Capacitor Filter

The pulsating DC from a rectifier is useless for most electronics. A capacitor filter placed across the load smooths those pulses into steady DC by storing charge during peaks and releasing it during the gaps. Every DC power supply from a 5 V USB charger to a 48 V server rail uses this technique.

Core Concept

A capacitor in parallel with the load charges to the peak voltage during each conduction pulse. When the rectified voltage falls below the capacitor voltage, the diodes stop conducting and the capacitor discharges slowly through the load. The result is a nearly flat DC output with a small ripple voltage superimposed on it.

The size of the ripple depends on two things: the capacitance C and the load current I_L. A larger capacitor discharges more slowly, giving less ripple. A heavier load draws more current, discharging the capacitor faster and giving more ripple. This is why low-ripple supplies use large electrolytic capacitors, such as a 4700 µF unit in a 1 A audio amplifier supply.

The ripple factor r is the ratio of ripple RMS voltage to DC output voltage. Without a filter it is 0.482 for a bridge rectifier. With a 1000 µF capacitor and a 100 Ω load on a 100 Hz ripple source, r drops to below 0.05, a tenfold improvement. This is the main reason capacitor filters appear in virtually every power supply.

Key Equations

Ripple voltage (peak-to-peak): Vr = I_L / (f * C) = V_dc / (f * C * RL) where f = ripple frequency (100 Hz for bridge, 50 Hz for half-wave).

DC output voltage with filter: V_dc = Vm - Vr/2 (approximately).

Ripple factor with capacitor: r = 1 / (2*sqrt(3) * f * C * RL) — valid when Vr << Vm.

Capacitor selection: C = I_L / (Vr_allowed * f)

Given:
Bridge rectifier, 50 Hz AC supply
Ripple frequency f = 100 Hz (full-wave)
Peak output voltage Vm = 15 V
Load current I_L = 100 mA = 0.1 A
Filter capacitor C = 1000 µF = 1000e-6 F
Load RL = 150 Ω

Why this formula:
Capacitor discharges linearly between peaks; approximation gives ripple as I/fC.

Formula:
Vr = I_L / (f * C)

Substitution:
Vr = 0.1 / (100 * 1000e-6)
Vr = 0.1 / 0.1

Calculation:
Vr = 1 V (peak-to-peak ripple)

DC output voltage:
V_dc = Vm - Vr/2 = 15 - 0.5 = 14.5 V

Ripple factor:
r = Vr / (2*sqrt(3) * V_dc)
r = 1 / (2 * 1.732 * 14.5)
r = 1 / 50.2 = 0.0199 ≈ 2%

Final Answer:
Vr = 1 V, V_dc = 14.5 V, ripple factor r = 0.02 (2%)

Exam Tip: Always use the correct ripple frequency: 100 Hz for a bridge rectifier on a 50 Hz supply, and 50 Hz for a half-wave rectifier. Using 50 Hz in a bridge circuit doubles your calculated ripple. Also, the formula Vr = I_L/(f*C) gives peak-to-peak ripple. When asked for ripple factor r, use Vr_rms = Vr_pp / (2*sqrt(3)) for a sawtooth-like ripple. Forgetting the sqrt(3) factor is a common mistake.

Key Properties

• Ripple voltage Vr = I_L / (f*C). Doubling C halves the ripple. This is why large electrolytic capacitors (1000 µF to 10000 µF) are used.
• Bridge rectifier ripple frequency is 100 Hz (for 50 Hz mains). Half-wave rectifier ripple is 50 Hz. The bridge needs a smaller C for the same ripple.
• The capacitor charges quickly through the diode forward resistance (very low) and discharges slowly through RL (much larger). This asymmetry creates the smooth top on the waveform.
• Diode peak current with a capacitor filter is much higher than average DC current. For a 100 mA load, peak diode current can reach 1 to 2 A during the brief charging pulse.
• Electrolytic capacitors have voltage ratings. A 25 V rated capacitor must not be used on a supply with Vm above 22 V (allow 20% margin).
• ESR (equivalent series resistance) of the capacitor adds to ripple at high frequencies. Low-ESR capacitors rated for 105°C are used in switching supplies.

Quick Revision

• Capacitor charges to Vm during rectifier peak; discharges through RL between peaks.
• Ripple Vr = I_L / (f * C). f = 100 Hz for bridge rectifier.
• V_dc ≈ Vm - Vr/2 with capacitor filter.
• Larger C and larger RL both reduce ripple.
• Ripple factor r = 1 / (2*sqrt(3)*f*C*RL) for bridge with capacitor.
• Peak diode current >> average load current due to short charging pulse.
• Exam trap: Using f = 50 Hz instead of f = 100 Hz in the ripple formula for a bridge rectifier. This doubles the calculated ripple and gives a wrong filter capacitor size.