Rectifiers
Use diode state logic to derive AC-to-unidirectional conversion rather than memorizing waveforms. Analyze the resistive-load half-wave rectifier, derive its average and RMS output, extend the same reasoning to a full-wave bridge and its two-diode conduction paths, then add a reservoir capacitor and derive the small-ripple approximation ΔV≈Iload/(fripple C) with explicit assumptions. Separate output ripple from diode/source pulse current, voltage stress and safety limits.
- Draw and label the half-wave rectifier with explicit vin, vo, diode current and load-current references before writing its piecewise equations.
- Derive the ideal half-wave output vo(t)=max[0,vin(t)] and extend it to a constant-drop diode model.
- Derive the average and RMS values of an ideal half-wave-rectified sinusoid.
- Trace the two alternating conduction paths of a four-diode bridge while proving that load current keeps the same polarity.
- Derive the ideal bridge output vo(t)=|vin(t)| and the first-order constant-drop approximation max[0,|vin|−2Vγ].
- Derive the average and RMS values of an ideal full-wave-rectified sinusoid.
- Explain how a reservoir capacitor charges near waveform peaks and discharges through the load between peaks.
- Derive ΔV≈Iload/(fripple C) and VDC≈Vpeak−NdropVγ−ΔV/2 under a declared small-ripple/approximately constant-load-current assumption.
- Distinguish ripple frequency for half-wave and full-wave rectification and explain why diode/source current occurs in short pulses rather than equal to the load current waveform.
- State topology-specific reverse-voltage/current stress and low-voltage isolation/safety checks before turning a rectifier calculation into hardware.
A rectifier uses diode state changes to enforce one load-current polarity. Its raw output still varies strongly with time; a reservoir capacitor adds memory and pulsed charging current.
03.1Half-wave state solution
Figure 3.1 — Half-wave rectifier
The one question is: which references define vin, vo, iD and iL?
During the conducting half-cycle, iD=iL and vo follows the input minus diode drop; during the blocked half-cycle, iD=0.
Thus ideal vo=max(0,vin). With a constant drop, conduction requires vin>Vγ and vo≈vin−Vγ while ON.
03.2Half-wave average and RMS
Average is the DC component; RMS determines resistor heating. With Vγ and θ0=asin(Vγ/Vm), conduction is θ0 to π−θ0 and
Subtracting Vγ from the ideal average is therefore not exact because conduction angle also changes.
03.3Full-wave bridge paths
Figure 3.2 — Graetz bridge: two paths, one load polarity
The one question is: which pair conducts when A and then B is the positive AC terminal?
D1 and D2 meet the + node at their cathodes; D3 and D4 leave the − node at their anodes. Whichever AC terminal is instantaneously positive, current crosses RL from + to −.
Two bridge diodes conduct in series during either half-cycle, while load-current polarity stays fixed.
03.4Full-wave average and RMS
Ideal full-wave rectification changes the sine average from zero to 2Vm/π while preserving its RMS magnitude across a resistor.
03.5Reservoir capacitor adds a state
Figure 3.3 — Half-wave rectifier with reservoir capacitor
The one question is: where do recharge current and load current flow?
The capacitor is in parallel with the load. The diode conducts only when the source rises enough above stored capacitor voltage, so diode current becomes pulsed.
C charges near input peaks when the source exceeds the stored capacitor voltage plus diode drop. After the source falls, the diode turns OFF and C supplies the load. Diode current is therefore concentrated into recharge pulses.
03.6Small-ripple derivation
Figure 3.4 — Small-ripple approximation
The one question is: how does ΔV follow from discharge time between recharge events?
Between recharge events separated by Δt≈1/fripple, the capacitor supplies approximately Iload and loses ΔV≈IloadΔt/C.
For small approximately triangular ripple,
ND=1 for the simple half-wave path and 2 for a bridge path.
03.7Load coupling and self-consistency
Example: bridge, Vpeak=10 V, Vγ=0.7 V, RL=1 kΩ, C=470 μF, fin=50 Hz. Solving the approximation self-consistently gives VDC≈8.51 V and ΔVpp≈0.181 V.
03.8Low ripple can mean high charging pulses
Increasing C reduces ripple but usually narrows the recharge interval. Peak current then depends on source/transformer impedance, diode dynamic resistance, capacitor ESR and waveform shape. The ripple formula alone does not predict peak diode current.
03.9Reverse voltage and hardware safety
Reverse stress is topology-specific. A resistive half-wave diode sees about Vm peak reverse voltage; with a charged reservoir capacitor it can approach roughly 2Vm in the idealized light-load limit. In a conventional ideal bridge, the reverse stress of one diode is instead typically of order the input peak; derive it from the actual topology rather than importing the half-wave result. Current ratings must include repetitive charging pulses and inrush.
03.10Mastery exercises
- Ideal half-wave, Vm=12 V: Vavg and Vrms?
- Ideal bridge, Vm=12 V: Vavg and Vrms?
- Vm=5 V,Vγ=0.7 V half-wave: θ0 and conduction interval?
- Why two diode drops in a bridge path?
- Full-wave bridge at 60 Hz: ripple frequency?
- Iload=20 mA,C=1000 μF,full-wave 50 Hz: ΔVpp?
- What happens to ripple and charging pulses when C doubles?
- Why can a capacitor-filtered half-wave diode require nearly 2Vm reverse rating?
Solutions
1. 12/π≈3.82 V; 6 V.
2. 24/π≈7.64 V; 12/√2≈8.49 V.
3. θ0=asin(0.14)≈8.05°; about 8.05° to 171.95°.
4. Two diodes conduct in series.
5. 120 Hz.
6. 0.20 V.
7. Ripple roughly halves, while charging pulses can become narrower/higher.
8. At negative input peak, source voltage adds to the still-positive stored capacitor voltage across the blocked diode.