩ. The diode allows conduction during positive half cycles,
producing an output that follows the input minus the diode drop (0.7 V) when forward
biased, and zero during negative cycles. This behavior results in a clipped sinusoidal
output known as a half-wave rectified signal.
Schematic in EasyEDA:

• Source: VAC, 10 V amplitude, 50 Hz frequency.

• Diode: 1N4001 or 1N4148 (cathode connected to the load resistor).

• Load resistor: 10 kΩ connected between diode cathode and ground.

Simulation Procedure: Run a transient analysis in EasyEDA for 0–100 ms. The
expected output is the positive half sine cycles reduced by the diode drop.

Vout
D1
Vin RL




Figure 1: Half-wave rectifier circuit using a diode and resistive load.


V (t) Input Vin
Output Vout
t




Figure 2: Waveforms for Q1.1: Input sine and half-wave rectified output.




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