The DC Power Supply — Rectifiers and Filters

By the end you should be able to: Draw the block diagram of a DC power supply, describe half wave, full wave and bridge rectifiers, and calculate ripple factor with and without a capacitor filter.

Rectifier circuits are treated in detail in Physics (GAPHT121) Module 4, including the derivations of ripple factor and efficiency. This topic gives the systems view the Electrical syllabus asks for — the whole supply as a chain of blocks — and the filter arithmetic.

The block diagram

Every linear DC supply is the same four stages:

Transformer → Rectifier → Filter → Regulator

Transformer steps 230 V mains down to the working voltage, and — just as importantly — provides isolation from the mains. A supply without a transformer has live parts referenced to the mains neutral, which is why transformerless designs are confined to sealed appliances.
Rectifier converts AC to unidirectional but heavily pulsating DC.
Filter smooths the pulsations towards steady DC.
Regulator holds the output constant against changes in load and input, and rejects what ripple remains.

Each stage fixes what the previous one could not. Drawing this diagram and naming the function of each block is a standard exam question.

The three rectifier circuits

	Half wave	Centre-tapped FW	Bridge
Diodes	1	2	4
$V_{dc}$	$V_m/\pi = 0.318V_m$	$2V_m/\pi = 0.636V_m$	$0.636V_m$
Ripple factor	1.21	0.48	0.48
Efficiency	40.6%	81.2%	81.2%
Ripple frequency	$f$	$2f$	$2f$
PIV	$V_m$	$2V_m$	$V_m$
Centre tap needed	no	yes	no

The bridge dominates in practice. Its four diodes buy a lower PIV (so cheaper diodes), better transformer utilisation (so a smaller, cheaper transformer), and no centre tap. The cost is two diode drops in the conduction path instead of one — about 1.4 V, negligible at 24 V but a 14% loss at 5 V. Low-voltage high-current supplies therefore still use the centre-tapped circuit.

Ripple factor

r = \frac{\text{rms of the AC component}}{\text{DC value}} = \sqrt{\left(\frac{V_{rms}}{V_{dc}}\right)^{2} - 1}

Note $r > 1$ for the half wave circuit: its output contains more AC than DC. That is why nothing uses an unfiltered half wave rectifier for a supply.

The capacitor filter

A capacitor across the load charges to near $V_m$ at each peak and discharges into the load between peaks. Output becomes near-constant with a small sawtooth ripple:

r = \frac{1}{4\sqrt{3}\,f\,R_L\,C}

Read it as instructions: ripple falls with larger $C$ , larger $R_L$ (that is, smaller load current), and higher ripple frequency $f$ . Full wave rectification therefore helps twice — better ripple before filtering, and 100 Hz rather than 50 Hz to filter afterwards.

Two things the formula does not tell you.

First, with the capacitor fitted the DC output rises to near $V_m$ rather than $0.636V_m$ , because the capacitor holds near the peak instead of averaging over the waveform. A "12 V" supply often measures 17 V unloaded.

Second, and more practically: the capacitor recharges only during the brief interval near each peak, so the diodes conduct in short, tall current spikes. Peak diode current is far above the average load current, and the diode must be rated for the peak, not the average. This is also why a large filter capacitor causes a heavy inrush surge at switch-on.