Calculate peak or DC voltage in a half-wave rectifier from the other value using Vdc = Vpeak/π or Vpeak = Vdc×π, with results in V, mV, or kV.

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Ideal diode, unfiltered sinusoidal input and resistive load. Voltages are nonnegative magnitudes.

Result unit

Half Wave Rectifier Voltage Formula

The calculator uses the ideal average DC output voltage formula for a half wave rectifier with a sinusoidal input, ideal diode, resistive load, and no filter capacitor. Voltages are nonnegative magnitudes; zero gives zero output.

Vdc = Vₚₑₐₖ / π
Vₚₑₐₖ = Vdc × π
  • Vdc = average DC output voltage of the half wave rectifier
  • Vpeak = peak value of the input AC voltage
  • π = pi, approximately 3.1416

Select average DC voltage or required peak voltage, then enter the known voltage. The calculator divides peak voltage by π or multiplies average DC voltage by π. Input and result units support millivolts, volts, and kilovolts.

Half Wave Rectifier Voltage Reference

Peak Voltage Average DC Voltage Calculation
5 V 1.5915 V 5 ÷ π
12 V 3.8197 V 12 ÷ π
24 V 7.6394 V 24 ÷ π
100 V 31.8310 V 100 ÷ π

The values below show the unit conversions used before the voltage formula is applied.

Unit Meaning Value in Volts
mV millivolt 0.001 V
V volt 1 V
kV kilovolt 1000 V

Example Problems

Example 1: Find DC voltage from peak voltage

You have a half wave rectifier with a peak voltage of 12 V.

Vdc = 12 / π = 3.8197 V

The average DC voltage is approximately 3.8197 V.

Example 2: Find peak voltage from DC voltage

You need an average DC voltage of 10 V from an ideal half wave rectifier.

Vₚₑₐₖ = 10 × π = 31.4159 V

The required peak voltage is approximately 31.4159 V.

FAQ

Is this the same as RMS voltage?

No. This calculator uses average DC voltage for a half wave rectified sine wave. RMS voltage is different. For an ideal half wave rectifier, the RMS output voltage is Vpeak ÷ 2, while the average DC voltage is Vpeak ÷ π.

Does the formula include diode voltage drop?

No. The formula is for an ideal diode. In a real circuit, the diode forward voltage drop reduces the output voltage. A silicon diode often drops about 0.7 V when conducting, but the exact value depends on the diode and current.

Why is the DC voltage lower than the peak voltage?

A half wave rectifier only passes one half of the AC sine wave and blocks the other half. Since the output is zero for half of each cycle, the average DC value is much lower than the peak value.