ECD2601
Portfolio
Examination
Due 7 October 2025
, Question 1: Half-Wave Rectifier Design and Redesign
1.1 Half-Wave Rectifier Using EasyEDA
Answer: The half-wave rectifier was designed using EasyEDA by connecting a sinusoidal
AC input source to a diode and a resistive load. The circuit components included:
• Input: 5 V peak, 1 kHz AC source
• Diode: 1N4148 (standard EasyEDA diode)
• Load: RL = 1 kΩ
In operation, the diode conducts during the positive half-cycles, producing a positive half-
sine output across the resistor. During negative half-cycles, the diode blocks current, and
the output remains near zero.
Waveform: The simulated rectified waveform shows the clipped negative half-cycles.
The average DC component obtained is approximately 1.578 V.
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.
1
Portfolio
Examination
Due 7 October 2025
, Question 1: Half-Wave Rectifier Design and Redesign
1.1 Half-Wave Rectifier Using EasyEDA
Answer: The half-wave rectifier was designed using EasyEDA by connecting a sinusoidal
AC input source to a diode and a resistive load. The circuit components included:
• Input: 5 V peak, 1 kHz AC source
• Diode: 1N4148 (standard EasyEDA diode)
• Load: RL = 1 kΩ
In operation, the diode conducts during the positive half-cycles, producing a positive half-
sine output across the resistor. During negative half-cycles, the diode blocks current, and
the output remains near zero.
Waveform: The simulated rectified waveform shows the clipped negative half-cycles.
The average DC component obtained is approximately 1.578 V.
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.
1
