〰️▷ Half-Wave Rectifier Calculator
Calculate DC output voltage, ripple frequency, and efficiency for half-wave rectifier circuits. Essential for power supply design.
⚡ Calculate Half-Wave Rectifier
RMS voltage of AC input
50Hz (Europe) or 60Hz (US)
0.7V for silicon diodes
Load resistor value
Smoothing capacitor (0 for no filter)
📊 Rectifier Output
Peak Voltage
16.3 V
DC Output (avg)
5.0 V
Ripple Frequency
50 Hz
Ripple Voltage
0.32 V
DC Current
5 mA
Efficiency
40.6%
📝 Detailed Calculation Steps
1
Peak Voltage (Vpeak)
Vpeak = √2 × Vrms − Vf
2
Ripple Frequency (fripple)
fripple = finput (same as input for
half-wave)
Without Capacitor Filter
3a
DC Output Voltage (No Filter)
VDC = Vpeak / π ≈ 0.318 ×
Vpeak
Without a filter capacitor, the DC output is the average of
the half-wave pulses.
4a
Ripple Voltage (No Filter)
Vripple = γ × VDC (γ = 1.21 for
half-wave)
Ripple factor (γ) = 1.21 is a constant for unfiltered
half-wave rectification.
5a
DC Current (No Filter)
IDC = VDC / Rload
With Capacitor Filter
3b
Ripple Voltage (With Filter)
Vripple = Iload / (f × C)
The capacitor charges to peak voltage and discharges through
the load between cycles.
4b
DC Output Voltage (With Filter)
VDC = Vpeak − (Vripple /
2)
The DC output is approximately the peak voltage minus half
the ripple.
5b
DC Current (With Filter)
IDC = VDC / Rload
6
Maximum Efficiency (η)
ηmax = 40.6% (theoretical maximum for
half-wave)
The half-wave rectifier utilizes only one half of the AC
cycle, resulting in lower efficiency compared to full-wave rectifiers.
Half-Wave Rectifier Circuit
The half-wave rectifier uses a single diode to convert AC to pulsating DC. Only the positive half of the AC cycle passes through, while the negative half is blocked.
📈 Input & Output Waveforms
The blue dashed line shows the AC input signal. The green solid line shows the rectified output - notice only the positive halves pass through.
📐 Half-Wave Rectifier Formulas
VDC = Vpeak/π ≈ 0.318×Vpeak
Average DC output voltage (without filter capacitor)
Peak Voltage
Vp = √2×Vrms - Vf
Ripple Frequency
fr = fin
Ripple Voltage
Vr = I / (f × C)
Efficiency
η = 40.6% (max)
Key Characteristics
- Simplest Design: Uses only one diode, making it the cheapest rectifier option
- Low Efficiency: Maximum theoretical efficiency is only 40.6%
- High Ripple: Output has significant AC ripple at the same frequency as input
- Poor Utilization: Only uses half of the AC cycle, wasting 50% of the input power
- Large Filter Required: Needs larger filter capacitor compared to full-wave rectifiers