Half-Wave Rectifiers Notes

Half-Wave Rectifiers Overview

Sections Covered:
- 3.1/3.2: Single-phase Half-wave Rectifiers
- 3.3: Resistive and inductive load
- 3.7: Free-wheeling diode
- 3.8: Half-wave Rectifiers with a capacitor filter
- 3.9: The Controlled Half-wave Rectifier

Half-Wave Rectifier with Resistive Load

Average Output Voltage ( $V<em>{avg}$ ):
$V</em>{avg} = \frac{V_m}{ heta}$ where $ heta$ is the angle of conduction.
RMS Voltage ( $V<em>{rms}$ ):
$V</em>{rms} = \frac{V_m}{2}$
Equations:
- $I<em>{avg} = \frac{V</em>{avg}}{R}$
- $P<em>{avg} = I</em>{avg} imes V_{avg}$

Half-Wave Rectifier with Inductive Load

Current Response:
$V<em>m\frac{di(t)}{dt} + Ri(t) + Li(t) = V</em>m ext{ sin }( heta)$
Forced Response: $i(t) = i<em>f(t) + i</em>n(t)$ (total current)
- Where $Z = R + j\frac{1}{ heta}$
Natural Response:
$i(t) = A e^{-t/ au} + ext{term from forced response}$
Extinction Angle ( $\beta$ ):
- Current $i(t)$ goes to zero at extinction angle $\beta$ .

Free-Wheeling Diode

Prevents back EMF from causing negative voltages.
Configuration:
- When $V_s > 0$ : D1 conducts;
- When $V_s < 0$ : D2 conducts.
Maintains output voltage ( $V_o$ ) positive.

Example Problem from Free-Wheeling Diode Section

Given:
- Source: 240 V rms at 60 Hz
- Load Resistance ( $R = 8 \, ext{Ω}$ )
- Steps to Solve:
- (a) Calculate average current, $I_o$ .
- (b) Find the average power delivered.
- (c) Determine the power factor of the source.
- (d) Find the current in each diode.
- (e) Use a 1st harmonic approximation to calculate the inductance $L$ to limit ripple.

Capacitor Filter in Half-Wave Rectifiers

For small changes in output voltage ( $ΔV<em>o$ ): $\frac{V</em>m}{R C} < f$
Used to smoothen the output and reduces ripple.

General Notes

Half-wave rectifiers are simpler but less efficient than full-wave rectifiers.
Output voltage/current waveforms can be further analyzed using Fourier series to estimate ripple levels.
Phase control techniques allow for different conduction times.