Rectifier

A rectifier converts an alternating input into a unidirectional output. Diode orientation selects which direction of electric current is transmitted.^[ref]

An electrical circuit that uses diodes and capacitors to produce an output electric potential difference larger in magnitude than the input peak. The attainable output depends on the circuit arrangement, electric current drawn by the load, capacitance and input frequency; an ideal unloaded result does not specify the output under load.

Half-wave rectifier ( $v_{L} (t) = max (0, v_{S} (t))$ )

 0  0

A half-wave rectifier with one diode and a resistor load passes one half of each input cycle.^[ref]

For an original ideal example, let an ideal voltage source supply the sinusoid

v_{S} (t) = V_{p} sin (ω t)

, with

V_{p} = 5 V

and angular frequency

ω > 0

. Connect its positive reference terminal to the diode anode, the diode cathode to the upper terminal of a load resistor with electrical resistance

R_{L} = 1 kΩ

, and the lower terminal to the source return. Let

v_{L}

be the upper-minus-lower load electric potential difference, and

i_{L}

the downward load electric current. Assume the ideal diode model, zero source electrical resistance, and no filter capacitor, inductor or switching delay.

Figure 1.
An ideal half-wave rectifier selects the positive half-cycle
. Original schematic and calculated load electric potential difference for $v_{S} = 5 sin (ω t) V$ and $R_{L} = 1 kΩ$ . The diode cathode is the bar at $K$ ; positive load electric current flows downward through the resistor. Dashed orange: source sinusoid. Blue: load output, zero during the negative half-cycle. Green: its cycle-average electric potential difference $5/ π V$ . No smoothing capacitor is present; these are ideal model predictions. **Details**.

The Kirchhoff cycle law gives

v_{S} = V_{D} + v_{L}

, and Ohm's law gives

v_{L} = R_{L} i_{L}

. If

v_{S} > 0

, the blocking assumption would give

V_{D} = v_{S} > 0

, contradicting the ideal diode model, so the diode conducts with

V_{D} = 0

. If

v_{S} < 0

, conduction would give

i_{L} = v_{S} / R_{L} < 0

, also forbidden, so the diode blocks. At the crossings both values vanish. Thus

v_{L} (t) = max {0, V_{p} sin (ω t)}, i_{L} (t) = \frac{v _{L} ( t )}{R _{L}} .

(1)

The peak load electric current is

5 mA

. The output is unidirectional but still varies in time, reaching zero throughout every negative half-cycle. Over a period of a function

2 π / ω

, write the wave phase as

θ = ω t

. Its average value of a function and root mean square follow from two integrals:

\overline{v}_{L} = \frac{1}{2 π} \int_{0}^{π} V_{p} sin θ d θ = \frac{V _{p}}{π} = 1.59155 V, v_{L, rms} = (\frac{1}{2 π} \int_{0}^{π} V_{p}^{2} sin^{2} θ d θ)^{1/2} = \frac{V _{p}}{2} = 2.500 V .

(2)

Consequently the average load power is

v_{L, rms}^{2} / R_{L} = 6.25 mW

, not

\overline{v}_{L}^{2} / R_{L}

. The ideal diode has zero loss; the resistor converts supplied energy into Joule heating. The same switching argument predicts

v_{L} = min (0, v_{S})

when the diode is reversed. Directional selection changes the waveform and its average value of a function; it does not by itself produce a constant output.