Half-Wave Rectifier with Filter Capacitor

Calculate ripple voltage and charging voltage of half-wave rectifier

Half-Wave Rectification with Capacitor

Capacitor Smoothing

The filter capacitor reduces ripple voltage and stabilizes the output voltage. Internal resistance, diode forward voltage and frequency are considered.

Input Voltage

Diode Voltage U_D

Filter Capacitor C_L

µF

Load Resistance R_L

Internal Resistance R_i

Frequency f

Decimal Places

Results

Input Values

Input U_RMS:

Input U_Peak:

Output Voltages

No-Load Voltage:

Voltage Under Load:

Ripple Voltage (V_PP):

Additional Values

Maximum Voltage:

Minimum Voltage:

Load Current:

Diode Reverse Voltage:

Voltage Diagram

Voltage Comparison with Ripple Voltage

The right blue bar shows the minimum output voltage. The red area corresponds to the ripple voltage (V_PP). The total height corresponds to the maximum output voltage.

Circuit Type

Half-wave rectifier with filter capacitor

Capacitor charges during positive half-wave
Discharge through load resistance during blocking phase
Significantly reduced ripple voltage

Circuit diagram: Half-wave rectifier with filter capacitor

Practical Calculation Example

Example: 12V Power Supply with Half-Wave Rectifier

Given: Transformer secondary voltage U_RMS = 15V, Filter capacitor C = 1000µF, Load resistance R_L = 100Ω, Internal resistance R_i = 2Ω, Frequency f = 50Hz

Step 1: Calculate no-load voltage

Peak voltage of secondary winding:

\[U_{Peak} = \sqrt{2} \times 15V = 21.21V\]

No-load voltage (minus diode forward voltage):

\[U_{No-load} = 21.21V - 0.7V = 20.51V\]

Step 2: Output voltage under load

Voltage drop due to internal resistance:

\[U_{Load} = U_{No-load} \times \left(1 - \sqrt{\frac{R_i}{R_L}}\right) = 20.51V \times \left(1 - \sqrt{\frac{2Ω}{100Ω}}\right)\] \[U_{Load} = 20.51V \times (1 - 0.141) = 20.51V \times 0.859 = 17.62V\]

Step 3: Calculate ripple voltage

Ripple voltage (peak-to-peak):

\[U_{RipplePP} = \frac{U_{Load}}{C \times R_L \times f} \times \left(1 - \sqrt[4]{\frac{R_i}{R_L}}\right)\] \[U_{RipplePP} = \frac{17.62V}{1000 \times 10^{-6}F \times 100Ω \times 50Hz} \times (1 - 0.377)\] \[U_{RipplePP} = \frac{17.62V}{5} \times 0.623 = 2.20V\]

Step 4: Min/Max voltages

Maximum voltage: \[U_{max} = U_{Load} + \frac{U_{RipplePP}}{2} = 17.62V + 1.10V = 18.72V\]

Minimum voltage: \[U_{min} = U_{Load} - \frac{U_{RipplePP}}{2} = 17.62V - 1.10V = 16.52V\]

Step 5: Load current and reverse voltage

Load current: \[I_{Load} = \frac{U_{Load}}{R_L} = \frac{17.62V}{100Ω} = 176.2mA\]

Diode reverse voltage: \[U_{Reverse} = 2\sqrt{2} \times U_{RMS} = 2.83 \times 15V = 42.4V\]

Result: The 12V power supply delivers an output voltage of 17.6V ± 1.1V with 2.2V ripple voltage. For a stable 12V power supply, a voltage regulator would be required.

Capacitor Dimensioning

Rules of thumb:

• 1000µF per 1A load current at 50Hz

• 2000µF per 1A load current at 100Hz

• Larger capacitor → less ripple voltage

• Voltage rating > 1.5 × U_Peak

Practical Notes

• Consider inrush current (Ri or series resistor)

• Electrolytic capacitor: observe polarity

• Diode must handle peak currents

• Reverse voltage of diode > 2.83 × U_RMS

Theory of Half-Wave Rectifier with Capacitor

Operating Principle

The half-wave rectifier with filter capacitor significantly improves the smoothing of the output voltage. During the positive half-wave, the diode conducts and the capacitor charges to the peak voltage. During the negative half-wave, the capacitor discharges through the load resistance.

Detailed Operation

Charging phase: Diode conducts, capacitor charges quickly
Discharging phase: Diode blocks, capacitor discharges slowly through R_L
Smoothing: Capacitor bridges the blocking time of the diode
Ripple voltage: Remains due to incomplete smoothing

Mathematical Relationships

No-load voltage:

\[U_{No-load} = \sqrt{2} \cdot U_{RMS} - U_D\]

Output voltage under load:

\[U_{Load} = U_{No-load} \left(1 - \sqrt{\frac{R_i}{R_L}}\right)\]

Ripple voltage:

\[U_{RipplePP} = \frac{U_{Load}}{C \cdot R_L \cdot f} \left(1 - \sqrt[4]{\frac{R_i}{R_L}}\right)\]

Reverse voltage:

\[U_{Reverse} = 2\sqrt{2} \cdot U_{RMS}\]

Component Influence

Capacitor C: Larger capacitor → lower ripple voltage
Load resistance R_L: Higher resistance → lower ripple voltage
Internal resistance R_i: Higher resistance → lower output voltage
Frequency f: Higher frequency → lower ripple voltage

Disadvantages

High inrush current
Poor transformer utilization
Voltage dependency on load
Remaining ripple voltage
High diode reverse voltage required

Advantages

Significantly reduced ripple voltage
Higher output voltage
Simple construction
Cost-effective solution
Proven technology

Typical Applications

Simple DC power supplies
Battery chargers
Amplifier power supply
LED lighting
DC motors

Voltage Waveform Under Load

Voltage waveform shows charging during positive half-wave and discharge through load resistance

Symbol Directory

U_RMS	RMS value of input voltage [V]
U_No-load	No-load voltage without load resistance [V]
U_Load	Output voltage at load resistance [V]
U_RipplePP	Ripple voltage (peak-to-peak) [V]
U_D	Diode forward voltage (typ. 0.7V) [V]
R_i	Internal resistance of current source [Ω]
R_L	Load resistance [Ω]
C_L	Filter capacitor [F]
f	Mains frequency [Hz]
U_Reverse	Maximum reverse voltage at diode [V]

Other electronics functions

LED resistor • Zener diode resistor (variable) • Zener diode resistor (fix) • Half-Wave rectification • Half-Wave rectification with capacitor • Full-Wave rectification • Full-Wave rectification with capacitor •